Find the ratio between :
(i) 60 paise and ₹ 1.35
(ii) 1.8 m and 75 cm
(iii) 1.5 kg and 600 g
(iv) 35 min and 1 3 4 1\dfrac{3}{4} 1 4 3
Answer
(i) Given,
60 paise and ₹ 1.35
₹ 1 = 100 paise
₹ 1.35 = 135 paise
Required ratio = 60 135 = 4 9 \dfrac{60}{135} = \dfrac{4}{9} 135 60 = 9 4
= 4 : 9
Hence, ratio = 4 : 9.
(ii) Given,
1.8 m and 75 cm
1 m = 100 cm
1.8 m = 180 cm
Required ratio = 180 75 = 12 5 \dfrac{180}{75} = \dfrac{12}{5} 75 180 = 5 12
= 12 : 5.
Hence, ratio = 12 : 5.
(iii) Given,
1.5 kg and 600 g
1 kg = 1000 g
1.5 kg = 1500 g
Required ratio = 1500 600 = 5 2 \dfrac{1500}{600} = \dfrac{5}{2} 600 1500 = 2 5
= 5 : 2.
Hence, ratio = 5 : 2.
(iv) Given,
35 min and 1 3 4 1\dfrac{3}{4} 1 4 3
1 hour = 60 min
⇒ 1 3 4 = 7 4 \Rightarrow 1\dfrac{3}{4} = \dfrac{7}{4} ⇒ 1 4 3 = 4 7
= 7 4 × 60 \dfrac{7}{4} \times 60 4 7 × 60
Required ratio = 35 105 = 1 3 \dfrac{35}{105} = \dfrac{1}{3} 105 35 = 3 1
Hence, ratio = 1 : 3.
If A : B = 5 : 6 and B : C = 9 : 11, find (i) A : C (ii) A : B : C.
Answer
Given,
A : B = 5 : 6
B : C = 9 : 11
Taking L.C.M of two values of B i.e. 6 and 9 = 18.
So, A B = 5 × 3 6 × 3 = 15 18 \dfrac{A}{B} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18} B A = 6 × 3 5 × 3 = 18 15
and B C = 9 × 2 11 × 2 = 18 22 \dfrac{B}{C} = \dfrac{9 \times 2}{11 \times 2} = \dfrac{18}{22} C B = 11 × 2 9 × 2 = 22 18
∴ A : B : C = 15 : 18 : 22
⇒ A : C = 15 : 22.
(i)
Hence, A : C = 15 : 22.
(ii)
Hence, A : B : C = 15 : 18 : 22.
If P : Q = 7 : 4 and Q : R = 5 : 14, find (i) R : P (ii) P : Q : R.
Answer
Given,
P : Q = 7 : 4
Q : R = 5 : 14
To find R : P and P : Q : R, we will make Q same in both cases.
Taking L.C.M of two values of Q i.e. 4 and 5 = 20
So, P Q = 7 × 5 4 × 5 = 35 20 \dfrac{P}{Q} = \dfrac{7 \times 5}{4 \times 5} = \dfrac{35}{20} Q P = 4 × 5 7 × 5 = 20 35
and Q R = 5 × 4 14 × 4 = 20 56 \dfrac{Q}{R} = \dfrac{5 \times 4}{14 \times 4} = \dfrac{20}{56} R Q = 14 × 4 5 × 4 = 56 20
⇒ P Q × Q R = 35 20 × 20 56 ⇒ P R = 35 56 ⇒ P R = 5 8 ⇒ R P = 8 5 . \Rightarrow \dfrac{P}{Q} \times \dfrac{Q}{R} = \dfrac{35}{20} \times \dfrac{20}{56} \\[1em] \Rightarrow \dfrac{P}{R} = \dfrac{35}{56} \\[1em] \Rightarrow \dfrac{P}{R} = \dfrac{5}{8} \\[1em] \Rightarrow \dfrac{R}{P} = \dfrac{8}{5}. ⇒ Q P × R Q = 20 35 × 56 20 ⇒ R P = 56 35 ⇒ R P = 8 5 ⇒ P R = 5 8 .
Hence, R : P = 8 : 5.
∴ P : Q : R = 35 : 20 : 56
Hence, ratio of P : Q : R = 35 : 20 : 56.
If 3A = 5B = 6C, find A : B : C.
Answer
Let 3A = 5B = 6C = k.
A = k 3 , B = k 5 , C = k 6 A = \dfrac{k}{3}, B = \dfrac{k}{5}, C = \dfrac{k}{6} A = 3 k , B = 5 k , C = 6 k
⇒ A : B : C = k 3 : k 5 : k 6 ⇒ A : B : C = 1 3 : 1 5 : 1 6 \Rightarrow A : B : C = \dfrac{k}{3} : \dfrac{k}{5} : \dfrac{k}{6} \\[1em] \Rightarrow A : B : C = \dfrac{1}{3} : \dfrac{1}{5} : \dfrac{1}{6} ⇒ A : B : C = 3 k : 5 k : 6 k ⇒ A : B : C = 3 1 : 5 1 : 6 1
Taking L.C.M of values of 3, 5, 6 = 30.
⇒ A : B : C = 1 3 × 30 : 1 5 × 30 : 1 6 × 30 ⇒ A : B : C = 10 : 6 : 5. \Rightarrow A : B : C = \dfrac{1}{3} \times 30 : \dfrac{1}{5} \times 30 : \dfrac{1}{6} \times 30 \\[1em] \Rightarrow A : B : C = 10 : 6 : 5. ⇒ A : B : C = 3 1 × 30 : 5 1 × 30 : 6 1 × 30 ⇒ A : B : C = 10 : 6 : 5.
Hence, A : B : C = 10 : 6 : 5.
If A 2 = B 3 = C 6 \dfrac{A}{2} = \dfrac{B}{3} = \dfrac{C}{6} 2 A = 3 B = 6 C
Answer
Let, A 2 = B 3 = C 6 \dfrac{A}{2} = \dfrac{B}{3} = \dfrac{C}{6} 2 A = 3 B = 6 C
∴ A = 2k, B = 3k, C = 6k.
A : B : C = 2k : 3k : 6k
A : B : C = 2 : 3 : 6.
Hence, A : B : C = 2 : 3 : 6.
If a : b = 8 : 5, find (7a + 5b) : (8a − 9b).
Answer
Given,
a : b = 8 : 5
Let a = 8x and b = 5x.
Substituting values in 7 a + 5 b 8 a − 9 b \dfrac{7a + 5b}{8a − 9b} 8 a − 9 b 7 a + 5 b
⇒ 7 ( 8 x ) + 5 ( 5 x ) 8 ( 8 x ) − 9 ( 5 x ) ⇒ 56 x + 25 x 64 x − 45 x ⇒ 81 x 19 x ⇒ 81 19 . \Rightarrow \dfrac{7(8x) + 5(5x)}{8(8x) - 9(5x)} \\[1em] \Rightarrow \dfrac{56x + 25x}{64x - 45x} \\[1em] \Rightarrow \dfrac{81x}{19x} \\[1em] \Rightarrow \dfrac{81}{19}. ⇒ 8 ( 8 x ) − 9 ( 5 x ) 7 ( 8 x ) + 5 ( 5 x ) ⇒ 64 x − 45 x 56 x + 25 x ⇒ 19 x 81 x ⇒ 19 81 .
Hence, 7a + 5b : 8a − 9b = 81 : 19.
If x : y = 3 : 2, find (5x − 3y) : (7x + 2y).
Answer
Given,
x : y = 3 : 2
Let x = 3a and y = 2a.
Substituting values in 5 x − 3 y 7 x + 2 y \dfrac{5x − 3y}{7x + 2y} 7 x + 2 y 5 x − 3 y
⇒ 5 ( 3 a ) − 3 ( 2 a ) 7 ( 3 a ) + 2 ( 2 a ) ⇒ 15 a − 6 a 21 a + 4 a ⇒ 9 a 25 a ⇒ 9 25 . \Rightarrow \dfrac{5(3a) - 3(2a)}{7(3a) + 2(2a)} \\[1em] \Rightarrow \dfrac{15a - 6a}{21a + 4a} \\[1em] \Rightarrow \dfrac{9a}{25a} \\[1em] \Rightarrow \dfrac{9}{25}. ⇒ 7 ( 3 a ) + 2 ( 2 a ) 5 ( 3 a ) − 3 ( 2 a ) ⇒ 21 a + 4 a 15 a − 6 a ⇒ 25 a 9 a ⇒ 25 9 .
Hence, 5x − 3y : 7x + 2y = 9 : 25.
If x : y = 10 : 3, find (3x2 + 2y2 ) : (3x2 − 2y2 ).
Answer
Given,
x : y = 10 : 3
Let x = 10a and y = 3a.
Substituting values in 3 x 2 + 2 y 2 3 x 2 − 2 y 2 \dfrac{3x^2 + 2y^2}{3x^2 − 2y^2} 3 x 2 − 2 y 2 3 x 2 + 2 y 2
⇒ 3 ( 10 a ) 2 + 2 ( 3 a ) 2 3 ( 10 a ) 2 − 2 ( 3 a ) 2 ⇒ 3 ( 100 a 2 ) + 2 ( 9 a 2 ) 3 ( 100 a 2 ) − 2 ( 9 a 2 ) ⇒ 300 a 2 + 18 a 2 300 a 2 − 18 a 2 ⇒ 318 a 2 282 a 2 ⇒ 53 47 . \Rightarrow \dfrac{3(10a)^2 + 2(3a)^2}{3(10a)^2 - 2(3a)^2} \\[1em] \Rightarrow \dfrac{3(100a^2) + 2(9a^2)}{3(100a^2) - 2(9a^2)} \\[1em] \Rightarrow \dfrac{300a^2 + 18a^2}{300a^2 - 18a^2} \\[1em] \Rightarrow \dfrac{318a^2}{282a^2} \\[1em] \Rightarrow \dfrac{53}{47}. ⇒ 3 ( 10 a ) 2 − 2 ( 3 a ) 2 3 ( 10 a ) 2 + 2 ( 3 a ) 2 ⇒ 3 ( 100 a 2 ) − 2 ( 9 a 2 ) 3 ( 100 a 2 ) + 2 ( 9 a 2 ) ⇒ 300 a 2 − 18 a 2 300 a 2 + 18 a 2 ⇒ 282 a 2 318 a 2 ⇒ 47 53 .
Hence, (3x2 + 2y2 ) : (3x2 − 2y2 ) = 53 : 47.
If a : b = 2 : 5, find (3a2 − 2ab + 5b2 ) : (a2 + 7ab − 2b2 ).
Answer
Given,
a : b = 2 : 5
Let a = 2x, then b = 5x.
Substituting values in 3 a 2 − 2 a b + 5 b 2 a 2 + 7 a b − 2 b 2 \dfrac{3a^2 − 2ab + 5b^2}{a^2 + 7ab − 2b^2} a 2 + 7 ab − 2 b 2 3 a 2 − 2 ab + 5 b 2
⇒ 3 × ( 2 x ) 2 − 2 × ( 2 x ) × ( 5 x ) + 5 × ( 5 x ) 2 ( 2 x ) 2 + 7 × ( 2 x ) × ( 5 x ) − 2 × ( 5 x ) 2 ⇒ 12 x 2 − 20 x 2 + 125 x 2 4 x 2 + 70 x 2 − 50 x 2 ⇒ 137 x 2 − 20 x 2 74 x 2 − 50 x 2 ⇒ 117 x 2 24 x 2 ⇒ 39 8 . \Rightarrow \dfrac{3 \times (2x)^2 − 2 \times (2x) \times (5x) + 5 \times (5x)^2}{(2x)^2 + 7 \times (2x) \times (5x) − 2 \times (5x)^2} \\[1em] \Rightarrow \dfrac{12x^2 − 20x^2 + 125x^2}{4x^2 + 70x^2 − 50x^2} \\[1em] \Rightarrow \dfrac{137x^2 − 20x^2}{74x^2 − 50x^2} \\[1em] \Rightarrow \dfrac{117x^2}{24x^2} \\[1em] \Rightarrow \dfrac{39}{8}. ⇒ ( 2 x ) 2 + 7 × ( 2 x ) × ( 5 x ) − 2 × ( 5 x ) 2 3 × ( 2 x ) 2 − 2 × ( 2 x ) × ( 5 x ) + 5 × ( 5 x ) 2 ⇒ 4 x 2 + 70 x 2 − 50 x 2 12 x 2 − 20 x 2 + 125 x 2 ⇒ 74 x 2 − 50 x 2 137 x 2 − 20 x 2 ⇒ 24 x 2 117 x 2 ⇒ 8 39 .
Hence, 3a2 − 2ab + 5b2 : a2 + 7ab − 2b2 = 39 : 8 .
If (5x + 2y) : (7x + 4y) = 13 : 20, find x : y.
Answer
Given,
(5x + 2y) : (7x + 4y) = 13 : 20,
⇒ 5 x + 2 y 7 x + 4 y = 13 20 ⇒ 20 ( 5 x + 2 y ) = 13 ( 7 x + 4 y ) ⇒ 100 x + 40 y = 91 x + 52 y ⇒ 100 x − 91 x = 52 y − 40 y ⇒ 9 x = 12 y ⇒ x y = 12 9 ⇒ x y = 4 3 . \Rightarrow \dfrac{5x + 2y}{7x + 4y} = \dfrac{13}{20} \\[1em] \Rightarrow 20(5x + 2y) = 13(7x + 4y) \\[1em] \Rightarrow 100x + 40y = 91x + 52y \\[1em] \Rightarrow 100x - 91x = 52y - 40y \\[1em] \Rightarrow 9x = 12y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{12}{9} \\[1em] \Rightarrow \dfrac{x}{y} =\dfrac{4}{3}. ⇒ 7 x + 4 y 5 x + 2 y = 20 13 ⇒ 20 ( 5 x + 2 y ) = 13 ( 7 x + 4 y ) ⇒ 100 x + 40 y = 91 x + 52 y ⇒ 100 x − 91 x = 52 y − 40 y ⇒ 9 x = 12 y ⇒ y x = 9 12 ⇒ y x = 3 4 .
Hence, x : y = 4 : 3.
If (3x + 5y) : (3x − 5y) = 7 : 3, find x : y.
Answer
Given,
(3x + 5y) : (3x − 5y) = 7 : 3,
⇒ 3 x + 5 y 3 x − 5 y = 7 3 ⇒ 3 ( 3 x + 5 y ) = 7 ( 3 x − 5 y ) ⇒ 9 x + 15 y = 21 x − 35 y ⇒ 15 y + 35 y = 21 x − 9 x ⇒ 50 y = 12 x ⇒ x y = 50 12 ⇒ x y = 25 6 . \Rightarrow \dfrac{3x + 5y}{3x - 5y} = \dfrac{7}{3} \\[1em] \Rightarrow 3(3x + 5y) = 7(3x - 5y) \\[1em] \Rightarrow 9x + 15y = 21x - 35y \\[1em] \Rightarrow 15y + 35y = 21x - 9x \\[1em] \Rightarrow 50y = 12x \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{50}{12} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{25}{6}. ⇒ 3 x − 5 y 3 x + 5 y = 3 7 ⇒ 3 ( 3 x + 5 y ) = 7 ( 3 x − 5 y ) ⇒ 9 x + 15 y = 21 x − 35 y ⇒ 15 y + 35 y = 21 x − 9 x ⇒ 50 y = 12 x ⇒ y x = 12 50 ⇒ y x = 6 25 .
Hence, x : y = 25 : 6.
If (6x2 − xy) : (2xy − y2 ) = 6 : 1, find x : y.
Answer
Given,
(6x2 − xy) : (2xy − y2 ) = 6 : 1,
Dividing numerator and denominator by y2 , we get :
⇒ 6 x 2 − x y y 2 2 x y − y 2 y 2 = 6 1 ⇒ 6 ( x y ) 2 − x y 2 x y − 1 = 6 \Rightarrow \dfrac{\dfrac{6x^2 − xy}{y^2}}{\dfrac{2xy − y^2}{y^2}} = \dfrac{6}{1} \\[1em] \Rightarrow \dfrac{6\Big(\dfrac{x}{y}\Big)^2 - \dfrac{x}{y}}{2\dfrac{x}{y} - 1} = 6 ⇒ y 2 2 x y − y 2 y 2 6 x 2 − x y = 1 6 ⇒ 2 y x − 1 6 ( y x ) 2 − y x = 6
Let, x y \dfrac{x}{y} y x
⇒ 6 t 2 − t 2 t − 1 = 6 ⇒ 6 t 2 − t = ( 2 t − 1 ) × 6 ⇒ 6 t 2 − t = 12 t − 6 ⇒ 6 t 2 − t − 12 t + 6 = 0 ⇒ 6 t 2 − 13 t + 6 = 0 ⇒ 6 t 2 − 4 t − 9 t + 6 = 0 ⇒ 2 t ( 3 t − 2 ) − 3 ( 3 t − 2 ) = 0 ⇒ ( 2 t − 3 ) ( 3 t − 2 ) = 0 ⇒ ( 2 t − 3 ) = 0 or ( 3 t − 2 ) = 0 [Using zero - product rule] ⇒ 2 t = 3 or 3 t = 2 ⇒ t = 3 2 or t = 2 3 . \Rightarrow \dfrac{6t^2 - t}{2t - 1} = 6 \\[1em] \Rightarrow 6t^2 - t = (2t - 1) \times 6 \\[1em] \Rightarrow 6t^2 - t = 12t - 6 \\[1em] \Rightarrow 6t^2 - t - 12t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 13t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 4t - 9t + 6 = 0 \\[1em] \Rightarrow 2t(3t - 2) - 3(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3)(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3) = 0 \text{ or }(3t - 2) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow 2t = 3 \text{ or } 3t = 2 \\[1em] \Rightarrow t = \dfrac{3}{2} \text{ or } t = \dfrac{2}{3} . ⇒ 2 t − 1 6 t 2 − t = 6 ⇒ 6 t 2 − t = ( 2 t − 1 ) × 6 ⇒ 6 t 2 − t = 12 t − 6 ⇒ 6 t 2 − t − 12 t + 6 = 0 ⇒ 6 t 2 − 13 t + 6 = 0 ⇒ 6 t 2 − 4 t − 9 t + 6 = 0 ⇒ 2 t ( 3 t − 2 ) − 3 ( 3 t − 2 ) = 0 ⇒ ( 2 t − 3 ) ( 3 t − 2 ) = 0 ⇒ ( 2 t − 3 ) = 0 or ( 3 t − 2 ) = 0 [Using zero - product rule] ⇒ 2 t = 3 or 3 t = 2 ⇒ t = 2 3 or t = 3 2 .
Thus,
x y = 3 2 or x y = 2 3 \dfrac{x}{y} = \dfrac{3}{2} \text{ or } \dfrac{x}{y} = \dfrac{2}{3} y x = 2 3 or y x = 3 2
Hence, x : y = 3 : 2 or 2 : 3.
If (4x2 − 3y2 ) : (2x2 + 5y2 ) = 12 : 19, find x : y.
Answer
Given,
(4x2 − 3y2 ) : (2x2 + 5y2 ) = 12 : 19,
Dividing numerator and denominator both by y2 we get,
⇒ 4 x 2 − 3 y 2 y 2 2 x 2 + 5 y 2 y 2 = 12 19 ⇒ 4 ( x y ) 2 − 3 y 2 y 2 2 ( x y ) 2 + 5 y 2 y 2 = 12 19 ⇒ 4 ( x y ) 2 − 3 2 ( x y ) 2 + 5 = 12 19 \Rightarrow \dfrac{\dfrac{4x^2 − 3y^2}{y^2}}{\dfrac{2x^2 + 5y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3\dfrac{y^2}{y^2}}{2\Big(\dfrac{x}{y}\Big)^2 + 5\dfrac{y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3}{2\Big(\dfrac{x}{y}\Big)^2 + 5} = \dfrac{12}{19} ⇒ y 2 2 x 2 + 5 y 2 y 2 4 x 2 − 3 y 2 = 19 12 ⇒ 2 ( y x ) 2 + 5 y 2 y 2 4 ( y x ) 2 − 3 y 2 y 2 = 19 12 ⇒ 2 ( y x ) 2 + 5 4 ( y x ) 2 − 3 = 19 12
Let, x y \dfrac{x}{y} y x
⇒ 4 t 2 − 3 2 t 2 + 5 = 12 19 ⇒ ( 4 t 2 − 3 ) × 19 = ( 2 t 2 + 5 ) × 12 ⇒ 76 t 2 − 57 = 24 t 2 + 60 ⇒ 76 t 2 − 24 t 2 = 60 + 57 ⇒ 52 t 2 = 117 ⇒ t 2 = 117 52 ⇒ t 2 = 9 4 ⇒ t = 9 4 ⇒ t = 3 2 . \Rightarrow \dfrac{4t^2 - 3}{2t^2 + 5} = \dfrac{12}{19} \\[1em] \Rightarrow (4t^2 - 3) \times 19 = (2t^2 + 5) \times 12 \\[1em] \Rightarrow 76t^2 - 57 = 24t^2 + 60 \\[1em] \Rightarrow 76t^2 - 24t^2 = 60 + 57 \\[1em] \Rightarrow 52t^2 = 117 \\[1em] \Rightarrow t^2 = \dfrac{117}{52} \\[1em] \Rightarrow t^2 = \dfrac{9}{4} \\[1em] \Rightarrow t = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow t = \dfrac{3}{2}. ⇒ 2 t 2 + 5 4 t 2 − 3 = 19 12 ⇒ ( 4 t 2 − 3 ) × 19 = ( 2 t 2 + 5 ) × 12 ⇒ 76 t 2 − 57 = 24 t 2 + 60 ⇒ 76 t 2 − 24 t 2 = 60 + 57 ⇒ 52 t 2 = 117 ⇒ t 2 = 52 117 ⇒ t 2 = 4 9 ⇒ t = 4 9 ⇒ t = 2 3 .
Hence, x : y = 3 : 2.
If x2 + 4y2 = 4xy, find x : y.
Answer
Given,
x2 + 4y2 = 4xy
Dividing both sides by xy we get,
⇒ x 2 + 4 y 2 x y = 4 x y x y ⇒ x y + 4 y x = 4 \Rightarrow \dfrac{x^2 + 4y^2}{xy} = \dfrac{4xy}{xy} \\[1em] \Rightarrow \dfrac{x}{y} + 4\dfrac{y}{x} = 4 ⇒ x y x 2 + 4 y 2 = x y 4 x y ⇒ y x + 4 x y = 4
Let x y \dfrac{x}{y} y x
⇒ t + 4 1 t = 4 ⇒ t 2 + 4 t = 4 ⇒ t 2 + 4 = 4 t ⇒ t 2 − 4 t + 4 = 0 ⇒ t 2 − 2 t − 2 t + 4 = 0 ⇒ t ( t − 2 ) − 2 ( t − 2 ) = 0 ⇒ ( t − 2 ) ( t − 2 ) = 0 ⇒ t − 2 = 0 or t − 2 = 0 ⇒ t = 2 ⇒ x y = 2. \Rightarrow t + 4\dfrac{1}{t} = 4 \\[1em] \Rightarrow \dfrac{t^2 + 4}{t} = 4 \\[1em] \Rightarrow t^2 + 4 = 4t \\[1em] \Rightarrow t^2 - 4t + 4 = 0 \\[1em] \Rightarrow t^2 - 2t - 2t + 4 = 0 \\[1em] \Rightarrow t(t - 2) - 2(t - 2) = 0 \\[1em] \Rightarrow (t - 2)(t - 2) = 0 \\[1em] \Rightarrow t - 2 = 0 \text{ or } t - 2 = 0 \\[1em] \Rightarrow t = 2 \\[1em] \Rightarrow \dfrac{x}{y} = 2. ⇒ t + 4 t 1 = 4 ⇒ t t 2 + 4 = 4 ⇒ t 2 + 4 = 4 t ⇒ t 2 − 4 t + 4 = 0 ⇒ t 2 − 2 t − 2 t + 4 = 0 ⇒ t ( t − 2 ) − 2 ( t − 2 ) = 0 ⇒ ( t − 2 ) ( t − 2 ) = 0 ⇒ t − 2 = 0 or t − 2 = 0 ⇒ t = 2 ⇒ y x = 2.
Hence, x : y = 2 : 1.
If 10x2 − 23xy + 9y2 = 0, find x : y.
Answer
Given,
10x2 − 23xy + 9y2 = 0
Dividing both sides by xy we get,
⇒ 10 x 2 − 23 x y + 9 y 2 x y = 0 ⇒ 10 x y − 23 + 9 y x = 0 ⇒ 10 x y + 9 y x = 23 \Rightarrow \dfrac{10x^2 − 23xy + 9y^2}{xy} = 0 \\[1em] \Rightarrow 10\dfrac{x}{y} - 23 + 9\dfrac{y}{x} = 0 \\[1em] \Rightarrow 10\dfrac{x}{y} + 9\dfrac{y}{x} = 23 ⇒ x y 10 x 2 − 23 x y + 9 y 2 = 0 ⇒ 10 y x − 23 + 9 x y = 0 ⇒ 10 y x + 9 x y = 23
Let, x y \dfrac{x}{y} y x
⇒ 10 t + 9 1 t = 23 ⇒ 10 t 2 + 9 t = 23 ⇒ 10 t 2 + 9 = 23 t ⇒ 10 t 2 − 23 t + 9 = 0 ⇒ 10 t 2 − 5 t − 18 t + 9 = 0 ⇒ 5 t ( 2 t − 1 ) − 9 ( 2 t − 1 ) = 0 ⇒ ( 5 t − 9 ) ( 2 t − 1 ) = 0 ⇒ 5 t − 9 = 0 or 2 t − 1 = 0 ⇒ 5 t = 9 or 2 t = 1 ⇒ t = 9 5 or t = 1 2 ⇒ x y = 9 5 or x y = 1 2 \Rightarrow 10t + 9\dfrac{1}{t} = 23 \\[1em] \Rightarrow \dfrac{10t^2 + 9}{t} = 23 \\[1em] \Rightarrow 10t^2 + 9 = 23t \\[1em] \Rightarrow 10t^2 - 23t + 9 = 0 \\[1em] \Rightarrow 10t^2 - 5t - 18t + 9 = 0 \\[1em] \Rightarrow 5t(2t - 1) - 9(2t - 1) = 0 \\[1em] \Rightarrow (5t - 9)(2t - 1) = 0 \\[1em] \Rightarrow 5t - 9 = 0 \text{ or } 2t - 1 = 0 \\[1em] \Rightarrow 5t = 9 \text{ or } 2t = 1 \\[1em] \Rightarrow t = \dfrac{9}{5} \text{ or } t = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{9}{5} \text{ or } \dfrac{x}{y} = \dfrac{1}{2} ⇒ 10 t + 9 t 1 = 23 ⇒ t 10 t 2 + 9 = 23 ⇒ 10 t 2 + 9 = 23 t ⇒ 10 t 2 − 23 t + 9 = 0 ⇒ 10 t 2 − 5 t − 18 t + 9 = 0 ⇒ 5 t ( 2 t − 1 ) − 9 ( 2 t − 1 ) = 0 ⇒ ( 5 t − 9 ) ( 2 t − 1 ) = 0 ⇒ 5 t − 9 = 0 or 2 t − 1 = 0 ⇒ 5 t = 9 or 2 t = 1 ⇒ t = 5 9 or t = 2 1 ⇒ y x = 5 9 or y x = 2 1
Hence, x : y = 9 : 5 or 1 : 2.
A ratio is equal to 3 : 4. If its consequent is 144, what is its antecedent?
Answer
Let the antecedent be x. Then,
3 : 4 = x : 144
⇒ 3 4 = x 144 ⇒ 3 × 144 = 4 x ⇒ 432 = 4 x ⇒ x = 432 4 ⇒ x = 108 \Rightarrow \dfrac{3}{4} = \dfrac{x}{144} \\[1em] \Rightarrow 3 \times 144 = 4x \\[1em] \Rightarrow 432 = 4x \\[1em] \Rightarrow x = \dfrac{432}{4} \\[1em] \Rightarrow x = 108 ⇒ 4 3 = 144 x ⇒ 3 × 144 = 4 x ⇒ 432 = 4 x ⇒ x = 4 432 ⇒ x = 108
Hence, the antecedent = 108.
Two numbers are in the ratio 8 : 13. If 14 is added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.
Answer
Let the numbers be 8x and 13x.
Given,
If 14 is added to each of the numbers, the ratio becomes 2 : 3.
⇒ 8 x + 14 13 x + 14 = 2 3 ⇒ ( 8 x + 14 ) × 3 = ( 13 x + 14 ) × 2 ⇒ 24 x + 42 = 26 x + 28 ⇒ 26 x − 24 x = 42 − 28 ⇒ 2 x = 14 ⇒ x = 14 2 ⇒ x = 7. \Rightarrow \dfrac{8x + 14}{13x + 14} = \dfrac{2}{3} \\[1em] \Rightarrow (8x + 14) \times 3 = (13x + 14) \times 2 \\[1em] \Rightarrow 24x + 42 = 26x + 28 \\[1em] \Rightarrow 26x - 24x = 42 - 28 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \dfrac{14}{2} \\[1em] \Rightarrow x = 7. ⇒ 13 x + 14 8 x + 14 = 3 2 ⇒ ( 8 x + 14 ) × 3 = ( 13 x + 14 ) × 2 ⇒ 24 x + 42 = 26 x + 28 ⇒ 26 x − 24 x = 42 − 28 ⇒ 2 x = 14 ⇒ x = 2 14 ⇒ x = 7.
The two numbers are 8x and 13x
= 8 × 7, 13 × 7
= 56, 91.
Hence, the numbers are 56 and 91.
Two numbers are in the ratio 5 : 7. If 8 is subtracted from each, the ratio becomes 3 : 5. Find the numbers.
Answer
Let the numbers be 5x and 7x.
Given,
If 8 is subtracted from each, the ratio becomes 3 : 5.
⇒ 5 x − 8 7 x − 8 = 3 5 ⇒ ( 5 x − 8 ) × 5 = ( 7 x − 8 ) × 3 ⇒ 25 x − 40 = 21 x − 24 ⇒ 25 x − 21 x = 40 − 24 ⇒ 4 x = 16 ⇒ x = 16 4 ⇒ x = 4. \Rightarrow \dfrac{5x - 8}{7x - 8} = \dfrac{3}{5} \\[1em] \Rightarrow (5x - 8) \times 5 = (7x - 8) \times 3 \\[1em] \Rightarrow 25x - 40 = 21x - 24\\[1em] \Rightarrow 25x - 21x = 40 - 24 \\[1em] \Rightarrow 4x = 16 \\[1em] \Rightarrow x = \dfrac{16}{4} \\[1em] \Rightarrow x = 4. ⇒ 7 x − 8 5 x − 8 = 5 3 ⇒ ( 5 x − 8 ) × 5 = ( 7 x − 8 ) × 3 ⇒ 25 x − 40 = 21 x − 24 ⇒ 25 x − 21 x = 40 − 24 ⇒ 4 x = 16 ⇒ x = 4 16 ⇒ x = 4.
The two numbers are 5x and 7x
= 5 × 4, 7 × 4
= 20, 28.
Hence, the numbers are 20 and 28.
What least number must be added to each term of the ratio 5 : 7 to make it 8 : 9?
Answer
Let x be the least number must be added to each term of the ratio 5 : 7 to make it 8 : 9.
⇒ 5 + x 7 + x = 8 9 ⇒ ( 5 + x ) × 9 = ( 7 + x ) × 8 ⇒ 45 + 9 x = 56 + 8 x ⇒ 9 x − 8 x = 56 − 45 ⇒ x = 11. \Rightarrow \dfrac{5 + x}{7 + x} = \dfrac{8}{9} \\[1em] \Rightarrow (5 + x) \times 9 = (7 + x) \times 8 \\[1em] \Rightarrow 45 + 9x = 56 + 8x \\[1em] \Rightarrow 9x - 8x = 56 - 45 \\[1em] \Rightarrow x = 11. ⇒ 7 + x 5 + x = 9 8 ⇒ ( 5 + x ) × 9 = ( 7 + x ) × 8 ⇒ 45 + 9 x = 56 + 8 x ⇒ 9 x − 8 x = 56 − 45 ⇒ x = 11.
Hence, the least number to be added is 11.
Out of the monthly income of ₹ 45,000, Rahul spends ₹ 31,500 and the rest he saves. Find the ratio of his
(i) income to expenditure
(ii) income to savings
(iii) savings to expenditure.
Answer
Given,
Income = ₹ 45,000
Expenditure = ₹ 31,500
Total savings = Income - Expenditure = ₹ 45,000 - 31,500
= ₹ 13,500.
(i) Required ratio,
Income : Expenditure = 45000 : 31500
= 45000 31500 = 10 7 = \dfrac{45000}{31500} = \dfrac{10}{7} = 31500 45000 = 7 10
= 10 : 7.
Hence, required ratio = 10 : 7.
(ii) Required ratio,
Income : Savings = 45000 : 13500
= 45000 13500 = 10 3 = \dfrac{45000}{13500} = \dfrac{10}{3} = 13500 45000 = 3 10
= 10 : 3.
Hence, required ratio = 10 : 3.
(iii) Required ratio is savings to expenditure.
Savings : expenditure = 13500 : 31500
= 13500 31500 = 3 7 =\dfrac{13500}{31500} = \dfrac{3}{7} = 31500 13500 = 7 3
= 3 : 7.
Hence, required ratio = 3 : 7.
The cost of making an umbrella is divided between material, labour and overheads in the ratio 6 : 4 : 1. If the material costs ₹ 132, find the cost of production of an umbrella.
Answer
Given,
The ratio of material : labour : overheads = 6 : 4 : 1
Let the cost of material be 6x, labour be 4x and overheads be x.
Given,
Cost of material = ₹ 132
⇒ 6x = 132
⇒ x = 132 6 \dfrac{132}{6} 6 132
⇒ x = 22.
The total cost of production of an umbrella = 6x + 4x + 1x = 11x
= 11 × 22
= ₹ 242.
Hence, cost of production of an umbrella = ₹ 242.
Divide ₹ 6,720 in the ratio 5 : 3.
Answer
Given,
Let ₹ 6,720 be divided in two parts A and B.
A = 5a and B = 3a
To find A's part,
⇒ 5 a 5 a + 3 a × 6 , 720 ⇒ 5 a 8 a × 6 , 720 ⇒ 5 8 × 6 , 720 ⇒ ₹ 4 , 200. \Rightarrow \dfrac{5a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \dfrac{5a}{8a} \times 6,720 \\[1em] \Rightarrow \dfrac{5}{8} \times 6,720 \\[1em] \Rightarrow ₹ 4,200. ⇒ 5 a + 3 a 5 a × 6 , 720 ⇒ 8 a 5 a × 6 , 720 ⇒ 8 5 × 6 , 720 ⇒ ₹4 , 200.
To find B's part,
⇒ 3 a 5 a + 3 a × 6 , 720 ⇒ 3 a 8 a × 6 , 720 ⇒ 3 8 × 6 , 720 ⇒ ₹ 2 , 520 \Rightarrow \dfrac{3a}{5a + 3a} \times 6,720 \\[1em] \Rightarrow \dfrac{3a}{8a} \times 6,720 \\[1em] \Rightarrow \dfrac{3}{8} \times 6,720 \\[1em] \Rightarrow ₹ 2,520 ⇒ 5 a + 3 a 3 a × 6 , 720 ⇒ 8 a 3 a × 6 , 720 ⇒ 8 3 × 6 , 720 ⇒ ₹2 , 520
Hence, ₹ 6,720 can be divided into ₹ 4,200 and ₹ 2,520.
Divide ₹ 11,620 among A, B and C in the ratio 35 : 28 : 20.
Answer
Given,
Let A = 35a and B = 28a and C = 20a,
To find A's part,
⇒ 35 a 35 a + 28 a + 20 a × 11 , 620 ⇒ 35 a 83 a × 11 , 620 ⇒ 35 83 × 11 , 620 ⇒ ₹ 4 , 900. \Rightarrow \dfrac{35a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{35a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{35}{83} \times 11,620 \\[1em] \Rightarrow ₹ 4,900. ⇒ 35 a + 28 a + 20 a 35 a × 11 , 620 ⇒ 83 a 35 a × 11 , 620 ⇒ 83 35 × 11 , 620 ⇒ ₹4 , 900.
To find B's part,
⇒ 28 a 35 a + 28 a + 20 a × 11 , 620 ⇒ 28 a 83 a × 11 , 620 ⇒ 28 83 × 11 , 620 ⇒ ₹ 3 , 920. \Rightarrow \dfrac{28a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{28a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{28}{83} \times 11,620 \\[1em] \Rightarrow ₹ 3,920. ⇒ 35 a + 28 a + 20 a 28 a × 11 , 620 ⇒ 83 a 28 a × 11 , 620 ⇒ 83 28 × 11 , 620 ⇒ ₹3 , 920.
To find C's part,
⇒ 20 a 35 a + 28 a + 20 a × 11 , 620 ⇒ 20 a 83 a × 11 , 620 ⇒ 20 83 × 11 , 620 ⇒ ₹ 2 , 800 \Rightarrow \dfrac{20a}{35a + 28a + 20a} \times 11,620 \\[1em] \Rightarrow \dfrac{20a}{83a} \times 11,620 \\[1em] \Rightarrow \dfrac{20}{83} \times 11,620 \\[1em] \Rightarrow ₹ 2,800 ⇒ 35 a + 28 a + 20 a 20 a × 11 , 620 ⇒ 83 a 20 a × 11 , 620 ⇒ 83 20 × 11 , 620 ⇒ ₹2 , 800
Hence, A = ₹ 4,900, B = ₹ 3,920 and C = ₹ 2,800.
Divide ₹ 782 among P, Q and R in the ratio 1 2 : 2 3 : 3 4 \dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4} 2 1 : 3 2 : 4 3
Answer
Given,
P : Q : R = 1 2 : 2 3 : 3 4 \dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4} 2 1 : 3 2 : 4 3
Multiply each ratio by 12 (LCM of denominators) to clear fractions :
= 1 2 × 12 : 2 3 × 12 : 3 4 × 12 \dfrac{1}{2} \times 12 : \dfrac{2}{3} \times 12 : \dfrac{3}{4} \times 12 2 1 × 12 : 3 2 × 12 : 4 3 × 12
= 6 : 4 : 9.
Let P = 6a and Q = 8a and R = 9a.
To find P's part,
⇒ 6 a 6 a + 8 a + 9 a × 782 ⇒ 6 a 23 a × 782 ⇒ 6 23 × 782 ⇒ 204 \Rightarrow \dfrac{6a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{6a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{6}{23} \times 782 \\[1em] \Rightarrow 204 ⇒ 6 a + 8 a + 9 a 6 a × 782 ⇒ 23 a 6 a × 782 ⇒ 23 6 × 782 ⇒ 204
To find Q's part,
⇒ 8 a 6 a + 8 a + 9 a × 782 ⇒ 8 a 23 a × 782 ⇒ 8 23 × 782 ⇒ 272 \Rightarrow \dfrac{8a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{8a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{8}{23} \times 782 \\[1em] \Rightarrow 272 ⇒ 6 a + 8 a + 9 a 8 a × 782 ⇒ 23 a 8 a × 782 ⇒ 23 8 × 782 ⇒ 272
To find Q's part,
⇒ 9 a 6 a + 8 a + 9 a × 782 ⇒ 9 a 23 a × 782 ⇒ 9 23 × 782 ⇒ 306 \Rightarrow \dfrac{9a}{6a + 8a + 9a} \times 782 \\[1em] \Rightarrow \dfrac{9a}{23a} \times 782 \\[1em] \Rightarrow \dfrac{9}{23} \times 782 \\[1em] \Rightarrow 306 ⇒ 6 a + 8 a + 9 a 9 a × 782 ⇒ 23 a 9 a × 782 ⇒ 23 9 × 782 ⇒ 306
Hence, ₹ 782 can be divided into P = ₹ 204, Q = ₹ 272 and R = ₹ 306.
If ₹ 5,100 be divided among A, B, C in such a way that A gets 2 3 \dfrac{2}{3} 3 2 1 4 \dfrac{1}{4} 4 1
Answer
Given,
A = 2 3 \dfrac{2}{3} 3 2 1 4 \dfrac{1}{4} 4 1
Express all in terms of B:
A = 2 3 B \dfrac{2}{3}B 3 2 B
Total amount divided among A, B and C = ₹ 5,100
So,
⇒ 2 3 B + B + 4 B = 5100 ⇒ 2 B + 3 B + 12 B 3 = 5100 ⇒ 17 B 3 = 5100 ⇒ B = 5100 × 3 17 ⇒ B = ₹ 900. \Rightarrow \dfrac{2}{3}B + B + 4B = 5100 \\[1em] \Rightarrow \dfrac{2B + 3B + 12B}{3} = 5100 \\[1em] \Rightarrow \dfrac{17B}{3} = 5100 \\[1em] \Rightarrow B = \dfrac{5100 \times 3}{17} \\[1em] \Rightarrow B = ₹ 900. ⇒ 3 2 B + B + 4 B = 5100 ⇒ 3 2 B + 3 B + 12 B = 5100 ⇒ 3 17 B = 5100 ⇒ B = 17 5100 × 3 ⇒ B = ₹900.
B's share = ₹ 900
Therefore,
A's share = 2 3 × 900 \dfrac{2}{3} \times 900 3 2 × 900
C's share = 4 × 900 = ₹ 3,600
Hence, A = ₹ 600, B = ₹ 900 and C = ₹ 3,600.
Divide ₹ 8,300 among A, B and C such that 4 times A’s share, 5 times B’s share and 7 times C’s share may all be equal.
Answer
Given,
Let, 4A = 5B = 7C = k
Then, A = k 4 \dfrac{k}{4} 4 k k 5 \dfrac{k}{5} 5 k k 7 \dfrac{k}{7} 7 k
Total amount divided among A, B and C = ₹ 8,300
So,
⇒ k 4 + k 5 + k 7 = 8300 ⇒ k ( 1 4 + 1 5 + 1 7 ) = 8300 ⇒ k ( 35 + 28 + 20 140 ) = 8300 ⇒ k ( 83 140 ) = 8300 ⇒ k = 8300 × 140 83 ⇒ k = 14000 \Rightarrow \dfrac{k}{4} + \dfrac{k}{5} + \dfrac{k}{7} = 8300 \\[1em] \Rightarrow k\Big(\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{7}\Big) = 8300 \\[1em] \Rightarrow k\Big(\dfrac{35 + 28 + 20}{140}\Big) = 8300 \\[1em] \Rightarrow k\Big(\dfrac{83}{140}\Big) = 8300 \\[1em] \Rightarrow k = \dfrac{8300 \times 140}{83} \\[1em] \Rightarrow k = 14000 ⇒ 4 k + 5 k + 7 k = 8300 ⇒ k ( 4 1 + 5 1 + 7 1 ) = 8300 ⇒ k ( 140 35 + 28 + 20 ) = 8300 ⇒ k ( 140 83 ) = 8300 ⇒ k = 83 8300 × 140 ⇒ k = 14000
Therefore,
A' share = 14000 4 \dfrac{14000}{4} 4 14000
B's share = 14000 5 \dfrac{14000}{5} 5 14000
C"s share = 14000 7 \dfrac{14000}{7} 7 14000
Hence, A = ₹ 3,500, B = ₹ 2,800 and C = ₹ 2,000.
A sum of money is divided between A and B in the ratio 6 : 11. If B’s share is ₹ 7,315, find (i) A’s share (ii) the total amount of money.
Answer
(i) Let A's share be ₹ x.
Then,
⇒ 6 11 = x 7315 ⇒ x = 6 11 × 7315 ⇒ x = ₹ 3 , 990. \Rightarrow \dfrac{6}{11} = \dfrac{x}{7315} \\[1em] \Rightarrow x = \dfrac{6}{11} \times 7315 \\[1em] \Rightarrow x = ₹ 3,990. ⇒ 11 6 = 7315 x ⇒ x = 11 6 × 7315 ⇒ x = ₹3 , 990.
Hence, A' share of money = ₹ 3,990.
(ii) Total sum of money = A's share + B's share
= ₹ 3,990 + ₹ 7,315
= ₹ 11,305.
Hence, total sum of money = ₹ 11,305.
The ages of Tanvy and Divya are in the ratio 5 : 7. Five years hence, their ages will be in the ratio 3 : 4. Find their present ages.
Answer
Given,
The ages of Tanvy and Divya are in the ratio 5 : 7.
Let the present age of Tanvy be 5x and the present age of Divya be 7x.
Five years from now :
Tanvy's age will be 5x + 5.
Divya's age will be 7x + 5.
Given,
The ratio of their ages in five years will be 3 : 4.
⇒ 5 x + 5 7 x + 5 = 3 4 \Rightarrow \dfrac{5x + 5}{7x + 5} = \dfrac{3}{4} ⇒ 7 x + 5 5 x + 5 = 4 3
⇒ 4(5x + 5) = 3(7x + 5)
⇒ 20x + 20 = 21x + 15
⇒ 21x - 20x = 20 - 15
⇒ x = 20 - 15
⇒ x = 5.
Tanvy's present age = 5x = 5(5) = 25 years.
Divya's present age = 7x = 7(5) = 35 years.
Hence, present age of Tnavy and Divya are 25 years and 35 years respectively.
One year ago, the ratio of Amit’s and Arun’s ages was 6 : 7 respectively. Four years hence, their ages will be in the ratio 7 : 8. How old is Amit?
Answer
Given,
The ratio of Amit’s and Arun’s ages was 6 : 7.
Let Amit's age one year ago be 6x and Arun's age one year ago be 7x.
Their present ages are :
Amit's present age = 6x + 1
Arun's present age = 7x + 1
After four years their ages will be:
Amit's age will be (6x + 1) + 4 = 6x + 5
Arun's age will be (7x + 1) + 4 = 7x + 5
Given,
The ratio of their ages four years hence will be 7 : 8.
∴ 6 x + 5 7 x + 5 = 7 8 \therefore \dfrac{6x + 5}{7x + 5} = \dfrac{7}{8} ∴ 7 x + 5 6 x + 5 = 8 7
⇒ 8(6x + 5) = 7(7x + 5)
⇒ 48x + 40 = 49x + 35
⇒ 40 - 35 = 49x - 48x
⇒ x = 5
Amit's present age is 6x + 1.
= 6(5) + 1 = 30 + 1
= 31 years.
Hence, Amit's present age = 31 years.
Reena reduces her weight in the ratio 5 : 4. What is her weight now, if originally it was 70 kg?
Answer
Given,
Reena reduces her weight in the ratio 5 : 4
Let her orignal weight be 5x and new weight be 4x.
Reena's orignal weight = 70 kg
⇒ 5x = 70
⇒ x = 70 5 \dfrac{70}{5} 5 70
⇒ x = 14 kg
Reena's new weight = 4x
= 4 × 14
= 56 kg.
Hence, Reena's present weight = 56 kg.
68 kg of a mixture contains milk and water in the ratio 27 : 7. How much more water is to be added to this mixture to get a new mixture containing milk and water in the ratio 3 : 1?
Answer
Given,
Milk : Water = 27 : 7 and total = 68 kg
Milk = 27 27 + 7 × 68 = 27 34 × 68 = 54 kg \dfrac{27}{27 + 7} \times 68 = \dfrac{27}{34} \times 68 = 54\text{ kg} 27 + 7 27 × 68 = 34 27 × 68 = 54 kg
Water = 7 27 + 7 × 68 = 7 34 × 68 = 14 kg \dfrac{7}{27 + 7} \times 68 = \dfrac{7}{34} \times 68 = 14\text{ kg} 27 + 7 7 × 68 = 34 7 × 68 = 14 kg
Let x kg of water be added to make the ratio of milk to water as 3 : 1.
⇒ 54 14 + x = 3 1 ⇒ 54 = 3 ( 14 + x ) ⇒ 54 = 42 + 3 x ⇒ 54 − 42 = 3 x ⇒ 12 = 3 x ⇒ x = 4. \Rightarrow \dfrac{54}{14 + x} = \dfrac{3}{1} \\[1em] \Rightarrow 54 = 3(14 + x) \\[1em] \Rightarrow 54 = 42 + 3x \\[1em] \Rightarrow 54 - 42 = 3x \\[1em] \Rightarrow 12 = 3x \\[1em] \Rightarrow x = 4. ⇒ 14 + x 54 = 1 3 ⇒ 54 = 3 ( 14 + x ) ⇒ 54 = 42 + 3 x ⇒ 54 − 42 = 3 x ⇒ 12 = 3 x ⇒ x = 4.
Hence, 4 kg of water must be added.
A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk to water becomes 5 : 2. Find the quantity of milk in the original mixture.
Answer
Given,
Milk : Water = 5 : 1
Let Milk = 5x and Water = x
After adding 5 litres of water, the ratio of milk to water becomes 5 : 2.
⇒ 5 x x + 5 = 5 2 ⇒ 2 ( 5 x ) = 5 ( x + 5 ) ⇒ 10 x = 5 x + 25 ⇒ 10 x − 5 x = 25 ⇒ 5 x = 25 ⇒ x = 5. \Rightarrow \dfrac{5x}{x + 5} = \dfrac{5}{2} \\[1em] \Rightarrow 2(5x) = 5(x + 5) \\[1em] \Rightarrow 10x = 5x + 25 \\[1em] \Rightarrow 10x - 5x = 25 \\[1em] \Rightarrow 5x = 25 \\[1em] \Rightarrow x = 5. ⇒ x + 5 5 x = 2 5 ⇒ 2 ( 5 x ) = 5 ( x + 5 ) ⇒ 10 x = 5 x + 25 ⇒ 10 x − 5 x = 25 ⇒ 5 x = 25 ⇒ x = 5.
Milk in the original mixture = 5x = 5 × 5 = 25 litres.
Hence, the quantity of milk in the original mixture = 25 litres.
In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination?
Answer
Given,
Ratio of Passes to Failures = 4 : 1
Let students who Pass = 4k and Fail = k
Total students appeared for examination = 4k + k = 5k
Given,
30 students didn't appear for examination.
Now total students appeared for exam = 5k − 30
If 20 less students passed the exam, then students that pass the exam now = 4k − 20
Number of students that fail in exam = (5k − 30) − (4k − 20) = k − 10.
The new ratio of passes to failures = 5 : 1
∴ 4 k − 20 k − 10 = 5 1 ⇒ 4 k − 20 = 5 ( k − 10 ) ⇒ 4 k − 20 = 5 k − 50 ⇒ − 20 + 50 = 5 k − 4 k ⇒ k = 30. \therefore \dfrac{4k - 20}{k - 10} = \dfrac{5}{1} \\[1em] \Rightarrow 4k - 20 = 5(k - 10) \\[1em] \Rightarrow 4k - 20 = 5k - 50 \\[1em] \Rightarrow -20 + 50 = 5k - 4k \\[1em] \Rightarrow k = 30. ∴ k − 10 4 k − 20 = 1 5 ⇒ 4 k − 20 = 5 ( k − 10 ) ⇒ 4 k − 20 = 5 k − 50 ⇒ − 20 + 50 = 5 k − 4 k ⇒ k = 30.
Total students who appeared exam = 5k = 5(30) = 150.
Hence, total number of students appeared for the examination = 150.
Find the angles of a triangle which are in the ratio 5 : 4 : 3.
Answer
Given,
Let the angles of triangle be 5x, 4x and 3x.
We know that,
Sum of angles of a triangle = 180°
⇒ 5 x + 4 x + 3 x = 180 ⇒ 12 x = 180 ⇒ x = 180 12 = 15. \Rightarrow 5x + 4x + 3x = 180 \\[1em] \Rightarrow 12x = 180 \\[1em] \Rightarrow x = \dfrac{180}{12} = 15. ⇒ 5 x + 4 x + 3 x = 180 ⇒ 12 x = 180 ⇒ x = 12 180 = 15.
⇒ 5x = 5(15) = 75°
⇒ 4x = 4(15) = 60°
⇒ 3x = 3(15) = 45°
Hence, the angles of the triangle are 75°, 60° and 45°.
The sides of a triangle are in the ratio 1 2 : 1 3 : 1 4 \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4} 2 1 : 3 1 : 4 1
Answer
Given,
Side1 : Side2 : Side3 = 1 2 : 1 3 : 1 4 \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4} 2 1 : 3 1 : 4 1
= 1 2 × 12 : 1 3 × 12 : 1 4 × 12 \dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 2 1 × 12 : 3 1 × 12 : 4 1 × 12
= 6 : 4 : 3.
Let Side1 = 6a and Side2 = 4a and Side3 = 3a
Sum of sides of triangle = 6a + 4a + 3a = 13a
Given,
Perimeter of triangle = 91 cm
⇒ 13a = 91
⇒ a = 91 13 \dfrac{91}{13} 13 91
⇒ a = 7.
Therefore,
Length of Side1 = 6a = 6(7) = 42 cm
Length of Side2 = 4a = 4(7) = 28 cm
Length of Side3 = 3a = 3(7) = 21 cm
Hence, the lengths of the sides are 42 cm, 28 cm and 21 cm.
In a school, the boys and girls are in the ratio 9 : 5. If there are 425 girls, what is the total number of students in the school?
Answer
Given,
Boys : Girls = 9 : 5
Let number of Boys = 9x and Girls = 5x.
Total number of girls in school = 425
So,
⇒ 5x = 425
⇒ x = 425 5 \dfrac{425}{5} 5 425
⇒ x = 85.
Therefore,
Number of Boys in school = 9x = 9(85) = 765.
Total number of students in school = 765 + 425 = 1190.
Hence, the total number of students in the school = 1190.
Compare the following ratios :
(i) (7 : 9) and (11 : 16)
(ii) (19 : 25) and (17 : 20)
(iii) ( 1 2 : 1 5 ) \Big(\dfrac{1}{2} : \dfrac{1}{5}\Big) ( 2 1 : 5 1 )
Answer
(i) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.
Given,
A : B = 7 : 9 and C : D = 11: 16
L.C.M. of 9 and 16 is 144.
⇒ A B = 7 × 16 9 × 16 = 112 144 ⇒ C D = 11 × 9 16 × 9 = 99 144 ⇒ 112 144 > 99 144 ⇒ 7 9 > 11 16 \Rightarrow \dfrac{A}{B} = \dfrac{7 \times 16}{9 \times 16} = \dfrac{112}{144} \\[1em] \Rightarrow \dfrac{C}{D} = \dfrac{11 \times 9}{16 \times 9} = \dfrac{99}{144} \\[1em] \Rightarrow \dfrac{112}{144} \gt \dfrac{99}{144} \\[1em] \Rightarrow \dfrac{7}{9} \gt \dfrac{11}{16} ⇒ B A = 9 × 16 7 × 16 = 144 112 ⇒ D C = 16 × 9 11 × 9 = 144 99 ⇒ 144 112 > 144 99 ⇒ 9 7 > 16 11
Hence, 7 : 9 > 11 : 16.
(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.
Let A : B = 19 : 25 and C : D = 17 : 20.
L.C.M. of 25 and 20 is 100.
⇒ A B = 19 × 4 25 × 4 = 76 100 ⇒ C D = 17 × 5 20 × 5 = 85 100 ⇒ 76 100 < 85 100 ⇒ 19 25 < 17 20 \Rightarrow \dfrac{A}{B} = \dfrac{19 \times 4}{25 \times 4} = \dfrac{76}{100} \\[1em] \Rightarrow \dfrac{C}{D} = \dfrac{17 \times 5}{20 \times 5} = \dfrac{85}{100} \\[1em] \Rightarrow \dfrac{76}{100} \lt \dfrac{85}{100} \\[1em] \Rightarrow \dfrac{19}{25} \lt \dfrac{17}{20} ⇒ B A = 25 × 4 19 × 4 = 100 76 ⇒ D C = 20 × 5 17 × 5 = 100 85 ⇒ 100 76 < 100 85 ⇒ 25 19 < 20 17
Hence, 19 : 25 < 17 : 20.
(ii) To compare 2 ratios, the consequent of the first ratio and 2nd ratio must be made equal.
Let A : B = ( 1 2 : 1 5 ) \Big(\dfrac{1}{2} : \dfrac{1}{5}\Big) ( 2 1 : 5 1 )
First simplifying A : B,
L.C.M of 2 and 5 is 10.
⇒ ( 1 2 × 10 : 1 5 × 10 ) ⇒ 5 : 2 \Rightarrow \Big(\dfrac{1}{2} \times 10 : \dfrac{1}{5} \times 10\Big) \\[1em] \Rightarrow 5:2 ⇒ ( 2 1 × 10 : 5 1 × 10 ) ⇒ 5 : 2
Since both ratios are same.
Hence, ( 1 2 : 1 5 ) \Big(\dfrac{1}{2} : \dfrac{1}{5}\Big) ( 2 1 : 5 1 )
Arrange the following ratios in descending order of magnitudes :
(i) (5 : 6), (8 : 9), (13 : 18) and (19 : 24)
(ii) (6 : 7), (13 : 14), (19 : 21) and (23 : 28)
(iii) (7 : 12), (9 : 16), (13 : 20) and (5 : 8)
Answer
(i) Given,
(5 : 6), (8 : 9), (13 : 18) and (19 : 24)
We convert them into equivalent like fractions.
L.C.M of 6, 9, 18, 24 is 72.
⇒ 5 × 12 6 × 12 = 60 72 ⇒ 8 × 8 9 × 8 = 64 72 ⇒ 13 × 4 18 × 4 = 52 72 ⇒ 19 × 3 24 × 3 = 57 72 ⇒ 64 72 > 60 72 > 57 72 > 52 72 \Rightarrow \dfrac{5 \times 12}{6 \times 12} = \dfrac{60}{72} \\[1em] \Rightarrow \dfrac{8 \times 8}{9 \times 8} = \dfrac{64}{72} \\[1em] \Rightarrow \dfrac{13 \times 4}{18 \times 4} = \dfrac{52}{72} \\[1em] \Rightarrow \dfrac{19 \times 3}{24 \times 3} = \dfrac{57}{72} \\[1em] \Rightarrow \dfrac{64}{72} \gt \dfrac{60}{72} \gt \dfrac{57}{72} \gt \dfrac{52}{72} ⇒ 6 × 12 5 × 12 = 72 60 ⇒ 9 × 8 8 × 8 = 72 64 ⇒ 18 × 4 13 × 4 = 72 52 ⇒ 24 × 3 19 × 3 = 72 57 ⇒ 72 64 > 72 60 > 72 57 > 72 52
8 : 9 > 5 : 6 > 19 : 24 > 13 : 18
Hence, the ratios in descending order are (8 : 9) > (5 : 6) > (19 : 24) > (13 : 18).
(ii) Given,
(6 : 7), (13 : 14), (19 : 21) and (23 : 28)
We convert them into equivalent like fractions.
L.C.M of 7, 14, 21, 28 is 84
⇒ 6 × 12 7 × 12 = 72 84 ⇒ 13 × 6 14 × 6 = 78 84 ⇒ 19 × 4 21 × 4 = 76 84 ⇒ 23 × 3 28 × 3 = 69 84 ⇒ 78 84 > 76 84 > 72 84 > 69 84 \Rightarrow \dfrac{6 \times 12}{7 \times 12} = \dfrac{72}{84} \\[1em] \Rightarrow \dfrac{13 \times 6}{14 \times 6} = \dfrac{78}{84} \\[1em] \Rightarrow \dfrac{19 \times 4}{21 \times 4} = \dfrac{76}{84} \\[1em] \Rightarrow \dfrac{23 \times 3}{28 \times 3} = \dfrac{69}{84} \\[1em] \Rightarrow \dfrac{78}{84} \gt \dfrac{76}{84} \gt \dfrac{72}{84} \gt \dfrac{69}{84} ⇒ 7 × 12 6 × 12 = 84 72 ⇒ 14 × 6 13 × 6 = 84 78 ⇒ 21 × 4 19 × 4 = 84 76 ⇒ 28 × 3 23 × 3 = 84 69 ⇒ 84 78 > 84 76 > 84 72 > 84 69
(13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).
Hence, the ratios in descending order are (13 : 14) > (19 : 21) > (6 : 7) > (23 : 28).
(iii) Given,
(7 : 12), (9 : 16), (13 : 20) and (5 : 8)
We convert them into equivalent like fractions.
L.C.M of 12, 16, 20, 8 is 240.
⇒ 7 × 20 12 × 20 = 140 240 ⇒ 9 × 15 16 × 15 = 135 240 ⇒ 13 × 12 20 × 12 = 156 240 ⇒ 5 × 30 8 × 30 = 150 240 ⇒ 156 240 > 150 240 > 140 240 > 135 240 \Rightarrow \dfrac{7 \times 20}{12 \times 20} = \dfrac{140}{240} \\[1em] \Rightarrow \dfrac{9 \times 15}{16 \times 15} = \dfrac{135}{240} \\[1em] \Rightarrow \dfrac{13 \times 12}{20 \times 12} = \dfrac{156}{240} \\[1em] \Rightarrow \dfrac{5 \times 30}{8 \times 30} = \dfrac{150}{240} \\[1em] \Rightarrow \dfrac{156}{240} \gt \dfrac{150}{240} \gt \dfrac{140}{240} \gt \dfrac{135}{240} ⇒ 12 × 20 7 × 20 = 240 140 ⇒ 16 × 15 9 × 15 = 240 135 ⇒ 20 × 12 13 × 12 = 240 156 ⇒ 8 × 30 5 × 30 = 240 150 ⇒ 240 156 > 240 150 > 240 140 > 240 135
(13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).
Hence, the ratios in descending order are (13 : 20) > (5 : 8) > (7 : 12) > (9 : 16).
Arrange the following ratios in ascending order of magnitudes :
(i) (4 : 9), (6 : 11), (7 : 13) and (27 : 50)
(ii) (2 : 3), (8 : 15), (11 : 12) and (7 : 16)
(iii) (3 : 5), (4 : 9), (5 : 11) and (10 : 17)
Answer
(i) Given,
(4 : 9), (6 : 11), (7 : 13) and (27 : 50)
We convert them into decimals.
⇒ 4 9 \dfrac{4}{9} 9 4
⇒ 6 11 \dfrac{6}{11} 11 6
⇒ 7 13 \dfrac{7}{13} 13 7
⇒ 27 50 \dfrac{27}{50} 50 27
⇒ 0.444 < 0.5385 < 0.5400 < 0.5455
⇒ (4 : 9) < (7 : 13) < (27 : 50) < (6 : 11).
Hence, the ratios in ascending order are (4 : 9), (7 : 13), (27 : 50), (6 : 11).
(ii) Given,
(2 : 3), (8 : 15), (11 : 12) and (7 : 16)
We convert them into decimals.
2 3 \dfrac{2}{3} 3 2
8 15 \dfrac{8}{15} 15 8
11 12 \dfrac{11}{12} 12 11
7 16 \dfrac{7}{16} 16 7
0.4375 < 0.533 < 0.667 < 0.917
(7 : 16) < (8 : 15) < (2 : 3) < (11 : 12)
Hence, the ratios in ascending order are (7 : 16), (8 : 15), (2 : 3), (11 : 12).
(iii) Given,
(3 : 5), (4 : 9), (5 : 11) and (10 : 17)
We convert them into decimals.
3 5 \dfrac{3}{5} 5 3
4 9 \dfrac{4}{9} 9 4
5 11 \dfrac{5}{11} 11 5
10 17 \dfrac{10}{17} 17 10
⇒ 0.444 < 0.455 < 0.588 < 0.600
⇒ (4 : 9) < (5 : 11) < (10 : 17) < (3 : 5)
Hence, the ratios in ascending order are (4 : 9), (5 : 11), (10 : 17), (3 : 5).
If (3a + 2b) : (5a + 3b) = 18 : 29, find (a : b).
Answer
Given,
(3a + 2b) : (5a + 3b) = 18 : 29
Solving,
3 a + 2 b 5 a + 3 b = 18 29 \dfrac{3a + 2b}{5a + 3b} = \dfrac{18}{29} 5 a + 3 b 3 a + 2 b = 29 18
⇒ 29(3a + 2b) = 18(5a + 3b)
⇒ 87a + 58b = 90a + 54b
⇒ 58b - 54b = 90a - 87a
⇒ 4b = 3a
⇒ a b = 4 3 \dfrac{a}{b} = \dfrac{4}{3} b a = 3 4
⇒ a : b = 4 : 3
Hence, a : b = 4 : 3.
Find x, when :
(i) 3 : 4 :: 2.4 : x
(ii) 1 : 3 :: x : 7
(iii) x : 1.5 :: 3 : 5
Answer
(i) Given,
3 : 4 :: 2.4 : x
Solving for x,
⇒ 3 4 = 2.4 x ⇒ 3 × x = 4 × 2.4 ⇒ 3 x = 9.6 ⇒ x = 9.6 3 ⇒ x = 3.2 \Rightarrow \dfrac{3}{4} = \dfrac{2.4}{x} \\[1em] \Rightarrow 3 \times x = 4 \times 2.4 \\[1em] \Rightarrow 3x = 9.6 \\[1em] \Rightarrow x = \dfrac{9.6}{3} \\[1em] \Rightarrow x = 3.2 ⇒ 4 3 = x 2.4 ⇒ 3 × x = 4 × 2.4 ⇒ 3 x = 9.6 ⇒ x = 3 9.6 ⇒ x = 3.2
Hence, x = 3.2
(ii) Given,
1 : 3 :: x : 7
Solving for x,
⇒ 1 3 = x 7 ⇒ x = 7 3 = 2 1 3 . \Rightarrow \dfrac{1}{3} = \dfrac{x}{7} \\[1em] \Rightarrow x = \dfrac{7}{3} = 2\dfrac{1}{3}. ⇒ 3 1 = 7 x ⇒ x = 3 7 = 2 3 1 .
Hence, x = 2 1 3 2\dfrac{1}{3} 2 3 1
(iii) Given,
x : 1.5 :: 3 : 5
Solving for x,
⇒ x 1.5 = 3 5 ⇒ 5 × x = 1.5 × 3 ⇒ x = 4.5 5 ⇒ x = 0.9 \Rightarrow \dfrac{x}{1.5} = \dfrac{3}{5} \\[1em] \Rightarrow 5 \times x = 1.5 \times 3 \\[1em] \Rightarrow x = \dfrac{4.5}{5} \\[1em] \Rightarrow x = 0.9 ⇒ 1.5 x = 5 3 ⇒ 5 × x = 1.5 × 3 ⇒ x = 5 4.5 ⇒ x = 0.9
Hence, x = 0.9
Find the fourth proportional to :
(i) 3, 8 and 21
(ii) 1.4, 3.2 and 7
(iii) 1.5, 4.5 and 3.6
(iv) a2 , ab and b2
(v) (a2 − ab + b2 ), (a3 + b3 ) and (a − b)
Answer
(i) Given,
3, 8 and 21
Let the fourth proportional to 3, 8 and 21 be x,
⇒ 3 : 8 = 21 : x
⇒ 3 8 = 21 x ⇒ x = 21 × 8 3 ⇒ x = 168 3 ⇒ x = 56. \Rightarrow \dfrac{3}{8} = \dfrac{21}{x} \\[1em] \Rightarrow x = \dfrac{21 \times 8}{3} \\[1em] \Rightarrow x = \dfrac{168}{3} \\[1em] \Rightarrow x = 56. ⇒ 8 3 = x 21 ⇒ x = 3 21 × 8 ⇒ x = 3 168 ⇒ x = 56.
Hence, the fourth proportional is 56.
(ii) Given,
1.4, 3.2 and 7
Let the fourth proportional to 1.4, 3.2 and 7 be x,
⇒ 1.4 : 3.2 = 7 : x
⇒ 1.4 3.2 = 7 x ⇒ x = 7 × 3.2 1.4 ⇒ x = 22.4 1.4 ⇒ x = 16. \Rightarrow \dfrac{1.4}{3.2} = \dfrac{7}{x} \\[1em] \Rightarrow x = \dfrac{7 \times 3.2}{1.4} \\[1em] \Rightarrow x = \dfrac{22.4}{1.4} \\[1em] \Rightarrow x = 16. ⇒ 3.2 1.4 = x 7 ⇒ x = 1.4 7 × 3.2 ⇒ x = 1.4 22.4 ⇒ x = 16.
Hence, the fourth proportional is 16.
(iii) Given,
1.5, 4.5 and 3.6
Let the fourth proportional to 1.5, 4.5 and 3.6 be x,
⇒ 1.5 : 4.5 = 3.6 : x
⇒ 1.5 4.5 = 3.6 x ⇒ x = 4.5 × 3.6 1.5 ⇒ x = 3 × 3.6 ⇒ x = 10.8. \Rightarrow \dfrac{1.5}{4.5} = \dfrac{3.6}{x} \\[1em] \Rightarrow x = \dfrac{4.5 \times 3.6}{1.5} \\[1em] \Rightarrow x = 3 \times 3.6 \\[1em] \Rightarrow x = 10.8. ⇒ 4.5 1.5 = x 3.6 ⇒ x = 1.5 4.5 × 3.6 ⇒ x = 3 × 3.6 ⇒ x = 10.8.
Hence, the fourth proportional is 10.8.
(iv) Given,
a2 , ab and b2
Let the fourth proportional to a2 , ab and b2 be x,
⇒ a2 : ab = b2 : x
⇒ a 2 a b = b 2 x ⇒ x = a b × b 2 a 2 ⇒ x = b 3 a . \Rightarrow \dfrac{a^2}{ab} = \dfrac{b^2}{x} \\[1em] \Rightarrow x = \dfrac{ab \times b^2}{a^2} \\[1em] \Rightarrow x = \dfrac{b^3}{a}. ⇒ ab a 2 = x b 2 ⇒ x = a 2 ab × b 2 ⇒ x = a b 3 .
Hence, the fourth proportional is b 3 a \dfrac{b^3}{a} a b 3
(v) Given,
(a2 − ab + b2 ), (a3 + b3 ) and (a − b)
Let the fourth proportional to (a2 − ab + b2 ), (a3 + b3 ) and (a − b) be x,
⇒ a2 : ab = b2 :x
⇒ a 2 − a b + b 2 a 3 + b 3 = a − b x ⇒ x = ( a 3 + b 3 ) ( a − b ) a 2 − a b + b 2 ⇒ x = ( a + b ) ( a 2 − a b + b 2 ) ( a − b ) a 2 − a b + b 2 ⇒ x = ( a + b ) ( a − b ) ⇒ x = a 2 − b 2 . \Rightarrow \dfrac{a^2 - ab + b^2}{a^3 + b^3} = \dfrac{a - b}{x} \\[1em] \Rightarrow x = \dfrac{(a^3 + b^3)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = \dfrac{(a + b)(a^2 - ab + b^2)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = (a + b)(a - b) \\[1em] \Rightarrow x = a^2 - b^2. ⇒ a 3 + b 3 a 2 − ab + b 2 = x a − b ⇒ x = a 2 − ab + b 2 ( a 3 + b 3 ) ( a − b ) ⇒ x = a 2 − ab + b 2 ( a + b ) ( a 2 − ab + b 2 ) ( a − b ) ⇒ x = ( a + b ) ( a − b ) ⇒ x = a 2 − b 2 .
Hence, the fourth proportional is a2 - b2 .
Find the third proportional to :
(i) 9 and 6
(ii) 2 2 3 2\dfrac{2}{3} 2 3 2
(iii) 1.6 and 2.4
(iv) (2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
(v) ( a b + b a ) \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) ( b a + a b ) a 2 + b 2 \sqrt{a^{2} + b^{2}} a 2 + b 2
Answer
(i) Given,
9 and 6
Let third proportional to 9 and 6 be x.
⇒ 9 : 6 = 6 : x
⇒ 9 6 = 6 x \dfrac{9}{6} = \dfrac{6}{x} 6 9 = x 6
⇒ x = 6 2 9 = 36 4 \dfrac{6^2}{9} = \dfrac{36}{4} 9 6 2 = 4 36
⇒ x = 4.
Hence, the third proportional is 4.
(ii) Given,
2 2 3 2\dfrac{2}{3} 2 3 2
Let third proportional to 8 3 \dfrac{8}{3} 3 8
8 3 \dfrac{8}{3} 3 8
⇒ 4 x = 8 3 4 ⇒ 4 x = 2 3 ⇒ x = 3 2 × 4 ⇒ x = 6. \Rightarrow \dfrac{4}{x} = \dfrac{\dfrac{8}{3}}{4} \\[1em] \Rightarrow \dfrac{4}{x} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow x = 6. ⇒ x 4 = 4 3 8 ⇒ x 4 = 3 2 ⇒ x = 2 3 × 4 ⇒ x = 6.
Hence, the third proportional is 6.
(iii) Given,
1.6 and 2.4
Let third proportional to 1.6 and 2.4 be x.
1.6 : 2.4 = 2.4 : x
⇒ 1.6 2.4 = 2.4 x \dfrac{1.6}{2.4} = \dfrac{2.4}{x} 2.4 1.6 = x 2.4
⇒ x = ( 2.4 ) 2 1.6 = 5.76 1.6 \dfrac{(2.4)^2}{1.6} = \dfrac{5.76}{1.6} 1.6 ( 2.4 ) 2 = 1.6 5.76
⇒ x = 3.6
Hence, the third proportional is 3.6.
(iv) Given,
(2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
Let third proportional to (2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
( 2 + 3 ) : ( 5 + 4 3 ) = ( 5 + 4 3 ) : x (2 + \sqrt{3}) : (5 + 4 \sqrt{3}) = (5 + 4\sqrt{3}) : x ( 2 + 3 ) : ( 5 + 4 3 ) = ( 5 + 4 3 ) : x
Thus,
⇒ ( 2 + 3 ) ( 5 + 4 3 ) = ( 5 + 4 3 ) x ⇒ x = ( 5 + 4 3 ) 2 ( 2 + 3 ) = 5 2 + 2 ( 5 ) ( 4 3 ) + ( 4 3 ) 2 ( 2 + 3 ) = 25 + 40 3 + 16 × 3 ( 2 + 3 ) = 25 + 40 3 + 48 ( 2 + 3 ) = 73 + 40 3 ( 2 + 3 ) \Rightarrow \dfrac{(2 + \sqrt{3})}{(5 + 4 \sqrt{3})} = \dfrac{(5 + 4 \sqrt{3})}{x} \\[1em] \Rightarrow x = \dfrac{(5 + 4 \sqrt{3})^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{5^2 + 2(5)(4\sqrt3) + (4\sqrt3)^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 16 \times 3}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 48}{(2 + \sqrt{3})} \\[1em] = \dfrac{73 + 40\sqrt3}{(2 + \sqrt{3})} ⇒ ( 5 + 4 3 ) ( 2 + 3 ) = x ( 5 + 4 3 ) ⇒ x = ( 2 + 3 ) ( 5 + 4 3 ) 2 = ( 2 + 3 ) 5 2 + 2 ( 5 ) ( 4 3 ) + ( 4 3 ) 2 = ( 2 + 3 ) 25 + 40 3 + 16 × 3 = ( 2 + 3 ) 25 + 40 3 + 48 = ( 2 + 3 ) 73 + 40 3
Multiplying numerator and denominator by ( 2 − 3 ) (2 - \sqrt{3}) ( 2 − 3 )
= ( 73 + 40 3 ) ( 2 − 3 ) ( 2 + 3 ) ( 2 − 3 ) = 146 − 73 3 + 80 3 − 40 ( 3 ) 2 2 2 − ( 3 ) 2 = 146 + 7 3 − 120 4 − 3 = 26 + 7 3 . = \dfrac{(73 + 40\sqrt3)(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \\[1em] = \dfrac{146 - 73\sqrt3 + 80\sqrt{3} - 40(\sqrt3)^2}{2^2 - (\sqrt{3})^2} \\[1em] = \dfrac{146 + 7\sqrt3 - 120}{4 - 3} \\[1em] = 26 + 7\sqrt3. = ( 2 + 3 ) ( 2 − 3 ) ( 73 + 40 3 ) ( 2 − 3 ) = 2 2 − ( 3 ) 2 146 − 73 3 + 80 3 − 40 ( 3 ) 2 = 4 − 3 146 + 7 3 − 120 = 26 + 7 3 .
Hence, the third proportional is 26 + 7 3 26 + 7\sqrt3 26 + 7 3
(v) Given,
( a b + b a ) \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) ( b a + a b ) a 2 + b 2 \sqrt{a^{2} + b^{2}} a 2 + b 2
Let third proportional to ( a b + b a ) and a 2 + b 2 \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) \text{ and } \sqrt{a^{2} + b^{2}} ( b a + a b ) and a 2 + b 2
( a b + b a ) : a 2 + b 2 = a 2 + b 2 : x \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big): \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + b^{2}}:x ( b a + a b ) : a 2 + b 2 = a 2 + b 2 : x
( a b + b a ) a 2 + b 2 = a 2 + b 2 x x = ( a 2 + b 2 ) 2 ( a b + b a ) = a 2 + b 2 a 2 + b 2 a b = ( a 2 + b 2 ) × a b a 2 + b 2 = a b . \dfrac{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)}{\sqrt{a^{2} + b^{2}}} = \dfrac{\sqrt{a^{2} + b^{2}}}{x} \\[1em] x = \dfrac{(\sqrt{a^2 + b^2})^2}{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)} \\[1em] = \dfrac{a^2 + b^2}{\dfrac{a^2 + b^2}{ab}} \\[1em] = (a^2 + b^2) \times \dfrac{ab}{a^2 + b^2} \\[1em] = ab. a 2 + b 2 ( b a + a b ) = x a 2 + b 2 x = ( b a + a b ) ( a 2 + b 2 ) 2 = ab a 2 + b 2 a 2 + b 2 = ( a 2 + b 2 ) × a 2 + b 2 ab = ab .
Hence, the third proportional is ab.
Find the mean proportion between :
(i) 28 and 63
(ii) 2.5 and 0.9
(iii) 6.25 and 1.6
(iv) ( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
(v) 6 + 3 3 \sqrt{3} 3 3 \sqrt{3} 3
Answer
(i) Given,
28 and 63
Let mean proportional between 28 and 63 be x.
28 : x :: x : 63
⇒ 28 x = x 63 ⇒ x 2 = 28 × 63 ⇒ x 2 = 1764 ⇒ x = 1764 = 42 \Rightarrow \dfrac{28}{x} = \dfrac{x}{63} \\[1em] \Rightarrow x^2 = 28 \times 63 \\[1em] \Rightarrow x^2 = 1764 \\[1em] \Rightarrow x = \sqrt{1764} = 42 ⇒ x 28 = 63 x ⇒ x 2 = 28 × 63 ⇒ x 2 = 1764 ⇒ x = 1764 = 42
Hence, the mean proportional is 42.
(ii) Given,
2.5 and 0.9
Let mean proportional between 2.5 and 0.9 be x.
2.5 : x :: x : 0.9
⇒ 2.5 x = x 0.9 ⇒ x 2 = 2.5 × 0.9 ⇒ x 2 = 2.25 ⇒ x = 2.25 = 1.5 \Rightarrow \dfrac{2.5}{x} = \dfrac{x}{0.9} \\[1em] \Rightarrow x^2 = 2.5 \times 0.9 \\[1em] \Rightarrow x^2 = 2.25 \\[1em] \Rightarrow x = \sqrt{2.25} = 1.5 ⇒ x 2.5 = 0.9 x ⇒ x 2 = 2.5 × 0.9 ⇒ x 2 = 2.25 ⇒ x = 2.25 = 1.5
Hence, the mean proportional is 1.5.
(iii) Given,
6.25 and 1.6
Let mean proportional between 6.25 and 1.6 be x.
6.25 : x :: x : 1.6
⇒ 6.25 x = x 1.6 ⇒ x 2 = 6.25 × 1.6 ⇒ x 2 = 10 ⇒ x = 10 \Rightarrow \dfrac{6.25}{x} = \dfrac{x}{1.6} \\[1em] \Rightarrow x^2 = 6.25 \times 1.6 \\[1em] \Rightarrow x^2 = 10 \\[1em] \Rightarrow x = \sqrt{10} ⇒ x 6.25 = 1.6 x ⇒ x 2 = 6.25 × 1.6 ⇒ x 2 = 10 ⇒ x = 10
Hence, the mean proportional is 10 \sqrt{10} 10
(iv) Given,
( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
Let mean proportional between ( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
( 26 − 17 ) : x : : x : ( 26 + 17 ) \Big(\sqrt{26} - \sqrt{17}\Big):x::x:\Big(\sqrt{26} + \sqrt{17}\Big) ( 26 − 17 ) : x :: x : ( 26 + 17 )
⇒ 26 − 17 x = x 26 + 17 ⇒ x 2 = ( 26 − 17 ) × ( 26 + 17 ) \Rightarrow \dfrac{\sqrt{26} - \sqrt{17}}{x} = \dfrac{x}{\sqrt{26} + \sqrt{17}} \\[1em] \Rightarrow x^2 = (\sqrt{26} - \sqrt{17}) \times (\sqrt{26} + \sqrt{17}) \\[1em] ⇒ x 26 − 17 = 26 + 17 x ⇒ x 2 = ( 26 − 17 ) × ( 26 + 17 )
Hence, the mean proportional is 3.
(v) Given,
6 + 3 3 3\sqrt{3} 3 3 4 3 4\sqrt{3} 4 3
Let mean proportion between 6 + 3 3 3\sqrt{3} 3 3 4 3 4\sqrt{3} 4 3
6 + 3 3 : x : : x : 8 − 4 3 \sqrt{3} : x :: x : 8 − 4 \sqrt{3} 3 : x :: x : 8 − 4 3
⇒ 6 + 3 3 x = x 8 − 4 3 ⇒ x 2 = ( 6 + 3 3 ) × ( 8 − 4 3 ) ⇒ x 2 = 48 − 24 3 + 24 3 − 36 ⇒ x 2 = 12 ⇒ x = 2 3 \Rightarrow \dfrac{6 + 3\sqrt{3}}{x} = \dfrac{x}{8 − 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3}) \times (8 − 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt3 ⇒ x 6 + 3 3 = 8 − 4 3 x ⇒ x 2 = ( 6 + 3 3 ) × ( 8 − 4 3 ) ⇒ x 2 = 48 − 24 3 + 24 3 − 36 ⇒ x 2 = 12 ⇒ x = 2 3
Hence, the mean proportional is 2 3 2\sqrt3 2 3
6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.
Answer
Let two numbers be x and y.
Given,
6 is mean proportion between x and y,
∴ x 6 = 6 y ⇒ x y = 36 . . . . . ( 1 ) \therefore \dfrac{x}{6} = \dfrac{6}{y} \\[1em] \Rightarrow xy = 36 \space .....(1) ∴ 6 x = y 6 ⇒ x y = 36 ..... ( 1 )
Given,
48 is third proportional to x and y,
∴ x y = y 48 ⇒ y 2 = 48 x ⇒ x = y 2 48 . . . . . ( 2 ) \therefore \dfrac{x}{y} = \dfrac{y}{48} \\[1em] \Rightarrow y^2 = 48x \\[1em] \Rightarrow x = \dfrac{y^2}{48} \space .....(2) ∴ y x = 48 y ⇒ y 2 = 48 x ⇒ x = 48 y 2 ..... ( 2 )
Substituting value of x from equation (2) in (1) we get,
⇒ y 2 48 . y = 36 ⇒ y 3 = 36 × 48 ⇒ y 3 = 1728 ⇒ y = 1728 3 ⇒ y = 12. \Rightarrow \dfrac{y^2}{48}.y = 36 \\[1em] \Rightarrow y^3 = 36 \times 48 \\[1em] \Rightarrow y^3 = 1728 \\[1em] \Rightarrow y = \sqrt[3]{1728} \\[1em] \Rightarrow y = 12. ⇒ 48 y 2 . y = 36 ⇒ y 3 = 36 × 48 ⇒ y 3 = 1728 ⇒ y = 3 1728 ⇒ y = 12.
Substituting value of y in equation (2), we get :
x = 12 2 48 = 144 48 x = \dfrac{12^2}{48} = \dfrac{144}{48} x = 48 1 2 2 = 48 144
Hence, numbers are 3 and 12.
What least number must be added to each of the numbers 5, 11, 19 and 37, so that the resulting numbers are proportional.
Answer
Let least number to be added to numbers be x.
∴ 5 + x : 11 + x :: 19 + x : 37 + x
⇒ 5 + x 11 + x = 19 + x 37 + x ⇒ ( 5 + x ) ( 37 + x ) = ( 19 + x ) ( 11 + x ) ⇒ 185 + 5 x + 37 x + x 2 = 209 + 19 x + 11 x + x 2 ⇒ x 2 + 42 x + 185 = x 2 + 30 x + 209 ⇒ x 2 − x 2 + 42 x − 30 x = 209 − 185 ⇒ 12 x = 24 ⇒ x = 2. \Rightarrow \dfrac{5 + x}{11 + x} = \dfrac{19 + x}{37 + x} \\[1em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[1em] \Rightarrow 185 + 5x + 37x + x^2 = 209 + 19x + 11x + x^2 \\[1em] \Rightarrow x^2 + 42x + 185 = x^2 + 30x + 209 \\[1em] \Rightarrow x^2 - x^2 + 42x - 30x = 209 - 185 \\[1em] \Rightarrow 12x = 24 \\[1em] \Rightarrow x = 2. ⇒ 11 + x 5 + x = 37 + x 19 + x ⇒ ( 5 + x ) ( 37 + x ) = ( 19 + x ) ( 11 + x ) ⇒ 185 + 5 x + 37 x + x 2 = 209 + 19 x + 11 x + x 2 ⇒ x 2 + 42 x + 185 = x 2 + 30 x + 209 ⇒ x 2 − x 2 + 42 x − 30 x = 209 − 185 ⇒ 12 x = 24 ⇒ x = 2.
Hence, least number to be added to make numbers proportional is 2.
What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion ?
Answer
Let the number to be added to numbers be x.
∴ 4 + x : 6 + x :: 8 + x : 11 + x
⇒ 4 + x 6 + x = 8 + x 11 + x ⇒ ( 4 + x ) ( 11 + x ) = ( 8 + x ) ( 6 + x ) ⇒ 44 + 4 x + 11 x + x 2 = 48 + 8 x + 6 x + x 2 ⇒ x 2 + 15 x + 44 = x 2 + 14 x + 48 ⇒ x 2 − x 2 + 15 x − 14 x = 48 − 44 ⇒ x = 4. \Rightarrow \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x)(11 + x) = (8 + x)(6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2 \\[1em] \Rightarrow x^2 + 15x + 44 = x^2 + 14x + 48 \\[1em] \Rightarrow x^2 - x^2 + 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4. ⇒ 6 + x 4 + x = 11 + x 8 + x ⇒ ( 4 + x ) ( 11 + x ) = ( 8 + x ) ( 6 + x ) ⇒ 44 + 4 x + 11 x + x 2 = 48 + 8 x + 6 x + x 2 ⇒ x 2 + 15 x + 44 = x 2 + 14 x + 48 ⇒ x 2 − x 2 + 15 x − 14 x = 48 − 44 ⇒ x = 4.
Hence, the number to be added to make numbers proportional is 4.
What least number must be subtracted from each of the numbers 23, 30, 57 and 78, so that the remainders are in proportion?
Answer
Let least number to be subtracted to numbers be x.
∴ 23 - x : 30 - x :: 57 - x : 78 - x
⇒ 23 − x 30 − x = 57 − x 78 − x ⇒ ( 23 − x ) ( 78 − x ) = ( 57 − x ) ( 30 − x ) ⇒ 1794 − 23 x − 78 x + x 2 = 1710 − 57 x − 30 x + x 2 ⇒ x 2 − 101 x + 1794 = x 2 − 87 x + 1710 ⇒ x 2 − x 2 − 101 x + 87 x = 1710 − 1794 ⇒ − 14 x = − 84 ⇒ x = − 84 − 14 ⇒ x = 6. \Rightarrow \dfrac{23 - x}{30 - x} = \dfrac{57 - x}{78 - x} \\[1em] \Rightarrow (23 - x)(78 - x) = (57 - x)(30 - x) \\[1em] \Rightarrow 1794 - 23x - 78x + x^2 = 1710 - 57x - 30x + x^2 \\[1em] \Rightarrow x^2 - 101x + 1794 = x^2 - 87x + 1710 \\[1em] \Rightarrow x^2 - x^2 - 101x + 87x = 1710 - 1794 \\[1em] \Rightarrow -14x = -84 \\[1em] \Rightarrow x = \dfrac{-84}{-14} \\[1em] \Rightarrow x = 6. ⇒ 30 − x 23 − x = 78 − x 57 − x ⇒ ( 23 − x ) ( 78 − x ) = ( 57 − x ) ( 30 − x ) ⇒ 1794 − 23 x − 78 x + x 2 = 1710 − 57 x − 30 x + x 2 ⇒ x 2 − 101 x + 1794 = x 2 − 87 x + 1710 ⇒ x 2 − x 2 − 101 x + 87 x = 1710 − 1794 ⇒ − 14 x = − 84 ⇒ x = − 14 − 84 ⇒ x = 6.
Hence, least number to be subtracted to make numbers proportional is 6.
If (x − 2), (x + 2), (2x + 1) and (2x + 19) are in proportion, find the value of x.
Answer
Let,
∴ x - 2 : x + 2 :: 2x + 1 : 2x + 19
⇒ x − 2 x + 2 = 2 x + 1 2 x + 19 ⇒ ( x − 2 ) ( 2 x + 19 ) = ( 2 x + 1 ) ( x + 2 ) ⇒ 2 x 2 + 19 x − 4 x − 38 = 2 x 2 + 4 x + x + 2 ⇒ 2 x 2 + 15 x − 38 = 2 x 2 + 5 x + 2 ⇒ 2 x 2 − 2 x 2 + 15 x − 5 x = 2 + 38 ⇒ 10 x = 40 ⇒ x = 4. \Rightarrow \dfrac{x - 2}{x + 2} = \dfrac{2x + 1}{2x + 19} \\[1em] \Rightarrow (x - 2)(2x + 19) = (2x + 1)(x + 2) \\[1em] \Rightarrow 2x^2 + 19x - 4x - 38 = 2x^2 + 4x + x + 2 \\[1em] \Rightarrow 2x^2 + 15x - 38 = 2x^2 + 5x + 2 \\[1em] \Rightarrow 2x^2 - 2x^2 + 15x - 5x = 2 + 38 \\[1em] \Rightarrow 10x = 40 \\[1em] \Rightarrow x = 4. ⇒ x + 2 x − 2 = 2 x + 19 2 x + 1 ⇒ ( x − 2 ) ( 2 x + 19 ) = ( 2 x + 1 ) ( x + 2 ) ⇒ 2 x 2 + 19 x − 4 x − 38 = 2 x 2 + 4 x + x + 2 ⇒ 2 x 2 + 15 x − 38 = 2 x 2 + 5 x + 2 ⇒ 2 x 2 − 2 x 2 + 15 x − 5 x = 2 + 38 ⇒ 10 x = 40 ⇒ x = 4.
Hence, the value of x = 4.
The following numbers, K + 3, K + 2, 3K − 7 and 2K − 3 are in proportion. Find the value of K.
Answer
Let,
∴ K + 3 : K + 2 :: 3K − 7 : 2K − 3
⇒ K + 3 K + 2 = 3 K − 7 2 K − 3 ⇒ ( K + 3 ) ( 2 K − 3 ) = ( 3 K − 7 ) ( K + 2 ) ⇒ 2 K 2 − 3 K + 6 k − 9 = 3 K 2 + 6 K − 7 K − 14 ⇒ 2 K 2 + 3 K − 9 = 3 K 2 − K − 14 ⇒ 0 = 3 K 2 − K − 14 − 2 K 2 − 3 K + 9 ⇒ K 2 − 4 K − 5 = 0 ⇒ K 2 + 1 K − 5 K − 5 = 0 ⇒ K ( K + 1 ) − 5 ( K + 1 ) = 0 ⇒ ( K − 5 ) ( K + 1 ) = 0 ⇒ ( K − 5 ) = 0 or ( K + 1 ) = 0 [Using Zero - product rule] ⇒ K = 5 or K = − 1 \Rightarrow \dfrac{K + 3}{K + 2} = \dfrac{3K - 7}{2K - 3} \\[1em] \Rightarrow (K + 3)(2K - 3) = (3K - 7)(K + 2) \\[1em] \Rightarrow 2K^2 - 3K + 6k - 9 = 3K^2 + 6K - 7K - 14 \\[1em] \Rightarrow 2K^2 + 3K - 9 = 3K^2 - K - 14 \\[1em] \Rightarrow 0 = 3K^2 - K - 14 - 2K^2 - 3K + 9 \\[1em] \Rightarrow K^2 - 4K - 5 = 0 \\[1em] \Rightarrow K^2 + 1K - 5K - 5 = 0 \\[1em] \Rightarrow K(K + 1) - 5(K + 1) = 0 \\[1em] \Rightarrow (K - 5)(K + 1) = 0 \\[1em] \Rightarrow (K - 5) = 0 \text{ or }(K + 1) = 0 \text{[Using Zero - product rule]}\\[1em] \Rightarrow K = 5 \text{ or } K = - 1 ⇒ K + 2 K + 3 = 2 K − 3 3 K − 7 ⇒ ( K + 3 ) ( 2 K − 3 ) = ( 3 K − 7 ) ( K + 2 ) ⇒ 2 K 2 − 3 K + 6 k − 9 = 3 K 2 + 6 K − 7 K − 14 ⇒ 2 K 2 + 3 K − 9 = 3 K 2 − K − 14 ⇒ 0 = 3 K 2 − K − 14 − 2 K 2 − 3 K + 9 ⇒ K 2 − 4 K − 5 = 0 ⇒ K 2 + 1 K − 5 K − 5 = 0 ⇒ K ( K + 1 ) − 5 ( K + 1 ) = 0 ⇒ ( K − 5 ) ( K + 1 ) = 0 ⇒ ( K − 5 ) = 0 or ( K + 1 ) = 0 [Using Zero - product rule] ⇒ K = 5 or K = − 1
Hence, K = 5 or K = −1.
If (x + 5) is the geometric mean between (x + 2) and (x + 9), find the value of x.
Answer
Given, x + 5 is G.M. between x + 2 and x + 9.
∴ x + 2 x + 5 = x + 5 x + 9 \therefore \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} ∴ x + 5 x + 2 = x + 9 x + 5
⇒ (x + 5)2 = (x + 2)(x + 9)
⇒ x2 + 10x + 25 = x2 + 9x + 2x + 18
⇒ x2 + 10x + 25 = x2 + 11x + 18
⇒ x2 - x2 + 10x - 11x = 18 - 25
⇒ -x = -7
⇒ x = 7.
Hence, the value of x = 7.
Find two numbers whose mean proportion is 36 and the third proportional is 288.
Answer
Let the two numbers be x and y.
Thus, 36 is the mean proportion between x and y.
⇒ x 36 = 36 y ⇒ x y = 36 2 ⇒ x y = 1296 ⇒ x = 1296 y .....(1) \Rightarrow \dfrac{x}{36} = \dfrac{36}{y} \\[1em] \Rightarrow xy = 36^2 \\[1em] \Rightarrow xy = 1296 \\[1em] \Rightarrow x = \dfrac{1296}{y}\text{.....(1)} ⇒ 36 x = y 36 ⇒ x y = 3 6 2 ⇒ x y = 1296 ⇒ x = y 1296 .....(1)
The third proportional for x and y is 288.
⇒ x y = y 288 ⇒ y 2 = 288 x ..........(2) \Rightarrow \dfrac{x}{y} = \dfrac{y}{288} \\[1em] \Rightarrow y^2 = 288x \text{ ..........(2)} ⇒ y x = 288 y ⇒ y 2 = 288 x ..........(2)
Substituting value of x from equation (1) in (2), we get :
⇒ y 2 = 288 × 1296 y ⇒ y 3 = 373248 ⇒ y = 373248 3 = 72. \Rightarrow y^2 = 288 \times \dfrac{1296}{y} \\[1em] \Rightarrow y^3 = 373248 \\[1em] \Rightarrow y = \sqrt[3]{373248} = 72. ⇒ y 2 = 288 × y 1296 ⇒ y 3 = 373248 ⇒ y = 3 373248 = 72.
Substituting the value of y in equation (1), we get :
⇒ x = 1296 y ⇒ x = 1296 72 ⇒ x = 18 \Rightarrow x = \dfrac{1296}{y} \\[1em] \Rightarrow x = \dfrac{1296}{72} \\[1em] \Rightarrow x = 18 ⇒ x = y 1296 ⇒ x = 72 1296 ⇒ x = 18
Hence, the two numbers are 18 and 72.
If a : b :: c : d, prove that :
(i) (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd)
(ii) (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
(iii) (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 )
Answer
(i) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd), we get :
⇒ a 2 + a b c 2 + c d ⇒ ( c k ) 2 + c k . d k c 2 + c d ⇒ k 2 c 2 + k 2 c d c 2 + c d ⇒ k 2 ( c 2 + c d ) c 2 + c d ⇒ k 2 . \Rightarrow \dfrac{a^2 + ab}{c^2 + cd} \\[1em] \Rightarrow \dfrac{(ck)^2 + ck.dk}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd)}{c^2 + cd} \\[1em] \Rightarrow k^2. ⇒ c 2 + c d a 2 + ab ⇒ c 2 + c d ( c k ) 2 + c k . d k ⇒ c 2 + c d k 2 c 2 + k 2 c d ⇒ c 2 + c d k 2 ( c 2 + c d ) ⇒ k 2 .
Substituting value of a and b in R.H.S. of (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd), we get :
⇒ b 2 − 2 a b d 2 − 2 c d ⇒ ( k d ) 2 − 2 × c k × d k d 2 − 2 c d ⇒ k 2 d 2 − 2 k 2 c d d 2 − 2 c d ⇒ k 2 ( d 2 − 2 c d ) d 2 − 2 c d ⇒ k 2 . \Rightarrow \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{(kd)^2 - 2 \times ck \times dk}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2 d^2 - 2k^2 cd}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2(d^2 - 2cd)}{d^2 - 2cd} \\[1em] \Rightarrow k^2. ⇒ d 2 − 2 c d b 2 − 2 ab ⇒ d 2 − 2 c d ( k d ) 2 − 2 × c k × d k ⇒ d 2 − 2 c d k 2 d 2 − 2 k 2 c d ⇒ d 2 − 2 c d k 2 ( d 2 − 2 c d ) ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd).
(ii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
⇒ a 2 + b 2 c 2 + d 2 ⇒ ( k c ) 2 + ( k d ) 2 c 2 + d 2 ⇒ k 2 c 2 + k 2 d 2 c 2 + d 2 ⇒ k 2 ( c 2 + d 2 ) c 2 + d 2 ⇒ k 2 . \Rightarrow \dfrac{a^2 + b^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kd)^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{k^2c^2 + k^2d^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{k^2(c^2 + d^2)}{c^2 + d^2} \\[1em] \Rightarrow k^2. ⇒ c 2 + d 2 a 2 + b 2 ⇒ c 2 + d 2 ( k c ) 2 + ( k d ) 2 ⇒ c 2 + d 2 k 2 c 2 + k 2 d 2 ⇒ c 2 + d 2 k 2 ( c 2 + d 2 ) ⇒ k 2 .
Substituting value of a and b in R.H.S. of (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
⇒ a b + a d − b c c d − a d + b c ⇒ ( k c ) ( k d ) + ( k c ) d − ( k d ) c c d − ( k c ) d + ( k d ) c ⇒ k 2 c d + k c d − k c d c d − k c d + k c d ⇒ k 2 ( c d ) ( c d ) ⇒ k 2 . \Rightarrow \dfrac{ab + ad - bc}{cd - ad + bc} \\[1em] \Rightarrow \dfrac{(kc)(kd) + (kc)d - (kd)c}{cd - (kc)d + (kd)c} \\[1em] \Rightarrow \dfrac{k^2 cd + kcd - kcd}{cd - kcd + kcd} \\[1em] \Rightarrow \dfrac{k^2 (cd)}{(cd)} \\[1em] \Rightarrow k^2. ⇒ c d − a d + b c ab + a d − b c ⇒ c d − ( k c ) d + ( k d ) c ( k c ) ( k d ) + ( k c ) d − ( k d ) c ⇒ c d − k c d + k c d k 2 c d + k c d − k c d ⇒ ( c d ) k 2 ( c d ) ⇒ k 2 .
Since. L.H.S. = R.H.S.
Hence, proved that (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc).
(iii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ),
⇒ a 2 + a c + c 2 a 2 − a c + c 2 ⇒ ( k c ) 2 + ( k c ) c + c 2 ( k c ) 2 − ( k c ) c + c 2 ⇒ k 2 c 2 + k c 2 + c 2 k 2 c 2 − k c 2 + c 2 ⇒ c 2 ( k 2 + k + 1 ) c 2 ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{a^2 + ac + c^2}{a^2 - ac + c^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)c + c^2}{(kc)^2 - (kc)c + c^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + kc^2 + c^2}{k^2 c^2 - kc^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)}{c^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a 2 − a c + c 2 a 2 + a c + c 2 ⇒ ( k c ) 2 − ( k c ) c + c 2 ( k c ) 2 + ( k c ) c + c 2 ⇒ k 2 c 2 − k c 2 + c 2 k 2 c 2 + k c 2 + c 2 ⇒ c 2 ( k 2 − k + 1 ) c 2 ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Substituting value of a and b in R.H.S. of (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ),
⇒ b 2 + b d + d 2 b 2 − b d + d 2 ⇒ ( k d ) 2 + ( k d ) d + d 2 ( k d ) 2 − ( k d ) d + d 2 ⇒ k 2 d 2 + k d 2 + d 2 k 2 d 2 − k d 2 + d 2 ⇒ d 2 ( k 2 + k + 1 ) d 2 ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{b^2 + bd + d^2}{b^2 - bd + d^2} \\[1em] \Rightarrow \dfrac{(kd)^2 + (kd)d + d^2}{(kd)^2 - (kd)d + d^2} \\[1em] \Rightarrow \dfrac{k^2 d^2 + kd^2 + d^2}{k^2 d^2 - kd^2 + d^2} \\[1em] \Rightarrow \dfrac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ b 2 − b d + d 2 b 2 + b d + d 2 ⇒ ( k d ) 2 − ( k d ) d + d 2 ( k d ) 2 + ( k d ) d + d 2 ⇒ k 2 d 2 − k d 2 + d 2 k 2 d 2 + k d 2 + d 2 ⇒ d 2 ( k 2 − k + 1 ) d 2 ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ).
If a : b :: c : d, show that :
(i) a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
(ii) m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
(iii) a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
(iv) ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
Answer
(i) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
⇒ a + b c + d ⇒ k c + k d c + d ⇒ k ( c + d ) ( c + d ) ⇒ k . \Rightarrow \dfrac{a + b}{c + d} \\[1em] \Rightarrow \dfrac{kc + kd}{c + d} \\[1em] \Rightarrow \dfrac{k(c + d)}{(c + d)} \\[1em] \Rightarrow k. ⇒ c + d a + b ⇒ c + d k c + k d ⇒ ( c + d ) k ( c + d ) ⇒ k .
Substituting value of a and b in R.H.S. of a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
⇒ 2 a 2 + 7 b 2 2 c 2 + 7 d 2 ⇒ 2 ( k c ) 2 + 7 ( d k ) 2 2 c 2 + 7 d 2 ⇒ k 2 ( 2 c 2 + 7 d 2 ) 2 c 2 + 7 d 2 ⇒ k 2 ⇒ k . \Rightarrow \sqrt{\dfrac{2a^2 + 7b^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{2(kc)^2 + 7(dk)^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{k^2(2c^2 + 7d^2)}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{k^2} \\[1em] \Rightarrow k. ⇒ 2 c 2 + 7 d 2 2 a 2 + 7 b 2 ⇒ 2 c 2 + 7 d 2 2 ( k c ) 2 + 7 ( d k ) 2 ⇒ 2 c 2 + 7 d 2 k 2 ( 2 c 2 + 7 d 2 ) ⇒ k 2 ⇒ k .
Since, L.H.S. = R.H.S.
Hence, proved that a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
(ii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
⇒ m a 2 + n c 2 m b 2 + n d 2 ⇒ m ( k c ) 2 + n c 2 m ( k d ) 2 + n d 2 ⇒ m k 2 c 2 + n c 2 m k 2 d 2 + n d 2 ⇒ c 2 ( m k 2 + n ) d 2 ( m k 2 + n ) ⇒ c 2 d 2 . \Rightarrow \dfrac{ma^2 + nc^2}{mb^2 + nd^2} \\[1em] \Rightarrow \dfrac{m(kc)^2 + nc^2}{m(kd)^2 + nd^2} \\[1em] \Rightarrow \dfrac{m k^2 c^2 + n c^2}{m k^2 d^2 + n d^2} \\[1em] \Rightarrow \dfrac{c^2(m k^2 + n)}{d^2(m k^2 + n)} \\[1em] \Rightarrow \dfrac{c^2}{d^2}. ⇒ m b 2 + n d 2 m a 2 + n c 2 ⇒ m ( k d ) 2 + n d 2 m ( k c ) 2 + n c 2 ⇒ m k 2 d 2 + n d 2 m k 2 c 2 + n c 2 ⇒ d 2 ( m k 2 + n ) c 2 ( m k 2 + n ) ⇒ d 2 c 2 .
Substituting value of a and b in R.H.S. of m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
⇒ a 4 + c 4 b 4 + d 4 ⇒ ( k c ) 4 + c 4 ( k d ) 4 + d 4 ⇒ k 4 c 4 + c 4 k 4 d 4 + d 4 ⇒ c 4 ( k 4 + 1 ) d 4 ( k 4 + 1 ) ⇒ c 4 d 4 ⇒ c 2 d 2 . \Rightarrow \sqrt{\dfrac{a^4 + c^4}{b^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{(kc)^4 + c^4}{(kd)^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{k^4 c^4 + c^4}{k^4 d^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4(k^4 + 1)}{d^4(k^4 + 1)}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4}{d^4}} \\[1em] \Rightarrow \dfrac{c^2}{d^2}. ⇒ b 4 + d 4 a 4 + c 4 ⇒ ( k d ) 4 + d 4 ( k c ) 4 + c 4 ⇒ k 4 d 4 + d 4 k 4 c 4 + c 4 ⇒ d 4 ( k 4 + 1 ) c 4 ( k 4 + 1 ) ⇒ d 4 c 4 ⇒ d 2 c 2 .
Since, L.H.S. = R.H.S.
Hence, proved that m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
(iii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting values of a and b in L.H.S. of a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
⇒ a 2 + a b + b 2 a 2 − a b + b 2 ⇒ ( k c ) 2 + ( k c ) ( k d ) + ( k d ) 2 ( k c ) 2 − ( k c ) ( k d ) + ( k d ) 2 ⇒ k 2 c 2 + k 2 c d + k 2 d 2 k 2 c 2 − k 2 c d + k 2 d 2 ⇒ k 2 ( c 2 + c d + d 2 ) k 2 ( c 2 − c d + d 2 ) ⇒ c 2 + c d + d 2 c 2 − c d + d 2 . \Rightarrow \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)(kd) + (kd)^2}{(kc)^2 - (kc)(kd) + (kd)^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd + k^2 d^2}{k^2 c^2 - k^2 cd + k^2 d^2} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd + d^2)}{k^2(c^2 - cd + d^2)} \\[1em] \Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}. ⇒ a 2 − ab + b 2 a 2 + ab + b 2 ⇒ ( k c ) 2 − ( k c ) ( k d ) + ( k d ) 2 ( k c ) 2 + ( k c ) ( k d ) + ( k d ) 2 ⇒ k 2 c 2 − k 2 c d + k 2 d 2 k 2 c 2 + k 2 c d + k 2 d 2 ⇒ k 2 ( c 2 − c d + d 2 ) k 2 ( c 2 + c d + d 2 ) ⇒ c 2 − c d + d 2 c 2 + c d + d 2 .
Substituting values of a and b in R.H.S. of a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
⇒ c 2 + c d + d 2 c 2 − c d + d 2 . \Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}. ⇒ c 2 − c d + d 2 c 2 + c d + d 2 .
Since, L.H.S. = R.H.S.
Hence, proved that a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
(iv) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting values of a and b in L.H.S. of ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
⇒ ( a + c ) 3 ( b + d ) 3 ⇒ ( k c + c ) 3 ( k d + d ) 3 ⇒ c 3 ( k + 1 ) 3 d 3 ( k + 1 ) 3 ⇒ c 3 d 3 . \Rightarrow \dfrac{(a + c)^3}{(b + d)^3} \\[1em] \Rightarrow \dfrac{(kc + c)^3}{(kd + d)^3} \\[1em] \Rightarrow \dfrac{c^3(k + 1)^3}{d^3(k + 1)^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}. ⇒ ( b + d ) 3 ( a + c ) 3 ⇒ ( k d + d ) 3 ( k c + c ) 3 ⇒ d 3 ( k + 1 ) 3 c 3 ( k + 1 ) 3 ⇒ d 3 c 3 .
Substituting values of a and b in R.H.S. of ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
⇒ a ( a − c ) 2 b ( b − d ) 2 ⇒ k c ( k c − c ) 2 k d ( k d − d ) 2 ⇒ k c ⋅ c 2 ( k − 1 ) 2 k d ⋅ d 2 ( k − 1 ) 2 ⇒ k c 3 k d 3 ⇒ c 3 d 3 . \Rightarrow \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \Rightarrow \dfrac{kc(kc - c)^2}{kd(kd - d)^2} \\[1em] \Rightarrow \dfrac{kc \cdot c^2(k - 1)^2}{kd \cdot d^2(k - 1)^2} \\[1em] \Rightarrow \dfrac{kc^3}{kd^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}. ⇒ b ( b − d ) 2 a ( a − c ) 2 ⇒ k d ( k d − d ) 2 k c ( k c − c ) 2 ⇒ k d ⋅ d 2 ( k − 1 ) 2 k c ⋅ c 2 ( k − 1 ) 2 ⇒ k d 3 k c 3 ⇒ d 3 c 3 .
Since, L.H.S. = R.H.S.
Hence, proved that ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
If x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
(i) x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
(ii) x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
(iii) x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
(iv) a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) = 3 \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3 ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x = 3
Answer
(i) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
⇒ x 2 + y 2 + z 2 a 2 + b 2 + c 2 ⇒ k 2 a 2 + k 2 b 2 + k 2 c 2 a 2 + b 2 + c 2 ⇒ k 2 ( a 2 + b 2 + c 2 ) a 2 + b 2 + c 2 ⇒ k 2 . \Rightarrow \dfrac{x^2 + y^2 + z^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 a^2 + k^2 b^2 + k^2 c^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 (a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \\[1em] \Rightarrow k^2. ⇒ a 2 + b 2 + c 2 x 2 + y 2 + z 2 ⇒ a 2 + b 2 + c 2 k 2 a 2 + k 2 b 2 + k 2 c 2 ⇒ a 2 + b 2 + c 2 k 2 ( a 2 + b 2 + c 2 ) ⇒ k 2 .
Substituting values of x, y and z in R.H.S. of the equation x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
⇒ ( p x + q y + r z p a + q b + r c ) 2 ⇒ ( p ( k a ) + q ( k b ) + r ( k c ) p a + q b + r c ) 2 ⇒ ( k ( p a + q b + r c ) p a + q b + r c ) 2 ⇒ k 2 . \Rightarrow \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{p(ka) + q(kb) + r(kc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{k(pa + qb + rc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow k^2. ⇒ ( p a + q b + rc p x + q y + rz ) 2 ⇒ ( p a + q b + rc p ( ka ) + q ( kb ) + r ( k c ) ) 2 ⇒ ( p a + q b + rc k ( p a + q b + rc ) ) 2 ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
(ii) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
⇒ x 3 a 3 + y 3 b 3 + z 3 c 3 ⇒ k 3 a 3 a 3 + k 3 b 3 b 3 + k 3 c 3 c 3 ⇒ k 3 + k 3 + k 3 ⇒ 3 k 3 . \Rightarrow \dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^3} + \dfrac{k^3 b^3}{b^3} + \dfrac{k^3 c^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3. ⇒ a 3 x 3 + b 3 y 3 + c 3 z 3 ⇒ a 3 k 3 a 3 + b 3 k 3 b 3 + c 3 k 3 c 3 ⇒ k 3 + k 3 + k 3 ⇒ 3 k 3 .
Substituting values of x, y and z in R.H.S. of the equation x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
⇒ 3 x y z a b c ⇒ 3 ( k a ) ( k b ) ( k c ) a b c ⇒ 3 k 3 a b c a b c ⇒ 3 k 3 . \Rightarrow \dfrac{3xyz}{abc} \\[1em] \Rightarrow \dfrac{3(ka)(kb)(kc)}{abc} \\[1em] \Rightarrow \dfrac{3k^3 abc}{abc} \\[1em] \Rightarrow 3k^3. ⇒ ab c 3 x yz ⇒ ab c 3 ( ka ) ( kb ) ( k c ) ⇒ ab c 3 k 3 ab c ⇒ 3 k 3 .
Since, L.H.S. = R.H.S.
Hence, proved that x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
(iii) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
⇒ x 3 a 2 + y 3 b 2 + z 3 c 2 ⇒ k 3 a 3 a 2 + k 3 b 3 b 2 + k 3 c 3 c 2 ⇒ k 3 a + k 3 b + k 3 c ⇒ k 3 ( a + b + c ) . \Rightarrow \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^2} + \dfrac{k^3 b^3}{b^2} + \dfrac{k^3 c^3}{c^2} \\[1em] \Rightarrow k^3 a + k^3 b + k^3 c \\[1em] \Rightarrow k^3 (a + b + c). ⇒ a 2 x 3 + b 2 y 3 + c 2 z 3 ⇒ a 2 k 3 a 3 + b 2 k 3 b 3 + c 2 k 3 c 3 ⇒ k 3 a + k 3 b + k 3 c ⇒ k 3 ( a + b + c ) .
Substituting values of x, y and z in R.H.S. of the equation x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
⇒ ( x + y + z ) 3 ( a + b + c ) 2 ⇒ ( k ( a + b + c ) ) 3 ( a + b + c ) 2 ⇒ k 3 ( a + b + c ) 3 ( a + b + c ) 2 ⇒ k 3 ( a + b + c ) . \Rightarrow \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{(k(a + b + c))^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{k^3 (a + b + c)^3}{(a + b + c)^2} \\[1em] \Rightarrow k^3 (a + b + c). ⇒ ( a + b + c ) 2 ( x + y + z ) 3 ⇒ ( a + b + c ) 2 ( k ( a + b + c ) ) 3 ⇒ ( a + b + c ) 2 k 3 ( a + b + c ) 3 ⇒ k 3 ( a + b + c ) .
Since, L.H.S. = R.H.S.
Hence, proved that x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
(iv) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in a x − b y ( a + b ) ( x − y ) \dfrac{ax - by}{(a + b)(x - y)} ( a + b ) ( x − y ) a x − b y
⇒ a x − b y ( a + b ) ( x − y ) ⇒ a ( k a ) − b ( k b ) ( a + b ) ( k a − k b ) ⇒ k ( a 2 − b 2 ) k ( a + b ) ( a − b ) ⇒ k ( a − b ) ( a + b ) k ( a + b ) ( a − b ) ⇒ 1.... ( 1 ) \Rightarrow \dfrac{ax - by}{(a + b)(x - y)} \\[1em] \Rightarrow \dfrac{a(ka) - b(kb)}{(a + b)(ka - kb)} \\[1em] \Rightarrow \dfrac{k(a^2 - b^2)}{k(a + b)(a - b)} \\[1em] \Rightarrow \dfrac{k(a - b)(a + b)}{k(a + b)(a - b)} \\[1em] \Rightarrow 1....(1) ⇒ ( a + b ) ( x − y ) a x − b y ⇒ ( a + b ) ( ka − kb ) a ( ka ) − b ( kb ) ⇒ k ( a + b ) ( a − b ) k ( a 2 − b 2 ) ⇒ k ( a + b ) ( a − b ) k ( a − b ) ( a + b ) ⇒ 1.... ( 1 )
Substituting values of x, y and z in b y − c z ( b + c ) ( y − z ) \dfrac{by - cz}{(b + c)(y - z)} ( b + c ) ( y − z ) b y − cz
⇒ b y − c z ( b + c ) ( y − z ) ⇒ b ( k b ) − c ( k c ) ( b + c ) ( k b − k c ) ⇒ k ( b 2 − c 2 ) k ( b + c ) ( b − c ) ⇒ k ( b − c ) ( b + c ) k ( b + c ) ( b − c ) ⇒ 1.... ( 2 ) \Rightarrow \dfrac{by - cz}{(b + c)(y - z)} \\[1em] \Rightarrow \dfrac{b(kb) - c(kc)}{(b + c)(kb - kc)} \\[1em] \Rightarrow \dfrac{k(b^2 - c^2)}{k(b + c)(b - c)} \\[1em] \Rightarrow \dfrac{k(b - c)(b + c)}{k(b + c)(b - c)} \\[1em] \Rightarrow 1....(2) ⇒ ( b + c ) ( y − z ) b y − cz ⇒ ( b + c ) ( kb − k c ) b ( kb ) − c ( k c ) ⇒ k ( b + c ) ( b − c ) k ( b 2 − c 2 ) ⇒ k ( b + c ) ( b − c ) k ( b − c ) ( b + c ) ⇒ 1.... ( 2 )
Substituting values of x, y and z in c z − a x ( c + a ) ( z − x ) \dfrac{cz - ax}{(c + a)(z - x)} ( c + a ) ( z − x ) cz − a x
⇒ c z − a x ( c + a ) ( z − x ) ⇒ c ( k c ) − a ( k a ) ( c + a ) ( k c − k a ) ⇒ k ( c 2 − a 2 ) k ( c + a ) ( c − a ) ⇒ k ( c − a ) ( c + a ) k ( c + a ) ( c − a ) ⇒ 1.... ( 3 ) \Rightarrow \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow \dfrac{c(kc) - a(ka)}{(c + a)(kc - ka)} \\[1em] \Rightarrow \dfrac{k(c^2 - a^2)}{k(c + a)(c - a)} \\[1em] \Rightarrow \dfrac{k(c - a)(c + a)}{k(c + a)(c - a)} \\[1em] \Rightarrow 1....(3) ⇒ ( c + a ) ( z − x ) cz − a x ⇒ ( c + a ) ( k c − ka ) c ( k c ) − a ( ka ) ⇒ k ( c + a ) ( c − a ) k ( c 2 − a 2 ) ⇒ k ( c + a ) ( c − a ) k ( c − a ) ( c + a ) ⇒ 1.... ( 3 )
Adding 1,2 and 3 we get,
⇒ a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) ⇒ 1 + 1 + 1 ⇒ 3. \Rightarrow \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3. ⇒ ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x ⇒ 1 + 1 + 1 ⇒ 3.
Since, L.H.S = R.H.S
Hence, proved that a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) = 3 \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3 ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x = 3
If a b = c d = e f \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} b a = d c = f e
(i) (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2
(ii) a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
(iii) ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
(iv) (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
Answer
(i) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 , we get:
⇒ ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) ⇒ ( b 2 + d 2 + f 2 ) ( k 2 b 2 + k 2 d 2 + k 2 f 2 ) ⇒ ( b 2 + d 2 + f 2 ) [ k 2 ( b 2 + d 2 + f 2 ) ] ⇒ k 2 ( b 2 + d 2 + f 2 ) 2 . \Rightarrow (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)(k^2 b^2 + k^2 d^2 + k^2 f^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)[k^2(b^2 + d^2 + f^2)] \\[1em] \Rightarrow k^2(b^2 + d^2 + f^2)^2. ⇒ ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) ⇒ ( b 2 + d 2 + f 2 ) ( k 2 b 2 + k 2 d 2 + k 2 f 2 ) ⇒ ( b 2 + d 2 + f 2 ) [ k 2 ( b 2 + d 2 + f 2 )] ⇒ k 2 ( b 2 + d 2 + f 2 ) 2 .
Substituting values of a,c and e in R.H.S. of (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 , we get:
⇒ a b + c d + e f ⇒ ( k b ) b + ( k d ) d + ( k f ) f ⇒ k ( b 2 + d 2 + f 2 ) ∴ ( a b + c d + e f ) 2 = k 2 ( b 2 + d 2 + f 2 ) 2 . \Rightarrow ab + cd + ef \\[1em] \Rightarrow (k b)b + (k d)d + (k f)f \\[1em] \Rightarrow k(b^2 + d^2 + f^2) \\[1em] \therefore (ab + cd + ef)^2 = k^2(b^2 + d^2 + f^2)^2. ⇒ ab + c d + e f ⇒ ( kb ) b + ( k d ) d + ( k f ) f ⇒ k ( b 2 + d 2 + f 2 ) ∴ ( ab + c d + e f ) 2 = k 2 ( b 2 + d 2 + f 2 ) 2 .
Since, L.H.S. = R.H.S.
Hence, proved that (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 .
(ii) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
⇒ a 3 + c 3 + e 3 b 3 + d 3 + f 3 ⇒ k 3 b 3 + k 3 d 3 + k 3 f 3 b 3 + d 3 + f 3 ⇒ k 3 ( b 3 + d 3 + f 3 ) b 3 + d 3 + f 3 ⇒ k 3 . \Rightarrow \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3 b^3 + k^3 d^3 + k^3 f^3}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3(b^3 + d^3 + f^3)}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow k^3. ⇒ b 3 + d 3 + f 3 a 3 + c 3 + e 3 ⇒ b 3 + d 3 + f 3 k 3 b 3 + k 3 d 3 + k 3 f 3 ⇒ b 3 + d 3 + f 3 k 3 ( b 3 + d 3 + f 3 ) ⇒ k 3 .
SSubstituting values of a,c and e in R.H.S. of a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
⇒ a c e b d f ⇒ ( k b ) ( k d ) ( k f ) b d f ⇒ k 3 . \Rightarrow \dfrac{ace}{bdf} \\[1em] \Rightarrow \dfrac{(k b)(k d)(k f)}{b d f} \\[1em] \Rightarrow k^3. ⇒ b df a ce ⇒ b df ( kb ) ( k d ) ( k f ) ⇒ k 3 .
Since L.H.S. = R.H.S.
Hence, proved that a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
(iii) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
⇒ a 2 b 2 + c 2 d 2 + e 2 f 2 ⇒ k 2 b 2 b 2 + k 2 d 2 d 2 + k 2 f 2 f 2 ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 . \Rightarrow \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \\[1em] \Rightarrow \dfrac{k^2 b^2}{b^2} + \dfrac{k^2 d^2}{d^2} + \dfrac{k^2 f^2}{f^2} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2. ⇒ b 2 a 2 + d 2 c 2 + f 2 e 2 ⇒ b 2 k 2 b 2 + d 2 k 2 d 2 + f 2 k 2 f 2 ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 .
Substituting values of a,c and e in R.H.S. of ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
⇒ a c b d + c e d f + a e b f ⇒ ( k b ) ( k d ) b d + ( k d ) ( k f ) d f + ( k b ) ( k f ) b f ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 . \Rightarrow \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\[1em] \Rightarrow \dfrac{(k b)(k d)}{b d} + \dfrac{(k d)(k f)}{d f} + \dfrac{(k b)(k f)}{b f} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2. ⇒ b d a c + df ce + b f a e ⇒ b d ( kb ) ( k d ) + df ( k d ) ( k f ) + b f ( kb ) ( k f ) ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
(iv) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a, c and d in L.H.S. of (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
⇒ b d f ⋅ ( a + b b + c + d d + e + f f ) 3 ⇒ b d f ⋅ ( k b + b b + k d + d d + k f + f f ) 3 ⇒ b d f ⋅ [ ( k + 1 ) + ( k + 1 ) + ( k + 1 ) ] ⇒ b d f ⋅ [ 3 ( k + 1 ) ] 3 ⇒ b d f ⋅ 27 ( k + 1 ) 3 . \Rightarrow bdf \cdot \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} \\[1em] \Rightarrow bdf \cdot \Big(\dfrac{kb + b}{b} + \dfrac{kd + d}{d} + \dfrac{kf + f}{f}\Big)^3 \\[1em] \Rightarrow bdf \cdot [(k + 1) + (k + 1) + (k + 1)] \\[1em] \Rightarrow bdf \cdot [3(k + 1)]^3 \\[1em] \Rightarrow bdf\cdot 27(k + 1)^3. ⇒ b df ⋅ ( b a + b + d c + d + f e + f ) 3 ⇒ b df ⋅ ( b kb + b + d k d + d + f k f + f ) 3 ⇒ b df ⋅ [( k + 1 ) + ( k + 1 ) + ( k + 1 )] ⇒ b df ⋅ [ 3 ( k + 1 ) ] 3 ⇒ b df ⋅ 27 ( k + 1 ) 3 .
Substituting values of a, c and d in R.H.S. of (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
⇒ 27 ( a + b ) ( c + d ) ( e + f ) ⇒ 27 ( k b + b ) ( k d + d ) ( k f + f ) ⇒ 27 ( b ( k + 1 ) ) ( d ( k + 1 ) ) ( f ( k + 1 ) ) ⇒ 27 b d f ( k + 1 ) 3 . \Rightarrow 27(a + b)(c + d)(e + f) \\[1em] \Rightarrow 27(kb + b)(kd + d)(kf + f) \\[1em] \Rightarrow 27\big(b(k + 1)\big)\big(d(k + 1)\big)\big(f(k + 1)\big) \\[1em] \Rightarrow 27 bdf(k + 1)^3. ⇒ 27 ( a + b ) ( c + d ) ( e + f ) ⇒ 27 ( kb + b ) ( k d + d ) ( k f + f ) ⇒ 27 ( b ( k + 1 ) ) ( d ( k + 1 ) ) ( f ( k + 1 ) ) ⇒ 27 b df ( k + 1 ) 3 .
Since, L.H.S. = R.H.S.
Hence, proved that (bdf).( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
If a, b, c are in continued proportion, prove that :
(i) a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
(ii) a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
(iii) a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
(iv) (a + b + c)(a − b + c) = (a2 + b2 + c2 )
(v) a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 )
Answer
(i) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
⇒ a + b b + c ⇒ c k 2 + c k c k + c ⇒ c k ( k + 1 ) c ( k + 1 ) ⇒ k . \Rightarrow \dfrac{a + b}{b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck}{ck + c} \\[1em] \Rightarrow \dfrac{c k(k + 1)}{c(k + 1)} \\[1em] \Rightarrow k. ⇒ b + c a + b ⇒ c k + c c k 2 + c k ⇒ c ( k + 1 ) c k ( k + 1 ) ⇒ k .
Substituting values of a and b in R.H.S. of equation a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
⇒ a 2 ( b − c ) b 2 ( a − b ) ⇒ ( c k 2 ) 2 ( c k − c ) ( c k ) 2 ( c k 2 − c k ) ⇒ c 2 k 4 ( c k − c ) c 2 k 2 ( c k 2 − c k ) ⇒ c 3 k 4 ( k − 1 ) c 3 k 3 ( k − 1 ) ⇒ k . \Rightarrow \dfrac{a^2(b - c)}{b^2(a - b)} \\[1em] \Rightarrow \dfrac{(ck^2)^2\big(ck - c\big)}{(ck)^2\big(ck^2 - ck\big)} \\[1em] \Rightarrow \dfrac{c^2k^4(ck - c)}{c^2k^2(ck^2 - ck)} \\[1em] \Rightarrow \dfrac{c^3 k^4 (k - 1)}{c^3k^3(k - 1)} \\[1em] \Rightarrow k. ⇒ b 2 ( a − b ) a 2 ( b − c ) ⇒ ( c k ) 2 ( c k 2 − c k ) ( c k 2 ) 2 ( c k − c ) ⇒ c 2 k 2 ( c k 2 − c k ) c 2 k 4 ( c k − c ) ⇒ c 3 k 3 ( k − 1 ) c 3 k 4 ( k − 1 ) ⇒ k .
Since, L.H.S. = R.H.S.
Hence, proved that a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
(ii) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
⇒ a + b + c a − b + c ⇒ c k 2 + c k + c c k 2 − c k + c ⇒ c ( k 2 + k + 1 ) c ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{a + b + c}{a - b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck + c}{ck^2 - ck + c} \\[1em] \Rightarrow \dfrac{c(k^2 + k + 1)}{c(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a − b + c a + b + c ⇒ c k 2 − c k + c c k 2 + c k + c ⇒ c ( k 2 − k + 1 ) c ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Substituting values of a and b in R.H.S. of equation a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
⇒ ( a + b + c ) 2 a 2 + b 2 + c 2 ⇒ ( c k 2 + c k + c ) 2 ( c k 2 ) 2 + ( c k ) 2 + c 2 ⇒ c 2 ( k 2 + k + 1 ) 2 c 2 ( k 4 + k 2 + 1 ) ⇒ ( k 2 + k + 1 ) 2 ( k 4 + k 2 + 1 ) ⇒ ( k 2 + k + 1 ) 2 ( k 2 + k + 1 ) ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{(a + b + c)^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{(ck^2 + ck + c)^2}{(ck^2)^2 + (ck)^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)^2}{c^2(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^2 + k + 1)(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a 2 + b 2 + c 2 ( a + b + c ) 2 ⇒ ( c k 2 ) 2 + ( c k ) 2 + c 2 ( c k 2 + c k + c ) 2 ⇒ c 2 ( k 4 + k 2 + 1 ) c 2 ( k 2 + k + 1 ) 2 ⇒ ( k 4 + k 2 + 1 ) ( k 2 + k + 1 ) 2 ⇒ ( k 2 + k + 1 ) ( k 2 − k + 1 ) ( k 2 + k + 1 ) 2 ⇒ k 2 − k + 1 k 2 + k + 1 .
Since, L.H.S. = R.H.S.
Hence, proved that a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
(iii) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
⇒ a 2 + a b + b 2 b 2 + b c + c 2 ⇒ ( c k 2 ) 2 + ( c k 2 ) ( c k ) + ( c k ) 2 ( c k ) 2 + ( c k ) c + c 2 ⇒ c 2 k 4 + c 2 k 3 + c 2 k 2 c 2 k 2 + c 2 k + c 2 ⇒ c 2 ( k 4 + k 3 + k 2 ) c 2 ( k 2 + k + 1 ) ⇒ k 2 ( k 2 + k + 1 ) ( k 2 + k + 1 ) ⇒ k 2 . \Rightarrow \dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} \\[1em] \Rightarrow \dfrac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \\[1em] \Rightarrow \dfrac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^4 + k^3 + k^2)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow \dfrac{k^2(k^2 + k + 1)}{(k^2 + k + 1)} \\[1em] \Rightarrow k^2. ⇒ b 2 + b c + c 2 a 2 + ab + b 2 ⇒ ( c k ) 2 + ( c k ) c + c 2 ( c k 2 ) 2 + ( c k 2 ) ( c k ) + ( c k ) 2 ⇒ c 2 k 2 + c 2 k + c 2 c 2 k 4 + c 2 k 3 + c 2 k 2 ⇒ c 2 ( k 2 + k + 1 ) c 2 ( k 4 + k 3 + k 2 ) ⇒ ( k 2 + k + 1 ) k 2 ( k 2 + k + 1 ) ⇒ k 2 .
Substituting values of a and b in R.H.S. of equation a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
⇒ a c ⇒ c k 2 c ⇒ k 2 . \Rightarrow \dfrac{a}{c} \\[1em] \Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2. ⇒ c a ⇒ c c k 2 ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
(iv) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of (a + b + c)(a − b + c) = (a2 + b2 + c2 ), we get:
⇒ (a + b + c)(a − b + c)
⇒ ((ck2 ) + (ck) + c)((ck2 ) − (ck) + c)
⇒ c(k2 + k + 1). c(k2 - k + 1)
⇒ c2 (k4 + k2 + 1)
Substituting values of a and b in R.H.S. of (a + b + c)(a − b + c) = (a2 + b2 + c2 ), we get:
⇒ (a2 + b2 + c2 )
⇒ (ck2 )2 + (ck)2 + c2
⇒ (c2 k4 + c2 k2 + c2 )
⇒ c2 (k4 + k2 + 1)
Since L.H.S. = R.H.S.
Hence, proved that (a + b + c)(a − b + c) = (a2 + b2 + c2 ).
(v) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ), we get:
⇒ a 2 b 2 c 2 ( a − 3 + b − 3 + c − 3 ) ⇒ c 6 k 6 ( 1 c 3 k 6 + 1 c 3 k 3 + 1 c 3 ) ⇒ c 3 ( 1 + k 3 + k 6 ) . \Rightarrow a^2 b^2 c^2\big(a^{ - 3} + b^{ - 3} + c^{ - 3}\big) \\[1em] \Rightarrow c^6 k^6\Big(\dfrac{1}{c^3 k^6} + \dfrac{1}{c^3 k^3} + \dfrac{1}{c^3}\Big) \\[1em] \Rightarrow c^3\big(1 + k^3 + k^6\big). ⇒ a 2 b 2 c 2 ( a − 3 + b − 3 + c − 3 ) ⇒ c 6 k 6 ( c 3 k 6 1 + c 3 k 3 1 + c 3 1 ) ⇒ c 3 ( 1 + k 3 + k 6 ) .
Substituting values of a and b in R.H.S. of a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ), we get:
⇒ a 3 + b 3 + c 3 ⇒ ( c k 2 ) 3 + ( c k ) 3 + c 3 ⇒ c 3 k 6 + c 3 k 3 + c 3 ⇒ c 3 ( k 6 + k 3 + 1 ) . \Rightarrow a^3 + b^3 + c^3 \\[1em] \Rightarrow (ck^2)^3 + (ck)^3 + c^3 \\[1em] \Rightarrow c^3k^6 + c^3k^3 + c^3 \\[1em] \Rightarrow c^3(k^6 + k^3 + 1). ⇒ a 3 + b 3 + c 3 ⇒ ( c k 2 ) 3 + ( c k ) 3 + c 3 ⇒ c 3 k 6 + c 3 k 3 + c 3 ⇒ c 3 ( k 6 + k 3 + 1 ) .
Since L.H.S. = R.H.S.
Hence, proved that a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ).
If a, b, c, d are in continued proportion, prove that :
(b + c)(b + d) = (c + a)(c + d)
Answer
Given,
⇒ a, b, c, d are in continued proportion
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :
⇒ (b + c)(b + d)
⇒ (d k2 + d k)(d k2 + d)
⇒ d2 (k2 + k)(k2 + 1)
⇒ d2 k(k + 1)(k2 + 1).
Substituting values of a, b and c in R.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :
⇒ (c + a)(c + d)
⇒ (dk + dk3 )(dk + d)
⇒ d2 (k + k3 )(k + 1)
⇒ d2 k(1 + k2 )(k + 1).
Since, L.H.S. = R.H.S.
Hence, (b + c)(b + d) = (c + a)(c + d).
If a, b, c, d are in continued proportion, prove that :
(a2 − b2 )(c2 − d2 ) = (b2 − c2 )2
Answer
Given a, b, c, d are in continued proportion.
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 , we get :
⇒ (a2 - b2 )(c2 - d2 )
⇒ [(dk3 )2 - (dk2 )2 ] [(dk)2 - d2 ]
⇒ [d2 k6 - d2 k4 ] [d2 k2 - d2 ]
⇒ d4 (k6 - k4 )(k2 - 1)
⇒ d4 k4 (k2 - 1)(k2 - 1)
⇒ d4 k4 (k2 - 1)2 .
Substituting values of a, b and c in R.H.S. of (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 , we get :
⇒ (b2 - c2 )2
⇒ [(dk2 )2 - (dk)2 ]2
⇒ [d2 k4 - d2 k2 ]2
⇒ [d2 k2 (k2 - 1)]2
⇒ d4 k4 (k2 - 1)2 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 .
If ax = by = cz, prove that x 2 y z + y 2 z x + z 2 x y = b c a 2 + c a b 2 + a b c 2 \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} = \dfrac{bc}{a^{2}} + \dfrac{ca}{b^{2}} + \dfrac{ab}{c^{2}} yz x 2 + z x y 2 + x y z 2 = a 2 b c + b 2 c a + c 2 ab
Answer
Given,
ax = by = cz
⇒ a x a b c = b y a b c = c z a b c \Rightarrow \dfrac{ax}{abc} = \dfrac{by}{abc} = \dfrac{cz}{abc} ⇒ ab c a x = ab c b y = ab c cz
⇒ x b c = y c a = z a b = k \Rightarrow \dfrac{x}{bc} = \dfrac{y}{ca} = \dfrac{z}{ab} = k ⇒ b c x = c a y = ab z = k
Thus,
x = kbc, y = kca, z = kab
Substitute values of x, y , z in L.H.S , we get :
⇒ x 2 y z + y 2 z x + z 2 x y ⇒ ( k b c ) 2 ( k c a ) ( k a b ) + ( k c a ) 2 ( k a b ) ( k b c ) + ( k a b ) 2 ( k b c ) ( k c a ) ⇒ k 2 b 2 c 2 k 2 a 2 b c + k 2 c 2 a 2 k 2 b 2 c a + k 2 a 2 b 2 k 2 c 2 a b ⇒ b 2 c 2 a 2 b c + c 2 a 2 b 2 c a + a 2 b 2 c 2 a b ⇒ b c a 2 + c a b 2 + a b c 2 . \Rightarrow \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} \\[1em] \Rightarrow \dfrac{(kbc)^{2}}{(kca)(kab)} + \dfrac{(kca)^{2}}{(kab)(kbc)} + \dfrac{(kab)^{2}}{(kbc)(kca)} \\[1em] \Rightarrow \dfrac{k^2b^2c^2}{k^2a^2bc} + \dfrac{k^2c^2a^2}{k^2b^2ca} + \dfrac{k^2a^2b^2}{k^2c^2ab} \\[1em] \Rightarrow \dfrac{b^2c^2}{a^2bc} + \dfrac{c^2a^2}{b^2ca} + \dfrac{a^2b^2}{c^2ab} \\[1em] \Rightarrow \dfrac{bc}{a^2} + \dfrac{ca}{b^2} + \dfrac{ab}{c^2}. ⇒ yz x 2 + z x y 2 + x y z 2 ⇒ ( k c a ) ( kab ) ( kb c ) 2 + ( kab ) ( kb c ) ( k c a ) 2 + ( kb c ) ( k c a ) ( kab ) 2 ⇒ k 2 a 2 b c k 2 b 2 c 2 + k 2 b 2 c a k 2 c 2 a 2 + k 2 c 2 ab k 2 a 2 b 2 ⇒ a 2 b c b 2 c 2 + b 2 c a c 2 a 2 + c 2 ab a 2 b 2 ⇒ a 2 b c + b 2 c a + c 2 ab .
Hence, proved that x 2 y z + y 2 z x + z 2 x y = b c a 2 + c a b 2 + a b c 2 \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} = \dfrac{bc}{a^{2}} + \dfrac{ca}{b^{2}} + \dfrac{ab}{c^{2}} yz x 2 + z x y 2 + x y z 2 = a 2 b c + b 2 c a + c 2 ab
If, x b + c − a = y c + a − b = z a + b − c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} b + c − a x = c + a − b y = a + b − c z x + y + z a + b + c \dfrac{x + y + z}{a + b + c} a + b + c x + y + z
Also, show that (b − c)x + (c − a)y + (a − b)z = 0.
Answer
Given,
x b + c − a = y c + a − b = z a + b − c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} b + c − a x = c + a − b y = a + b − c z
Let the common value of the given ratios be k.
x b + c − a = y c + a − b = z a + b − c = k \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k b + c − a x = c + a − b y = a + b − c z = k
Therefore,
x = k(b + c - a), y = k(c + a - b), z = k(a + b - c)
Adding x,y and z, we get:
⇒ x + y + z = k[(b + c − a) + (c + a − b) + (a + b − c)]
⇒ x + y + z = k(a + b + c)
⇒ k = x + y + z a + b + c \dfrac{x + y + z}{a + b + c} a + b + c x + y + z
Therefore,
x b + c − a = y c + a − b = z a + b − c = k = x + y + z a + b + c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k = \dfrac{x + y + z}{a + b + c} b + c − a x = c + a − b y = a + b − c z = k = a + b + c x + y + z
Given,
(b − c)x + (c − a)y + (a − b)z = 0.
Substituting value of x, y, z in L.H.S of above equation, we get :
⇒ (b − c)[k(b + c - a)] + (c − a)[ k(c + a - b)] + (a − b)[k(a + b - c)]
⇒ k[(b − c)(b + c - a) + (c − a)(c + a - b) + (a − b)(a + b - c)]
⇒ k[(b2 - c2 ) - a(b - c) + (c2 - a2 ) - b(c - a) + (a2 - b2 ) - c(a - b)]
⇒ k[b2 - c2 - ab + ac + c2 - a2 - bc + ab + a2 - b2 - ca + bc]
⇒ k[b2 - b2 + c2 - c2 + a2 - a2 - ab + ab + ac - ac - bc + bc]
⇒ k(0)
⇒ 0.
Hence, proved that x b + c − a = y c + a − b = z a + b − c = x + y + z a + b + c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = \dfrac{x + y + z}{a + b + c} b + c − a x = c + a − b y = a + b − c z = a + b + c x + y + z
If b is the mean proportion between a and c, show that:
a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4 = c 2 a 2
Answer
Given,
Since b is the mean proportional between a and c, we have
⇒ a : b :: b : c
⇒ a b = b c \dfrac{a}{b} = \dfrac{b}{c} b a = c b
⇒ b2 = ac
Substituting value of b2 in a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4
⇒ a 4 + a 2 b 2 + ( b 2 ) 2 ( b 2 ) 2 + b 2 c 2 + c 4 ⇒ a 4 + a 2 . a c + ( a c ) 2 ( a c ) 2 + ( a c ) . c 2 + c 4 ⇒ a 2 ( a 2 + a c + c 2 ) c 2 ( a 2 + a c + c 2 ) ⇒ a 2 c 2 . \Rightarrow \dfrac{a^{4} + a^{2}b^{2} + (b^{2})^2}{(b^{2})^2 + b^{2}c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^{4} + a^{2}.ac + (ac)^2}{(ac)^2 + (ac).c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^2(a^{2} + ac + c^2)}{c^2(a^2 + ac + c^2)} \\[1em] \Rightarrow \dfrac{a^2}{c^2}. ⇒ ( b 2 ) 2 + b 2 c 2 + c 4 a 4 + a 2 b 2 + ( b 2 ) 2 ⇒ ( a c ) 2 + ( a c ) . c 2 + c 4 a 4 + a 2 . a c + ( a c ) 2 ⇒ c 2 ( a 2 + a c + c 2 ) a 2 ( a 2 + a c + c 2 ) ⇒ c 2 a 2 .
Hence, proved that a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4 = c 2 a 2
If a : b = c : d, prove that (9a + 13b) : (9a − 13b) = (9c + 13d) : (9c − 13d).
Answer
Given,
a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
Multiplying both sides by 9 13 \dfrac{9}{13} 13 9
9 a 13 b = 9 c 13 d \dfrac{9a}{13b} = \dfrac{9c}{13d} 13 b 9 a = 13 d 9 c
Applying componendo and dividendo:
9 a + 13 b 9 a − 13 b = 9 c + 13 d 9 c − 13 d \dfrac{9a + 13b}{9a - 13b} = \dfrac{9c + 13d}{9c - 13d} 9 a − 13 b 9 a + 13 b = 9 c − 13 d 9 c + 13 d
Hence, proved that (9a + 13b) : (9a − 13b) = (9c + 13d) : (9c − 13d).
If a : b = c : d, prove that (3a + 2b) : (3a − 2b) = (3c + 2d) : (3c − 2d).
Answer
Given,
a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
Multiplying both sides by 3 2 \dfrac{3}{2} 2 3
3 a 2 b = 3 c 2 d \dfrac{3a}{2b} = \dfrac{3c}{2d} 2 b 3 a = 2 d 3 c
Applying componendo and dividendo:
3 a + 2 b 3 a − 2 b = 3 c + 2 d 3 c − 2 d \dfrac{3a + 2b}{3a - 2b} = \dfrac{3c + 2d}{3c - 2d} 3 a − 2 b 3 a + 2 b = 3 c − 2 d 3 c + 2 d
Hence, proved that (3a + 2b) : (3a − 2b) = (3c + 2d) : (3c − 2d).
If (3a + 5b) : (3a − 5b) = (3c + 5d) : (3c − 5d), prove that a : b = c : d.
Answer
Given,
3 a + 5 b 3 a − 5 b = 3 c + 5 d 3 c − 5 d \dfrac{3a + 5b}{3a - 5b} = \dfrac{3c + 5d}{3c - 5d} 3 a − 5 b 3 a + 5 b = 3 c − 5 d 3 c + 5 d
Applying componendo and dividendo, we get :
⇒ ( 3 a + 5 b ) + ( 3 a − 5 b ) ( 3 a + 5 b ) − ( 3 a − 5 b ) = ( 3 c + 5 d ) + ( 3 c − 5 d ) ( 3 c + 5 d ) − ( 3 c − 5 d ) ⇒ 3 a + 5 b + 3 a − 5 b 3 a + 5 b − 3 a + 5 b = 3 c + 5 d + 3 c − 5 d 3 c + 5 d − 3 c + 5 d ⇒ 6 a 10 b = 6 c 10 d ⇒ a b = c d \Rightarrow \dfrac{(3a + 5b) + (3a - 5b)}{(3a + 5b) - (3a - 5b)} = \dfrac{(3c + 5d) + (3c - 5d)}{(3c + 5d) - (3c - 5d)} \\[1em] \Rightarrow \dfrac{3a + 5b + 3a - 5b}{3a + 5b - 3a + 5b} = \dfrac{3c + 5d + 3c - 5d}{3c +5d - 3c + 5d} \\[1em] \Rightarrow \dfrac{6a}{10b} = \dfrac{6c}{10d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}\\[1em] ⇒ ( 3 a + 5 b ) − ( 3 a − 5 b ) ( 3 a + 5 b ) + ( 3 a − 5 b ) = ( 3 c + 5 d ) − ( 3 c − 5 d ) ( 3 c + 5 d ) + ( 3 c − 5 d ) ⇒ 3 a + 5 b − 3 a + 5 b 3 a + 5 b + 3 a − 5 b = 3 c + 5 d − 3 c + 5 d 3 c + 5 d + 3 c − 5 d ⇒ 10 b 6 a = 10 d 6 c ⇒ b a = d c
Hence, proved that a : b = c : d.
If (4a2 + 7b2 ) : (4a2 − 7b2 ) = (4c2 + 7d2 ) : (4c2 − 7d2 ), prove that a : b = c : d.
Answer
Given,
(4a2 + 7b2 ) : (4a2 − 7b2 ) = (4c2 + 7d2 ) : (4c2 − 7d2 )
⇒ 4 a 2 + 7 b 2 4 a 2 − 7 b 2 = 4 c 2 + 7 d 2 4 c 2 − 7 d 2 \Rightarrow \dfrac{4a^2 + 7b^2}{4a^2 - 7b^2} = \dfrac{4c^2 + 7d^2}{4c^2 - 7d^2} ⇒ 4 a 2 − 7 b 2 4 a 2 + 7 b 2 = 4 c 2 − 7 d 2 4 c 2 + 7 d 2
Applying componendo and dividendo:
⇒ 4 a 2 + 7 b 2 + 4 a 2 − 7 b 2 4 a 2 + 7 b 2 − ( 4 a 2 − 7 b 2 ) = 4 c 2 + 7 d 2 + 4 c 2 − 7 d 2 4 c 2 + 7 d 2 − ( 4 c 2 − 7 d 2 ) ⇒ 8 a 2 4 a 2 + 7 b 2 − 4 a 2 + 7 b 2 = 8 c 2 4 c 2 + 7 d 2 − 4 c 2 + 7 d 2 ⇒ 8 a 2 14 b 2 = 8 c 2 14 d 2 ⇒ a 2 b 2 = c 2 d 2 ⇒ a 2 b 2 = c 2 d 2 ⇒ a b = c d . \Rightarrow \dfrac{4a^2 + 7b^2 + 4a^2 - 7b^2}{4a^2 + 7b^2 - (4a^2 - 7b^2)} = \dfrac{4c^2 + 7d^2 + 4c^2 - 7d^2}{4c^2 + 7d^2 - (4c^2 - 7d^2)} \\[1em] \Rightarrow \dfrac{8a^2}{4a^2 + 7b^2 - 4a^2 + 7b^2} = \dfrac{8c^2}{4c^2 + 7d^2 - 4c^2 + 7d^2} \\[1em] \Rightarrow \dfrac{8a^2}{14b^2} = \dfrac{8c^2}{14d^2} \\[1em] \Rightarrow \dfrac{a^2}{b^2} = \dfrac{c^2}{d^2} \\[1em] \Rightarrow \sqrt{\dfrac{a^2}{b^2}} = \sqrt{\dfrac{c^2}{d^2}} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}. ⇒ 4 a 2 + 7 b 2 − ( 4 a 2 − 7 b 2 ) 4 a 2 + 7 b 2 + 4 a 2 − 7 b 2 = 4 c 2 + 7 d 2 − ( 4 c 2 − 7 d 2 ) 4 c 2 + 7 d 2 + 4 c 2 − 7 d 2 ⇒ 4 a 2 + 7 b 2 − 4 a 2 + 7 b 2 8 a 2 = 4 c 2 + 7 d 2 − 4 c 2 + 7 d 2 8 c 2 ⇒ 14 b 2 8 a 2 = 14 d 2 8 c 2 ⇒ b 2 a 2 = d 2 c 2 ⇒ b 2 a 2 = d 2 c 2 ⇒ b a = d c .
Hence, proved that a : b = c : d.
If (ma + nb) : (mc + nd) = (ma − nb) : (mc − nd), prove that a : b = c : d.
Answer
Given,
(ma + nb) : (mc + nd) = (ma − nb) : (mc − nd)
⇒ m a + n b m c + n d = m a − n b m c − n d \Rightarrow \dfrac{ma + nb}{mc + nd} = \dfrac{ma - nb}{mc - nd} ⇒ m c + n d ma + nb = m c − n d ma − nb
Apply Alternendo,
m a + n b m a − n b = m c + n d m c − n d \dfrac{ma + nb}{ma - nb} = \dfrac{mc + nd}{mc - nd} ma − nb ma + nb = m c − n d m c + n d
Applying componendo and dividendo:
⇒ ( m a + n b ) + ( m a − n b ) ( m a + n b ) − ( m a − n b ) = ( m c + n d ) + ( m c − n d ) ( m c + n d ) − ( m c − n d ) ⇒ m a + n b + m a − n b m a + n b − m a + n b = m c + n d + m c − n d m c + n d − m c + n d ⇒ 2 m a 2 n b = 2 m c 2 n d ⇒ a b = c d . \Rightarrow \dfrac{(ma + nb) + (ma - nb)}{(ma + nb) - (ma - nb)} = \dfrac{(mc + nd) + (mc - nd)}{(mc + nd) - (mc - nd)} \\[1em] \Rightarrow \dfrac{ma + nb + ma - nb}{ma + nb - ma + nb} = \dfrac{mc + nd + mc - nd}{mc + nd - mc + nd} \\[1em] \Rightarrow \dfrac{2ma}{2nb} = \dfrac{2mc}{2nd} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}. ⇒ ( ma + nb ) − ( ma − nb ) ( ma + nb ) + ( ma − nb ) = ( m c + n d ) − ( m c − n d ) ( m c + n d ) + ( m c − n d ) ⇒ ma + nb − ma + nb ma + nb + ma − nb = m c + n d − m c + n d m c + n d + m c − n d ⇒ 2 nb 2 ma = 2 n d 2 m c ⇒ b a = d c .
Hence, proved that a : b = c : d.
If 5 x + 6 y 5 x − 6 y = 5 u + 6 v 5 u − 6 v \dfrac{5x + 6y}{5x - 6y} = \dfrac{5u + 6v}{5u - 6v} 5 x − 6 y 5 x + 6 y = 5 u − 6 v 5 u + 6 v x y = u v \dfrac{x}{y} = \dfrac{u}{v} y x = v u
Answer
Apply Componendo & Dividendo :
⇒ ( 5 x + 6 y ) + ( 5 x − 6 y ) ( 5 x + 6 y ) − ( 5 x − 6 y ) = ( 5 u + 6 v ) + ( 5 u − 6 v ) ( 5 u + 6 v ) − ( 5 u − 6 v ) ⇒ 5 x + 6 y + 5 x − 6 y 5 x + 6 y − 5 x + 6 y = 5 u + 6 v + 5 u − 6 v 5 u + 6 v − 5 u + 6 v ⇒ 10 x 12 y = 10 u 12 v ⇒ x y = u v . \Rightarrow \dfrac{(5x + 6y) + (5x - 6y)}{(5x + 6y) - (5x - 6y)} = \dfrac{(5u + 6v) + (5u - 6v)}{(5u + 6v) - (5u - 6v)} \\[1em] \Rightarrow \dfrac{5x + 6y + 5x - 6y}{5x + 6y - 5x + 6y} = \dfrac{5u + 6v + 5u - 6v}{5u + 6v - 5u + 6v} \\[1em] \Rightarrow \dfrac{10x}{12y} = \dfrac{10u}{12v} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{u}{v}. ⇒ ( 5 x + 6 y ) − ( 5 x − 6 y ) ( 5 x + 6 y ) + ( 5 x − 6 y ) = ( 5 u + 6 v ) − ( 5 u − 6 v ) ( 5 u + 6 v ) + ( 5 u − 6 v ) ⇒ 5 x + 6 y − 5 x + 6 y 5 x + 6 y + 5 x − 6 y = 5 u + 6 v − 5 u + 6 v 5 u + 6 v + 5 u − 6 v ⇒ 12 y 10 x = 12 v 10 u ⇒ y x = v u .
Hence, proved that x y = u v \dfrac{x}{y} = \dfrac{u}{v} y x = v u
If x = 6 a b a + b x = \dfrac{6ab}{a + b} x = a + b 6 ab ( x + 3 a x − 3 a + x + 3 b x − 3 b ) = 2 \Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2 ( x − 3 a x + 3 a + x − 3 b x + 3 b ) = 2
Answer
Given,
x = 6 a b a + b x = 3 a × 2 b a + b x 3 a = 2 b a + b x = \dfrac{6ab}{a + b} \\[1em] x = \dfrac{3a \times 2b}{a + b} \\[1em] \dfrac{x}{3a} = \dfrac{2b}{a + b} x = a + b 6 ab x = a + b 3 a × 2 b 3 a x = a + b 2 b
Applying componendo and dividendo rule,
⇒ x + 3 a x − 3 a = 2 b + ( a + b ) 2 b − ( a + b ) ⇒ x + 3 a x − 3 a = a + 3 b 2 b − a − b ⇒ x + 3 a x − 3 a = a + 3 b b − a ...(1) \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{2b + (a + b)}{2b - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{a + 3b}{2b - a - b} \\[1em] \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{a + 3b}{b - a} \text{ ...(1)} ⇒ x − 3 a x + 3 a = 2 b − ( a + b ) 2 b + ( a + b ) ⇒ x − 3 a x + 3 a = 2 b − a − b a + 3 b ⇒ x − 3 a x + 3 a = b − a a + 3 b ...(1)
Again solving x,
⇒ x = 6 a b a + b ⇒ x = 3 b × 2 a a + b ⇒ x 3 b = 2 a a + b \Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow x = \dfrac{3b \times 2a}{a + b} \\[1em] \Rightarrow \dfrac{x}{3b} = \dfrac{2a}{a + b} ⇒ x = a + b 6 ab ⇒ x = a + b 3 b × 2 a ⇒ 3 b x = a + b 2 a
Apply the Componendo and Dividendo rule,
⇒ x + 3 b x − 3 b = 2 a + ( a + b ) 2 a − ( a + b ) ⇒ x + 3 b x − 3 b = 3 a + b 2 a − a − b ⇒ x + 3 b x − 3 b = 3 a + b a − b .....(2) \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{2a + (a + b)}{2a - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{3a + b}{2a - a - b} \\[1em] \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{3a + b}{a - b} \text{ .....(2)} ⇒ x − 3 b x + 3 b = 2 a − ( a + b ) 2 a + ( a + b ) ⇒ x − 3 b x + 3 b = 2 a − a − b 3 a + b ⇒ x − 3 b x + 3 b = a − b 3 a + b .....(2)
Adding equations (1) and (2), we get :
⇒ ( x + 3 a x − 3 a ) + ( x + 3 b x − 3 b ) = ( a + 3 b b − a ) + ( 3 a + b a − b ) = ( a + 3 b b − a ) + ( 3 a + b − ( b − a ) ) = ( a + 3 b b − a ) − ( 3 a + b b − a ) = ( a + 3 b − ( 3 a + b ) b − a ) = ( a + 3 b − 3 a − b b − a ) = ( 2 b − 2 a b − a ) = ( 2 ( b − a ) b − a ) = 2. \Rightarrow \Big(\dfrac{x + 3a}{x - 3a}\Big) + \Big(\dfrac{x + 3b}{x - 3b}\Big) = \Big(\dfrac{a + 3b}{b - a}\Big) + \Big(\dfrac{3a + b}{a - b}\Big) \\[1em] = \Big(\dfrac{a + 3b}{b - a}\Big) + \Big(\dfrac{3a + b}{ - (b - a)}\Big) \\[1em] = \Big(\dfrac{a + 3b}{b - a}\Big) - \Big(\dfrac{3a + b}{b - a}\Big) \\[1em] = \Big(\dfrac{a + 3b - (3a + b)}{b - a}\Big) \\[1em] = \Big(\dfrac{a + 3b - 3a - b}{b - a}\Big) \\[1em] = \Big(\dfrac{2b - 2a}{b - a}\Big) \\[1em] = \Big(\dfrac{2(b - a)}{b - a}\Big) \\[1em] = 2. ⇒ ( x − 3 a x + 3 a ) + ( x − 3 b x + 3 b ) = ( b − a a + 3 b ) + ( a − b 3 a + b ) = ( b − a a + 3 b ) + ( − ( b − a ) 3 a + b ) = ( b − a a + 3 b ) − ( b − a 3 a + b ) = ( b − a a + 3 b − ( 3 a + b ) ) = ( b − a a + 3 b − 3 a − b ) = ( b − a 2 b − 2 a ) = ( b − a 2 ( b − a ) ) = 2.
Hence, proved that ( x + 3 a x − 3 a + x + 3 b x − 3 b ) = 2 \Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2 ( x − 3 a x + 3 a + x − 3 b x + 3 b ) = 2
If, x 3 + 3 x 3 x 2 + 1 = 341 91 \dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91} 3 x 2 + 1 x 3 + 3 x = 91 341
Answer
Given,
x 3 + 3 x 3 x 2 + 1 = 341 91 \dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91} 3 x 2 + 1 x 3 + 3 x = 91 341
Solving L.H.S:
Applying Componendo and Dividendo, we get :
⇒ ( x 3 + 3 x ) + ( 3 x 2 + 1 ) ( x 3 + 3 x ) − ( 3 x 2 + 1 ) ⇒ ( x 3 + 3 x + 3 x 2 + 1 ) ( x 3 + 3 x − 3 x 2 − 1 ) ⇒ ( x + 1 ) 3 ( x − 1 ) 3 ⇒ ( x + 1 x − 1 ) 3 . \Rightarrow \dfrac{(x^{3} + 3x) + (3x^{2} + 1)}{(x^{3} + 3x) - (3x^{2} + 1)} \\[1em] \Rightarrow \dfrac{(x^{3} + 3x + 3x^{2} + 1)}{(x^{3} + 3x - 3x^{2} - 1)} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3. ⇒ ( x 3 + 3 x ) − ( 3 x 2 + 1 ) ( x 3 + 3 x ) + ( 3 x 2 + 1 ) ⇒ ( x 3 + 3 x − 3 x 2 − 1 ) ( x 3 + 3 x + 3 x 2 + 1 ) ⇒ ( x − 1 ) 3 ( x + 1 ) 3 ⇒ ( x − 1 x + 1 ) 3 .
Solving R.H.S:
Apply Componendo and Dividendo:
⇒ 341 + 91 341 − 91 ⇒ 432 250 . \Rightarrow \dfrac{341 + 91}{341 - 91} \\[1em] \Rightarrow \dfrac{432}{250}. ⇒ 341 − 91 341 + 91 ⇒ 250 432 .
Equating L.H.S. and R.H.S.,
⇒ ( x + 1 x − 1 ) 3 = 432 250 ⇒ ( x + 1 x − 1 ) 3 = 216 125 ⇒ ( x + 1 x − 1 ) 3 = ( 6 5 ) 3 ⇒ ( x + 1 x − 1 ) = ( 6 5 ) ⇒ 5 ( x + 1 ) = 6 ( x − 1 ) ⇒ 5 x + 5 = 6 x − 6 ⇒ 6 x − 5 x = 6 + 5 ⇒ x = 11. \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{432}{250} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{216}{125} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{6}{5}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big) = \Big(\dfrac{6}{5}\Big) \\[1em] \Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 6x - 5x = 6 + 5 \\[1em] \Rightarrow x = 11. ⇒ ( x − 1 x + 1 ) 3 = 250 432 ⇒ ( x − 1 x + 1 ) 3 = 125 216 ⇒ ( x − 1 x + 1 ) 3 = ( 5 6 ) 3 ⇒ ( x − 1 x + 1 ) = ( 5 6 ) ⇒ 5 ( x + 1 ) = 6 ( x − 1 ) ⇒ 5 x + 5 = 6 x − 6 ⇒ 6 x − 5 x = 6 + 5 ⇒ x = 11.
Hence, proved that x = 11.
If x + 2 + x − 3 x + 2 − x − 3 = 5 \dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5 x + 2 − x − 3 x + 2 + x − 3 = 5
Answer
Given,
x + 2 + x − 3 x + 2 − x − 3 = 5 \dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5 x + 2 − x − 3 x + 2 + x − 3 = 5
Applying Componendo and Dividendo, we get :
⇒ ( x + 2 + x − 3 ) + ( x + 2 − x − 3 ) ( x + 2 + x − 3 ) − ( x + 2 − x − 3 ) = 5 + 1 5 − 1 ⇒ 2 x + 2 2 x − 3 = 6 4 ⇒ x + 2 x − 3 = 3 2 ⇒ ( x + 2 x − 3 ) 2 = ( 3 2 ) 2 \Rightarrow \dfrac{(\sqrt{x + 2} + \sqrt{x - 3}) + (\sqrt{x + 2} - \sqrt{x - 3})}{(\sqrt{x + 2} + \sqrt{x - 3}) - (\sqrt{x + 2} - \sqrt{x - 3})} = \dfrac{5 + 1}{5 - 1}\\[1em] \Rightarrow \dfrac{2\sqrt{x + 2}}{2\sqrt{x - 3}} = \dfrac{6}{4}\\[1em] \Rightarrow \dfrac{\sqrt{x + 2}}{\sqrt{x - 3}} = \dfrac{3}{2}\\[1em] \Rightarrow \Big(\dfrac{\sqrt{x + 2}}{\sqrt{x - 3}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2 ⇒ ( x + 2 + x − 3 ) − ( x + 2 − x − 3 ) ( x + 2 + x − 3 ) + ( x + 2 − x − 3 ) = 5 − 1 5 + 1 ⇒ 2 x − 3 2 x + 2 = 4 6 ⇒ x − 3 x + 2 = 2 3 ⇒ ( x − 3 x + 2 ) 2 = ( 2 3 ) 2
Squaring both sides, we get :
⇒ x + 2 x − 3 = ( 9 4 ) ⇒ 4 ( x + 2 ) = 9 ( x − 3 ) ⇒ 4 x + 8 = 9 x − 27 ⇒ 9 x − 4 x = 27 + 8 ⇒ 5 x = 35 ⇒ x = 35 5 = 7. \Rightarrow \dfrac{x + 2}{x - 3} = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(x + 2) = 9(x - 3) \\[1em] \Rightarrow 4x + 8 = 9x - 27 \\[1em] \Rightarrow 9x - 4x = 27 + 8 \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \dfrac{35}{5} = 7. ⇒ x − 3 x + 2 = ( 4 9 ) ⇒ 4 ( x + 2 ) = 9 ( x − 3 ) ⇒ 4 x + 8 = 9 x − 27 ⇒ 9 x − 4 x = 27 + 8 ⇒ 5 x = 35 ⇒ x = 5 35 = 7.
Hence, proved that x = 7.
If x + 5 + x − 16 x + 5 − x − 16 = 7 3 \dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3} x + 5 − x − 16 x + 5 + x − 16 = 3 7
Answer
Given,
x + 5 + x − 16 x + 5 − x − 16 = 7 3 \dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3} x + 5 − x − 16 x + 5 + x − 16 = 3 7
Applying Componendo and Dividendo, we get :
⇒ ( x + 5 + x − 16 ) + ( x + 5 − x − 16 ) ( x + 5 + x − 16 ) − ( x + 5 − x − 16 ) = 7 + 3 7 − 3 ⇒ ( x + 5 + x − 16 + x + 5 − x − 16 ) ( x + 5 + x − 16 − x + 5 + x − 16 ) = 10 4 ⇒ 2 x + 5 2 x − 16 = 5 2 ⇒ x + 5 x − 16 = 5 2 \Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16}) + (\sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16}) - (\sqrt{x + 5} - \sqrt{x - 16})} = \dfrac{7 + 3}{7 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16} + \sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16} - \sqrt{x + 5} + \sqrt{x - 16})} = \dfrac{10}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{\sqrt{x + 5}}{\sqrt{x - 16}} = \dfrac{5}{2} \\[1em] ⇒ ( x + 5 + x − 16 ) − ( x + 5 − x − 16 ) ( x + 5 + x − 16 ) + ( x + 5 − x − 16 ) = 7 − 3 7 + 3 ⇒ ( x + 5 + x − 16 − x + 5 + x − 16 ) ( x + 5 + x − 16 + x + 5 − x − 16 ) = 4 10 ⇒ 2 x − 16 2 x + 5 = 2 5 ⇒ x − 16 x + 5 = 2 5
Squaring both sides, we get :
⇒ ( x + 5 x − 16 ) 2 = ( 5 2 ) 2 ⇒ ( x + 5 x − 16 ) = ( 25 4 ) ⇒ 4 ( x + 5 ) = 25 ( x − 16 ) ⇒ 4 x + 20 = 25 x − 400 ⇒ 25 x − 4 x = 400 + 20 ⇒ 21 x = 420 ⇒ x = 420 21 ⇒ x = 20. \Rightarrow \Big(\dfrac{\sqrt{x + 5}}{\sqrt{x - 16}}\Big)^2 = \Big(\dfrac{5}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{x + 5}{x - 16}\Big) = \Big(\dfrac{25}{4}\Big) \\[1em] \Rightarrow 4(x + 5) = 25(x - 16) \\[1em] \Rightarrow 4x + 20 = 25x - 400 \\[1em] \Rightarrow 25x - 4x = 400 + 20 \\[1em] \Rightarrow 21x = 420 \\[1em] \Rightarrow x = \dfrac{420}{21} \\[1em] \Rightarrow x = 20. ⇒ ( x − 16 x + 5 ) 2 = ( 2 5 ) 2 ⇒ ( x − 16 x + 5 ) = ( 4 25 ) ⇒ 4 ( x + 5 ) = 25 ( x − 16 ) ⇒ 4 x + 20 = 25 x − 400 ⇒ 25 x − 4 x = 400 + 20 ⇒ 21 x = 420 ⇒ x = 21 420 ⇒ x = 20.
Hence, proved that x = 20.
If 3 x + 2 x − 1 3 x − 2 x − 1 = 5 \dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5 3 x − 2 x − 1 3 x + 2 x − 1 = 5 3 2 \dfrac{3}{2} 2 3
Answer
(i) Given,
3 x + 2 x − 1 3 x − 2 x − 1 = 5 \dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5 3 x − 2 x − 1 3 x + 2 x − 1 = 5
Applying Componendo and Dividendo, we get :
⇒ 3 x + 2 x − 1 + 3 x − 2 x − 1 3 x + 2 x − 1 − ( 3 x − 2 x − 1 ) = 5 + 1 5 − 1 ⇒ 2 3 x 3 x + 2 x − 1 − 3 x + 2 x − 1 = 6 4 ⇒ 2 3 x 2 2 x − 1 = 3 2 ⇒ 3 x 2 x − 1 = 3 2 \Rightarrow \dfrac{\sqrt{3x} + \sqrt{2x - 1} + \sqrt{3x} - \sqrt{2x - 1}}{\sqrt{3x} + \sqrt{2x - 1} - (\sqrt{3x} - \sqrt{2x - 1})} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{\sqrt{3x} + \sqrt{2x - 1} - \sqrt{3x} + \sqrt{2x - 1}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{2 \sqrt{2x - 1}} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3x}}{ \sqrt{2x - 1}} = \dfrac{3}{2} ⇒ 3 x + 2 x − 1 − ( 3 x − 2 x − 1 ) 3 x + 2 x − 1 + 3 x − 2 x − 1 = 5 − 1 5 + 1 ⇒ 3 x + 2 x − 1 − 3 x + 2 x − 1 2 3 x = 4 6 ⇒ 2 2 x − 1 2 3 x = 2 3 ⇒ 2 x − 1 3 x = 2 3
Squaring both sides, we get :
⇒ ( 3 x 2 x − 1 ) 2 = ( 3 2 ) 2 ⇒ ( 3 x 2 x − 1 ) = ( 9 4 ) ⇒ 4 ( 3 x ) = 9 ( 2 x − 1 ) ⇒ 12 x = 18 x − 9 ⇒ 18 x − 12 x = 9 ⇒ 6 x = 9 ⇒ x = 9 6 = 3 2 \Rightarrow \Big(\dfrac{\sqrt{3x}}{ \sqrt{2x - 1}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{3x}{2x - 1}\Big) = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(3x) = 9(2x - 1) \\[1em] \Rightarrow 12x = 18x - 9 \\[1em] \Rightarrow 18x - 12x = 9 \\[1em] \Rightarrow 6x = 9 \\[1em] \Rightarrow x = \dfrac{9}{6} = \dfrac{3}{2} ⇒ ( 2 x − 1 3 x ) 2 = ( 2 3 ) 2 ⇒ ( 2 x − 1 3 x ) = ( 4 9 ) ⇒ 4 ( 3 x ) = 9 ( 2 x − 1 ) ⇒ 12 x = 18 x − 9 ⇒ 18 x − 12 x = 9 ⇒ 6 x = 9 ⇒ x = 6 9 = 2 3
Hence, proved that x = 3 2 \dfrac{3}{2} 2 3
Using properties of proportion, solve for x. Given that x is positive :
2 x + 4 x 2 − 1 2 x − 4 x 2 − 1 = 4 \dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 = 4
Answer
Given,
2 x + 4 x 2 − 1 2 x − 4 x 2 − 1 = 4 \dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 = 4
Applying Componendo and Dividendo, we get :
⇒ 2 x + 4 x 2 − 1 + 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 − ( 2 x − 4 x 2 − 1 ) = 4 + 1 4 − 1 ⇒ 4 x 2 x + 4 x 2 − 1 − 2 x + 4 x 2 − 1 = 5 3 ⇒ 4 x 2 ( 4 x 2 − 1 ) = 5 3 ⇒ 2 x 4 x 2 − 1 = 5 3 \Rightarrow \dfrac{2x + \sqrt{4x^{2} - 1} + 2x - \sqrt{4x^{2} - 1}}{2x + \sqrt{4x^{2} - 1} - (2x - \sqrt{4x^{2} - 1})} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{4x}{2x + \sqrt{4x^{2} - 1} - 2x + \sqrt{4x^{2} - 1}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{4x}{2\sqrt{(4x^2 - 1)}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{4x^2 - 1}} = \dfrac{5}{3} \\[1em] ⇒ 2 x + 4 x 2 − 1 − ( 2 x − 4 x 2 − 1 ) 2 x + 4 x 2 − 1 + 2 x − 4 x 2 − 1 = 4 − 1 4 + 1 ⇒ 2 x + 4 x 2 − 1 − 2 x + 4 x 2 − 1 4 x = 3 5 ⇒ 2 ( 4 x 2 − 1 ) 4 x = 3 5 ⇒ 4 x 2 − 1 2 x = 3 5
Squaring both sides, we get :
⇒ ( 2 x 4 x 2 − 1 ) 2 = ( 5 3 ) 2 ⇒ ( 4 x 2 4 x 2 − 1 ) = ( 25 9 ) ⇒ 9 ( 4 x 2 ) = 25 ( 4 x 2 − 1 ) ⇒ 36 x 2 = 100 x 2 − 25 ⇒ 100 x 2 − 36 x 2 = 25 ⇒ 64 x 2 = 25 ⇒ x 2 = 25 64 ⇒ x = 25 64 ⇒ x = 5 8 \Rightarrow \Big(\dfrac{2x}{\sqrt{4x^2 - 1}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{4x^2}{4x^2 - 1}\Big) = \Big(\dfrac{25}{9}\Big) \\[1em] \Rightarrow 9(4x^2) = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 100x^2 - 36x^2 = 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \dfrac{25}{64} \\[1em] \Rightarrow x = \sqrt{\dfrac{25}{64}} \\[1em] \Rightarrow x = \dfrac{5}{8} ⇒ ( 4 x 2 − 1 2 x ) 2 = ( 3 5 ) 2 ⇒ ( 4 x 2 − 1 4 x 2 ) = ( 9 25 ) ⇒ 9 ( 4 x 2 ) = 25 ( 4 x 2 − 1 ) ⇒ 36 x 2 = 100 x 2 − 25 ⇒ 100 x 2 − 36 x 2 = 25 ⇒ 64 x 2 = 25 ⇒ x 2 = 64 25 ⇒ x = 64 25 ⇒ x = 8 5
Hence, x = 5 8 \dfrac{5}{8} 8 5
Using properties of proportion solve for x, given :
5 x + 2 x − 6 5 x − 2 x − 6 = 4 \dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 5 x − 2 x − 6 5 x + 2 x − 6 = 4
Answer
Given,
5 x + 2 x − 6 5 x − 2 x − 6 = 4 \dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 5 x − 2 x − 6 5 x + 2 x − 6 = 4
Applying Componendo and Dividendo, we get :
⇒ 5 x + 2 x − 6 + 5 x − 2 x − 6 5 x + 2 x − 6 − ( 5 x − 2 x − 6 ) = 4 + 1 4 − 1 ⇒ 2 5 x 5 x + 2 x − 6 − 5 x + 2 x − 6 = 5 3 ⇒ 2 5 x 2 2 x − 6 = 5 3 ⇒ 5 x 2 x − 6 = 5 3 \Rightarrow \dfrac{\sqrt{5x} + \sqrt{2x - 6} + \sqrt{5x} - \sqrt{2x - 6}}{\sqrt{5x} + \sqrt{2x - 6} - (\sqrt{5x} - \sqrt{2x - 6})} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{5x}}{\sqrt{5x} + \sqrt{2x - 6} - \sqrt{5x} + \sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{\sqrt{5x}}{\sqrt{2x - 6}} = \dfrac{5}{3} ⇒ 5 x + 2 x − 6 − ( 5 x − 2 x − 6 ) 5 x + 2 x − 6 + 5 x − 2 x − 6 = 4 − 1 4 + 1 ⇒ 5 x + 2 x − 6 − 5 x + 2 x − 6 2 5 x = 3 5 ⇒ 2 2 x − 6 2 5 x = 3 5 ⇒ 2 x − 6 5 x = 3 5
Squaring both sides, we get :
⇒ ( 5 x 2 x − 6 ) 2 = ( 5 3 ) 2 ⇒ ( 5 x 2 x − 6 ) = ( 25 9 ) ⇒ 9 ( 5 x ) = 25 ( 2 x − 6 ) ⇒ 45 x = 50 x − 150 ⇒ 50 x − 45 x = 150 ⇒ 5 x = 150 ⇒ x = 150 5 = 30 \Rightarrow \Big(\dfrac{\sqrt{5x}}{\sqrt{2x - 6}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{5x}{2x - 6}\Big) = \Big(\dfrac{25}{9}\Big) \\[1em] \Rightarrow 9(5x) = 25(2x - 6) \\[1em] \Rightarrow 45x = 50x - 150 \\[1em] \Rightarrow 50x - 45x = 150 \\[1em] \Rightarrow 5x = 150 \\[1em] \Rightarrow x = \dfrac{150}{5} = 30 ⇒ ( 2 x − 6 5 x ) 2 = ( 3 5 ) 2 ⇒ ( 2 x − 6 5 x ) = ( 9 25 ) ⇒ 9 ( 5 x ) = 25 ( 2 x − 6 ) ⇒ 45 x = 50 x − 150 ⇒ 50 x − 45 x = 150 ⇒ 5 x = 150 ⇒ x = 5 150 = 30
Hence, x = 30.
Using Componendo and Dividendo solve for x :
2 x + 2 + 2 x − 1 2 x + 2 − 2 x − 1 = 3 \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3 2 x + 2 − 2 x − 1 2 x + 2 + 2 x − 1 = 3
Answer
Given,
2 x + 2 + 2 x − 1 2 x + 2 − 2 x − 1 = 3 \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3 2 x + 2 − 2 x − 1 2 x + 2 + 2 x − 1 = 3
Applying Componendo and Dividendo, we get :
⇒ 2 x + 2 + 2 x − 1 + 2 x + 2 − 2 x − 1 2 x + 2 + 2 x − 1 − ( 2 x + 2 − 2 x − 1 ) = 3 + 1 3 − 1 ⇒ 2 2 x + 2 2 x + 2 + 2 x − 1 − 2 x + 2 + 2 x − 1 = 4 2 ⇒ 2 2 x + 2 2 2 x − 1 = 2 \Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{\sqrt{2x + 2} + \sqrt{2x - 1} - \sqrt{2x + 2} + \sqrt{2x - 1}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = 2 ⇒ 2 x + 2 + 2 x − 1 − ( 2 x + 2 − 2 x − 1 ) 2 x + 2 + 2 x − 1 + 2 x + 2 − 2 x − 1 = 3 − 1 3 + 1 ⇒ 2 x + 2 + 2 x − 1 − 2 x + 2 + 2 x − 1 2 2 x + 2 = 2 4 ⇒ 2 2 x − 1 2 2 x + 2 = 2
Squaring both sides, we get :
⇒ ( 2 x + 2 2 x − 1 ) 2 = 2 2 ⇒ ( 2 x + 2 2 x − 1 ) = 4 ⇒ 2 x + 2 = 4 ( 2 x − 1 ) ⇒ 2 x + 2 = 8 x − 4 ⇒ 8 x − 2 x = 4 + 2 ⇒ 6 x = 6 ⇒ x = 6 6 = 1. \Rightarrow \Big(\dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}}\Big)^2 = 2^2 \\[1em] \Rightarrow \Big(\dfrac{2x + 2}{2x - 1}\Big) = 4 \\[1em] \Rightarrow 2x + 2 = 4(2x - 1) \\[1em] \Rightarrow 2x + 2 = 8x - 4 \\[1em] \Rightarrow 8x - 2x = 4 + 2 \\[1em] \Rightarrow 6x = 6 \\[1em] \Rightarrow x = \dfrac{6}{6} = 1. ⇒ ( 2 x − 1 2 x + 2 ) 2 = 2 2 ⇒ ( 2 x − 1 2 x + 2 ) = 4 ⇒ 2 x + 2 = 4 ( 2 x − 1 ) ⇒ 2 x + 2 = 8 x − 4 ⇒ 8 x − 2 x = 4 + 2 ⇒ 6 x = 6 ⇒ x = 6 6 = 1.
Hence, x = 1.
If 16 ( a − x a + x ) 3 = ( a + x a − x ) 16\Big(\dfrac{a - x}{a + x}\Big)^{3} = \Big(\dfrac{a + x}{a - x}\Big) 16 ( a + x a − x ) 3 = ( a − x a + x ) a 3 \dfrac{a}{3} 3 a
Answer
Given,
⇒ 16 ( a − x a + x ) 3 = ( a + x a − x ) \Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big) ⇒ 16 ( a + x a − x ) 3 = ( a − x a + x )
Let, r = a + x a − x , 1 r = a − x a + x . r = \dfrac{a + x}{a - x} , \dfrac{1}{r} = \dfrac{a - x}{a + x}. r = a − x a + x , r 1 = a + x a − x .
Substituting value of r and 1 r \dfrac{1}{r} r 1 16 ( a − x a + x ) 3 = ( a + x a − x ) 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big) 16 ( a + x a − x ) 3 = ( a − x a + x )
⇒ 16 × ( 1 r ) 3 = r ⇒ 16 = r 4 ⇒ r 4 = 2 4 ⇒ r = 2. ⇒ a + x a − x = 2 ⇒ a + x = 2 a − 2 x ⇒ 3 x = a ⇒ x = a 3 . \Rightarrow 16 \times \Big(\dfrac{1}{r}\Big)^3 = r \\[1em] \Rightarrow 16 = r^4 \\[1em] \Rightarrow r^4 = 2^4 \\[1em] \Rightarrow r = 2. \\[1em] \Rightarrow \dfrac{a + x}{a - x} = 2 \\[1em] \Rightarrow a + x = 2a - 2x \\[1em] \Rightarrow 3x = a \\[1em] \Rightarrow x = \dfrac{a}{3}. ⇒ 16 × ( r 1 ) 3 = r ⇒ 16 = r 4 ⇒ r 4 = 2 4 ⇒ r = 2. ⇒ a − x a + x = 2 ⇒ a + x = 2 a − 2 x ⇒ 3 x = a ⇒ x = 3 a .
Hence, proved that x = a 3 \dfrac{a}{3} 3 a
If x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} x = 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 2 - 4ax + 1 = 0.
Answer
Given,
⇒ x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 \Rightarrow x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} ⇒ x = 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1
Applying componendo and dividendo, we get :
⇒ x + 1 x − 1 = 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 − ( 2 a + 1 − 2 a − 1 ) ⇒ x + 1 x − 1 = 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 − 2 a + 1 + 2 a − 1 ⇒ x + 1 x − 1 = 2 2 a + 1 2 2 a − 1 ⇒ x + 1 x − 1 = 2 a + 1 2 a − 1 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a + 1} - \sqrt{2a - 1})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - \sqrt{2a + 1} + \sqrt{2a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}} ⇒ x − 1 x + 1 = 2 a + 1 + 2 a − 1 − ( 2 a + 1 − 2 a − 1 ) 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 ⇒ x − 1 x + 1 = 2 a + 1 + 2 a − 1 − 2 a + 1 + 2 a − 1 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 ⇒ x − 1 x + 1 = 2 2 a − 1 2 2 a + 1 ⇒ x − 1 x + 1 = 2 a − 1 2 a + 1
Squaring both sides, we get :
⇒ ( x + 1 ) 2 ( x − 1 ) 2 = 2 a + 1 2 a − 1 ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = 2 a + 1 2 a − 1 ⇒ ( x 2 + 1 + 2 x ) ( 2 a − 1 ) = ( x 2 + 1 − 2 x ) ( 2 a + 1 ) ⇒ 2 a x 2 − x 2 + 2 a − 1 + 4 a x − 2 x = 2 a x 2 + x 2 + 2 a + 1 − 4 a x − 2 x ⇒ 2 a x 2 − 2 a x 2 + x 2 + x 2 + 2 a − 2 a + 1 + 1 − 4 a x − 4 a x − 2 x + 2 x = 0 ⇒ 2 x 2 − 8 a x + 2 = 0 ⇒ 2 ( x 2 − 4 a x + 1 ) = 0 ⇒ x 2 − 4 a x + 1 = 0. \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 1}{2a - 1} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 1}{2a - 1} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 1) = (x^2 + 1 - 2x)(2a + 1) \\[1em] \Rightarrow 2ax^2 - x^2 + 2a - 1 + 4ax - 2x = 2ax^2 + x^2 + 2a + 1 - 4ax - 2x \\[1em] \Rightarrow 2ax^2 - 2ax^2 + x^2 + x^2 + 2a - 2a + 1 + 1 - 4ax - 4ax - 2x + 2x = 0 \\[1em] \Rightarrow 2x^2 - 8ax + 2 = 0 \\[1em] \Rightarrow 2(x^2 - 4ax + 1) = 0 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0. ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = 2 a − 1 2 a + 1 ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = 2 a − 1 2 a + 1 ⇒ ( x 2 + 1 + 2 x ) ( 2 a − 1 ) = ( x 2 + 1 − 2 x ) ( 2 a + 1 ) ⇒ 2 a x 2 − x 2 + 2 a − 1 + 4 a x − 2 x = 2 a x 2 + x 2 + 2 a + 1 − 4 a x − 2 x ⇒ 2 a x 2 − 2 a x 2 + x 2 + x 2 + 2 a − 2 a + 1 + 1 − 4 a x − 4 a x − 2 x + 2 x = 0 ⇒ 2 x 2 − 8 a x + 2 = 0 ⇒ 2 ( x 2 − 4 a x + 1 ) = 0 ⇒ x 2 − 4 a x + 1 = 0.
Hence, proved that x2 - 4ax + 1 = 0.
If x = b + 3 a + b − 3 a b + 3 a − b − 3 a x = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a}}{\sqrt{b + 3a} - \sqrt{b - 3a}} x = b + 3 a − b − 3 a b + 3 a + b − 3 a 2 - 2bx + 3a = 0.
Answer
Given,
⇒ x = b + 3 a + b − 3 a b + 3 a − b − 3 a \Rightarrow x = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a}}{\sqrt{b + 3a} - \sqrt{b - 3a}} ⇒ x = b + 3 a − b − 3 a b + 3 a + b − 3 a
Applying componendo and dividendo,
⇒ x + 1 x − 1 = b + 3 a + b − 3 a + b + 3 a − b − 3 a b + 3 a + b − 3 a − ( b + 3 a − b − 3 a ) ⇒ x + 1 x − 1 = b + 3 a + b − 3 a + b + 3 a − b − 3 a b + 3 a + b − 3 a − b + 3 a + b − 3 a ⇒ x + 1 x − 1 = 2 b + 3 a 2 b − 3 a ⇒ x + 1 x − 1 = b + 3 a b − 3 a \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - (\sqrt{b + 3a} - \sqrt{b - 3a})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - \sqrt{b + 3a} + \sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{b + 3a}}{2\sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a}}{\sqrt{b - 3a}} ⇒ x − 1 x + 1 = b + 3 a + b − 3 a − ( b + 3 a − b − 3 a ) b + 3 a + b − 3 a + b + 3 a − b − 3 a ⇒ x − 1 x + 1 = b + 3 a + b − 3 a − b + 3 a + b − 3 a b + 3 a + b − 3 a + b + 3 a − b − 3 a ⇒ x − 1 x + 1 = 2 b − 3 a 2 b + 3 a ⇒ x − 1 x + 1 = b − 3 a b + 3 a
Squaring both sides:
⇒ ( x + 1 ) 2 ( x − 1 ) 2 = b + 3 a b − 3 a ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = b + 3 a b − 3 a ⇒ ( x 2 + 1 + 2 x ) ( b − 3 a ) = ( x 2 + 1 − 2 x ) ( b + 3 a ) ⇒ b x 2 + b + 2 b x − 3 a x 2 − 3 a − 6 a x = b x 2 + b − 2 b x + 3 a x 2 + 3 a − 6 a x ⇒ b x 2 − b x 2 + b − b + 2 b x + 2 b x − 3 a x 2 − 3 a x 2 − 3 a − 3 a − 6 a x + 6 a x = 0 ⇒ 4 b x − 6 a x 2 − 6 a = 0 ⇒ 6 a x 2 − 4 b x + 6 a = 0 ⇒ 2 ( 3 a x 2 − 2 b x + 3 a ) = 0 ⇒ 3 a x 2 − 2 b x + 3 a = 0. \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{b + 3a}{b - 3a} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{b + 3a}{b - 3a} \\[1em] \Rightarrow (x^2 + 1 + 2x)(b - 3a) = (x^2 + 1 - 2x)(b + 3a) \\[1em] \Rightarrow bx^2 + b + 2bx - 3ax^2 - 3a - 6ax = bx^2 + b - 2bx + 3ax^2 + 3a - 6ax \\[1em] \Rightarrow bx^2 - bx^2 + b - b + 2bx + 2bx - 3ax^2 - 3ax^2 - 3a - 3a - 6ax + 6ax = 0 \\[1em] \Rightarrow 4bx - 6ax^2 - 6a = 0 \\[1em] \Rightarrow 6ax^2 - 4bx + 6a = 0 \\[1em] \Rightarrow 2(3ax^2 - 2bx + 3a) = 0 \\[1em] \Rightarrow 3ax^2 - 2bx + 3a = 0. ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = b − 3 a b + 3 a ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = b − 3 a b + 3 a ⇒ ( x 2 + 1 + 2 x ) ( b − 3 a ) = ( x 2 + 1 − 2 x ) ( b + 3 a ) ⇒ b x 2 + b + 2 b x − 3 a x 2 − 3 a − 6 a x = b x 2 + b − 2 b x + 3 a x 2 + 3 a − 6 a x ⇒ b x 2 − b x 2 + b − b + 2 b x + 2 b x − 3 a x 2 − 3 a x 2 − 3 a − 3 a − 6 a x + 6 a x = 0 ⇒ 4 b x − 6 a x 2 − 6 a = 0 ⇒ 6 a x 2 − 4 b x + 6 a = 0 ⇒ 2 ( 3 a x 2 − 2 b x + 3 a ) = 0 ⇒ 3 a x 2 − 2 b x + 3 a = 0.
Hence, proved that 3ax2 - 2bx + 3a = 0.
If x = 2 a + 3 b + 2 a − 3 b 2 a + 3 b − 2 a − 3 b x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}} x = 2 a + 3 b − 2 a − 3 b 2 a + 3 b + 2 a − 3 b 2 - 4ax + 3b = 0.
Answer
Given,
⇒ x = 2 a + 3 b + 2 a − 3 b 2 a + 3 b − 2 a − 3 b \Rightarrow x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}} ⇒ x = 2 a + 3 b − 2 a − 3 b 2 a + 3 b + 2 a − 3 b
Applying componendo and dividendo, we get :
⇒ x + 1 x − 1 = 2 a + 3 b + 2 a − 3 b + 2 a + 3 b − 2 a − 3 b 2 a + 3 b + 2 a − 3 b − ( 2 a + 3 b − 2 a − 3 b ) ⇒ x + 1 x − 1 = 2 a + 3 b + 2 a − 3 b + 2 a + 3 b − 2 a − 3 b 2 a + 3 b + 2 a − 3 b − 2 a + 3 b + 2 a − 3 b ⇒ x + 1 x − 1 = 2 2 a + 3 b 2 2 a − 3 b ⇒ x + 1 x − 1 = 2 a + 3 b 2 a − 3 b \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - (\sqrt{2a + 3b} - \sqrt{2a - 3b})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - \sqrt{2a + 3b} + \sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 3b}}{2\sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b}}{\sqrt{2a - 3b}} ⇒ x − 1 x + 1 = 2 a + 3 b + 2 a − 3 b − ( 2 a + 3 b − 2 a − 3 b ) 2 a + 3 b + 2 a − 3 b + 2 a + 3 b − 2 a − 3 b ⇒ x − 1 x + 1 = 2 a + 3 b + 2 a − 3 b − 2 a + 3 b + 2 a − 3 b 2 a + 3 b + 2 a − 3 b + 2 a + 3 b − 2 a − 3 b ⇒ x − 1 x + 1 = 2 2 a − 3 b 2 2 a + 3 b ⇒ x − 1 x + 1 = 2 a − 3 b 2 a + 3 b
Squaring both sides, we get :
⇒ ( x + 1 ) 2 ( x − 1 ) 2 = 2 a + 3 b 2 a − 3 b ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = 2 a + 3 b 2 a − 3 b ⇒ ( x 2 + 1 + 2 x ) ( 2 a − 3 b ) = ( x 2 + 1 − 2 x ) ( 2 a + 3 b ) ⇒ 2 a x 2 + 2 a + 4 a x − 3 b x 2 − 3 b − 6 b x = 2 a x 2 + 2 a − 4 a x + 3 b x 2 + 3 b − 6 b x ⇒ 2 a x 2 − 2 a x 2 + 2 a − 2 a + 4 a x + 4 a x − 3 b x 2 − 3 b x 2 − 3 b − 3 b − 6 b x + 6 b x = 0 ⇒ 8 a x − 6 b x 2 − 6 b = 0 ⇒ 6 b x 2 − 8 a x + 6 b = 0 ⇒ 2 ( 3 b x 2 − 4 a x + 3 b ) = 0 ⇒ 3 b x 2 − 4 a x + 3 b = 0. \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 3b) = (x^2 + 1 - 2x)(2a + 3b) \\[1em] \Rightarrow 2ax^2 + 2a + 4ax - 3bx^2 - 3b - 6bx = 2ax^2 + 2a - 4ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow 2ax^2 - 2ax^2 + 2a - 2a + 4ax + 4ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 8ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 8ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 4ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 4ax + 3b = 0. ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = 2 a − 3 b 2 a + 3 b ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = 2 a − 3 b 2 a + 3 b ⇒ ( x 2 + 1 + 2 x ) ( 2 a − 3 b ) = ( x 2 + 1 − 2 x ) ( 2 a + 3 b ) ⇒ 2 a x 2 + 2 a + 4 a x − 3 b x 2 − 3 b − 6 b x = 2 a x 2 + 2 a − 4 a x + 3 b x 2 + 3 b − 6 b x ⇒ 2 a x 2 − 2 a x 2 + 2 a − 2 a + 4 a x + 4 a x − 3 b x 2 − 3 b x 2 − 3 b − 3 b − 6 b x + 6 b x = 0 ⇒ 8 a x − 6 b x 2 − 6 b = 0 ⇒ 6 b x 2 − 8 a x + 6 b = 0 ⇒ 2 ( 3 b x 2 − 4 a x + 3 b ) = 0 ⇒ 3 b x 2 − 4 a x + 3 b = 0.
Hence, proved that 3bx2 - 4ax + 3b = 0.
If x = m + 1 3 + m − 1 3 m + 1 3 − m − 1 3 x = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}} x = 3 m + 1 − 3 m − 1 3 m + 1 + 3 m − 1 3 - 3x2 m + 3x - m = 0.
Answer
Given,
⇒ x = m + 1 3 + m − 1 3 m + 1 3 − m − 1 3 \Rightarrow x = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}} ⇒ x = 3 m + 1 − 3 m − 1 3 m + 1 + 3 m − 1
Applying componendo and dividendo, we get :
⇒ x + 1 x − 1 = m + 1 3 + m − 1 3 + m + 1 3 − m − 1 3 m + 1 3 + m − 1 3 − ( m + 1 3 − m − 1 3 ) ⇒ x + 1 x − 1 = m + 1 3 + m − 1 3 + m + 1 3 − m − 1 3 m + 1 3 + m − 1 3 − m + 1 3 + m − 1 3 ⇒ x + 1 x − 1 = 2 m + 1 3 2 m − 1 3 ⇒ x + 1 x − 1 = m + 1 3 m − 1 3 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} + \sqrt[3]{m + 1} - \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} - (\sqrt[3]{m + 1} - \sqrt[3]{m - 1})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} + \sqrt[3]{m + 1} - \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} - \sqrt[3]{m + 1} + \sqrt[3]{m - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt[3]{m + 1}}{2\sqrt[3]{m - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1}}{\sqrt[3]{m - 1}} ⇒ x − 1 x + 1 = 3 m + 1 + 3 m − 1 − ( 3 m + 1 − 3 m − 1 ) 3 m + 1 + 3 m − 1 + 3 m + 1 − 3 m − 1 ⇒ x − 1 x + 1 = 3 m + 1 + 3 m − 1 − 3 m + 1 + 3 m − 1 3 m + 1 + 3 m − 1 + 3 m + 1 − 3 m − 1 ⇒ x − 1 x + 1 = 2 3 m − 1 2 3 m + 1 ⇒ x − 1 x + 1 = 3 m − 1 3 m + 1
Cubing both sides, we get :
⇒ ( x + 1 x − 1 ) 3 = m + 1 m − 1 ⇒ ( x + 1 ) 3 ( m − 1 ) = ( x − 1 ) 3 ( m + 1 ) ⇒ ( x 3 + 3 x 2 + 3 x + 1 ) ( m − 1 ) = ( x 3 − 3 x 2 + 3 x − 1 ) ( m + 1 ) ⇒ m x 3 + 3 m x 2 + 3 m x + m − x 3 − 3 x 2 − 3 x − 1 = m x 3 − 3 m x 2 + 3 m x − m + x 3 − 3 x 2 + 3 x − 1 ⇒ m x 3 + 3 m x 2 + 3 m x + m − x 3 − 3 x 2 − 3 x − 1 − ( m x 3 − 3 m x 2 + 3 m x − m + x 3 − 3 x 2 + 3 x − 1 ) = 0 ⇒ m x 3 − m x 3 + 3 m x 2 + 3 m x 2 + 3 m x − 3 m x + m + m − x 3 − x 3 − 3 x 2 + 3 x 2 − 3 x − 3 x − 1 + 1 = 0 ⇒ 6 m x 2 + 2 m − 2 x 3 − 6 x = 0 ⇒ 2 ( 3 m x 2 + m − x 3 − 3 x ) = 0 ⇒ 3 m x 2 + m − x 3 − 3 x = 0 ⇒ x 3 − 3 m x 2 + 3 x − m = 0. \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{m + 1}{m - 1} \\[1em] \Rightarrow (x + 1)^3 (m - 1) = (x - 1)^3 (m + 1) \\[1em] \Rightarrow (x^3 + 3x^2 + 3x + 1)(m - 1) = (x^3 - 3x^2 + 3x - 1)(m + 1) \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 = m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1 \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 -( m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1) = 0 \\[1em] \Rightarrow m x^3 - m x^3 + 3m x^2 + 3m x^2 + 3m x - 3m x + m + m - x^3 - x^3 - 3x^2 + 3x^2 - 3x - 3x - 1 + 1 = 0 \\[1em] \Rightarrow 6m x^2 + 2m - 2x^3 - 6x = 0 \\[1em] \Rightarrow 2(3mx^2 + m - x^3 - 3x) = 0 \\[1em] \Rightarrow 3mx^2 + m - x^3 - 3x = 0 \\[1em] \Rightarrow x^3 - 3mx^2 + 3x - m = 0. ⇒ ( x − 1 x + 1 ) 3 = m − 1 m + 1 ⇒ ( x + 1 ) 3 ( m − 1 ) = ( x − 1 ) 3 ( m + 1 ) ⇒ ( x 3 + 3 x 2 + 3 x + 1 ) ( m − 1 ) = ( x 3 − 3 x 2 + 3 x − 1 ) ( m + 1 ) ⇒ m x 3 + 3 m x 2 + 3 m x + m − x 3 − 3 x 2 − 3 x − 1 = m x 3 − 3 m x 2 + 3 m x − m + x 3 − 3 x 2 + 3 x − 1 ⇒ m x 3 + 3 m x 2 + 3 m x + m − x 3 − 3 x 2 − 3 x − 1 − ( m x 3 − 3 m x 2 + 3 m x − m + x 3 − 3 x 2 + 3 x − 1 ) = 0 ⇒ m x 3 − m x 3 + 3 m x 2 + 3 m x 2 + 3 m x − 3 m x + m + m − x 3 − x 3 − 3 x 2 + 3 x 2 − 3 x − 3 x − 1 + 1 = 0 ⇒ 6 m x 2 + 2 m − 2 x 3 − 6 x = 0 ⇒ 2 ( 3 m x 2 + m − x 3 − 3 x ) = 0 ⇒ 3 m x 2 + m − x 3 − 3 x = 0 ⇒ x 3 − 3 m x 2 + 3 x − m = 0.
Hence, proved that x3 - 3x2 m + 3x - m = 0.
What quantity must be added to each term of the ratio a : b to make it c : d ?
Answer
Let x be added.
∴ a + x b + x = c d \therefore \dfrac{a + x}{b + x} = \dfrac{c}{d} ∴ b + x a + x = d c
⇒ d(a + x) = c(b + x)
⇒ ad + dx = cb + cx
⇒ cx - dx = ad - cb
⇒ x(c - d) = ad - bc
⇒ x = a d − b c c − d \dfrac{ad - bc}{c - d} c − d a d − b c
Hence, quantity that must be added = a d − b c c − d \dfrac{ad - bc}{c - d} c − d a d − b c
If a + 3 b + 2 c + 6 d a − 3 b + 2 c − 6 d = a + 3 b − 2 c − 6 d a − 3 b − 2 c + 6 d \dfrac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \dfrac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d} a − 3 b + 2 c − 6 d a + 3 b + 2 c + 6 d = a − 3 b − 2 c + 6 d a + 3 b − 2 c − 6 d a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
Answer
Given,
⇒ a + 3 b + 2 c + 6 d a − 3 b + 2 c − 6 d = a + 3 b − 2 c − 6 d a − 3 b − 2 c + 6 d \Rightarrow \dfrac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \dfrac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d} ⇒ a − 3 b + 2 c − 6 d a + 3 b + 2 c + 6 d = a − 3 b − 2 c + 6 d a + 3 b − 2 c − 6 d
Applying componendo and dividendo, we get :
⇒ ( a + 3 b + 2 c + 6 d ) + ( a − 3 b + 2 c − 6 d ) ( a + 3 b + 2 c + 6 d ) − ( a − 3 b + 2 c − 6 d ) = ( a + 3 b − 2 c − 6 d ) + ( a − 3 b − 2 c + 6 d ) ( a + 3 b − 2 c − 6 d ) − ( a − 3 b − 2 c + 6 d ) ⇒ ( a + 3 b + 2 c + 6 d ) + ( a − 3 b + 2 c − 6 d ) ( a + 3 b + 2 c + 6 d ) − a + 3 b − 2 c + 6 d = ( a + 3 b − 2 c − 6 d ) + ( a − 3 b − 2 c + 6 d ) ( a + 3 b − 2 c − 6 d ) − a + 3 b + 2 c − 6 d ⇒ 2 a + 4 c 6 b + 12 d = 2 a − 4 c 6 b − 12 d ⇒ 2 ( a + 2 c ) 6 ( b + 2 d ) = 2 ( a − 2 c ) 6 ( b − 2 d ) ⇒ a + 2 c b + 2 d = a − 2 c b − 2 d ⇒ a + 2 c a − 2 c = b + 2 d b − 2 d \Rightarrow \dfrac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - (a - 3b + 2c - 6d)} = \dfrac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - (a - 3b - 2c + 6d)} \\[1em] \Rightarrow \dfrac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - a + 3b - 2c + 6d} = \dfrac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - a + 3b + 2c - 6d} \\[1em] \Rightarrow \dfrac{2a + 4c}{6b + 12d} = \dfrac{2a - 4c}{6b - 12d} \\[1em] \Rightarrow \dfrac{2(a + 2c)}{6(b + 2d)} = \dfrac{2(a - 2c)}{6(b - 2d)} \\[1em] \Rightarrow \dfrac{a + 2c}{b + 2d} = \dfrac{a - 2c}{b - 2d} \\[1em] \Rightarrow \dfrac{a + 2c}{a - 2c} = \dfrac{b + 2d}{b - 2d} ⇒ ( a + 3 b + 2 c + 6 d ) − ( a − 3 b + 2 c − 6 d ) ( a + 3 b + 2 c + 6 d ) + ( a − 3 b + 2 c − 6 d ) = ( a + 3 b − 2 c − 6 d ) − ( a − 3 b − 2 c + 6 d ) ( a + 3 b − 2 c − 6 d ) + ( a − 3 b − 2 c + 6 d ) ⇒ ( a + 3 b + 2 c + 6 d ) − a + 3 b − 2 c + 6 d ( a + 3 b + 2 c + 6 d ) + ( a − 3 b + 2 c − 6 d ) = ( a + 3 b − 2 c − 6 d ) − a + 3 b + 2 c − 6 d ( a + 3 b − 2 c − 6 d ) + ( a − 3 b − 2 c + 6 d ) ⇒ 6 b + 12 d 2 a + 4 c = 6 b − 12 d 2 a − 4 c ⇒ 6 ( b + 2 d ) 2 ( a + 2 c ) = 6 ( b − 2 d ) 2 ( a − 2 c ) ⇒ b + 2 d a + 2 c = b − 2 d a − 2 c ⇒ a − 2 c a + 2 c = b − 2 d b + 2 d
Applying componendo and dividendo again,
⇒ ( a + 2 c ) + ( a − 2 c ) ( a + 2 c ) − ( a − 2 c ) = ( b + 2 d ) + ( b − 2 d ) ( b + 2 d ) − ( b − 2 d ) ⇒ ( a + 2 c ) + ( a − 2 c ) ( a + 2 c ) − a + 2 c = ( b + 2 d ) + ( b − 2 d ) ( b + 2 d ) − b + 2 d ⇒ 2 a 4 c = 2 b 4 d ⇒ a c = b d ⇒ a b = c d . \Rightarrow \dfrac{(a + 2c) + (a - 2c)}{(a + 2c) - (a - 2c)} = \dfrac{(b + 2d) + (b - 2d)}{(b + 2d) - (b - 2d)} \\[1em] \Rightarrow \dfrac{(a + 2c) + (a - 2c)}{(a + 2c) - a + 2c} = \dfrac{(b + 2d) + (b - 2d)}{(b + 2d) - b + 2d} \\[1em] \Rightarrow \dfrac{2a}{4c} = \dfrac{2b}{4d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}. ⇒ ( a + 2 c ) − ( a − 2 c ) ( a + 2 c ) + ( a − 2 c ) = ( b + 2 d ) − ( b − 2 d ) ( b + 2 d ) + ( b − 2 d ) ⇒ ( a + 2 c ) − a + 2 c ( a + 2 c ) + ( a − 2 c ) = ( b + 2 d ) − b + 2 d ( b + 2 d ) + ( b − 2 d ) ⇒ 4 c 2 a = 4 d 2 b ⇒ c a = d b ⇒ b a = d c .
Hence, proved that a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
If, 2 a + 2 b − 3 c − 3 d 2 a − 2 b − 3 c + 3 d = a + b − 4 c − 4 d a − b − 4 c + 4 d \dfrac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \dfrac{a + b - 4c - 4d}{a - b - 4c + 4d} 2 a − 2 b − 3 c + 3 d 2 a + 2 b − 3 c − 3 d = a − b − 4 c + 4 d a + b − 4 c − 4 d a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
Answer
Given,
⇒ 2 a + 2 b − 3 c − 3 d 2 a − 2 b − 3 c + 3 d = a + b − 4 c − 4 d a − b − 4 c + 4 d \Rightarrow \dfrac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \dfrac{a + b - 4c - 4d}{a - b - 4c + 4d} ⇒ 2 a − 2 b − 3 c + 3 d 2 a + 2 b − 3 c − 3 d = a − b − 4 c + 4 d a + b − 4 c − 4 d
Applying componendo and dividendo, we get :
⇒ ( 2 a + 2 b − 3 c − 3 d ) + ( 2 a − 2 b − 3 c + 3 d ) ( 2 a + 2 b − 3 c − 3 d ) − ( 2 a − 2 b − 3 c + 3 d ) = ( a + b − 4 c − 4 d ) + ( a − b − 4 c + 4 d ) ( a + b − 4 c − 4 d ) − ( a − b − 4 c + 4 d ) ⇒ ( 2 a + 2 b − 3 c − 3 d ) + ( 2 a − 2 b − 3 c + 3 d ) ( 2 a + 2 b − 3 c − 3 d ) − 2 a + 2 b + 3 c − 3 d = ( a + b − 4 c − 4 d ) + ( a − b − 4 c + 4 d ) ( a + b − 4 c − 4 d ) − a + b + 4 c − 4 d ⇒ 4 a − 6 c 4 b − 6 d = 2 a − 8 c 2 b − 8 d ⇒ 2 ( 2 a − 3 c ) 2 ( 2 b − 3 d ) = 2 ( a − 4 c ) 2 ( b − 4 d ) ⇒ 2 a − 3 c 2 b − 3 d = a − 4 c b − 4 d ⇒ 2 a − 3 c a − 4 c = 2 b − 3 d b − 4 d \Rightarrow \dfrac{(2a + 2b - 3c - 3d) + (2a - 2b - 3c + 3d)}{(2a + 2b - 3c - 3d) - (2a - 2b - 3c + 3d)} = \dfrac{(a + b - 4c - 4d) + (a - b - 4c + 4d)}{(a + b - 4c - 4d) - (a - b - 4c + 4d)} \\[1em] \Rightarrow \dfrac{(2a + 2b - 3c - 3d) + (2a - 2b - 3c + 3d)}{(2a + 2b - 3c - 3d) - 2a + 2b + 3c - 3d} = \dfrac{(a + b - 4c - 4d) + (a - b - 4c + 4d)}{(a + b - 4c - 4d) - a + b + 4c - 4d} \\[1em] \Rightarrow \dfrac{4a - 6c}{4b - 6d} = \dfrac{2a - 8c}{2b - 8d} \\[1em] \Rightarrow \dfrac{2(2a - 3c)}{2(2b - 3d)} = \dfrac{2(a - 4c)}{2(b - 4d)} \\[1em] \Rightarrow \dfrac{2a - 3c}{2b - 3d} = \dfrac{a - 4c}{b - 4d} \\[1em] \Rightarrow \dfrac{2a - 3c}{a - 4c} = \dfrac{2b - 3d}{b - 4d} ⇒ ( 2 a + 2 b − 3 c − 3 d ) − ( 2 a − 2 b − 3 c + 3 d ) ( 2 a + 2 b − 3 c − 3 d ) + ( 2 a − 2 b − 3 c + 3 d ) = ( a + b − 4 c − 4 d ) − ( a − b − 4 c + 4 d ) ( a + b − 4 c − 4 d ) + ( a − b − 4 c + 4 d ) ⇒ ( 2 a + 2 b − 3 c − 3 d ) − 2 a + 2 b + 3 c − 3 d ( 2 a + 2 b − 3 c − 3 d ) + ( 2 a − 2 b − 3 c + 3 d ) = ( a + b − 4 c − 4 d ) − a + b + 4 c − 4 d ( a + b − 4 c − 4 d ) + ( a − b − 4 c + 4 d ) ⇒ 4 b − 6 d 4 a − 6 c = 2 b − 8 d 2 a − 8 c ⇒ 2 ( 2 b − 3 d ) 2 ( 2 a − 3 c ) = 2 ( b − 4 d ) 2 ( a − 4 c ) ⇒ 2 b − 3 d 2 a − 3 c = b − 4 d a − 4 c ⇒ a − 4 c 2 a − 3 c = b − 4 d 2 b − 3 d
Cross - multiplying and simplifying:
⇒ ( 2 a − 3 c ) ( b − 4 d ) = ( a − 4 c ) ( 2 b − 3 d ) ⇒ 2 a b − 8 a d − 3 b c + 12 c d = 2 a b − 3 a d − 8 b c + 12 c d ⇒ 2 a b − 2 a b − 8 a d + 3 a d − 3 b c + 8 b c + 12 c d − 12 c d = 0 ⇒ 5 b c − 5 a d = 0 ⇒ 5 b c = 5 a d ⇒ b c = a d ∴ a b = c d . \Rightarrow (2a - 3c)(b - 4d) = (a - 4c)(2b - 3d) \\[1em] \Rightarrow 2ab - 8ad - 3bc + 12cd = 2ab - 3ad - 8bc + 12cd \\[1em] \Rightarrow 2ab - 2ab - 8ad + 3ad - 3bc + 8bc + 12cd - 12cd = 0 \\[1em] \Rightarrow 5bc - 5ad = 0 \\[1em] \Rightarrow 5bc = 5ad \\[1em] \Rightarrow bc = ad \\[1em] \therefore \dfrac{a}{b} = \dfrac{c}{d}. ⇒ ( 2 a − 3 c ) ( b − 4 d ) = ( a − 4 c ) ( 2 b − 3 d ) ⇒ 2 ab − 8 a d − 3 b c + 12 c d = 2 ab − 3 a d − 8 b c + 12 c d ⇒ 2 ab − 2 ab − 8 a d + 3 a d − 3 b c + 8 b c + 12 c d − 12 c d = 0 ⇒ 5 b c − 5 a d = 0 ⇒ 5 b c = 5 a d ⇒ b c = a d ∴ b a = d c .
Hence, proved that a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
If (a + b + c + d) : (a + b − c − d) = (a − b + c − d) : (a − b − c + d), prove that a : b = c : d.
Answer
Given,
⇒ a + b + c + d a + b − c − d = a − b + c − d a − b − c + d \Rightarrow \dfrac{a + b + c + d}{a + b - c - d} = \dfrac{a - b + c - d}{a - b - c + d} ⇒ a + b − c − d a + b + c + d = a − b − c + d a − b + c − d
Applying componendo and dividendo, we get :
⇒ ( a + b + c + d ) + ( a + b − c − d ) ( a + b + c + d ) − ( a + b − c − d ) = ( a − b + c − d ) + ( a − b − c + d ) ( a − b + c − d ) − ( a − b − c + d ) ⇒ ( a + b + c + d ) + ( a + b − c − d ) ( a + b + c + d ) − a − b + c + d = ( a − b + c − d ) + ( a − b − c + d ) ( a − b + c − d ) − a + b + c − d ⇒ 2 ( a + b ) 2 ( c + d ) = 2 ( a − b ) 2 ( c − d ) ⇒ a + b c + d = a − b c − d ⇒ a + b a − b = c + d c − d \Rightarrow \dfrac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - (a + b - c - d)} = \dfrac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - (a - b - c + d)} \\[1em] \Rightarrow \dfrac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - a - b + c + d} = \dfrac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - a + b + c - d} \\[1em] \Rightarrow \dfrac{2(a + b)}{2(c + d)} = \dfrac{2(a - b)}{2(c - d)} \\[1em] \Rightarrow \dfrac{a + b}{c + d} = \dfrac{a - b}{c - d} \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} ⇒ ( a + b + c + d ) − ( a + b − c − d ) ( a + b + c + d ) + ( a + b − c − d ) = ( a − b + c − d ) − ( a − b − c + d ) ( a − b + c − d ) + ( a − b − c + d ) ⇒ ( a + b + c + d ) − a − b + c + d ( a + b + c + d ) + ( a + b − c − d ) = ( a − b + c − d ) − a + b + c − d ( a − b + c − d ) + ( a − b − c + d ) ⇒ 2 ( c + d ) 2 ( a + b ) = 2 ( c − d ) 2 ( a − b ) ⇒ c + d a + b = c − d a − b ⇒ a − b a + b = c − d c + d
Applying componendo and dividendo again:
⇒ a + b + ( a − b ) a + b − ( a − b ) = c + d + ( c − d ) c + d − ( c − d ) ⇒ a + b + ( a − b ) a + b − a + b = c + d + ( c − d ) c + d − c + d ⇒ 2 a 2 b = 2 c 2 d ⇒ a b = c d . \Rightarrow \dfrac{a + b + (a - b)}{a + b - (a - b)} = \dfrac{c + d + (c - d)}{c + d - (c - d)} \\[1em] \Rightarrow \dfrac{a + b + (a - b)}{a + b - a + b} = \dfrac{c + d + (c - d)}{c + d - c + d} \\[1em] \Rightarrow \dfrac{2a}{2b} = \dfrac{2c}{2d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}. ⇒ a + b − ( a − b ) a + b + ( a − b ) = c + d − ( c − d ) c + d + ( c − d ) ⇒ a + b − a + b a + b + ( a − b ) = c + d − c + d c + d + ( c − d ) ⇒ 2 b 2 a = 2 d 2 c ⇒ b a = d c .
Hence, proved that a : b = c : d.
Given : x 3 + 12 x 6 x 2 + 8 = y 3 + 27 y 9 y 2 + 27 \dfrac{x^{3} + 12x}{6x^{2} + 8} = \dfrac{y^{3} + 27y}{9y^{2} + 27} 6 x 2 + 8 x 3 + 12 x = 9 y 2 + 27 y 3 + 27 y
Answer
Given,
x 3 + 12 x 6 x 2 + 8 = y 3 + 27 y 9 y 2 + 27 \dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27} 6 x 2 + 8 x 3 + 12 x = 9 y 2 + 27 y 3 + 27 y
Applying componendo and dividendo we get,
⇒ x 3 + 12 x + 6 x 2 + 8 x 3 + 12 x − ( 6 x 2 + 8 ) = y 3 + 27 y + 9 y 2 + 27 y 3 + 27 y − ( 9 y 2 + 27 ) ⇒ x 3 + 12 x + 6 x 2 + 8 x 3 + 12 x − 6 x 2 − 8 = y 3 + 27 y + 9 y 2 + 27 y 3 + 27 y − 9 y 2 − 27 ⇒ ( x + 2 ) 3 ( x − 2 ) 3 = ( y + 3 ) 3 ( y − 3 ) 3 ⇒ x + 2 x − 2 = y + 3 y − 3 \Rightarrow \dfrac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - (6x^2 + 8)} = \dfrac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - (9y^2 + 27)} \\[1em] \Rightarrow \dfrac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \dfrac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \\[1em] \Rightarrow \dfrac{(x + 2)^3}{(x - 2)^3} = \dfrac{(y + 3)^3}{(y - 3)^3} \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3} ⇒ x 3 + 12 x − ( 6 x 2 + 8 ) x 3 + 12 x + 6 x 2 + 8 = y 3 + 27 y − ( 9 y 2 + 27 ) y 3 + 27 y + 9 y 2 + 27 ⇒ x 3 + 12 x − 6 x 2 − 8 x 3 + 12 x + 6 x 2 + 8 = y 3 + 27 y − 9 y 2 − 27 y 3 + 27 y + 9 y 2 + 27 ⇒ ( x − 2 ) 3 ( x + 2 ) 3 = ( y − 3 ) 3 ( y + 3 ) 3 ⇒ x − 2 x + 2 = y − 3 y + 3
Applying componendo and dividendo again we get,
⇒ x + 2 + x − 2 x + 2 − ( x − 2 ) = y + 3 + y − 3 y + 3 − ( y − 3 ) ⇒ x + 2 + x − 2 x + 2 − x + 2 = y + 3 + y − 3 y + 3 − y + 3 ⇒ 2 x 4 = 2 y 6 ⇒ x 2 = y 3 ⇒ x y = 2 3 ⇒ x : y = 2 : 3. \Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - (x - 2)} = \dfrac{y + 3 + y - 3}{y + 3 - (y - 3)} \\[1em] \Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - x + 2} = \dfrac{y + 3 + y - 3}{y + 3 - y + 3} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3. ⇒ x + 2 − ( x − 2 ) x + 2 + x − 2 = y + 3 − ( y − 3 ) y + 3 + y − 3 ⇒ x + 2 − x + 2 x + 2 + x − 2 = y + 3 − y + 3 y + 3 + y − 3 ⇒ 4 2 x = 6 2 y ⇒ 2 x = 3 y ⇒ y x = 3 2 ⇒ x : y = 2 : 3.
Hence, x : y = 2 : 3.
Using the properties of proportion, find x : y, given
x 2 + 2 x 2 x + 4 = y 2 + 3 y 3 y + 9 \dfrac{x^{2} + 2x}{2x + 4} = \dfrac{y^{2} + 3y}{3y + 9} 2 x + 4 x 2 + 2 x = 3 y + 9 y 2 + 3 y
Answer
Given,
⇒ x 2 + 2 x 2 x + 4 = y 2 + 3 y 3 y + 9 \Rightarrow \dfrac{x^{2} + 2x}{2x + 4} = \dfrac{y^{2} + 3y}{3y + 9} ⇒ 2 x + 4 x 2 + 2 x = 3 y + 9 y 2 + 3 y
Applying componendo and dividendo, we get :
⇒ x 2 + 2 x + 2 x + 4 x 2 + 2 x − ( 2 x + 4 ) = y 2 + 3 y + 3 y + 9 y 2 + 3 y − ( 3 y + 9 ) ⇒ x 2 + 2 x + 2 x + 4 x 2 + 2 x − 2 x − 4 = y 2 + 3 y + 3 y + 9 y 2 + 3 y − 3 y − 9 ⇒ x 2 + 4 x + 4 x 2 − 4 = y 2 + 6 y + 9 y 2 − 9 ⇒ ( x + 2 ) 2 ( x − 2 ) ( x + 2 ) = ( y + 3 ) 2 ( y − 3 ) ( y + 3 ) ⇒ x + 2 x − 2 = y + 3 y − 3 \Rightarrow \dfrac{x^{2} + 2x + 2x + 4}{x^{2} + 2x - (2x + 4)} = \dfrac{y^{2} + 3y + 3y + 9}{y^{2} + 3y - (3y + 9)} \\[1em] \Rightarrow \dfrac{x^{2} + 2x + 2x + 4}{x^{2} + 2x - 2x - 4} = \dfrac{y^{2} + 3y + 3y + 9}{y^{2} + 3y - 3y - 9} \\[1em] \Rightarrow \dfrac{x^{2} + 4x + 4}{x^{2} - 4} = \dfrac{y^{2} + 6y + 9}{y^{2} - 9} \\[1em] \Rightarrow \dfrac{(x + 2)^2}{(x - 2)(x + 2)} = \dfrac{(y + 3)^2}{(y - 3)(y + 3)} \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3} ⇒ x 2 + 2 x − ( 2 x + 4 ) x 2 + 2 x + 2 x + 4 = y 2 + 3 y − ( 3 y + 9 ) y 2 + 3 y + 3 y + 9 ⇒ x 2 + 2 x − 2 x − 4 x 2 + 2 x + 2 x + 4 = y 2 + 3 y − 3 y − 9 y 2 + 3 y + 3 y + 9 ⇒ x 2 − 4 x 2 + 4 x + 4 = y 2 − 9 y 2 + 6 y + 9 ⇒ ( x − 2 ) ( x + 2 ) ( x + 2 ) 2 = ( y − 3 ) ( y + 3 ) ( y + 3 ) 2 ⇒ x − 2 x + 2 = y − 3 y + 3
Applying componendo and dividendo again we get,
⇒ x + 2 + x − 2 x + 2 − ( x − 2 ) = y + 3 + y − 3 y + 3 − ( y − 3 ) ⇒ x + 2 + x − 2 x + 2 − x + 2 = y + 3 + y − 3 y + 3 − y + 3 ⇒ 2 x 4 = 2 y 6 ⇒ x 2 = y 3 ⇒ x y = 2 3 ⇒ x : y = 2 : 3. \Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - (x - 2)} = \dfrac{y + 3 + y - 3}{y + 3 - (y - 3)} \\[1em] \Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - x + 2} = \dfrac{y + 3 + y - 3}{y + 3 - y + 3} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3. ⇒ x + 2 − ( x − 2 ) x + 2 + x − 2 = y + 3 − ( y − 3 ) y + 3 + y − 3 ⇒ x + 2 − x + 2 x + 2 + x − 2 = y + 3 − y + 3 y + 3 + y − 3 ⇒ 4 2 x = 6 2 y ⇒ 2 x = 3 y ⇒ y x = 3 2 ⇒ x : y = 2 : 3.
Hence, x : y = 2 : 3.
Multiple Choice Questions
Which of the following is the ratio between a number and the number obtained by adding one-fifth of that number?
4 : 5
5 : 4
5 : 6
6 : 5
Answer
Let the number be x.
Given,
Number obtained by adding number and one-fifth of that number.
⇒ x + x 5 ⇒ x ( 1 + 1 5 ) ⇒ x ( 5 + 1 5 ) ⇒ ( 6 x 5 ) . \Rightarrow x + \dfrac{x}{5} \\[1em] \Rightarrow x\Big(1 + \dfrac{1}{5}\Big) \\[1em] \Rightarrow x\Big(\dfrac{5 + 1}{5}\Big) \\[1em] \Rightarrow \Big(\dfrac{6x}{5}\Big). ⇒ x + 5 x ⇒ x ( 1 + 5 1 ) ⇒ x ( 5 5 + 1 ) ⇒ ( 5 6 x ) .
Ratio = x : 6 5 \dfrac{6}{5} 5 6
= x 6 5 x \dfrac{x}{\dfrac{6}{5}x} 5 6 x x
= 5 x 6 x \dfrac{5x}{6x} 6 x 5 x
= 5 6 \dfrac{5}{6} 6 5
= 5 : 6.
Hence, option 3 is the correct option.
Govind spends ₹ 8,100 in buying some jeans at ₹ 1,200 each and some shirts at ₹ 300 each. The ratio of the number of shirts to that of jeans, when the maximum possible number of jeans is purchased is:
1 : 2
1 : 4
2 : 1
5 : 7
Answer
Let the number of jeans Gopal brought be x and shirts be y.
Given,
Cost of jeans = ₹ 1,200.
Cost of shirt = ₹ 300.
Total amount spent = ₹ 8,100.
⇒ 1200x + 300y = 8100
⇒ 1200x + 300y - 8100 = 0
⇒ 300(4x + y - 27) = 0
⇒ 4x + y - 27 = 0
To maximize x keep y ≥ 0
⇒ 4x = 27
Maximum value of x can be 6.
Substitute value of x to get y,
⇒ 4(6) + y - 27 = 0
⇒ 24 + y - 27 = 0
⇒ y - 3 = 0
⇒ y = 3.
Ratio of shirt(y) : jeans(x) = 3 : 6 = 1 : 2.
Hence, option 1 is the correct option.
which of the following represents xy = 64?
8 : x = 8 : y
x : 16 = y : 4
x : 8 = y : 8
32 : x = y : 2
Answer
If we consider,
⇒ 32 x = y 2 \Rightarrow \dfrac{32}{x} = \dfrac{y}{2} ⇒ x 32 = 2 y
⇒ 32(2) = xy
⇒ xy = 64.
Hence, option 4 is the correct option.
If a carton containing a dozen mirrors is dropped, then which of the following cannot be the ratio of broken mirrors to unbroken mirrors?
2 : 1
7 : 5
3 : 2
3 : 1
Answer
Let ratio of broken mirrors to unbroken mirrors = p : q
Number of mirrors in dozen = 12
Number of broken mirrors = p p + q × 12 \dfrac{p}{p + q} \times 12 p + q p × 12
Number of broken mirrors = q p + q × 12 \dfrac{q}{p + q} \times 12 p + q q × 12
If we consider 3 : 2,
3 + 2 = 5
3 5 × 12 = 7.2 \dfrac{3}{5} \times 12 = 7.2 5 3 × 12 = 7.2
Since 7.2 is not integer. Thus, 3 : 2 is the ratio that cannot exist.
Hence, option 3 is the correct option.
If (x + 1) : 8 = 3.75 : 7 then value of x is :
1 2 7 1\dfrac{2}{7} 1 7 2
2 2 7 2\dfrac{2}{7} 2 7 2
3 2 7 3\dfrac{2}{7} 3 7 2
4 2 7 4\dfrac{2}{7} 4 7 2
Answer
Given,
⇒ (x + 1) : 8 = 3.75 : 7
⇒ x + 1 8 = 3.75 7 ⇒ x + 1 8 = 3.75 7 ⇒ x + 1 8 = 375 700 ⇒ x + 1 8 = 15 28 ⇒ x + 1 = 15 28 × 8 ⇒ x + 1 = 30 7 ⇒ x = 30 7 − 1 ⇒ x = 30 − 7 7 ⇒ x = 23 7 ⇒ x = 3 2 7 . \Rightarrow \dfrac{x + 1}{8} = \dfrac{3.75}{7} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{3.75}{7} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{375}{700} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{15}{28} \\[1em] \Rightarrow x + 1 = \dfrac{15}{28} \times 8 \\[1em] \Rightarrow x + 1 = \dfrac{30}{7} \\[1em] \Rightarrow x = \dfrac{30}{7} - 1 \\[1em] \Rightarrow x = \dfrac{30 - 7}{7} \\[1em] \Rightarrow x = \dfrac{23}{7} \\[1em] \Rightarrow x = 3\dfrac{2}{7}. ⇒ 8 x + 1 = 7 3.75 ⇒ 8 x + 1 = 7 3.75 ⇒ 8 x + 1 = 700 375 ⇒ 8 x + 1 = 28 15 ⇒ x + 1 = 28 15 × 8 ⇒ x + 1 = 7 30 ⇒ x = 7 30 − 1 ⇒ x = 7 30 − 7 ⇒ x = 7 23 ⇒ x = 3 7 2 .
Hence, option 3 is the correct option.
If 2 : ( 1 + 3 ) : : 6 : x \sqrt2:(1 + \sqrt3)::\sqrt6:x 2 : ( 1 + 3 ) :: 6 : x
3 + 3 \sqrt3 + 3 3 + 3
1 − 3 1 - \sqrt3 1 − 3
3 − 3 \sqrt3 - 3 3 − 3
1 + 3 1 + \sqrt3 1 + 3
Answer
Given,
⇒ 2 : ( 1 + 3 ) : : 6 : x ⇒ 2 1 + 3 = 6 x ⇒ x = 6 ( 1 + 3 ) 2 ⇒ x = 3 ( 1 + 3 ) ⇒ x = 3 + 3. \Rightarrow \sqrt2:(1 + \sqrt3)::\sqrt6:x \\[1em] \Rightarrow \dfrac{\sqrt{2}}{1 + \sqrt3} = \dfrac{\sqrt6}{x} \\[1em] \Rightarrow x = \dfrac{\sqrt6(1 + \sqrt3)}{\sqrt2} \\[1em] \Rightarrow x = \sqrt3(1 + \sqrt3) \\[1em] \Rightarrow x = \sqrt3 + 3. ⇒ 2 : ( 1 + 3 ) :: 6 : x ⇒ 1 + 3 2 = x 6 ⇒ x = 2 6 ( 1 + 3 ) ⇒ x = 3 ( 1 + 3 ) ⇒ x = 3 + 3.
Hence, option 1 is the correct option.
0.5 of a number is equal to 0.07 of another. The ratio of the numbers is :
1 : 14
5 : 7
7 : 50
50 : 7
Answer
Let the numbers be x and y.
Given,
⇒ 0.5x = 0.07y
⇒ x y = 0.07 0.5 = 7 50 \dfrac{x}{y} = \dfrac{0.07}{0.5} = \dfrac{7}{50} y x = 0.5 0.07 = 50 7
⇒ x : y = 7 : 50
Hence, option 3 is the correct option.
25% of x is equal to 35% of y, then x : y equals to :
5 : 7
7 : 5
13 : 15
15 : 13
Answer
Given,
25% of x = 35% of y
⇒ 25 100 x = 35 100 y ⇒ 25 x = 35 y ⇒ x y = 35 25 ⇒ x y = 7 5 . \Rightarrow \dfrac{25}{100}x = \dfrac{35}{100}y \\[1em] \Rightarrow 25x = 35y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{35}{25} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{7}{5}. ⇒ 100 25 x = 100 35 y ⇒ 25 x = 35 y ⇒ y x = 25 35 ⇒ y x = 5 7 .
Thus,
x : y = 7 : 5
Hence, option 2 is the correct option.
If A = 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1
1 : 2 : 6
1 : 3 : 6
1 : 2 : 3
3 : 2 : 1
Answer
Given,
A = 1 3 \dfrac{1}{3} 3 1
B = 3A
B = 1 2 \dfrac{1}{2} 2 1
C = 2B
C = 2(3A)
C = 6A
A : B : C = A : 3A : 6A
= 1 : 3 : 6.
Hence, option 2 is the correct option.
If 2p = 3q = 4r then p : q : r is equal to:
2 : 3 : 4
3 : 4 : 6
4 : 3 : 2
6 : 4 : 3
Answer
Let, 2p = 3q = 4r = k
p = k 2 \dfrac{k}{2} 2 k
q = k 3 \dfrac{k}{3} 3 k
r = k 4 \dfrac{k}{4} 4 k
⇒ p : q : r = k 2 : k 3 : k 4 = k 2 × 12 : k 3 × 12 : k 4 × 12 = 6 k : 4 k : 3 k = 6 : 4 : 3. \Rightarrow p : q : r = \dfrac{k}{2}:\dfrac{k}{3}:\dfrac{k}{4} \\[1em] = \dfrac{k}{2} \times 12 : \dfrac{k}{3} \times 12 : \dfrac{k}{4} \times 12 \\[1em] = 6k : 4k : 3k \\[1em] = 6 : 4 : 3. ⇒ p : q : r = 2 k : 3 k : 4 k = 2 k × 12 : 3 k × 12 : 4 k × 12 = 6 k : 4 k : 3 k = 6 : 4 : 3.
Hence, option 4 is the correct option.
If a 3 = b 4 = c 7 \dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{7} 3 a = 4 b = 7 c a + b + c c \dfrac{a + b + c}{c} c a + b + c
5 2 \dfrac{5}{2} 2 5
1 7 \dfrac{1}{7} 7 1
2
7
Answer
Let, a 3 = b 4 = c 7 = k \dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{7} = k 3 a = 4 b = 7 c = k
We can express,
a = 3k, b = 4k, c = 7k
Substituting values in a + b + c c \dfrac{a + b + c}{c} c a + b + c
⇒ 3 k + 4 k + 7 k 7 k ⇒ 14 k 7 k ⇒ 2. \Rightarrow \dfrac{3k + 4k + 7k}{7k} \\[1em] \Rightarrow \dfrac{14k}{7k} \\[1em] \Rightarrow 2. ⇒ 7 k 3 k + 4 k + 7 k ⇒ 7 k 14 k ⇒ 2.
Hence, option 3 is the correct option.
If x, 5.4, 5, 9 are in proportion, then x is equal to:
3
9.72
25
25 3 \dfrac{25}{3} 3 25
Answer
Given,
x, 5.4, 5, 9 are in proportion.
⇒ x 5.4 = 5 9 \dfrac{x}{5.4} = \dfrac{5}{9} 5.4 x = 9 5
⇒ x = 5 9 × 5.4 x = \dfrac{5}{9} \times 5.4 x = 9 5 × 5.4
⇒ x = 0.6 × 5
⇒ x = 3.
Hence, option 1 is the correct option.
The fourth proportional to 5, 8, 15 is :
18
20
21
24
Answer
Let forth proportional be x.
⇒ 5 : 8 = 15 : x
⇒ 5 8 = 15 x \dfrac{5}{8} = \dfrac{15}{x} 8 5 = x 15
⇒ 5x = 8 × 15
⇒ x = 120 5 \dfrac{120}{5} 5 120
⇒ x = 24.
Hence, option 4 is the correct option.
The fourth proportional to 0.12, 0.21 and 8 is :
8.9
14
17
56
Answer
Let fourth proportional be x.
⇒ 0.12 : 0.21 = 8 : x
⇒ 0.12 0.21 = 8 x \dfrac{0.12}{0.21} = \dfrac{8}{x} 0.21 0.12 = x 8
⇒ 0.12x = 0.21 × 8
⇒ x = 1.68 0.12 \dfrac{1.68}{0.12} 0.12 1.68
⇒ x = 14.
Hence, option 2 is the correct option.
The third proportional to 38 and 15 is :
15 38 × 38 \dfrac{15}{38 \times 38} 38 × 38 15
38 × 38 15 \dfrac{38 \times 38}{15} 15 38 × 38
15 × 15 38 \dfrac{15 \times 15}{38} 38 15 × 15
38 × 15 2 \dfrac{38 \times 15}{2} 2 38 × 15
Answer
Let third proportional be x.
⇒ 38 : 15 = 15 : x
⇒ 38 15 = 15 x \dfrac{38}{15} = \dfrac{15}{x} 15 38 = x 15
⇒ 38x = 15 × 15
⇒ x = 15 × 15 38 \dfrac{15 \times 15}{38} 38 15 × 15
Hence, option 3 is the correct option.
The third proportional to (x2 − y2 ) and (x − y) is :
x − y x + y \dfrac{x - y}{x + y} x + y x − y
x + y x − y \dfrac{x + y}{x - y} x − y x + y
(x + y)
(x − y)
Answer
Let third proportional be p.
⇒ (x2 − y2 ) : (x − y) = (x − y) : p
⇒ x 2 − y 2 x − y = x − y p ⇒ p = ( x − y ) 2 x 2 − y 2 ⇒ p = ( x − y ) 2 ( x + y ) ( x − y ) ⇒ p = x − y x + y . \Rightarrow \dfrac{x^2 - y^2}{x - y} = \dfrac{x - y}{p} \\[1em] \Rightarrow p = \dfrac{(x - y)^2}{x^2 - y^2} \\[1em] \Rightarrow p = \dfrac{(x - y)^2}{(x + y)(x - y)} \\[1em] \Rightarrow p = \dfrac{x - y}{x + y}. ⇒ x − y x 2 − y 2 = p x − y ⇒ p = x 2 − y 2 ( x − y ) 2 ⇒ p = ( x + y ) ( x − y ) ( x − y ) 2 ⇒ p = x + y x − y .
Hence, option 1 is the correct option.
The mean proportion between 9 and 16 is:
7
12
25
144
Answer
Let mean proportional be x.
⇒ 9 : x = x : 16
⇒ 9 x = x 16 \dfrac{9}{x} = \dfrac{x}{16} x 9 = 16 x
⇒ x2 = 9(16)
⇒ x = 144 \sqrt{144} 144
⇒ x = 12.
Hence, option 2 is the correct option.
The mean proportion between 0.02 and 0.32 is :
0.08
0.16
0.3
0.34
Answer
Let mean proportion be x.
⇒ 0.02 : x = x : 0.32
⇒ 0.02 x = x 0.32 \dfrac{0.02}{x} = \dfrac{x}{0.32} x 0.02 = 0.32 x
⇒ x2 = 0.02 × 0.32
⇒ x2 = 0.0064
⇒ x = 0.0064 \sqrt{0.0064} 0.0064
⇒ x = 0.08
Hence, option 1 is the correct option.
The mean proportion between (3 + 2 \sqrt{2} 2 32 \sqrt{32} 32
6
2 7 2\sqrt{7} 2 7
7 \sqrt{7} 7
15 − 3 2 2 \dfrac{15 - 3\sqrt{2}}{2} 2 15 − 3 2
Answer
Let mean proportion be x.
(3 + 2 \sqrt{2} 2 32 \sqrt{32} 32
⇒ ( 3 + 2 ) x = x ( 12 − 32 ) ⇒ x 2 = ( 3 + 2 ) × ( 12 − 32 ) ⇒ x 2 = ( 3 + 2 ) × 4 ( 3 − 2 ) ⇒ x 2 = 4 [ ( 3 ) 2 − ( 2 ) ] 2 ⇒ x 2 = 4 ( 9 − 2 ) ⇒ x 2 = 4 × 7 ⇒ x 2 = 28 ⇒ x = 28 ⇒ x = 2 7 . \Rightarrow \dfrac{(3 + \sqrt{2})}{x} = \dfrac{x}{(12 − \sqrt{32})} \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times (12 − \sqrt{32}) \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times 4(3 − \sqrt{2}) \\[1em] \Rightarrow x^2 = 4[(3)^2 − (\sqrt{2})]^2 \\[1em] \Rightarrow x^2 = 4(9 - 2) \\[1em] \Rightarrow x^2 = 4 \times 7 \\[1em] \Rightarrow x^2 = 28 \\[1em] \Rightarrow x = \sqrt{28} \\[1em] \Rightarrow x = 2\sqrt7. ⇒ x ( 3 + 2 ) = ( 12 − 32 ) x ⇒ x 2 = ( 3 + 2 ) × ( 12 − 32 ) ⇒ x 2 = ( 3 + 2 ) × 4 ( 3 − 2 ) ⇒ x 2 = 4 [( 3 ) 2 − ( 2 ) ] 2 ⇒ x 2 = 4 ( 9 − 2 ) ⇒ x 2 = 4 × 7 ⇒ x 2 = 28 ⇒ x = 28 ⇒ x = 2 7 .
Hence, option 2 is the correct option.
The mean proportion between x and y is 6. The third proportional to x and y is 48. Then x : y equals :
1 : 3
1 : 4
1 : 6
1 : 9
Answer
Given,
Mean proportion between x and y is 6.
⇒ x : 6 = 6 : y
⇒ x 6 = 6 y \dfrac{x}{6} = \dfrac{6}{y} 6 x = y 6
⇒ y = 36 x \dfrac{36}{x} x 36
Given,
The third proportional to x and y is 48.
⇒ x : y = y : 48
⇒ x y = y 48 \dfrac{x}{y} = \dfrac{y}{48} y x = 48 y
⇒ y2 = 48x .......(2)
Substitute value of y from equation 1 in equation 2
⇒ 48 x = ( 36 x ) 2 ⇒ 48 x = ( 1296 x 2 ) ⇒ 48 x × x 2 = 1296 ⇒ 48 x 3 = 1296 ⇒ x 3 = 1296 48 ⇒ x 3 = 27 ⇒ x = 27 3 ⇒ x = 3. \Rightarrow 48x = \Big(\dfrac{36}{x}\Big)^2 \\[1em] \Rightarrow 48x = \Big(\dfrac{1296}{x^2}\Big) \\[1em] \Rightarrow 48x \times x^2 = 1296 \\[1em] \Rightarrow 48x^3 = 1296 \\[1em] \Rightarrow x^3 = \dfrac{1296}{48} \\[1em] \Rightarrow x^3 = 27 \\[1em] \Rightarrow x = \sqrt[3]{27} \\[1em] \Rightarrow x = 3. ⇒ 48 x = ( x 36 ) 2 ⇒ 48 x = ( x 2 1296 ) ⇒ 48 x × x 2 = 1296 ⇒ 48 x 3 = 1296 ⇒ x 3 = 48 1296 ⇒ x 3 = 27 ⇒ x = 3 27 ⇒ x = 3.
Substituting value of x in equation 1, we get :
⇒ y = 36 3 \dfrac{36}{3} 3 36
⇒ y = 12
⇒ x : y = 3 : 12
⇒ x : y = 1 : 4.
Hence, option 2 is the correct option.
The ratio between the third proportional to 12 and 30 and mean proportion of 9 and 25, is :
2 : 1
5 : 1
7 : 15
9 : 14
Answer
Let, third proportional to 12 and 30 be t.
If 12 : 30 :: 30 : t, then,
⇒ 12 30 = 30 t ⇒ t = 30 × 30 12 = 30 2 12 ⇒ t = 900 12 ⇒ t = 75. \Rightarrow \dfrac{12}{30} = \dfrac{30}{t} \\[1em] \Rightarrow t = \dfrac{30 \times 30}{12} = \dfrac{30^2}{12} \\[1em] \Rightarrow t = \dfrac{900}{12} \\[1em] \Rightarrow t = 75. ⇒ 30 12 = t 30 ⇒ t = 12 30 × 30 = 12 3 0 2 ⇒ t = 12 900 ⇒ t = 75.
Let Mean proportion between 9 and 25 be m.
⇒ 9 m = m 25 ⇒ m 2 = 9 × 25 ⇒ m = 9 × 25 ⇒ m = 225 ⇒ m = 15. \Rightarrow \dfrac{9}{m} = \dfrac{m}{25} \\[1em] \Rightarrow m^2 = 9 \times 25 \\[1em] \Rightarrow m = \sqrt{9 \times 25} \\[1em] \Rightarrow m = \sqrt{225} \\[1em] \Rightarrow m = 15. ⇒ m 9 = 25 m ⇒ m 2 = 9 × 25 ⇒ m = 9 × 25 ⇒ m = 225 ⇒ m = 15.
Required ratio = t : m = 75 : 15 = 5 : 1.
Hence, option 2 is the correct option.
When 30% of one number is subtracted from another number, the second number reduces to its four-fifths. What is the ratio of the first to the second number?
2 : 5
2 : 3
4 : 7
cannot be determined
Answer
Let the first number be x and the second number be y.
According to question,
⇒ y - 30% of (x) = 4 5 \dfrac{4}{5} 5 4
⇒ y − 30 100 x = 4 5 y ⇒ y − 3 10 x = 4 5 y ⇒ y − 4 5 y = 3 10 x ⇒ y 5 = 3 10 x ⇒ x y = 1 5 × 10 3 ⇒ x y = 10 15 ⇒ x y = 2 3 . \Rightarrow y - \dfrac{30}{100}x = \dfrac{4}{5}y \\[1em] \Rightarrow y - \dfrac{3}{10}x = \dfrac{4}{5}y \\[1em] \Rightarrow y - \dfrac{4}{5}y = \dfrac{3}{10}x \\[1em] \Rightarrow \dfrac{y}{5} = \dfrac{3}{10}x \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{5} \times \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{10}{15} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3}. ⇒ y − 100 30 x = 5 4 y ⇒ y − 10 3 x = 5 4 y ⇒ y − 5 4 y = 10 3 x ⇒ 5 y = 10 3 x ⇒ y x = 5 1 × 3 10 ⇒ y x = 15 10 ⇒ y x = 3 2 .
⇒ x : y = 2 : 3.
Hence, option 2 is the correct option.
Five mangoes and four guavas cost as much as three mangoes and seven guavas. What is the ratio of the cost of one mango to the cost of one guava?
1 : 3
3 : 2
4 : 3
5 : 2
Answer
Let the cost of a mangoe be M and the cost of a guava be G.
Given,
Five mangoes and four guavas cost as much as three mangoes and seven guavas.
⇒ 5 M + 4 G = 3 M + 7 G ⇒ 5 M − 3 M = 7 G − 4 G ⇒ 2 M = 3 G ⇒ M G = 3 2 . \Rightarrow 5M + 4G = 3M + 7G \\[1em] \Rightarrow 5M - 3M = 7G - 4G \\[1em] \Rightarrow 2M = 3G \\[1em] \Rightarrow \dfrac{M}{G} = \dfrac{3}{2}. ⇒ 5 M + 4 G = 3 M + 7 G ⇒ 5 M − 3 M = 7 G − 4 G ⇒ 2 M = 3 G ⇒ G M = 2 3 .
Hence, option 2 is the correct option.
Of 132 examinees of a certain class, the ratio of passed to failed students is 9 : 2. If 4 more students passed, what would have been the ratio of passed to failed students?
23 : 28
28 : 5
9 : 4
25 : 4
Answer
Given,
Total examinees of a class = 132 and ratio of passed to failed = 9 : 2.
Let number of passed students be 9x and failed students be 2x.
Then,
⇒ 9x + 2x = 132
⇒ 11x = 132
⇒ x = 132 11 \dfrac{132}{11} 11 132
The number of passed students = 9 × 12 = 108
The number of failed students = 2 × 12 = 24
If 4 more students passed, then
New number of passed students = 108 + 4 = 112
New number of failed students = 24 - 4 = 20
Then the new ratio of passed to failed = 112 : 20 = 28 : 5.
Hence, option 2 is the correct option.
There are three boxes – P, Q and R, containing marbles in the ratio 1 : 2 : 3. Total number of marbles is 60. The above ratio can be changed to 3 : 4 : 5 by transferring :
2 marbles from P to Q and 1 from R to Q
3 marbles from Q to R
4 marbles from R to Q
5 marbles from R to P
Answer
Given,
Initial ratio,
P : Q : R = 1 : 2 : 3
Let initial no. of marbles in P, Q and R be x, 2x and 3x respectively.
Total number of marbles = 60.
⇒ x + 2x + 3x = 60
⇒ 6x = 60
⇒ x = 60 6 \dfrac{60}{6} 6 60
⇒ x = 10
P = 1 × 10 = 10, Q = 2 × 10 = 20, R = 3 × 10 = 30.
Given,
New ratio of marbles = 3 : 4 : 5.
Let now the no. of marbles in box P, Q and R be 3x, 4x and 5x respectively.
⇒ 3x + 4x + 5x = 60
⇒ 12x = 60
⇒ x = 60 12 \dfrac{60}{12} 12 60
P = 3 × 5 = 15, Q = 4 × 5 = 20, R = 5 × 5 = 25.
Thus, if initially 5 marbles are transferred from R to P then the ratio changes from 1 : 2 : 3 to 3 : 4 : 5.
Hence, option 4 is the correct option.
If x2 + 4y2 = 4xy, then x : y is :
1 : 1
1 : 2
1 : 4
2 : 1
Answer
Given,
x2 + 4y2 = 4xy
Dividing the equation by y2 ,
∴ ( x y ) 2 + 4 = 4 ( x y ) \therefore \Big(\dfrac{x}{y}\Big)^2 + 4 = 4\Big(\dfrac{x}{y}\Big) ∴ ( y x ) 2 + 4 = 4 ( y x )
Let x y = k \dfrac{x}{y} = k y x = k
⇒ k2 + 4 = 4k
⇒ k2 - 4k + 4 = 0
⇒ k2 - 2k - 2k + 4 = 0
⇒ k(k - 2) - 2(k - 2) = 0
⇒ (k - 2)(k - 2) = 0
⇒ (k - 2)= 0 [Using Zero-product rule]
⇒ k = 2
Therefore, x y = k = 2 1 \dfrac{x}{y} = k = \dfrac{2}{1} y x = k = 1 2
Hence, option 4 is the correct option.
If 3A = 5B and 4B = 6C, then A : C is equal to :
2 : 5
3 : 5
4 : 5
5 : 2
Answer
Given,
3A = 5B
⇒ A B = 5 3 \Rightarrow \dfrac{A}{B} = \dfrac{5}{3} ⇒ B A = 3 5
Also, 4B = 6C
⇒ B C = 6 4 = 3 2 \Rightarrow \dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2} ⇒ C B = 4 6 = 2 3
From (1) and (2):
⇒ A C = A B × B C ⇒ A C = 5 3 × 3 2 ⇒ A C = 5 2 . \Rightarrow \dfrac{A}{C} = \dfrac{A}{B} \times \dfrac{B}{C} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{5}{3} \times \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{5}{2}. ⇒ C A = B A × C B ⇒ C A = 3 5 × 2 3 ⇒ C A = 2 5 .
Therefore, A : C = 5 : 2.
Hence, option 4 is the correct option.
If, A : B = 7 : 9 and B : C = 5 : 4, then A : B : C is:
7 : 45 : 36
35 : 45 : 36
28 : 36 : 35
none of these
Answer
Given,
A : B = 7 : 9 and B : C = 5 : 4
L.C.M of two values of B, that are 9 and 5 is 45.
⇒ A B = 7 × 5 9 × 5 = 35 45 ⇒ B C = 5 × 9 4 × 9 = 45 36 . \Rightarrow \dfrac{A}{B} = \dfrac{7 \times 5}{9\times 5} = \dfrac{35}{45} \\[1em] \Rightarrow \dfrac{B}{C} = \dfrac{5 \times 9}{4 \times 9} = \dfrac{45}{36}. ⇒ B A = 9 × 5 7 × 5 = 45 35 ⇒ C B = 4 × 9 5 × 9 = 36 45 .
A : B : C = 35 : 45 : 36.
Hence, option 2 is the correct option.
If 8a = 9b, then the ratio of a 9 \dfrac{a}{9} 9 a b 8 \dfrac{b}{8} 8 b
1 : 1
1 : 2
2 : 1
64 : 81
Answer
Given,
8a = 9b
∴ a b = 9 8 \therefore \dfrac{a}{b} = \dfrac{9}{8} ∴ b a = 8 9
Solving,
⇒ a 9 b 8 ⇒ a 9 × 8 b ⇒ 8 a 9 b ⇒ 8 9 × 9 8 ⇒ 9 9 ⇒ 1 1 ⇒ 1 : 1. \Rightarrow \dfrac{\dfrac{a}{9}}{\dfrac{b}{8}} \\[1em] \Rightarrow \dfrac{a}{9} \times \dfrac{8}{b} \\[1em] \Rightarrow \dfrac{8a}{9b} \\[1em] \Rightarrow \dfrac{8}{9} \times \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{9}{9} \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1 : 1. ⇒ 8 b 9 a ⇒ 9 a × b 8 ⇒ 9 b 8 a ⇒ 9 8 × 8 9 ⇒ 9 9 ⇒ 1 1 ⇒ 1 : 1.
Hence, option 1 is the correct option.
If a, b, c and d are proportional, then a + b a − b \dfrac{a + b}{a - b} a − b a + b
c d \dfrac{c}{d} d c
c − d c + d \dfrac{c-d}{c+d} c + d c − d
d c \dfrac{d}{c} c d
c + d c − d \dfrac{c+d}{c-d} c − d c + d
Answer
Given,
a, b, c and d are proportional.
∴ a : b = c : d
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} ⇒ b a = d c
Applying componendo and dividendo, we get :
⇒ a + b a − b = c + d c − d \Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} ⇒ a − b a + b = c − d c + d
Hence, option 4 is the correct option.
If x : y = 3 : 2, then the ratio (2x2 + 3y2 ) : (3x2 − 2y2 ) is equal to:
5 : 3
6 : 5
12 : 5
30 : 19
Answer
Given,
x : y = 3 : 2
Let x = 3k and y = 2k for some k.
For ratio (2x2 + 3y2 ) : (3x2 − 2y2 ),
Substituting value of x and y in the antecedent,
⇒ 2x2 + 3y2
⇒ 2(3k)2 + 3(2k)2
⇒ 2 × 9k2 + 3 × 4k2
⇒ 18k2 + 12k2
⇒ 30k2 .
Substituting value of x and y in the consequent,
⇒ 3x2 - 2y2
⇒ 3(3k)2 - 2(2k)2
⇒ 3 × 9k2 - 2 × 4k2
⇒ 27k2 - 8k2
⇒ 19k2 .
Therefore the required ratio is:
⇒ 2 x 2 + 3 y 2 3 x 2 − 2 y 2 ⇒ 30 k 2 19 k 2 ⇒ 30 19 ⇒ 30 : 19. \Rightarrow \dfrac{2x^2 + 3y^2}{3x^2 - 2y^2} \\[1em] \Rightarrow \dfrac{30k^2}{19k^2} \\[1em] \Rightarrow \dfrac{30}{19} \\[1em] \Rightarrow 30 : 19. ⇒ 3 x 2 − 2 y 2 2 x 2 + 3 y 2 ⇒ 19 k 2 30 k 2 ⇒ 19 30 ⇒ 30 : 19.
Hence, option 4 is the correct option.
If (5a + 3b) : (2a − 3b) = 23 : 5, then the value of a : b is:
1 : 2
1 : 4
2 : 1
4 : 1
Answer
Given,
∴ 5 a + 3 b 2 a − 3 b = 23 5 \therefore \dfrac{5a + 3b}{2a - 3b} = \dfrac{23}{5} ∴ 2 a − 3 b 5 a + 3 b = 5 23
Cross multiplying:
⇒ 5 ( 5 a + 3 b ) = 23 ( 2 a − 3 b ) ⇒ 25 a + 15 b = 46 a − 69 b ⇒ 25 a − 46 a = − 69 b − 15 b ⇒ − 21 a = − 84 b ⇒ 21 a = 84 b ⇒ a b = 84 21 = 4. \Rightarrow 5(5a + 3b) = 23(2a - 3b) \\[1em] \Rightarrow 25a + 15b = 46a - 69b \\[1em] \Rightarrow 25a - 46a = -69b - 15b \\[1em] \Rightarrow -21a = -84b \\[1em] \Rightarrow 21a = 84b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{84}{21} = 4. ⇒ 5 ( 5 a + 3 b ) = 23 ( 2 a − 3 b ) ⇒ 25 a + 15 b = 46 a − 69 b ⇒ 25 a − 46 a = − 69 b − 15 b ⇒ − 21 a = − 84 b ⇒ 21 a = 84 b ⇒ b a = 21 84 = 4.
Thus,
a : b = 4 : 1.
Hence, option 4 is the correct option.
If (4x2 − 3y2 ) : (2x2 + 5y2 ) = 12 : 19 then x : y is:
2 : 3
1 : 2
2 : 1
3 : 2
Answer
Given,
⇒ 4 x 2 − 3 y 2 2 x 2 + 5 y 2 = 12 19 ⇒ 19 ( 4 x 2 − 3 y 2 ) = 12 ( 2 x 2 + 5 y 2 ) ⇒ 76 x 2 − 57 y 2 = 24 x 2 + 60 y 2 ⇒ 76 x 2 − 24 x 2 = 60 y 2 + 57 y 2 ⇒ 52 x 2 = 117 y 2 ⇒ x 2 y 2 = 117 52 ⇒ x 2 y 2 = 9 4 ⇒ x y = 9 4 ⇒ x y = 3 2 . \Rightarrow \dfrac{4x^2 - 3y^2}{2x^2 + 5y^2} = \dfrac{12}{19} \\[1em] \Rightarrow 19(4x^2 - 3y^2) = 12(2x^2 + 5y^2) \\[1em] \Rightarrow 76x^2 - 57y^2 = 24x^2 + 60y^2 \\[1em] \Rightarrow 76x^2 - 24x^2 = 60y^2 + 57y^2 \\[1em] \Rightarrow 52x^2 = 117y^2 \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{117}{52} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{9}{4} \\[1em] \Rightarrow \dfrac{x}{y} = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{3}{2}. ⇒ 2 x 2 + 5 y 2 4 x 2 − 3 y 2 = 19 12 ⇒ 19 ( 4 x 2 − 3 y 2 ) = 12 ( 2 x 2 + 5 y 2 ) ⇒ 76 x 2 − 57 y 2 = 24 x 2 + 60 y 2 ⇒ 76 x 2 − 24 x 2 = 60 y 2 + 57 y 2 ⇒ 52 x 2 = 117 y 2 ⇒ y 2 x 2 = 52 117 ⇒ y 2 x 2 = 4 9 ⇒ y x = 4 9 ⇒ y x = 2 3 .
Thus,
x : y = 3 : 2.
Hence, option 4 is the correct option.
If, a : b = b : c then a4 : b4 would be equal to:
ac : b2
a2 : c2
b2 : ac
c2 : a2
Answer
Given,
a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b2 = ac
⇒ b4 = (ac)2
⇒ b4 = a2 c2 .
⇒ a 4 b 4 = a 4 a 2 c 2 = a 4 a 2 c 2 = a 4 − 2 c 2 = a 2 c 2 . \Rightarrow \dfrac{a^4}{b^4} = \dfrac{a^4}{a^2c^2} \\[1em] = \dfrac{a^4}{a^2c^2} \\[1em] = \dfrac{a^{4 - 2}}{c^2} \\[1em] = \dfrac{a^{2}}{c^2}. ⇒ b 4 a 4 = a 2 c 2 a 4 = a 2 c 2 a 4 = c 2 a 4 − 2 = c 2 a 2 .
⇒ a4 : b4 = a2 : c2
Hence, option 2 is the correct option.
If a : b : c = 2 : 3 : 4, then 1 a : 1 b : 1 c \dfrac{1}{a} : \dfrac{1}{b} : \dfrac{1}{c} a 1 : b 1 : c 1
1 4 : 1 3 : 1 2 \dfrac{1}{4} : \dfrac{1}{3} : \dfrac{1}{2} 4 1 : 3 1 : 2 1
4 : 3 : 2
6 : 4 : 3
none of these
Answer
Given,
a : b : c = 2 : 3 : 4
Let a = 2k, b = 3k, c = 4k.
Now,
⇒ 1 a : 1 b : 1 c = 1 2 k : 1 3 k : 1 4 k = 1 2 k × 12 k : 1 3 k × 12 k : 1 4 k × 12 k = 6 : 4 : 3. \Rightarrow \dfrac{1}{a} : \dfrac{1}{b} : \dfrac{1}{c} = \dfrac{1}{2k} : \dfrac{1}{3k} : \dfrac{1}{4k} \\[1em] = \dfrac{1}{2k} \times 12k : \dfrac{1}{3k} \times 12k : \dfrac{1}{4k} \times 12k \\[1em] = 6 : 4 : 3. ⇒ a 1 : b 1 : c 1 = 2 k 1 : 3 k 1 : 4 k 1 = 2 k 1 × 12 k : 3 k 1 × 12 k : 4 k 1 × 12 k = 6 : 4 : 3.
Hence, option 3 is the correct option.
If 1 x : 1 y : 1 z = 2 : 3 : 5 \dfrac{1}{x} : \dfrac{1}{y} : \dfrac{1}{z} = 2 : 3 : 5 x 1 : y 1 : z 1 = 2 : 3 : 5
2 : 3 : 5
5 : 3 : 2
15 : 10 : 6
6 : 10 : 15
Answer
Given,
1 x : 1 y : 1 z = 2 : 3 : 5 \dfrac{1}{x} : \dfrac{1}{y} : \dfrac{1}{z} = 2 : 3 : 5 x 1 : y 1 : z 1 = 2 : 3 : 5
Thus,
x : y : z = 1 2 : 1 3 : 1 5 \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{5} 2 1 : 3 1 : 5 1
Since, L.C.M of 2, 3, 5 is 30.
x : y : z = 1 2 × 30 : 1 3 × 30 : 1 5 × 30 x : y : z = \dfrac{1}{2} \times 30 : \dfrac{1}{3} \times 30 : \dfrac{1}{5} \times 30 x : y : z = 2 1 × 30 : 3 1 × 30 : 5 1 × 30
= 15 : 10 : 6.
Hence, option 3 is the correct option.
If (x + y) : (x − y) = 4 : 1, then (x2 + y2 ) : (x2 − y2 ) is :
8 : 17
17 : 8
16 : 1
25 : 9
Answer
Given,
⇒ (x + y) : (x − y) = 4 : 1
⇒ ( x + y ) ( x − y ) = 4 1 \Rightarrow \dfrac{(x + y)}{(x - y)} = \dfrac{4}{1} ⇒ ( x − y ) ( x + y ) = 1 4
Applying Componendo and Dividendo, we get :
⇒ ( x + y ) + ( x − y ) ( x + y ) − ( x − y ) = 4 + 1 4 − 1 ⇒ x + y + x − y x + y − x + y = 5 3 ⇒ 2 x 2 y = 5 3 ⇒ x y = 5 3 . \Rightarrow \dfrac{(x + y) + (x - y)}{(x + y) - (x - y)} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{x + y + x - y}{x + y - x + y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{2y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{5}{3}. ⇒ ( x + y ) − ( x − y ) ( x + y ) + ( x − y ) = 4 − 1 4 + 1 ⇒ x + y − x + y x + y + x − y = 3 5 ⇒ 2 y 2 x = 3 5 ⇒ y x = 3 5 .
Let x = 5k and y = 3k for some constant k.
Substituting value of x and y in x 2 + y 2 x 2 − y 2 \dfrac{x^2 + y^2}{x^2 - y^2} x 2 − y 2 x 2 + y 2
⇒ ( 5 k ) 2 + ( 3 k ) 2 ( 5 k ) 2 − ( 3 k ) 2 ⇒ 25 k 2 + 9 k 2 25 k 2 − 9 k 2 ⇒ 34 k 2 16 k 2 ⇒ 17 8 . \Rightarrow \dfrac{(5k)^2 + (3k)^2}{(5k)^2 - (3k)^2} \\[1em] \Rightarrow \dfrac{25k^2 + 9k^2}{25k^2 - 9k^2} \\[1em] \Rightarrow \dfrac{34k^2}{16k^2} \\[1em] \Rightarrow \dfrac{17}{8}. ⇒ ( 5 k ) 2 − ( 3 k ) 2 ( 5 k ) 2 + ( 3 k ) 2 ⇒ 25 k 2 − 9 k 2 25 k 2 + 9 k 2 ⇒ 16 k 2 34 k 2 ⇒ 8 17 .
Thus, (x2 + y2 ) : (x2 − y2 ) = 17 : 8.
Hence, option 2 is the correct option.
If a : b : c = 2 : 3 : 4 and 2a − 3b + 4c = 33, then the value of c is :
6
9
66 7 \dfrac{66}{7} 7 66
12
Answer
Given,
a : b : c = 2 : 3 : 4
Let a = 2k, b = 3k and c = 4k for some constant k.
Substituting value of a, b and c in 2a − 3b + 4c = 33, we get :
⇒ 2(2k) − 3(3k) + 4(4k) = 33
⇒ 4k - 9k + 16k = 33
⇒ 11k = 33
⇒ k = 33 11 \dfrac{33}{11} 11 33
⇒ k = 3.
⇒ 4k = 4(3) = 12.
Hence, option 4 is the correct option.
If x, y, z are in continued proportion, then (y2 + z2 ) : (x2 + y2 ) is equal to :
z : x
x : z
x : y
(y + z) : (x + y)
Answer
Given,
x, y, z are in continued proportion
Let x y = y z = k \dfrac{x}{y} = \dfrac{y}{z} = k y x = z y = k
⇒ x y = k , y z = k \dfrac{x}{y} = k, \dfrac{y}{z} = k y x = k , z y = k
⇒ y = zk and x = yk = (zk)k = zk2
Substitute value of x and y in y 2 + z 2 x 2 + y 2 \dfrac{y^2 + z^2}{x^2 + y^2} x 2 + y 2 y 2 + z 2
⇒ ( z k ) 2 + z 2 ( z k 2 ) 2 + ( z k ) 2 ⇒ z 2 k 2 + z 2 z 2 k 4 + z 2 k 2 ⇒ z 2 ( k 2 + 1 ) z 2 k 2 ( k 2 + 1 ) ⇒ 1 k 2 ⇒ 1 x y × y z ⇒ 1 x z ⇒ z x ⇒ z : x . \Rightarrow \dfrac{(zk)^2 + z^2}{(zk^2)^2 + (zk)^2} \\[1em] \Rightarrow \dfrac{z^2k^2 + z^2}{z^2k^4 + z^2k^2} \\[1em] \Rightarrow \dfrac{z^2(k^2 + 1)}{z^2k^2(k^2 + 1)} \\[1em] \Rightarrow \dfrac{1}{k^2} \\[1em] \Rightarrow \dfrac{1}{\dfrac{x}{y} \times \dfrac{y}{z}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{x}{z}} \\[1em] \Rightarrow \dfrac{z}{x} \\[1em] \Rightarrow z : x. ⇒ ( z k 2 ) 2 + ( z k ) 2 ( z k ) 2 + z 2 ⇒ z 2 k 4 + z 2 k 2 z 2 k 2 + z 2 ⇒ z 2 k 2 ( k 2 + 1 ) z 2 ( k 2 + 1 ) ⇒ k 2 1 ⇒ y x × z y 1 ⇒ z x 1 ⇒ x z ⇒ z : x .
Hence, option 1 is the correct option.
If a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c b 3 + c 3 + d 3 a 3 + b 3 + c 3 \dfrac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3} a 3 + b 3 + c 3 b 3 + c 3 + d 3
a b \dfrac{a}{b} b a
b c \dfrac{b}{c} c b
c d \dfrac{c}{d} d c
d a \dfrac{d}{a} a d
Answer
Let a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
Therefore,
c = dk
b = ck = (dk)k = dk2
a = bk = (dk2 )k = dk3
k3 = a d \dfrac{a}{d} d a
Substitute value of a,b and c in b 3 + c 3 + d 3 a 3 + b 3 + c 3 \dfrac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3} a 3 + b 3 + c 3 b 3 + c 3 + d 3
⇒ ( d k 2 ) 3 + ( d k ) 3 + d 3 ( d k 3 ) 3 + ( d k 2 ) 3 + ( d k ) 3 ⇒ d 3 k 6 + d 3 k 3 + d 3 d 3 k 9 + d 3 k 6 + d 3 k 3 ⇒ d 3 ( k 6 + k 3 + 1 ) d 3 k 3 ( k 6 + k 3 + 1 ) ⇒ 1 k 3 ⇒ 1 a d ⇒ d a . \Rightarrow \dfrac{(dk^2)^3 + (dk)^3 + d^3}{(dk^3)^3 + (dk^2)^3 + (dk)^3} \\[1em] \Rightarrow \dfrac{d^3k^6 + d^3k^3 + d^3}{d^3k^9 + d^3k^6 + d^3k^3} \\[1em] \Rightarrow \dfrac{d^3(k^6 + k^3 + 1)}{d^3k^3(k^6 + k^3 + 1)} \\[1em] \Rightarrow \dfrac{1}{k^3} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a}{d}} \\[1em] \Rightarrow \dfrac{d}{a}. ⇒ ( d k 3 ) 3 + ( d k 2 ) 3 + ( d k ) 3 ( d k 2 ) 3 + ( d k ) 3 + d 3 ⇒ d 3 k 9 + d 3 k 6 + d 3 k 3 d 3 k 6 + d 3 k 3 + d 3 ⇒ d 3 k 3 ( k 6 + k 3 + 1 ) d 3 ( k 6 + k 3 + 1 ) ⇒ k 3 1 ⇒ d a 1 ⇒ a d .
Hence, option 4 is the correct option.
If a : b = c : d, then m a + n c m b + n d \dfrac{ma + nc}{mb + nd} mb + n d ma + n c
m : n
dm : cn
an : mb
a : b
Answer
Given,
⇒ a : b = c : d
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} ⇒ b a = d c
Let, a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
Therefore, a = bk and c = dk
Substituting values of a and c in m a + n c m b + n d \dfrac{ma + nc}{mb + nd} mb + n d ma + n c
⇒ m ( b k ) + n ( d k ) m b + n d ⇒ k ( m b + n d ) m b + n d ⇒ k ⇒ a b ⇒ a : b . \Rightarrow \dfrac{m(bk) + n(dk)}{mb + nd} \\[1em] \Rightarrow \dfrac{k(mb + nd)}{mb + nd} \\[1em] \Rightarrow k \\[1em] \Rightarrow \dfrac{a}{b} \\[1em] \Rightarrow a : b. ⇒ mb + n d m ( bk ) + n ( d k ) ⇒ mb + n d k ( mb + n d ) ⇒ k ⇒ b a ⇒ a : b .
Hence, option 4 is the correct option.
In an alloy, the ratio of copper and zinc is 5 : 2. If 1.250 kg of zinc is mixed in 17 kg 500 g of alloy, then the ratio of copper and zinc in the alloy will be :
1 : 2
2 : 1
2 : 3
3 : 2
Answer
Given,
Total weight of alloy = 17 kg 500 g = 17.5 kg
Given,
Ratio of copper and zinc is 5 : 2.
Let initial quantity of copper be 5x and zinc be 2x.
Initial quantity of copper = 5 x 5 x + 2 x × 17.5 = 5 x 7 x × 17.5 = 5 × 2.5 = 12.5 kg Initial quantity of zinc = 2 x 5 x + 2 x × 17.5 = 2 x 7 x × 17.5 = 2 × 2.5 = 5 kg . \text{Initial quantity of copper} = \dfrac{5x}{5x + 2x} \times 17.5 \\[1em] = \dfrac{5x}{7x} \times 17.5 \\[1em] = 5 \times 2.5 \\[1em] = 12.5\text{ kg} \\[1em] \text{Initial quantity of zinc} = \dfrac{2x}{5x + 2x} \times 17.5 \\[1em] = \dfrac{2x}{7x} \times 17.5 \\[1em] = 2 \times 2.5 \\[1em] = 5\text{ kg}. Initial quantity of copper = 5 x + 2 x 5 x × 17.5 = 7 x 5 x × 17.5 = 5 × 2.5 = 12.5 kg Initial quantity of zinc = 5 x + 2 x 2 x × 17.5 = 7 x 2 x × 17.5 = 2 × 2.5 = 5 kg .
1.250 kg of zinc is mixed in 17 kg 500 g of alloy, so new quantity of zinc = 5 + 1.250 = 6.25 kg.
Ratio of copper to zinc = 12.5 : 6.25 = 2 : 1
Hence, option 2 is the correct option.
A sum of ₹ 6,400 is divided among three workers in the ratio 3 5 : 2 : 5 3 \dfrac{3}{5} : 2 : \dfrac{5}{3} 5 3 : 2 : 3 5
₹ 2,560
₹ 3,000
₹ 3,200
₹ 3,840
Answer
Wages ₹ 6,400 divided into 3 workers in ratio = 3 5 : 2 : 5 3 \dfrac{3}{5} : 2 : \dfrac{5}{3} 5 3 : 2 : 3 5
L.C.M of 5 and 3 is 15
⇒ 3 5 × 15 : 15 × 2 : 5 3 × 15 \Rightarrow \dfrac{3}{5} \times 15:15 \times 2: \dfrac{5}{3} \times 15 ⇒ 5 3 × 15 : 15 × 2 : 3 5 × 15
⇒ 9 : 30 : 25
Sum of the ratio = 9 + 30 + 25 = 64
The share of the second worker = 30 64 × 6400 \dfrac{30}{64} \times 6400 64 30 × 6400
= 30 × 100
= ₹ 3,000.
Hence, option 2 is the correct option.
If a sum of ₹ x is divided between A and B in the ratio a b : c d \dfrac{a}{b} : \dfrac{c}{d} b a : d c
₹( a b x a c + b d ) \Big(\dfrac{abx}{ac + bd}\Big) ( a c + b d ab x )
₹( a b x a d + b c ) \Big(\dfrac{abx}{ad + bc}\Big) ( a d + b c ab x )
₹( a d x a d + b c ) \Big(\dfrac{adx}{ad + bc}\Big) ( a d + b c a d x )
₹( a d x a b + c d ) \Big(\dfrac{adx}{ab + cd}\Big) ( ab + c d a d x )
Answer
The sum ₹ x is divided between A and B in the ratio :
A : B = a b : c d \dfrac{a}{b}:\dfrac{c}{d} b a : d c
L.C.M of denominators bd:
A : B = a b × b d : c d × b d \dfrac{a}{b} \times bd :\dfrac{c}{d} \times bd b a × b d : d c × b d
A : B = ad : bc
Sum of the ratio = ad + bc.
The sum of money A get = a d a d + b c × x = a d x a d + b c \dfrac{ad}{ad + bc} \times x = \dfrac{adx}{ad + bc} a d + b c a d × x = a d + b c a d x
Hence, option 3 is the correct option.
The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?
90
120
220
240
Answer
Given,
The total number of students is 720.
The initial ratio of boys (B) to girls (G) is 7 : 5.
The total number of ratio parts is 7 + 5 = 12.
Initial number of boys = 7 12 × 720 \dfrac{7}{12} \times 720 12 7 × 720
Initial number of girls = 5 12 × 720 \dfrac{5}{12} \times 720 12 5 × 720
In order to make the ratio 1 : 1, the number of girls and boys must be equal.
Thus, 420 - 300 = 120, more girls should be admitted.
Hence, option 2 is the correct option.
Two numbers are in the ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. The smaller number is:
39
49
66
77
Answer
Let the two numbers be 7x and 11x.
Given,
When 7 is added to each number, the new ratio is 2 : 3.
∴ 7 x + 7 11 x + 7 = 2 3 \therefore \dfrac{7x + 7}{11x + 7} = \dfrac{2}{3} ∴ 11 x + 7 7 x + 7 = 3 2
⇒ 3(7x + 7) = 2(11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 22x - 21x = 21 - 14
⇒ x = 7.
The smaller number is 7x = 7(7) = 49.
Hence, option 2 is the correct option.
What must be subtracted from each of 7, 9, 11 and 15, so that the resulting numbers are in proportion?
1
2
3
5
Answer
Let the number to be subtracted be x.
Thus, the numbers 7 - x, 9 - x, 11 - x and 15 - x are in proportion.
⇒ 7 − x 9 − x = 11 − x 15 − x \Rightarrow \dfrac{7 - x}{9 - x} = \dfrac{11 - x}{15 - x} ⇒ 9 − x 7 − x = 15 − x 11 − x
⇒ (7 - x)(15 - x) = (11 - x)(9 - x)
⇒ 105 - 7x - 15x + x2 = 99 - 11x - 9x + x2
⇒ 105 - 22x + x2 = 99 - 20x + x2
⇒ 105 - 99 = 22x - 20x
⇒ 6 = 2x
⇒ x = 6 2 \dfrac{6}{2} 2 6
⇒ x = 3.
Hence, option 3 is the correct option.
The ratio of milk to water in 80 litres of a mixture is 7 : 3. The quantity of water (in litres) to be added to it to make the ratio 2 : 1 is :
4
5
6
8
Answer
Let the initial quantities of milk and water be 7x and 3x respectively.
Given,
The total volume of the mixture is 80 litres.
⇒ 7x + 3x = 80
⇒ 10x = 80
⇒ x = 80 10 \dfrac{80}{10} 10 80
⇒ x = 8.
Initial Quantity of Milk : 7x = 7 × 8 = 56 litres
Initial Quantity of Water : 3x = 3 × 8 = 24 litres
Let W be the quantity of water added to the mixture.
The quantity of milk remains unchanged.
Water = 24 + W
The new ratio of milk to water is required to be 2 : 1.
⇒ 56 24 + W = 2 1 \dfrac{56}{24 + W} = \dfrac{2}{1} 24 + W 56 = 1 2
⇒ 56 = 2(24 + W)
⇒ 56 = 48 + 2W
⇒ 2W = 56 - 48
⇒ 2W = 8
⇒ W = 8 2 \dfrac{8}{2} 2 8
⇒ W = 4.
The quantity of water to be added is 4 litres.
Hence, option 1 is the correct option.
What number must be added to each of the numbers 7, 11 and 19 so that the resulting numbers may be in continued proportion?
−4
−3
3
4
Answer
Let the number to be added be x.
Thus, numbers 7 + x, 11 + x and 19 + x will be in continued proportion.
∴ 7 + x 11 + x = 11 + x 19 + x ⇒ ( 7 + x ) ( 19 + x ) = ( 11 + x ) ( 11 + x ) ⇒ 133 + 7 x + 19 x + x 2 = 121 + 11 x + 11 x + x 2 ⇒ x 2 + 26 x + 133 = x 2 + 22 x + 121 ⇒ x 2 − x 2 + 26 x − 22 x = 121 − 133 ⇒ 4 x = − 12 ⇒ x = − 12 4 ⇒ x = − 3. \therefore \dfrac{7 + x}{11 + x} = \dfrac{11 + x}{19 + x} \\[1em] \Rightarrow (7 + x)(19 + x) = (11 + x)(11 + x) \\[1em] \Rightarrow 133 + 7x + 19x + x^2 = 121 + 11x + 11x + x^2 \\[1em] \Rightarrow x^2 + 26x + 133 = x^2 + 22x + 121 \\[1em] \Rightarrow x^2 - x^2 + 26x - 22x = 121 - 133 \\[1em] \Rightarrow 4x = -12 \\[1em] \Rightarrow x = \dfrac{-12}{4} \\[1em] \Rightarrow x = -3. ∴ 11 + x 7 + x = 19 + x 11 + x ⇒ ( 7 + x ) ( 19 + x ) = ( 11 + x ) ( 11 + x ) ⇒ 133 + 7 x + 19 x + x 2 = 121 + 11 x + 11 x + x 2 ⇒ x 2 + 26 x + 133 = x 2 + 22 x + 121 ⇒ x 2 − x 2 + 26 x − 22 x = 121 − 133 ⇒ 4 x = − 12 ⇒ x = 4 − 12 ⇒ x = − 3.
Hence, option 2 is the correct option.
Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each, they are in the ratio of 12 : 23. The larger number is :
40
45
55
60
Answer
Given,
Let the two numbers be 3k and 5k.
After subtracting 9 from each number, the ratio changes to 12 : 23.
∴ 3 k − 9 5 k − 9 = 12 23 ⇒ 23 ( 3 k − 9 ) = 12 ( 5 k − 9 ) ⇒ 69 k − 207 = 60 k − 108 ⇒ 69 k − 60 k = 207 − 108 ⇒ 9 k = 99 ⇒ k = 99 9 ⇒ k = 11. \therefore \dfrac{3k - 9}{5k - 9} = \dfrac{12}{23} \\[1em] \Rightarrow 23(3k - 9) = 12(5k - 9) \\[1em] \Rightarrow 69k - 207 = 60k - 108 \\[1em] \Rightarrow 69k - 60k = 207 - 108 \\[1em] \Rightarrow 9k = 99 \\[1em] \Rightarrow k = \dfrac{99}{9} \\[1em] \Rightarrow k = 11. ∴ 5 k − 9 3 k − 9 = 23 12 ⇒ 23 ( 3 k − 9 ) = 12 ( 5 k − 9 ) ⇒ 69 k − 207 = 60 k − 108 ⇒ 69 k − 60 k = 207 − 108 ⇒ 9 k = 99 ⇒ k = 9 99 ⇒ k = 11.
The larger number = 5k = 5 × 11 = 55.
Hence, option 3 is the correct option.
What number must be added to the terms of 3 : 5 to make the ratio 5 : 6 ?
6
7
12
13
Answer
Let the number to be added be x.
∴ 3 + x 5 + x = 5 6 ⇒ 6 ( 3 + x ) = 5 ( 5 + x ) ⇒ 18 + 6 x = 25 + 5 x ⇒ 6 x − 5 x = 25 − 18 ⇒ x = 7. \therefore \dfrac{3 + x}{5 + x} = \dfrac{5}{6} \\[1em] \Rightarrow 6(3 + x) = 5(5 + x) \\[1em] \Rightarrow 18 + 6x = 25 + 5x \\[1em] \Rightarrow 6x - 5x = 25 - 18 \\[1em] \Rightarrow x = 7. ∴ 5 + x 3 + x = 6 5 ⇒ 6 ( 3 + x ) = 5 ( 5 + x ) ⇒ 18 + 6 x = 25 + 5 x ⇒ 6 x − 5 x = 25 − 18 ⇒ x = 7.
Hence, option 2 is the correct option.
Six numbers a, b, c, d, e, f are such that ab = 1, bc = 1 2 \dfrac{1}{2} 2 1 1 2 \dfrac{1}{2} 2 1
4 : 3 : 27
6 : 1 : 9
8 : 9 : 9
72 : 1 : 9
Answer
Given,
ab = 1, bc = 1 2 \dfrac{1}{2} 2 1 1 2 \dfrac{1}{2} 2 1
⇒ e f = 1 2 ⇒ e = 1 2 f ⇒ d e = 2 ⇒ d = 2 e = 2 1 2 f = 4 f ⇒ c d = 6 ⇒ c = 6 d = 6 4 f = 3 2 f ⇒ b c = 1 2 ⇒ b = 1 2 c = 1 2 × 3 2 f = f 3 ⇒ a b = 1 ⇒ a = 1 b = 1 f 3 = 3 f \Rightarrow ef = \dfrac{1}{2} \\[1em] \Rightarrow e = \dfrac{1}{2f} \\[1em] \Rightarrow de = 2 \\[1em] \Rightarrow d = \dfrac{2}{e} = \dfrac{2}{\dfrac{1}{2f}} = 4f \\[1em] \Rightarrow cd = 6 \\[1em] \Rightarrow c = \dfrac{6}{d} = \dfrac{6}{4f} = \dfrac{3}{2f} \\[1em] \Rightarrow bc = \dfrac{1}{2} \\[1em] \Rightarrow b = \dfrac{1}{2c} = \dfrac{1}{2 \times \dfrac{3}{2f}} = \dfrac{f}{3} \\[1em] \Rightarrow ab = 1 \\[1em] \Rightarrow a = \dfrac{1}{b} = \dfrac{1}{\dfrac{f}{3}} = \dfrac{3}{f} ⇒ e f = 2 1 ⇒ e = 2 f 1 ⇒ d e = 2 ⇒ d = e 2 = 2 f 1 2 = 4 f ⇒ c d = 6 ⇒ c = d 6 = 4 f 6 = 2 f 3 ⇒ b c = 2 1 ⇒ b = 2 c 1 = 2 × 2 f 3 1 = 3 f ⇒ ab = 1 ⇒ a = b 1 = 3 f 1 = f 3
ad : be : cf = 3 f × 4 f : f 3 × 1 2 f : 3 2 f × f \dfrac{3}{f} \times 4f : \dfrac{f}{3} \times \dfrac{1}{2f}: \dfrac{3}{2f} \times f f 3 × 4 f : 3 f × 2 f 1 : 2 f 3 × f
= 12 : 1 6 : 3 2 \dfrac{1}{6} : \dfrac{3}{2} 6 1 : 2 3
Since, L.C.M. of 2 and 6 is 6.
= 12 × 6 : 1 6 × 6 : 3 2 × 6 \dfrac{1}{6} \times 6 : \dfrac{3}{2} \times 6 6 1 × 6 : 2 3 × 6
⇒ 72 : 1 : 9.
Hence, option 4 is the correct option.
If a = 4 x y x + y \dfrac{4xy}{x+y} x + y 4 x y ( a + 2 x a − 2 x + a + 2 y a − 2 y ) \Big(\dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y}\Big) ( a − 2 x a + 2 x + a − 2 y a + 2 y )
1
2
1 2 \dfrac{1}{2} 2 1
none of these
Answer
Given,
a = 4 x y x + y \dfrac{4xy}{x+y} x + y 4 x y
Solving,
⇒ a + 2 x a − 2 x + a + 2 y a − 2 y = 4 x y x + y + 2 x 4 x y x + y − 2 x + 4 x y x + y + 2 y 4 x y x + y − 2 y ⇒ Multiply numerator and denominator by (x + y) ⇒ ( 4 x y x + y + 2 x ) × ( x + y ) ( 4 x y x + y − 2 x ) × ( x + y ) + ( 4 x y x + y + 2 y ) × ( x + y ) ( 4 x y x + y − 2 y ) × ( x + y ) ⇒ 4 x y + 2 x ( x + y ) 4 x y − 2 x ( x + y ) + 4 x y + 2 y ( x + y ) 4 x y − 2 y ( x + y ) ⇒ 4 x y + 2 x 2 + 2 x y 4 x y − 2 x 2 − 2 x y + 4 x y + 2 x y + 2 y 2 4 x y − 2 x y − 2 y 2 ⇒ 2 x 2 + 6 x y − 2 x 2 + 2 x y + 6 x y + 2 y 2 2 x y − 2 y 2 ⇒ 2 x ( x + 3 y ) 2 x ( y − x ) + 2 y ( 3 x + y ) 2 y ( x − y ) ⇒ x + 3 y y − x + 3 x + y x − y ⇒ x + 3 y − ( x − y ) + 3 x + y x − y ⇒ − ( x + 3 y ) x − y + 3 x + y x − y ⇒ − x − 3 y + 3 x + y x − y ⇒ − x + 3 x + y − 3 y x − y ⇒ 2 x − 2 y x − y ⇒ 2 ( x − y ) x − y ⇒ 2. \Rightarrow \dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y} = \dfrac{\dfrac{4xy}{x+y} + 2x}{\dfrac{4xy}{x+y} - 2x} + \dfrac{\dfrac{4xy}{x+y} + 2y}{\dfrac{4xy}{x+y} - 2y} \\[1em] \Rightarrow \text{Multiply numerator and denominator by (x + y)} \\[1em] \Rightarrow \dfrac{\Big(\dfrac{4xy}{x+y} + 2x\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2x\Big) \times (x + y)} + \dfrac{\Big(\dfrac{4xy}{x+y} + 2y\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2y\Big) \times (x + y)}\\[1em] \Rightarrow \dfrac{4xy + 2x(x+y)}{4xy - 2x(x+y)} + \dfrac{4xy + 2y(x+y)}{4xy - 2y(x+y)} \\[1em] \Rightarrow \dfrac{4xy + 2x^2 + 2xy}{4xy - 2x^2 - 2xy} + \dfrac{4xy + 2xy + 2y^2}{4xy - 2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x^2 + 6xy}{-2x^2 + 2xy} + \dfrac{6xy + 2y^2}{2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x(x + 3y)}{2x(y - x)} + \dfrac{2y(3x+y)}{2y(x-y)} \\[1em] \Rightarrow \dfrac{x + 3y}{y - x} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{x + 3y}{-(x - y)} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-(x + 3y)}{x - y} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x - 3y + 3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x + 3x + y - 3y}{x - y} \\[1em] \Rightarrow \dfrac{2x - 2y}{x - y} \\[1em] \Rightarrow \dfrac{2(x - y)}{x - y} \\[1em] \Rightarrow 2. ⇒ a − 2 x a + 2 x + a − 2 y a + 2 y = x + y 4 x y − 2 x x + y 4 x y + 2 x + x + y 4 x y − 2 y x + y 4 x y + 2 y ⇒ Multiply numerator and denominator by (x + y) ⇒ ( x + y 4 x y − 2 x ) × ( x + y ) ( x + y 4 x y + 2 x ) × ( x + y ) + ( x + y 4 x y − 2 y ) × ( x + y ) ( x + y 4 x y + 2 y ) × ( x + y ) ⇒ 4 x y − 2 x ( x + y ) 4 x y + 2 x ( x + y ) + 4 x y − 2 y ( x + y ) 4 x y + 2 y ( x + y ) ⇒ 4 x y − 2 x 2 − 2 x y 4 x y + 2 x 2 + 2 x y + 4 x y − 2 x y − 2 y 2 4 x y + 2 x y + 2 y 2 ⇒ − 2 x 2 + 2 x y 2 x 2 + 6 x y + 2 x y − 2 y 2 6 x y + 2 y 2 ⇒ 2 x ( y − x ) 2 x ( x + 3 y ) + 2 y ( x − y ) 2 y ( 3 x + y ) ⇒ y − x x + 3 y + x − y 3 x + y ⇒ − ( x − y ) x + 3 y + x − y 3 x + y ⇒ x − y − ( x + 3 y ) + x − y 3 x + y ⇒ x − y − x − 3 y + 3 x + y ⇒ x − y − x + 3 x + y − 3 y ⇒ x − y 2 x − 2 y ⇒ x − y 2 ( x − y ) ⇒ 2.
Hence, option 2 is the correct option.
If 1 + x + 1 − x 1 + x − 1 − x = a b \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b} 1 + x − 1 − x 1 + x + 1 − x = b a
a 2 a 2 + b 2 \dfrac{a^2}{a^2 + b^2} a 2 + b 2 a 2
b 2 a + b \dfrac{b^2}{a+b} a + b b 2
a b a 2 + b 2 \dfrac{ab}{a^2 + b^2} a 2 + b 2 ab
2 a b a 2 + b 2 \dfrac{2ab}{a^2 + b^2} a 2 + b 2 2 ab
Answer
Given,
⇒ 1 + x + 1 − x 1 + x − 1 − x = a b ⇒ b ( 1 + x + 1 − x ) = a ( 1 + x − 1 − x ) ⇒ b 1 + x + b 1 − x = a 1 + x − a 1 − x ⇒ b 1 + x − a 1 + x = − a 1 − x − b 1 − x ⇒ ( b − a ) 1 + x = − ( a + b ) 1 − x Squaring on Both Sides, ⇒ ( b − a ) 2 ( 1 + x ) = ( a + b ) 2 ( 1 − x ) ⇒ ( b 2 − 2 a b + a 2 ) ( 1 + x ) = ( a 2 + 2 a b + b 2 ) ( 1 − x ) ⇒ ( a 2 + b 2 − 2 a b ) ( 1 + x ) = ( a 2 + b 2 + 2 a b ) ( 1 − x ) ⇒ ( a 2 + b 2 − 2 a b ) + ( a 2 + b 2 − 2 a b ) x = ( a 2 + b 2 + 2 a b ) − ( a 2 + b 2 + 2 a b ) x ⇒ ( a 2 + b 2 − 2 a b ) − ( a 2 + b 2 + 2 a b ) = − ( a 2 + b 2 − 2 a b ) x − ( a 2 + b 2 + 2 a b ) x ⇒ a 2 + b 2 − 2 a b − a 2 − b 2 − 2 a b = − a 2 x − b 2 x + 2 a b x − a 2 x − b 2 x − 2 a b x ⇒ a 2 − a 2 − b 2 + b 2 − 2 a b − 2 a b = − a 2 x − a 2 x − b 2 x − b 2 x + 2 a b x − 2 a b x ⇒ − 4 a b = − 2 a 2 x − 2 b 2 x ⇒ − 4 a b = − 2 ( a 2 + b 2 ) x ⇒ x = − 4 a b − 2 ( a 2 + b 2 ) ⇒ x = 2 a b a 2 + b 2 . \Rightarrow \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b} \\[1em] \Rightarrow b\Big(\sqrt{1+x} + \sqrt{1-x}\Big) = a\Big(\sqrt{1+x} - \sqrt{1-x}\Big) \\[1em] \Rightarrow b\sqrt{1+x} + b\sqrt{1-x} = a\sqrt{1+x} - a\sqrt{1-x} \\[1em] \Rightarrow b\sqrt{1+x} - a\sqrt{1+x} = -a\sqrt{1-x} - b\sqrt{1-x} \\[1em] \Rightarrow (b-a)\sqrt{1+x} = -(a+b)\sqrt{1-x} \\[1em] \text{Squaring on Both Sides,} \\[1em] \Rightarrow (b-a)^2(1+x) = (a+b)^2(1-x) \\[1em] \Rightarrow (b^2 - 2ab + a^2)(1+x) = (a^2 + 2ab + b^2)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab)(1+x) = (a^2 + b^2 + 2ab)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab) + (a^2 + b^2 - 2ab)x = (a^2 + b^2 + 2ab) - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow (a^2 + b^2 - 2ab) - (a^2 + b^2 + 2ab) = -(a^2 + b^2 - 2ab)x - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow a^2 + b^2 - 2ab - a^2 - b^2 - 2ab = - a^2x - b^2x + 2abx - a^2x - b^2x - 2abx \\[1em] \Rightarrow a^2 - a^2 - b^2 + b^2 - 2ab - 2ab = - a^2x - a^2x - b^2x - b^2x + 2abx - 2abx \\[1em] \Rightarrow -4ab = - 2a^2x - 2b^2x \\[1em] \Rightarrow -4ab = -2(a^2 + b^2)x \\[1em] \Rightarrow x = \dfrac{-4ab}{-2(a^2 + b^2)} \\[1em] \Rightarrow x = \dfrac{2ab}{a^2 + b^2}. ⇒ 1 + x − 1 − x 1 + x + 1 − x = b a ⇒ b ( 1 + x + 1 − x ) = a ( 1 + x − 1 − x ) ⇒ b 1 + x + b 1 − x = a 1 + x − a 1 − x ⇒ b 1 + x − a 1 + x = − a 1 − x − b 1 − x ⇒ ( b − a ) 1 + x = − ( a + b ) 1 − x Squaring on Both Sides, ⇒ ( b − a ) 2 ( 1 + x ) = ( a + b ) 2 ( 1 − x ) ⇒ ( b 2 − 2 ab + a 2 ) ( 1 + x ) = ( a 2 + 2 ab + b 2 ) ( 1 − x ) ⇒ ( a 2 + b 2 − 2 ab ) ( 1 + x ) = ( a 2 + b 2 + 2 ab ) ( 1 − x ) ⇒ ( a 2 + b 2 − 2 ab ) + ( a 2 + b 2 − 2 ab ) x = ( a 2 + b 2 + 2 ab ) − ( a 2 + b 2 + 2 ab ) x ⇒ ( a 2 + b 2 − 2 ab ) − ( a 2 + b 2 + 2 ab ) = − ( a 2 + b 2 − 2 ab ) x − ( a 2 + b 2 + 2 ab ) x ⇒ a 2 + b 2 − 2 ab − a 2 − b 2 − 2 ab = − a 2 x − b 2 x + 2 ab x − a 2 x − b 2 x − 2 ab x ⇒ a 2 − a 2 − b 2 + b 2 − 2 ab − 2 ab = − a 2 x − a 2 x − b 2 x − b 2 x + 2 ab x − 2 ab x ⇒ − 4 ab = − 2 a 2 x − 2 b 2 x ⇒ − 4 ab = − 2 ( a 2 + b 2 ) x ⇒ x = − 2 ( a 2 + b 2 ) − 4 ab ⇒ x = a 2 + b 2 2 ab .
Hence, option 4 is the correct option.
If x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 \dfrac{\sqrt{2a+1} + \sqrt{2a-1}}{\sqrt{2a+1} - \sqrt{2a-1}} 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1
x2 + 2ax + 1 = 0
x2 − 2ax + 1 = 0
x2 + 4ax − 1 = 0
x2 − 4ax + 1 = 0
Answer
Given,
x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 x ( 2 a + 1 − 2 a − 1 ) = 2 a + 1 + 2 a − 1 x 2 a + 1 − x 2 a − 1 = 2 a + 1 + 2 a − 1 ( x − 1 ) 2 a + 1 = ( x + 1 ) 2 a − 1 Squaring on Both Sides, ( x − 1 ) 2 ( 2 a + 1 ) = ( x + 1 ) 2 ( 2 a − 1 ) ( 2 a + 1 ) ( x 2 − 2 x + 1 ) = ( 2 a − 1 ) ( x 2 + 2 x + 1 ) 2 a x 2 − 4 a x + 2 a + x 2 − 2 x + 1 = 2 a x 2 + 4 a x + 2 a − x 2 − 2 x − 1 2 a x 2 − 2 a x 2 + x 2 + x 2 − 4 a x − 4 a x + 2 a − 2 a − 2 x + 2 x + 1 + 1 = 0 2 x 2 − 8 a x + 2 = 0 2 ( x 2 − 4 a x + 1 ) = 0 x 2 − 4 a x + 1 = 0 x = \dfrac{\sqrt{2a+1} + \sqrt{2a-1}}{\sqrt{2a+1} - \sqrt{2a-1}} \\[1em] x\Big(\sqrt{2a+1} - \sqrt{2a-1}\Big) = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] x\sqrt{2a+1} - x\sqrt{2a-1} = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] (x-1)\sqrt{2a+1} = (x+1)\sqrt{2a-1} \\[1em] \text{Squaring on Both Sides,} \\[1em] (x-1)^2(2a+1) = (x+1)^2(2a-1) \\[1em] (2a+1)(x^2 - 2x + 1) = (2a-1)(x^2 + 2x + 1) \\[1em] 2ax^2 - 4ax + 2a + x^2 - 2x + 1 = 2ax^2 + 4ax + 2a - x^2 - 2x - 1 \\[1em] 2ax^2 - 2ax^2 + x^2 + x^2 - 4ax - 4ax + 2a - 2a - 2x + 2x + 1 + 1 = 0\\[1em] 2x^2 - 8ax + 2 = 0 \\[1em] 2(x^2 - 4ax + 1) = 0 \\[1em] x^2 - 4ax + 1 = 0 x = 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 x ( 2 a + 1 − 2 a − 1 ) = 2 a + 1 + 2 a − 1 x 2 a + 1 − x 2 a − 1 = 2 a + 1 + 2 a − 1 ( x − 1 ) 2 a + 1 = ( x + 1 ) 2 a − 1 Squaring on Both Sides, ( x − 1 ) 2 ( 2 a + 1 ) = ( x + 1 ) 2 ( 2 a − 1 ) ( 2 a + 1 ) ( x 2 − 2 x + 1 ) = ( 2 a − 1 ) ( x 2 + 2 x + 1 ) 2 a x 2 − 4 a x + 2 a + x 2 − 2 x + 1 = 2 a x 2 + 4 a x + 2 a − x 2 − 2 x − 1 2 a x 2 − 2 a x 2 + x 2 + x 2 − 4 a x − 4 a x + 2 a − 2 a − 2 x + 2 x + 1 + 1 = 0 2 x 2 − 8 a x + 2 = 0 2 ( x 2 − 4 a x + 1 ) = 0 x 2 − 4 a x + 1 = 0
Hence, option 4 is the correct option.
If a b + c = b c + a = c a + b \dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b} b + c a = c + a b = a + b c
1
−1
either 1 2 \dfrac{1}{2} 2 1
neither 1 2 \dfrac{1}{2} 2 1
Answer
Given,
a b + c = b c + a = c a + b = k \dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b} = k b + c a = c + a b = a + b c = k
From the first relation: a = k(b + c) ....(1)
From the second relation: b = k(c + a) ....(2)
From the third relation: c = k(a + b) ....(3)
Adding (1), (2) and (3):
a + b + c = k(b + c) + k(c + a) + k(a + b)
a + b + c = k((b + c) + (c + a) + (a + b))
a + b + c = k(b + c + c + a + a + b)
a + b + c = k(a + a + b + b + c + c)
a + b + c = k(2a + 2b + 2c)
a + b + c = 2k(a + b + c)
If a + b + c ≠ 0, then,
2 k = ( a + b + c ) ( a + b + c ) 2k =\dfrac{(a + b + c)}{(a + b + c)} 2 k = ( a + b + c ) ( a + b + c )
2k = 1
k = 1 2 . k = \dfrac{1}{2}. k = 2 1 .
If a + b + c = 0, then the equations will be,
a + b = -c ....(4)
b + c = -a ....(5)
c + a = -b ....(6)
Subtituting values in a b + c = b c + a = c a + b \dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b} b + c a = c + a b = a + b c
a − a = b − b = c − c = k \dfrac{a}{-a} = \dfrac{b}{-b} = \dfrac{c}{-c} = k − a a = − b b = − c c = k
k = -1
Hence, Option 3 is the correct option.
If b is the mean proportion between a and c, then a 2 − b 2 + c 2 a − 2 − b − 2 + c − 2 \dfrac{a^2 − b^2 + c^2}{a^{-2} − b^{-2} + c^{-2}} a − 2 − b − 2 + c − 2 a 2 − b 2 + c 2
a4
b4
a2
b2
Answer
Since b is the mean proportion between a and c,
∴ a b = b c \therefore \dfrac{a}{b} = \dfrac{b}{c} ∴ b a = c b
⇒ b 2 = a c \Rightarrow b^2 = ac ⇒ b 2 = a c
Solving, ⇒ a 2 − b 2 + c 2 a − 2 − b − 2 + c − 2 ⇒ a 2 − b 2 + c 2 1 a 2 − 1 b 2 + 1 c 2 ⇒ a 2 − b 2 + c 2 b 2 c 2 a 2 b 2 c 2 − a 2 c 2 a 2 b 2 c 2 + a 2 b 2 a 2 b 2 c 2 ⇒ a 2 − b 2 + c 2 b 2 c 2 a 2 b 2 c 2 − ( b 2 ) 2 a 2 b 2 c 2 + a 2 b 2 a 2 b 2 c 2 ⇒ a 2 − b 2 + c 2 b 2 ( c 2 − b 2 + a 2 ) a 2 b 2 c 2 ⇒ a 2 − b 2 + c 2 ( c 2 − b 2 + a 2 ) a 2 c 2 ⇒ a 2 − b 2 + c 2 × a 2 c 2 ( c 2 − b 2 + a 2 ) ⇒ a 2 c 2 = ( b 2 ) 2 = b 4 . \Rightarrow \dfrac{a^2 - b^2 + c^2}{a^{-2} - b^{-2} + c^{-2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{1}{a^{2}} - \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2c^2}{a^2b^2c^2} - \dfrac{a^2c^2}{a^2b^2c^2} + \dfrac{a^2b^2}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2c^2}{a^2b^2c^2} - \dfrac{(b^2)^2}{a^2b^2c^2} + \dfrac{a^2b^2}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{b^2(c^2 - b^2 + a^2)}{a^2b^2c^2}} \\[1em] \Rightarrow \dfrac{a^2 - b^2 + c^2}{\dfrac{(c^2 - b^2 + a^2)}{a^2c^2}} \\[1em] \Rightarrow a^2 - b^2 + c^2 \times {\dfrac{a^2c^2}{(c^2 - b^2 + a^2)}} \\[1em] \Rightarrow a^2c^2 = (b^2)^2 = b^4. ⇒ a − 2 − b − 2 + c − 2 a 2 − b 2 + c 2 ⇒ a 2 1 − b 2 1 + c 2 1 a 2 − b 2 + c 2 ⇒ a 2 b 2 c 2 b 2 c 2 − a 2 b 2 c 2 a 2 c 2 + a 2 b 2 c 2 a 2 b 2 a 2 − b 2 + c 2 ⇒ a 2 b 2 c 2 b 2 c 2 − a 2 b 2 c 2 ( b 2 ) 2 + a 2 b 2 c 2 a 2 b 2 a 2 − b 2 + c 2 ⇒ a 2 b 2 c 2 b 2 ( c 2 − b 2 + a 2 ) a 2 − b 2 + c 2 ⇒ a 2 c 2 ( c 2 − b 2 + a 2 ) a 2 − b 2 + c 2 ⇒ a 2 − b 2 + c 2 × ( c 2 − b 2 + a 2 ) a 2 c 2 ⇒ a 2 c 2 = ( b 2 ) 2 = b 4 .
Hence, option 2 is the correct option.
If b is the mean proportion between a and c, then the mean proportion between (a2 + b2 ) and (b2 + c2 ) is:
a(b + c)
b(a + c)
c(a + b)
none of these
Answer
Since b is the mean proportion between a and c,
⇒ a b = b c ⇒ b 2 = a c . \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow b^2 = ac. ⇒ b a = c b ⇒ b 2 = a c .
Let the required mean proportion be x. Then,
⇒ a 2 + b 2 x = x b 2 + c 2 ⇒ x 2 = ( a 2 + b 2 ) ( b 2 + c 2 ) . \Rightarrow \dfrac{a^2 + b^2}{x} = \dfrac{x}{b^2 + c^2} \\[1em] \Rightarrow x^2 = (a^2 + b^2)(b^2 + c^2). ⇒ x a 2 + b 2 = b 2 + c 2 x ⇒ x 2 = ( a 2 + b 2 ) ( b 2 + c 2 ) .
Now substitute b2 = ac:
⇒ x2 = (a2 + b2 )(b2 + c2 )
⇒ x2 = a2 b2 + a2 c2 + b4 + b2 c2
⇒ x2 = a2 b2 + (ac)2 + b4 + b2 c2
⇒ x2 = a2 b2 + (b)2 + b4 + b2 c2
⇒ x2 = a2 b2 + b4 + b4 + b2 c2
⇒ x2 = b2 (a2 + 2b2 + c2 )
⇒ x2 = b2 (a2 + 2ac + c2 )
⇒ x2 = b2 (a + c)2 .
Therefore,
⇒ x 2 = b 2 ( a + c ) 2 ⇒ x = b 2 ( a + c ) 2 ⇒ x = b ( a + c ) . \Rightarrow x^2 = b^2(a + c)^2 \\[1em] \Rightarrow x = \sqrt{b^2(a + c)^2} \\[1em] \Rightarrow x = b(a + c). ⇒ x 2 = b 2 ( a + c ) 2 ⇒ x = b 2 ( a + c ) 2 ⇒ x = b ( a + c ) .
Hence, option 2 is the correct option.
Assertion Reason Questions
Assertion (A) : If a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c a + c b + d = a b \dfrac{a+c}{b+d} = \dfrac{a}{b} b + d a + c = b a
Reason (R) : If two or more than two ratios are equal, then each ratio = sum of antecedents sum of consequents \dfrac{\text{sum of antecedents}}{\text{sum of consequents}} sum of consequents sum of antecedents
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
Let a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
a = kb, c = kd
Substituting value of a and c in a + c b + d \dfrac{a+c}{b+d} b + d a + c
⇒ k b + k d b + d ⇒ k ( b + d ) ( b + d ) ⇒ k ⇒ a b or c d . \Rightarrow \dfrac{kb + kd}{b + d} \\[1em] \Rightarrow \dfrac{k(b + d)}{(b + d)} \\[1em] \Rightarrow k \\[1em] \Rightarrow \dfrac{a}{b} \text{ or } \dfrac{c}{d}. ⇒ b + d kb + k d ⇒ ( b + d ) k ( b + d ) ⇒ k ⇒ b a or d c .
∴ Assertion (A) is true.
As proved in assertion,
Each ratio = sum of antecedents sum of consequents \dfrac{\text{sum of antecedents}}{\text{sum of consequents}} sum of consequents sum of antecedents
∴ Reason (R) is true.
Hence, option 3 is the correct option.
Assertion (A) : The mean proportion between a2 b and 1 b \dfrac{1}{b} b 1 a b \dfrac{a}{b} b a
Reason (R) : The mean proportion between x and y is given by x y \sqrt{xy} x y
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
We know that,
Mean proportion between two numbers
= First number × Second number \sqrt{\text{First number} \times \text{Second number}} First number × Second number
Thus,
The mean proportion between x and y is given by x y \sqrt{xy} x y
∴ Reason (R) is true.
Thus,
The mean proportion between a2 b and 1 b \dfrac{1}{b} b 1 a 2 b × 1 b \sqrt{a^2b \times \dfrac{1}{b}} a 2 b × b 1
= a 2 = a = \sqrt{a^2} = a = a 2 = a
∴ Assertion (A) is false.
Hence, option 2 is the correct option.
Assertion (A) : If 2x = 3y and 4y = 5z, then x : y : z is equal to 2 : 4 : 5.
Reason (R) : If x = y and y = z, then we cannot find the ratio x : y : z.
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
Given,
⇒ 2x = 3y
⇒ x y = 3 2 \dfrac{x}{y} = \dfrac{3}{2} y x = 2 3
⇒ x : y = 3 : 2
⇒ 4y = 5z
⇒ y z = 5 4 \dfrac{y}{z} = \dfrac{5}{4} z y = 4 5
⇒ y : z = 5 : 4
Since, x : y = 3 : 2 and y : z = 5 : 4. L.C.M of 2 and 5 is 10.
⇒ x : y = 3 : 2 = (3 × 5) : (2 × 5) = 15 : 10
⇒ y : z = 5 : 4 = (5 × 2) : (4 × 2) = 10 : 8
⇒ x : y : z = 15 : 10 : 8.
∴ Assertion (A) is false.
If x = y and y = z, then by the transitive property of equality, it follows that x = y = z.
x : y : z = x : x : x = 1 : 1 : 1.
∴ Reason (R) is false.
Both A and R are false.
Hence, option 4 is the correct option.
