If A : B = 7 : 5 and B : C = 7 : 5, then A : C is :
25 : 49
49 : 25
1
15 : 14
Answer
Given,
⇒ A : B = 7 : 5
⇒ A B = 7 5 \Rightarrow \dfrac{A}{B} = \dfrac{7}{5} ⇒ B A = 5 7
⇒ B : C = 7 : 5
⇒ B C = 7 5 \Rightarrow \dfrac{B}{C} = \dfrac{7}{5} ⇒ C B = 5 7
Multiplying equation (1) and (2), we get :
⇒ A B × B C = 7 5 × 7 5 ⇒ A C = 49 25 . \Rightarrow \dfrac{A}{B} \times \dfrac{B}{C} = \dfrac{7}{5} \times \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{49}{25}. ⇒ B A × C B = 5 7 × 5 7 ⇒ C A = 25 49 .
A : C = 49 : 25.
Hence, Option 2 is the correct option.
If x 2 − 1 x 2 + 1 = 3 5 \dfrac{x^2 - 1}{x^2 + 1} =\dfrac{3}{5} x 2 + 1 x 2 − 1 = 5 3
1 2 \dfrac{1}{2} 2 1
2
± 1 2 \pm \dfrac{1}{2} ± 2 1
± 2 \pm 2 ± 2
Answer
Given,
⇒ x 2 − 1 x 2 + 1 = 3 5 ⇒ 5 ( x 2 − 1 ) = 3 ( x 2 + 1 ) ⇒ 5 x 2 − 5 = 3 x 2 + 3 ⇒ 5 x 2 − 3 x 2 = 3 + 5 ⇒ 2 x 2 = 8 ⇒ x 2 = 4 ⇒ x = 4 ⇒ x = ± 2. \Rightarrow \dfrac{x^2 - 1}{x^2 + 1} =\dfrac{3}{5} \\[1em] \Rightarrow 5(x^2 - 1) = 3(x^2 + 1) \\[1em] \Rightarrow 5x^2 - 5 = 3x^2 + 3 \\[1em] \Rightarrow 5x^2 - 3x^2 = 3 + 5 \\[1em] \Rightarrow 2x^2 = 8 \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \sqrt{4} \\[1em] \Rightarrow x = \pm 2. ⇒ x 2 + 1 x 2 − 1 = 5 3 ⇒ 5 ( x 2 − 1 ) = 3 ( x 2 + 1 ) ⇒ 5 x 2 − 5 = 3 x 2 + 3 ⇒ 5 x 2 − 3 x 2 = 3 + 5 ⇒ 2 x 2 = 8 ⇒ x 2 = 4 ⇒ x = 4 ⇒ x = ± 2.
Hence, Option 4 is the correct option.
If (2x + 3y) : (3x + 2y) = 4 : 3; then x : y is :
6 : 1
2 : 3
1 : 6
3 : 2
Answer
Given,
⇒ ( 2 x + 3 y ) : ( 3 x + 2 y ) = 4 : 3 ⇒ 2 x + 3 y 3 x + 2 y = 4 3 ⇒ 3 ( 2 x + 3 y ) = 4 ( 3 x + 2 y ) ⇒ 6 x + 9 y = 12 x + 8 y ⇒ 12 x − 6 x = 9 y − 8 y ⇒ 6 x = y ⇒ x y = 1 6 ⇒ x : y = 1 : 6. \Rightarrow (2x + 3y) : (3x + 2y) = 4 : 3 \\[1em] \Rightarrow \dfrac{2x + 3y}{3x + 2y} = \dfrac{4}{3} \\[1em] \Rightarrow 3(2x + 3y) = 4(3x + 2y) \\[1em] \Rightarrow 6x + 9y = 12x + 8y \\[1em] \Rightarrow 12x - 6x = 9y - 8y \\[1em] \Rightarrow 6x = y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{6} \\[1em] \Rightarrow x : y = 1 : 6. ⇒ ( 2 x + 3 y ) : ( 3 x + 2 y ) = 4 : 3 ⇒ 3 x + 2 y 2 x + 3 y = 3 4 ⇒ 3 ( 2 x + 3 y ) = 4 ( 3 x + 2 y ) ⇒ 6 x + 9 y = 12 x + 8 y ⇒ 12 x − 6 x = 9 y − 8 y ⇒ 6 x = y ⇒ y x = 6 1 ⇒ x : y = 1 : 6.
Hence, Option 3 is the correct option.
Two numbers are in the ratio 5 : 8. If 10 is subtracted from each number, the ratio becomes 4 : 7. The numbers are :
80 and 40
50 and 80
25 and 40
40 and 25
Answer
Let two numbers be x and y.
Given,
Ratio of two numbers = 5 : 8.
⇒ x y = 5 8 ⇒ x = 5 8 y . . . . . ( 1 ) \Rightarrow \dfrac{x}{y} = \dfrac{5}{8} \\[1em] \Rightarrow x = \dfrac{5}{8}y \space .....(1) ⇒ y x = 8 5 ⇒ x = 8 5 y ..... ( 1 )
Given,
If 10 is subtracted from each number, the ratio becomes 4 : 7.
⇒ x − 10 y − 10 = 4 7 ⇒ 7 ( x − 10 ) = 4 ( y − 10 ) ⇒ 7 x − 70 = 4 y − 40 \Rightarrow \dfrac{x - 10}{y - 10} = \dfrac{4}{7} \\[1em] \Rightarrow 7(x - 10) = 4(y - 10) \\[1em] \Rightarrow 7x - 70 = 4y - 40 ⇒ y − 10 x − 10 = 7 4 ⇒ 7 ( x − 10 ) = 4 ( y − 10 ) ⇒ 7 x − 70 = 4 y − 40
Substituting value of x from equation (1), in above equation we get :
⇒ 7 × 5 8 y − 70 = 4 y − 40 ⇒ 35 y 8 − 70 = 4 y − 40 ⇒ 35 y 8 − 4 y = 70 − 40 ⇒ 35 y − 32 y 8 = 30 ⇒ 3 y 8 = 30 ⇒ y = 30 × 8 3 ⇒ y = 80. ⇒ x = 5 8 y ⇒ x = 5 8 × 80 = 50. \Rightarrow 7 \times \dfrac{5}{8}y - 70 = 4y - 40 \\[1em] \Rightarrow \dfrac{35y}{8} - 70 = 4y - 40 \\[1em] \Rightarrow \dfrac{35y}{8} - 4y = 70 - 40 \\[1em] \Rightarrow \dfrac{35y - 32y}{8} = 30 \\[1em] \Rightarrow \dfrac{3y}{8} = 30 \\[1em] \Rightarrow y = \dfrac{30 \times 8}{3} \\[1em] \Rightarrow y = 80. \\[1em] \Rightarrow x = \dfrac{5}{8}y \\[1em] \Rightarrow x = \dfrac{5}{8} \times 80 = 50. ⇒ 7 × 8 5 y − 70 = 4 y − 40 ⇒ 8 35 y − 70 = 4 y − 40 ⇒ 8 35 y − 4 y = 70 − 40 ⇒ 8 35 y − 32 y = 30 ⇒ 8 3 y = 30 ⇒ y = 3 30 × 8 ⇒ y = 80. ⇒ x = 8 5 y ⇒ x = 8 5 × 80 = 50.
Numbers = 50 and 80.
Hence, Option 2 is the correct option.
The compounded ratio of x - y : x + y and (x + y)2 : x2 - y2 is :
2 : 1
1 : (x + y)2
1 : 1
1 : 2
Answer
Compounded ratio of x - y : x + y and (x + y)2 : x2 - y2 :
⇒ x − y x + y × ( x + y ) 2 x 2 − y 2 ⇒ x − y x + y × ( x + y ) 2 ( x + y ) ( x − y ) ⇒ ( x − y ) ( x + y ) 2 ( x + y ) 2 ( x − y ) ⇒ 1 1 ⇒ 1 : 1. \Rightarrow \dfrac{x - y}{x + y} \times \dfrac{(x + y)^2}{x^2 - y^2} \\[1em] \Rightarrow \dfrac{x - y}{x + y} \times \dfrac{(x + y)^2}{(x + y)(x - y)} \\[1em] \Rightarrow \dfrac{(x - y)(x + y)^2}{(x + y)^2(x - y)} \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1 : 1. ⇒ x + y x − y × x 2 − y 2 ( x + y ) 2 ⇒ x + y x − y × ( x + y ) ( x − y ) ( x + y ) 2 ⇒ ( x + y ) 2 ( x − y ) ( x − y ) ( x + y ) 2 ⇒ 1 1 ⇒ 1 : 1.
Hence, Option 3 is the correct option.
The sub-duplicate ratio of 6x2 : 24y2 is :
x : 2y
2x : y
x : y
y : x
Answer
Sub-duplicate ratio of 6x2 : 24y2 is :
⇒ 6 x 2 24 y 2 ⇒ 6 x 2 6 y ⇒ x 2 y ⇒ x : 2 y . \Rightarrow \dfrac{\sqrt{6x^2}}{\sqrt{24y^2}} \\[1em] \Rightarrow \dfrac{\sqrt{6}x}{2\sqrt{6}y} \\[1em] \Rightarrow \dfrac{x}{2y} \\[1em] \Rightarrow x : 2y. ⇒ 24 y 2 6 x 2 ⇒ 2 6 y 6 x ⇒ 2 y x ⇒ x : 2 y .
Hence, Option 1 is the correct option.
If m + n m + 3 n = 2 3 \dfrac{m + n}{m + 3n} = \dfrac{2}{3} m + 3 n m + n = 3 2 2 n 2 3 m 2 + m n \dfrac{2n^2}{3m^2 + mn} 3 m 2 + mn 2 n 2
Answer
Given,
⇒ m + n m + 3 n = 2 3 ⇒ 3 ( m + n ) = 2 ( m + 3 n ) ⇒ 3 m + 3 n = 2 m + 6 n ⇒ 3 m − 2 m = 6 n − 3 n ⇒ m = 3 n . \phantom{\Rightarrow} \dfrac{m + n}{m + 3n} = \dfrac{2}{3} \\[1em] \Rightarrow 3(m + n) = 2(m + 3n) \\[1em] \Rightarrow 3m + 3n = 2m + 6n \\[1em] \Rightarrow 3m - 2m = 6n - 3n \\[1em] \Rightarrow m = 3n. ⇒ m + 3 n m + n = 3 2 ⇒ 3 ( m + n ) = 2 ( m + 3 n ) ⇒ 3 m + 3 n = 2 m + 6 n ⇒ 3 m − 2 m = 6 n − 3 n ⇒ m = 3 n .
Substituting value of m in 2 n 2 3 m 2 + m n \dfrac{2n^2}{3m^2 + mn} 3 m 2 + mn 2 n 2
⇒ 2 n 2 3 ( 3 n ) 2 + ( 3 n ) n ⇒ 2 n 2 3 ( 9 n 2 ) + 3 n 2 ⇒ 2 n 2 27 n 2 + 3 n 2 ⇒ 2 n 2 30 n 2 ⇒ 1 15 . \Rightarrow \dfrac{2n^2}{3(3n)^2 + (3n)n} \\[1em] \Rightarrow \dfrac{2n^2}{3(9n^2) + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{27n^2 + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{30n^2} \\[1em] \Rightarrow \dfrac{1}{15}. ⇒ 3 ( 3 n ) 2 + ( 3 n ) n 2 n 2 ⇒ 3 ( 9 n 2 ) + 3 n 2 2 n 2 ⇒ 27 n 2 + 3 n 2 2 n 2 ⇒ 30 n 2 2 n 2 ⇒ 15 1 .
Hence, 2 n 2 3 m 2 + m n = 1 15 . \dfrac{2n^2}{3m^2 + mn} = \dfrac{1}{15}. 3 m 2 + mn 2 n 2 = 15 1 .
If the ratio between 8 and 11 is same as the ratio of 2x - y to x + 2y, find the value of 7 x 9 y \dfrac{7x}{9y} 9 y 7 x
Answer
According to question,
⇒ 2 x − y x + 2 y = 8 11 ⇒ 11 ( 2 x − y ) = 8 ( x + 2 y ) ⇒ 22 x − 11 y = 8 x + 16 y 22 x − 8 x = 16 y + 11 y 14 x = 27 y x = 27 y 14 . \phantom{\Rightarrow} \dfrac{2x - y}{x + 2y} = \dfrac{8}{11} \\[1em] \Rightarrow 11(2x - y) = 8(x + 2y) \\[1em] \Rightarrow 22x - 11y = 8x + 16y \\[1em] 22x - 8x = 16y + 11y \\[1em] 14x = 27y \\[1em] x = \dfrac{27y}{14}. ⇒ x + 2 y 2 x − y = 11 8 ⇒ 11 ( 2 x − y ) = 8 ( x + 2 y ) ⇒ 22 x − 11 y = 8 x + 16 y 22 x − 8 x = 16 y + 11 y 14 x = 27 y x = 14 27 y .
Substituting value of x in 7 x 9 y \dfrac{7x}{9y} 9 y 7 x
7 x 9 y = 7 9 y × 27 y 14 = 3 2 . \dfrac{7x}{9y} = \dfrac{7}{9y} \times \dfrac{27y}{14} \\[1em] = \dfrac{3}{2}. 9 y 7 x = 9 y 7 × 14 27 y = 2 3 .
Hence, 7 x 9 y = 3 2 \dfrac{7x}{9y} = \dfrac{3}{2} 9 y 7 x = 2 3
Divide ₹1,290 into A, B and C such that A is 2 5 \dfrac{2}{5} 5 2
Answer
Given, B : C = 4 : 3
If B = 4a, then C = 3a
Given, A is 2 5 \dfrac{2}{5} 5 2
∴ A = 2 5 × 4 a = 8 a 5 \dfrac{2}{5} \times 4a = \dfrac{8a}{5} 5 2 × 4 a = 5 8 a
Share of A,
= A A + B + C × 1290 = 8 a 5 8 a 5 + 4 a + 3 a × 1290 = 8 a 5 8 a + 20 a + 15 a 5 × 1290 = 8 a 43 a × 1290 = 8 43 × 1290 = 240. = \dfrac{A}{A + B + C} \times 1290 \\[1em] = \dfrac{\dfrac{8a}{5}}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{\dfrac{8a}{5}}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{8a}{43a} \times 1290 \\[1em] = \dfrac{8}{43} \times 1290 \\[1em] = 240. = A + B + C A × 1290 = 5 8 a + 4 a + 3 a 5 8 a × 1290 = 5 8 a + 20 a + 15 a 5 8 a × 1290 = 43 a 8 a × 1290 = 43 8 × 1290 = 240.
Share of B,
= B A + B + C × 1290 = 4 a 8 a 5 + 4 a + 3 a × 1290 = 4 a 8 a + 20 a + 15 a 5 × 1290 = 20 a 43 a × 1290 = 20 43 × 1290 = 600. = \dfrac{B}{A + B + C} \times 1290 \\[1em] = \dfrac{4a}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{4a}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{20a}{43a} \times 1290 \\[1em] = \dfrac{20}{43} \times 1290 \\[1em] = 600. = A + B + C B × 1290 = 5 8 a + 4 a + 3 a 4 a × 1290 = 5 8 a + 20 a + 15 a 4 a × 1290 = 43 a 20 a × 1290 = 43 20 × 1290 = 600.
Share of C,
= C A + B + C × 1290 = 3 a 8 a 5 + 4 a + 3 a × 1290 = 3 a 8 a + 20 a + 15 a 5 × 1290 = 15 a 43 a × 1290 = 15 43 × 1290 = 450. = \dfrac{C}{A + B + C} \times 1290 \\[1em] = \dfrac{3a}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{3a}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{15a}{43a} \times 1290 \\[1em] = \dfrac{15}{43} \times 1290 \\[1em] = 450. = A + B + C C × 1290 = 5 8 a + 4 a + 3 a 3 a × 1290 = 5 8 a + 20 a + 15 a 3 a × 1290 = 43 a 15 a × 1290 = 43 15 × 1290 = 450.
Hence, the amount of money with A = ₹ 240, B = ₹ 600 and C = ₹ 450.
A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
Answer
Ratio of boys to girls = 3 : 2.
No. of boys = 3 3 + 2 × 630 = 3 5 × 630 = 378. \dfrac{3}{3 + 2} \times 630 = \dfrac{3}{5} \times 630 = 378. 3 + 2 3 × 630 = 5 3 × 630 = 378.
No. of girls = 2 3 + 2 × 630 = 2 5 × 630 = 252. \dfrac{2}{3 + 2} \times 630 = \dfrac{2}{5} \times 630 = 252. 3 + 2 2 × 630 = 5 2 × 630 = 252.
Let no. of newly admitted boys be x and so girls = 90 - x.
According to question on admission of 90 new students ratio changes to 7 : 5.
∴ 378 + x 252 + 90 − x = 7 5 ⇒ 378 + x 342 − x = 7 5 ⇒ 5 ( 378 + x ) = 7 ( 342 − x ) ⇒ 5 ( 378 + x ) = 7 ( 342 − x ) ⇒ 1890 + 5 x = 2394 − 7 x ⇒ 5 x + 7 x = 2394 − 1890 ⇒ 12 x = 504 ⇒ x = 504 12 ⇒ x = 42. \therefore \dfrac{378 + x}{252 + 90 - x} = \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{378 + x}{342 - x} = \dfrac{7}{5} \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 1890 + 5x = 2394 - 7x \\[1em] \Rightarrow 5x + 7x = 2394 - 1890 \\[1em] \Rightarrow 12x = 504 \\[1em] \Rightarrow x = \dfrac{504}{12} \\[1em] \Rightarrow x = 42. ∴ 252 + 90 − x 378 + x = 5 7 ⇒ 342 − x 378 + x = 5 7 ⇒ 5 ( 378 + x ) = 7 ( 342 − x ) ⇒ 5 ( 378 + x ) = 7 ( 342 − x ) ⇒ 1890 + 5 x = 2394 − 7 x ⇒ 5 x + 7 x = 2394 − 1890 ⇒ 12 x = 504 ⇒ x = 12 504 ⇒ x = 42.
Hence, there are 42 newly admitted boys.
What quantity must be subtracted from each term of ratio 9 : 17 to make it equal to 1 : 3 ?
Answer
Let number to be subtracted be x,
∴ 9 − x 17 − x = 1 3 ⇒ 3 ( 9 − x ) = 1 ( 17 − x ) ⇒ 27 − 3 x = 17 − x ⇒ 3 x − x = 27 − 17 ⇒ 2 x = 10 ⇒ x = 5. \therefore \dfrac{9 - x}{17 - x} = \dfrac{1}{3} \\[1em] \Rightarrow 3(9 - x) = 1(17 - x) \\[1em] \Rightarrow 27 - 3x = 17 - x \\[1em] \Rightarrow 3x - x = 27 - 17 \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = 5. ∴ 17 − x 9 − x = 3 1 ⇒ 3 ( 9 − x ) = 1 ( 17 − x ) ⇒ 27 − 3 x = 17 − x ⇒ 3 x − x = 27 − 17 ⇒ 2 x = 10 ⇒ x = 5.
Hence, quantity to be subtracted = 5.
The work done by (x - 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x - 3) days are in the ratio 3 : 8. Find the value of x.
Answer
Amount of work done by (x - 2) men in (4x + 1) days = (x - 2)(4x + 1),
Amount of work done by (4x + 1) men in (2x - 3) days = (4x + 1)(2x - 3)
According to question,
⇒ ( x − 2 ) ( 4 x + 1 ) ( 4 x + 1 ) ( 2 x − 3 ) = 3 8 ⇒ 4 x 2 + x − 8 x − 2 8 x 2 − 12 x + 2 x − 3 = 3 8 ⇒ 4 x 2 − 7 x − 2 8 x 2 − 10 x − 3 = 3 8 ⇒ 8 ( 4 x 2 − 7 x − 2 ) = 3 ( 8 x 2 − 10 x − 3 ) ⇒ 32 x 2 − 56 x − 16 = 24 x 2 − 30 x − 9 ⇒ 32 x 2 − 24 x 2 − 56 x + 30 x − 16 + 9 = 0 ⇒ 8 x 2 − 26 x − 7 = 0 ⇒ 8 x 2 − 28 x + 2 x − 7 = 0 ⇒ 4 x ( 2 x − 7 ) + 1 ( 2 x − 7 ) = 0 ⇒ ( 4 x + 1 ) ( 2 x − 7 ) = 0 ⇒ 4 x + 1 = 0 or 2 x − 7 = 0 ⇒ x = − 1 4 or x = 7 2 . \Rightarrow \dfrac{(x - 2)(4x + 1)}{(4x + 1)(2x - 3)} = \dfrac{3}{8} \\[1em] \Rightarrow \dfrac{4x^2 + x - 8x - 2}{8x^2 - 12x + 2x - 3} = \dfrac{3}{8} \\[1em] \Rightarrow \dfrac{4x^2 - 7x - 2}{8x^2 - 10x - 3} = \dfrac{3}{8} \\[1em] \Rightarrow 8(4x^2 - 7x - 2) = 3(8x^2 - 10x - 3) \\[1em] \Rightarrow 32x^2 - 56x - 16 = 24x^2 - 30x - 9 \\[1em] \Rightarrow 32x^2 - 24x^2 - 56x + 30x - 16 + 9 = 0 \\[1em] \Rightarrow 8x^2 - 26x - 7 = 0 \\[1em] \Rightarrow 8x^2 - 28x + 2x - 7 = 0 \\[1em] \Rightarrow 4x(2x - 7) + 1(2x - 7) = 0 \\[1em] \Rightarrow (4x + 1)(2x - 7) = 0 \\[1em] \Rightarrow 4x + 1 = 0 \text{ or } 2x - 7 = 0 \\[1em] \Rightarrow x = -\dfrac{1}{4} \text{ or } x = \dfrac{7}{2}. ⇒ ( 4 x + 1 ) ( 2 x − 3 ) ( x − 2 ) ( 4 x + 1 ) = 8 3 ⇒ 8 x 2 − 12 x + 2 x − 3 4 x 2 + x − 8 x − 2 = 8 3 ⇒ 8 x 2 − 10 x − 3 4 x 2 − 7 x − 2 = 8 3 ⇒ 8 ( 4 x 2 − 7 x − 2 ) = 3 ( 8 x 2 − 10 x − 3 ) ⇒ 32 x 2 − 56 x − 16 = 24 x 2 − 30 x − 9 ⇒ 32 x 2 − 24 x 2 − 56 x + 30 x − 16 + 9 = 0 ⇒ 8 x 2 − 26 x − 7 = 0 ⇒ 8 x 2 − 28 x + 2 x − 7 = 0 ⇒ 4 x ( 2 x − 7 ) + 1 ( 2 x − 7 ) = 0 ⇒ ( 4 x + 1 ) ( 2 x − 7 ) = 0 ⇒ 4 x + 1 = 0 or 2 x − 7 = 0 ⇒ x = − 4 1 or x = 2 7 .
x ≠ − 1 4 -\dfrac{1}{4} − 4 1
Hence, x = 7 2 \dfrac{7}{2} 2 7
If 3A = 4B = 6C, find : A : B : C.
Answer
3A = 4B
A B = 4 3 \dfrac{A}{B} = \dfrac{4}{3} B A = 3 4
⇒ A : B = 4 : 3
4B = 6C
B C = 6 4 = 3 2 \dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2} C B = 4 6 = 2 3
⇒ B : C = 3 : 2
∴ A : B : C = 4 : 3 : 2.
Hence, A : B : C = 4 : 3 : 2.
If 2a = 3b and 4b = 5c, find : a : c.
Answer
2a = 3b
a b = 3 2 \dfrac{a}{b} = \dfrac{3}{2} b a = 2 3
⇒ a : b = 3 : 2
4b = 5c
b c = 5 4 \dfrac{b}{c} = \dfrac{5}{4} c b = 4 5
⇒ b : c = 5 : 4
Multiplying a : b by 5 and b : c by 2 we get,
a : b = 15 : 10
b : c = 10 : 8
⇒ a : b : c = 15 : 10 : 8
Hence, a : c = 15 : 8.
Find the compound ratio of 2 : 3, 9 : 14 and 14 : 27.
Answer
Compound ratio of 2 : 3, 9 : 14 and 14 : 27,
= 2 3 × 9 14 × 14 27 = 252 1134 = 2 9 = 2 : 9. = \dfrac{2}{3} \times \dfrac{9}{14} \times \dfrac{14}{27} \\[1em] = \dfrac{252}{1134} \\[1em] = \dfrac{2}{9} = 2 : 9. = 3 2 × 14 9 × 27 14 = 1134 252 = 9 2 = 2 : 9.
Hence, compound ratio of 2 : 3, 9 : 14 and 14 : 27 is 2 : 9.
Find the compound ratio of 2a : 3b, mn : x2 and x : n
Answer
Compound ratio of 2a : 3b, mn : x2 and x : n,
= 2 a 3 b × m n x 2 × x n = 2 a m n x 3 b x 2 n = 2 a m 3 b x = 2 a m : 3 b x . = \dfrac{2a}{3b} \times \dfrac{mn}{x^2} \times \dfrac{x}{n} \\[1em] = \dfrac{2amnx}{3bx^2n} \\[1em] = \dfrac{2am}{3bx} = 2am : 3bx. = 3 b 2 a × x 2 mn × n x = 3 b x 2 n 2 amn x = 3 b x 2 am = 2 am : 3 b x .
Hence, compound ratio of 2a : 3b, mn : x2 and x : n is 2am : 3bx.
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Answer
According to question,
⇒ x + 3 4 x + 1 = 3 × 3 5 × 5 ⇒ x + 3 4 x + 1 = 9 25 ⇒ 25 ( x + 3 ) = 9 ( 4 x + 1 ) ⇒ 25 x + 75 = 36 x + 9 ⇒ 36 x − 25 x = 75 − 9 ⇒ 11 x = 66 ⇒ x = 6. \Rightarrow \dfrac{x + 3}{4x + 1} = \dfrac{3 \times 3}{5 \times 5} \\[1em] \Rightarrow \dfrac{x + 3}{4x + 1} = \dfrac{9}{25} \\[1em] \Rightarrow 25(x + 3) = 9(4x + 1) \\[1em] \Rightarrow 25x + 75 = 36x + 9 \\[1em] \Rightarrow 36x - 25x = 75 - 9 \\[1em] \Rightarrow 11x = 66 \\[1em] \Rightarrow x = 6. ⇒ 4 x + 1 x + 3 = 5 × 5 3 × 3 ⇒ 4 x + 1 x + 3 = 25 9 ⇒ 25 ( x + 3 ) = 9 ( 4 x + 1 ) ⇒ 25 x + 75 = 36 x + 9 ⇒ 36 x − 25 x = 75 − 9 ⇒ 11 x = 66 ⇒ x = 6.
Hence, x = 6.
If m : n is the duplicate ratio of m + x : n + x; show that : x2 = mn.
Answer
According to question,
⇒ m n = ( m + x ) 2 ( n + x ) 2 ⇒ m n = m 2 + x 2 + 2 m x n 2 + x 2 + 2 n x ⇒ m ( n 2 + x 2 + 2 n x ) = n ( m 2 + x 2 + 2 m x ) ⇒ m n 2 + m x 2 + 2 m n x = n m 2 + n x 2 + 2 m n x ⇒ m x 2 − n x 2 = n m 2 − m n 2 + 2 m n x − 2 m n x ⇒ x 2 ( m − n ) = m n ( m − n ) ⇒ x 2 = m n . \Rightarrow \dfrac{m}{n} = \dfrac{(m + x)^2}{(n + x)^2} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{m^2 + x^2 + 2mx}{n^2 + x^2 + 2nx} \\[1em] \Rightarrow m(n^2 + x^2 + 2nx) = n(m^2 + x^2 + 2mx) \\[1em] \Rightarrow mn^2 + mx^2 + 2mnx = nm^2 + nx^2 + 2mnx \\[1em] \Rightarrow mx^2 - nx^2 = nm^2 - mn^2 + 2mnx - 2mnx \\[1em] \Rightarrow x^2(m - n) = mn(m - n) \\[1em] \Rightarrow x^2 = mn. ⇒ n m = ( n + x ) 2 ( m + x ) 2 ⇒ n m = n 2 + x 2 + 2 n x m 2 + x 2 + 2 m x ⇒ m ( n 2 + x 2 + 2 n x ) = n ( m 2 + x 2 + 2 m x ) ⇒ m n 2 + m x 2 + 2 mn x = n m 2 + n x 2 + 2 mn x ⇒ m x 2 − n x 2 = n m 2 − m n 2 + 2 mn x − 2 mn x ⇒ x 2 ( m − n ) = mn ( m − n ) ⇒ x 2 = mn .
Hence, proved that x2 = mn.
If (3x - 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Answer
According to question,
⇒ 3 x − 9 5 x + 4 = 3 3 4 3 ⇒ 3 x − 9 5 x + 4 = 27 64 ⇒ 64 ( 3 x − 9 ) = 27 ( 5 x + 4 ) ⇒ 192 x − 576 = 135 x + 108 ⇒ 192 x − 135 x = 108 + 576 ⇒ 57 x = 684 ⇒ x = 12. \Rightarrow \dfrac{3x - 9}{5x + 4} = \dfrac{3^3}{4^3} \\[1em] \Rightarrow \dfrac{3x - 9}{5x + 4} = \dfrac{27}{64} \\[1em] \Rightarrow 64(3x - 9) = 27(5x + 4) \\[1em] \Rightarrow 192x - 576 = 135x + 108 \\[1em] \Rightarrow 192x - 135x = 108 + 576 \\[1em] \Rightarrow 57x = 684 \\[1em] \Rightarrow x = 12. ⇒ 5 x + 4 3 x − 9 = 4 3 3 3 ⇒ 5 x + 4 3 x − 9 = 64 27 ⇒ 64 ( 3 x − 9 ) = 27 ( 5 x + 4 ) ⇒ 192 x − 576 = 135 x + 108 ⇒ 192 x − 135 x = 108 + 576 ⇒ 57 x = 684 ⇒ x = 12.
Hence, x = 12.
If x, 2, 10 and y are in proportion, the values of x and y are respectively :
0.2 and 0.25
0.2 and 50
0.4 and 50
0.4 and 25
Answer
Given,
x, 2, 10 and y are in proportion.
∴ x 2 = 2 10 = 10 y \therefore \dfrac{x}{2} = \dfrac{2}{10} = \dfrac{10}{y} ∴ 2 x = 10 2 = y 10
Considering L.H.S. of the equation (1) :
⇒ x 2 = 2 10 ⇒ x = 2 × 2 10 ⇒ x = 4 10 = 0.4 \Rightarrow \dfrac{x}{2} = \dfrac{2}{10} \\[1em] \Rightarrow x = \dfrac{2 \times 2}{10} \\[1em] \Rightarrow x = \dfrac{4}{10} = 0.4 ⇒ 2 x = 10 2 ⇒ x = 10 2 × 2 ⇒ x = 10 4 = 0.4
Considering R.H.S. of the equation (1) :
⇒ 2 10 = 10 y ⇒ y = 10 × 10 2 ⇒ y = 100 2 ⇒ y = 50. \Rightarrow \dfrac{2}{10} = \dfrac{10}{y} \\[1em] \Rightarrow y = \dfrac{10 \times 10}{2} \\[1em] \Rightarrow y = \dfrac{100}{2} \\[1em] \Rightarrow y = 50. ⇒ 10 2 = y 10 ⇒ y = 2 10 × 10 ⇒ y = 2 100 ⇒ y = 50.
Hence, Option 3 is the correct option.
If a : b = 2 : 1, the value of (7a + 4b) : (5a - 2b) is :
9 : 4
4 : 9
11 : 3
3 : 11
Answer
Given,
⇒ a b = 2 1 ⇒ a = 2 b . \Rightarrow \dfrac{a}{b} = \dfrac{2}{1} \Rightarrow a = 2b. ⇒ b a = 1 2 ⇒ a = 2 b .
Substituting value of a in (7a + 4b) : (5a - 2b), we get :
⇒ 7 a + 4 b 5 a − 2 b ⇒ 7 × 2 b + 4 b 5 × 2 b − 2 b ⇒ 14 b + 4 b 10 b − 2 b ⇒ 18 b 8 b ⇒ 9 4 ⇒ 9 : 4. \Rightarrow \dfrac{7a + 4b}{5a - 2b} \\[1em] \Rightarrow \dfrac{7 \times 2b + 4b}{5 \times 2b - 2b} \\[1em] \Rightarrow \dfrac{14b + 4b}{10b - 2b} \\[1em] \Rightarrow \dfrac{18b}{8b} \\[1em] \Rightarrow \dfrac{9}{4} \\[1em] \Rightarrow 9 : 4. ⇒ 5 a − 2 b 7 a + 4 b ⇒ 5 × 2 b − 2 b 7 × 2 b + 4 b ⇒ 10 b − 2 b 14 b + 4 b ⇒ 8 b 18 b ⇒ 4 9 ⇒ 9 : 4.
Hence, Option 1 is the correct option.
If x : y = y : z, then x2 : y2 is :
1 : x
x : y
x : z
z : x
Answer
Given,
⇒ x : y = y : z
⇒ x y = y z ⇒ y 2 = x z . \Rightarrow \dfrac{x}{y} = \dfrac{y}{z} \\[1em] \Rightarrow y^2 = xz. ⇒ y x = z y ⇒ y 2 = x z .
Substituting value of y2 in x2 : y2 , we get :
⇒ x2 : xz
⇒ x : z.
Hence, Option 3 is the correct option.
The mean proportion between 3 + 2 2 and 3 − 2 2 3 + 2\sqrt{2} \text{ and } 3 - 2\sqrt{2} 3 + 2 2 and 3 − 2 2
1
-1
2 2 2\sqrt{2} 2 2
3
Answer
Let mean proportion be x.
⇒ 3 + 2 2 x = x 3 − 2 2 ⇒ x 2 = ( 3 + 2 2 ) ( 3 − 2 2 ) ⇒ x 2 = 3 2 − ( 2 2 ) 2 ⇒ x 2 = 9 − ( 4 × 2 ) ⇒ x 2 = 9 − 8 ⇒ x 2 = 1 ⇒ x = 1 = ± 1. \Rightarrow \dfrac{3 + 2\sqrt{2}}{x} = \dfrac{x}{3 - 2\sqrt{2}} \\[1em] \Rightarrow x^2 = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \\[1em] \Rightarrow x^2 = 3^2 - (2\sqrt{2})^2 \\[1em] \Rightarrow x^2 = 9 - (4 \times 2) \\[1em] \Rightarrow x^2 = 9 - 8 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1. ⇒ x 3 + 2 2 = 3 − 2 2 x ⇒ x 2 = ( 3 + 2 2 ) ( 3 − 2 2 ) ⇒ x 2 = 3 2 − ( 2 2 ) 2 ⇒ x 2 = 9 − ( 4 × 2 ) ⇒ x 2 = 9 − 8 ⇒ x 2 = 1 ⇒ x = 1 = ± 1.
Since, geometrical mean cannot be negative,
∴ x = 1.
Hence, Option 1 is the correct option.
If 2x, 9 and 18 are in continued proportion, the value of x is :
2 1 4 2\dfrac{1}{4} 2 4 1
4 9 \dfrac{4}{9} 9 4
1
9
Answer
Given,
2x, 9 and 18 are in continued proportion.
∴ 2 x 9 = 9 18 ⇒ x = 9 × 9 18 × 2 ⇒ x = 9 4 = 2 1 4 . \therefore \dfrac{2x}{9} = \dfrac{9}{18} \\[1em] \Rightarrow x = \dfrac{9 \times 9}{18 \times 2} \\[1em] \Rightarrow x = \dfrac{9}{4} = 2\dfrac{1}{4}. ∴ 9 2 x = 18 9 ⇒ x = 18 × 2 9 × 9 ⇒ x = 4 9 = 2 4 1 .
Hence, Option 1 is the correct option.
Find the fourth proportional to 1.5, 4.5 and 3.5
Answer
Let fourth proportional to 1.5, 4.5 and 3.5 be x
⇒ 1.5 : 4.5 = 3.5 : x ⇒ 1.5 4.5 = 3.5 x ⇒ x = 3.5 × 4.5 1.5 ⇒ x = 3.5 × 3 ⇒ x = 10.5 \Rightarrow 1.5 : 4.5 = 3.5 : x \\[1em] \Rightarrow \dfrac{1.5}{4.5} = \dfrac{3.5}{x} \\[1em] \Rightarrow x = \dfrac{3.5 \times 4.5}{1.5} \\[1em] \Rightarrow x = 3.5 \times 3 \\[1em] \Rightarrow x = 10.5 ⇒ 1.5 : 4.5 = 3.5 : x ⇒ 4.5 1.5 = x 3.5 ⇒ x = 1.5 3.5 × 4.5 ⇒ x = 3.5 × 3 ⇒ x = 10.5
Hence, fourth proportional to 1.5, 4.5 and 3.5 is 10.5
Find the fourth proportional to 3a, 6a2 and 2ab2
Answer
Let fourth proportional to 3a, 6a2 and 2ab2 be x
⇒ 3 a : 6 a 2 = 2 a b 2 : x ⇒ 3 a 6 a 2 = 2 a b 2 x ⇒ x = 2 a b 2 × 6 a 2 3 a ⇒ x = 4 a 2 b 2 \Rightarrow 3a : 6a^2 = 2ab^2 : x \\[1em] \Rightarrow \dfrac{3a}{6a^2} = \dfrac{2ab^2}{x} \\[1em] \Rightarrow x = \dfrac{2ab^2 \times 6a^2}{3a} \\[1em] \Rightarrow x = 4a^2b^2 ⇒ 3 a : 6 a 2 = 2 a b 2 : x ⇒ 6 a 2 3 a = x 2 a b 2 ⇒ x = 3 a 2 a b 2 × 6 a 2 ⇒ x = 4 a 2 b 2
Hence, fourth proportional to 3a, 6a2 and 2ab2 is 4a2 b2 .
Find the third proportional to 2 2 3 2\dfrac{2}{3} 2 3 2
Answer
Let the third proportion to 2 2 3 2\dfrac{2}{3} 2 3 2
⇒ 2 2 3 : 4 = 4 : x ⇒ 8 3 : 4 = 4 : x ⇒ 8 3 4 = 4 x ⇒ 8 12 = 4 x ⇒ x = 4 × 12 8 ⇒ x = 6. \Rightarrow 2\dfrac{2}{3} : 4 = 4 : x \\[1em] \Rightarrow \dfrac{8}{3} : 4 = 4 : x \\[1em] \Rightarrow \dfrac{\dfrac{8}{3}}{4} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{8}{12} = \dfrac{4}{x} \\[1em] \Rightarrow x = \dfrac{4 \times 12}{8} \\[1em] \Rightarrow x = 6. ⇒ 2 3 2 : 4 = 4 : x ⇒ 3 8 : 4 = 4 : x ⇒ 4 3 8 = x 4 ⇒ 12 8 = x 4 ⇒ x = 8 4 × 12 ⇒ x = 6.
Hence, third proportional to 2 2 3 2\dfrac{2}{3} 2 3 2
Find the third proportional to a - b and a2 - b2 .
Answer
Let the third proportion to a - b and a2 - b2 be x,
⇒ a − b : a 2 − b 2 = a 2 − b 2 : x ⇒ a − b a 2 − b 2 = a 2 − b 2 x ⇒ x = ( a 2 − b 2 ) × ( a 2 − b 2 ) ( a − b ) ⇒ x = ( a + b ) ( a − b ) × ( a 2 − b 2 ) ( a − b ) ⇒ x = ( a + b ) ( a 2 − b 2 ) \Rightarrow a - b : a^2 - b^2 = a^2 - b^2 : x \\[1em] \Rightarrow \dfrac{a - b}{a^2 - b^2} = \dfrac{a^2 - b^2}{x} \\[1em] \Rightarrow x = \dfrac{(a^2 - b^2) \times (a^2 - b^2)}{(a - b)} \\[1em] \Rightarrow x = \dfrac{(a + b)(a - b) \times (a^2 - b^2)}{(a - b)} \\[1em] \Rightarrow x = (a + b)(a^2 - b^2) ⇒ a − b : a 2 − b 2 = a 2 − b 2 : x ⇒ a 2 − b 2 a − b = x a 2 − b 2 ⇒ x = ( a − b ) ( a 2 − b 2 ) × ( a 2 − b 2 ) ⇒ x = ( a − b ) ( a + b ) ( a − b ) × ( a 2 − b 2 ) ⇒ x = ( a + b ) ( a 2 − b 2 )
Hence, third proportional to a - b and a2 - b2 is (a + b)(a2 - b2 ).
Find the mean proportional between 6 + 3 3 and 8 − 4 3 3\sqrt{3} \text{ and } 8 - 4\sqrt{3} 3 3 and 8 − 4 3
Answer
Let mean proportional between 6 + 3 3 and 8 − 4 3 3\sqrt{3} \text{ and } 8 - 4\sqrt{3} 3 3 and 8 − 4 3
∴ 6 + 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 = 48 − 36 ⇒ x 2 = 12 ⇒ x = 2 3 . \therefore 6 + 3\sqrt{3} : x = x : 8 - 4\sqrt{3} \\[1em] \Rightarrow \dfrac{6 + 3\sqrt{3}}{x} = \dfrac{x}{8 - 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3})(8 - 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 48 - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt{3}. ∴ 6 + 3 3 : x = x : 8 − 4 3 ⇒ 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 = 48 − 36 ⇒ x 2 = 12 ⇒ x = 2 3 .
Hence, x = 2 3 2\sqrt{3} 2 3
Find the mean proportional between a - b and a3 - a2 b
Answer
Let mean proportional between a - b and a3 - a2 b be x
⇒ a − b x = x a 3 − a 2 b ⇒ x 2 = ( a − b ) ( a 3 − a 2 b ) ⇒ x 2 = a 4 − a 3 b − a 3 b + a 2 b 2 ⇒ x 2 = a 4 − 2 a 3 b + a 2 b 2 ⇒ x 2 = a 2 ( a 2 + b 2 − 2 a b ) ⇒ x 2 = a 2 ( a − b ) 2 ⇒ x = a ( a − b ) . \Rightarrow \dfrac{a - b}{x} = \dfrac{x}{a^3 - a^2b} \\[1em] \Rightarrow x^2 = (a - b)(a^3 - a^2b) \\[1em] \Rightarrow x^2 = a^4 - a^3b - a^3b + a^2b^2 \\[1em] \Rightarrow x^2 = a^4 - 2a^3b + a^2b^2 \\[1em] \Rightarrow x^2 = a^2(a^2 + b^2 - 2ab) \\[1em] \Rightarrow x^2 = a^2(a - b)^2 \\[1em] \Rightarrow x = a(a - b). ⇒ x a − b = a 3 − a 2 b x ⇒ x 2 = ( a − b ) ( a 3 − a 2 b ) ⇒ x 2 = a 4 − a 3 b − a 3 b + a 2 b 2 ⇒ x 2 = a 4 − 2 a 3 b + a 2 b 2 ⇒ x 2 = a 2 ( a 2 + b 2 − 2 ab ) ⇒ x 2 = a 2 ( a − b ) 2 ⇒ x = a ( a − b ) .
Hence, mean proportional between a - b and a3 - a2 b = a(a - b).
If x + 5 is the mean proportion between x + 2 and x + 9; find the value of x.
Answer
Given,
x + 5 is the mean proportion between x + 2 and x + 9
∴ x + 2 : x + 5 = x + 5 : x + 9 ⇒ x + 2 x + 5 = x + 5 x + 9 ⇒ ( x + 2 ) ( x + 9 ) = ( x + 5 ) ( x + 5 ) ⇒ x 2 + 9 x + 2 x + 18 = x 2 + 5 x + 5 x + 25 ⇒ x 2 + 11 x + 18 = x 2 + 10 x + 25 ⇒ 11 x − 10 x = 25 − 18 ⇒ x = 7. \therefore x + 2 : x + 5 = x + 5 : x + 9 \\[1em] \Rightarrow \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} \\[1em] \Rightarrow (x + 2)(x + 9) = (x + 5)(x + 5) \\[1em] \Rightarrow x^2 + 9x + 2x + 18 = x^2 + 5x + 5x + 25 \\[1em] \Rightarrow x^2 + 11x + 18 = x^2 + 10x + 25 \\[1em] \Rightarrow 11x - 10x = 25 - 18 \\[1em] \Rightarrow x = 7. ∴ x + 2 : x + 5 = x + 5 : x + 9 ⇒ x + 5 x + 2 = x + 9 x + 5 ⇒ ( x + 2 ) ( x + 9 ) = ( x + 5 ) ( x + 5 ) ⇒ x 2 + 9 x + 2 x + 18 = x 2 + 5 x + 5 x + 25 ⇒ x 2 + 11 x + 18 = x 2 + 10 x + 25 ⇒ 11 x − 10 x = 25 − 18 ⇒ x = 7.
Hence, x = 7.
If x2 , 4 and 9 are in continued proportion, find x.
Answer
Given,
x2 , 4 and 9 are in continued proportion.
∴ x 2 4 = 4 9 ⇒ x 2 = 16 9 ⇒ x = 16 9 ⇒ x = 4 3 . \therefore \dfrac{x^2}{4} = \dfrac{4}{9} \\[1em] \Rightarrow x^2 = \dfrac{16}{9} \\[1em] \Rightarrow x = \sqrt{\dfrac{16}{9}} \\[1em] \Rightarrow x = \dfrac{4}{3}. ∴ 4 x 2 = 9 4 ⇒ x 2 = 9 16 ⇒ x = 9 16 ⇒ x = 3 4 .
Hence, x = 4 3 . \dfrac{4}{3}. 3 4 .
If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x2 + y2 and y2 + z2 .
Answer
Given,
y is the mean proportional between x and z
∴ x y = y z ⇒ y 2 = x z . \therefore \dfrac{x}{y} = \dfrac{y}{z} \Rightarrow y^2 = xz. ∴ y x = z y ⇒ y 2 = x z .
To prove,
xy + yz is the mean proportional between x2 + y2 and y2 + z2
∴ x 2 + y 2 x y + y z = x y + y z y 2 + z 2 ⇒ ( x y + y z ) ( x y + y z ) = ( x 2 + y 2 ) ( y 2 + z 2 ) ⇒ x 2 y 2 + x y 2 z + x y 2 z + y 2 z 2 = x 2 y 2 + x 2 z 2 + y 4 + y 2 z 2 . . . . [ i ] \therefore \dfrac{x^2 + y^2}{xy + yz} = \dfrac{xy + yz}{y^2 + z^2} \\[1em] \Rightarrow (xy + yz)(xy + yz) = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow x^2y^2 + xy^2z + xy^2z + y^2z^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 ....[\text{i}] ∴ x y + yz x 2 + y 2 = y 2 + z 2 x y + yz ⇒ ( x y + yz ) ( x y + yz ) = ( x 2 + y 2 ) ( y 2 + z 2 ) ⇒ x 2 y 2 + x y 2 z + x y 2 z + y 2 z 2 = x 2 y 2 + x 2 z 2 + y 4 + y 2 z 2 .... [ i ]
Substituting y2 = xz in L.H.S. of (i)
⇒ x2 (xz) + x(xz)z + x(xz)z + (xz)z2
⇒ x3 z + x2 z2 + x2 z2 + xz3
⇒ x3 z + 2x2 z2 + xz3 .........(ii)
Substituting y2 = xz in R.H.S. of (i)
⇒ x2 (xz) + x2 z2 + (xz)2 + (xz)z2
⇒ x3 z + x2 z2 + x2 z2 + xz3
⇒ x3 z + 2x2 z2 + xz3 .........(iii)
Since, (ii) = (iii)
Hence, proved that xy + yz is the mean proportional between x2 + y2 and y2 + z2 .
If q is the mean proportional between p and r, show that :
pqr(p + q + r)3 = (pq + qr + pr)3 .
Answer
Since, q is the mean proportional between p and r
∴ p q = q r ⇒ q 2 = p r . \therefore \dfrac{p}{q} = \dfrac{q}{r} \Rightarrow q^2 = pr. ∴ q p = r q ⇒ q 2 = p r .
Substituting pr = q2 in L.H.S. of pqr(p + q + r)3 = (pq + qr + pr)3
⇒ q.q2 (p + q + r)3
⇒ q3 (p + q + r)3
⇒ [q(p + q + r)]3
⇒ (pq + q2 + qr)3
⇒ (pq + pr + qr)3 = R.H.S. (As q2 = pr).
Hence, proved that pqr(p + q + r)3 = (pq + qr + pr)3 .
If three quantities are in continued proportion; show that the ratio of the first to third is duplicate ratio of first to the second.
Answer
Let x, y and z be in continued proportion.
∴ x : y = y : z
⇒ x y = y z ⇒ y 2 = x z \Rightarrow \dfrac{x}{y} = \dfrac{y}{z} \Rightarrow y^2 = xz ⇒ y x = z y ⇒ y 2 = x z
To prove :
x z = x 2 y 2 \dfrac{x}{z} = \dfrac{x^2}{y^2} z x = y 2 x 2
Substituting y2 = xz in R.H.S. of above equation we get,
x 2 x z = x z \dfrac{x^2}{xz} = \dfrac{x}{z} x z x 2 = z x
Hence, proved that ratio of the first to third is duplicate ratio of first to the second.
Find the third proportional to x y + y x and x 2 + y 2 \dfrac{x}{y} + \dfrac{y}{x} \text{ and } \sqrt{x^2 + y^2} y x + x y and x 2 + y 2
Answer
Let third proportional be p.
∴ x y + y x x 2 + y 2 = x 2 + y 2 p ⇒ ( x 2 + y 2 ) 2 = p ( x y + y x ) ⇒ x 2 + y 2 = p × x 2 + y 2 x y ⇒ p = ( x 2 + y 2 ) ( x y ) x 2 + y 2 ⇒ p = x y . \therefore \dfrac{\dfrac{x}{y} + \dfrac{y}{x}}{\sqrt{x^2 + y^2}} = \dfrac{\sqrt{x^2 + y^2}}{p} \\[1em] \Rightarrow \Big(\sqrt{x^2 + y^2}\Big)^2 = p\Big(\dfrac{x}{y} + \dfrac{y}{x}\Big) \\[1em] \Rightarrow x^2 + y^2 = p \times \dfrac{x^2 + y^2}{xy} \\[1em] \Rightarrow p = \dfrac{(x^2 + y^2)(xy)}{x^2 + y^2} \\[1em] \Rightarrow p = xy. ∴ x 2 + y 2 y x + x y = p x 2 + y 2 ⇒ ( x 2 + y 2 ) 2 = p ( y x + x y ) ⇒ x 2 + y 2 = p × x y x 2 + y 2 ⇒ p = x 2 + y 2 ( x 2 + y 2 ) ( x y ) ⇒ p = x y .
Hence, third proportional to x y + y x and x 2 + y 2 \dfrac{x}{y} + \dfrac{y}{x} \text{ and } \sqrt{x^2 + y^2} y x + x y and x 2 + y 2
If p : q = r : s; then show that:
mp + nq : q = mr + ns : s
Answer
Given,
⇒ p q = r s \Rightarrow \dfrac{p}{q} = \dfrac{r}{s} ⇒ q p = s r
Multiplying both sides by m:
⇒ m p q = m r s \Rightarrow \dfrac{mp}{q} = \dfrac{mr}{s} ⇒ q m p = s m r
Adding n on both sides:
⇒ m p q + n = m r s + n ⇒ m p + n q q = m r + n s s \Rightarrow \dfrac{mp}{q} + n = \dfrac{mr}{s} + n \\[1em] \Rightarrow \dfrac{mp + nq}{q}= \dfrac{mr + ns}{s} ⇒ q m p + n = s m r + n ⇒ q m p + n q = s m r + n s
Hence, proved that mp + nq : q = mr + ns : s.
If p + r = mq and 1 q + 1 s = m r ; \dfrac{1}{q} + \dfrac{1}{s} = \dfrac{m}{r}; q 1 + s 1 = r m ;
Answer
Given,
⇒ 1 q + 1 s = m r ⇒ s + q q s = m r ⇒ s + q s = m q r ⇒ s + q s = p + r r (As mq = p + r) ⇒ 1 + q s = p r + 1 ⇒ q s = p r ⇒ p q = r s . \Rightarrow \dfrac{1}{q} + \dfrac{1}{s} = \dfrac{m}{r} \\[1em] \Rightarrow \dfrac{s + q}{qs} = \dfrac{m}{r} \\[1em] \Rightarrow \dfrac{s + q}{s} = \dfrac{mq}{r} \\[1em] \Rightarrow \dfrac{s + q}{s} = \dfrac{p + r}{r} \text{ (As mq = p + r)} \\[1em] \Rightarrow 1 + \dfrac{q}{s} = \dfrac{p}{r} + 1 \\[1em] \Rightarrow \dfrac{q}{s} = \dfrac{p}{r} \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{r}{s}. ⇒ q 1 + s 1 = r m ⇒ q s s + q = r m ⇒ s s + q = r m q ⇒ s s + q = r p + r (As mq = p + r) ⇒ 1 + s q = r p + 1 ⇒ s q = r p ⇒ q p = s r .
Hence, proved that p : q = r : s.
If x + y + z = 0, the value of x y + z \dfrac{x}{y + z} y + z x
2
1 2 \dfrac{1}{2} 2 1
1
-1
Answer
Given,
⇒ x + y + z = 0
⇒ x = -(y + z).
Substituting value of x in x y + z \dfrac{x}{y + z} y + z x
⇒ x y + z = − ( y + z ) y + z \Rightarrow \dfrac{x}{y + z} = \dfrac{-(y + z)}{y + z} ⇒ y + z x = y + z − ( y + z )
Hence, Option 4 is the correct option.
If a, b, c and d are in proportion, the value of 8 a 2 − 5 b 2 8 c 2 − 5 d 2 \dfrac{8a^2 - 5b^2}{8c^2 - 5d^2} 8 c 2 − 5 d 2 8 a 2 − 5 b 2
a2 : b2
a2 : c2
a2 : d2
c2 : d2
Answer
Given,
a, b, c and d are in proportion.
∴ a b = c d \therefore \dfrac{a}{b} = \dfrac{c}{d} ∴ b a = d c
∴ a = bk and c = dk
Solving,
⇒ 8 a 2 − 5 b 2 8 c 2 − 5 d 2 ⇒ 8 ( b k ) 2 − 5 b 2 8 ( d k ) 2 − 5 d 2 ⇒ 8 b 2 k 2 − 5 b 2 8 d 2 k 2 − 5 d 2 ⇒ b 2 ( 8 k 2 − 5 ) d 2 ( 8 k 2 − 5 ) ⇒ b 2 d 2 . . . . . ( 1 ) \Rightarrow \dfrac{8a^2 - 5b^2}{8c^2 - 5d^2} \\[1em] \Rightarrow \dfrac{8(bk)^2 - 5b^2}{8(dk)^2 - 5d^2} \\[1em] \Rightarrow \dfrac{8b^2k^2 - 5b^2}{8d^2k^2 - 5d^2} \\[1em] \Rightarrow \dfrac{b^2(8k^2 - 5)}{d^2(8k^2 - 5)} \\[1em] \Rightarrow \dfrac{b^2}{d^2} \space .....(1) ⇒ 8 c 2 − 5 d 2 8 a 2 − 5 b 2 ⇒ 8 ( d k ) 2 − 5 d 2 8 ( bk ) 2 − 5 b 2 ⇒ 8 d 2 k 2 − 5 d 2 8 b 2 k 2 − 5 b 2 ⇒ d 2 ( 8 k 2 − 5 ) b 2 ( 8 k 2 − 5 ) ⇒ d 2 b 2 ..... ( 1 )
As,
⇒ a b = c d ⇒ b d = a c \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{b}{d} = \dfrac{a}{c} ⇒ b a = d c ⇒ d b = c a
Substituting value of b d \dfrac{b}{d} d b
⇒ b 2 d 2 = a 2 c 2 \Rightarrow \dfrac{b^2}{d^2} = \dfrac{a^2}{c^2} ⇒ d 2 b 2 = c 2 a 2 2 : c2 .
Hence, Option 2 is the correct option.
x + y z = y + z x = z + x y \dfrac{x + y}{z} = \dfrac{y + z}{x} = \dfrac{z + x}{y} z x + y = x y + z = y z + x
0
1
2
-2
Answer
Given,
x + y z = y + z x = z + x y \dfrac{x + y}{z} = \dfrac{y + z}{x} = \dfrac{z + x}{y} z x + y = x y + z = y z + x
Applying componendo on each side we get :
⇒ x + y z + 1 = y + z x + 1 = z + x y + 1 ⇒ x + y + z z = y + z + x x = z + x + y y . \Rightarrow \dfrac{x + y}{z} + 1 = \dfrac{y + z}{x} + 1 = \dfrac{z + x}{y} + 1 \\[1em] \Rightarrow \dfrac{x + y + z}{z} = \dfrac{y + z + x}{x} = \dfrac{z + x + y}{y}. ⇒ z x + y + 1 = x y + z + 1 = y z + x + 1 ⇒ z x + y + z = x y + z + x = y z + x + y .
Since, above fractions are equal.
∴ We can conclude that,
x = y = z = a (let)
Substituting values of x, y and z in given equation we get :
⇒ x + y z ⇒ a + a a ⇒ 2 a a ⇒ 2. \Rightarrow \dfrac{x + y}{z} \\[1em] \Rightarrow \dfrac{a + a}{a} \\[1em] \Rightarrow \dfrac{2a}{a} \\[1em] \Rightarrow 2. ⇒ z x + y ⇒ a a + a ⇒ a 2 a ⇒ 2.
Hence, Option 3 is the correct option.
If x 2 − 4 x 2 + 4 = 3 5 \dfrac{x^2 - 4}{x^2 + 4} = \dfrac{3}{5} x 2 + 4 x 2 − 4 = 5 3
4
± 4 \pm 4 ± 4
1 4 \dfrac{1}{4} 4 1
± 1 4 \pm \dfrac{1}{4} ± 4 1
Answer
Given,
⇒ x 2 − 4 x 2 + 4 = 3 5 ⇒ 5 ( x 2 − 4 ) = 3 ( x 2 + 4 ) ⇒ 5 x 2 − 20 = 3 x 2 + 12 ⇒ 5 x 2 − 3 x 2 = 12 + 20 ⇒ 2 x 2 = 32 ⇒ x 2 = 16 ⇒ x = 16 ⇒ x = ± 4. \Rightarrow \dfrac{x^2 - 4}{x^2 + 4} = \dfrac{3}{5} \\[1em] \Rightarrow 5(x^2 - 4) = 3(x^2 + 4) \\[1em] \Rightarrow 5x^2 - 20 = 3x^2 + 12 \\[1em] \Rightarrow 5x^2 - 3x^2 = 12 + 20 \\[1em] \Rightarrow 2x^2 = 32 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x= \sqrt{16} \\[1em] \Rightarrow x = \pm 4. ⇒ x 2 + 4 x 2 − 4 = 5 3 ⇒ 5 ( x 2 − 4 ) = 3 ( x 2 + 4 ) ⇒ 5 x 2 − 20 = 3 x 2 + 12 ⇒ 5 x 2 − 3 x 2 = 12 + 20 ⇒ 2 x 2 = 32 ⇒ x 2 = 16 ⇒ x = 16 ⇒ x = ± 4.
Hence, Option 2 is the correct option.
If x a + b − c = y b + c − a = z c + a − b \dfrac{x}{a + b - c} = \dfrac{y}{b + c - a} = \dfrac{z}{c + a - b} a + b − c x = b + c − a y = c + a − b z
35
7 5 \dfrac{7}{5} 5 7
5 7 \dfrac{5}{7} 7 5
42
Answer
Given,
⇒ a + b + c = 7
⇒ a + b = 7 - c
⇒ b + c = 7 - a
⇒ c + a = 7 - b
Given,
⇒ x a + b − c = 5 ⇒ x 7 − c − c = 5 ⇒ x = 5 ( 7 − 2 c ) . ⇒ y b + c − a = 5 ⇒ y 7 − a − a = 5 ⇒ y = 5 ( 7 − 2 a ) . ⇒ z c + a − b = 5 ⇒ z 7 − b − b = 5 ⇒ z = 5 ( 7 − 2 b ) . \phantom{\Rightarrow} \dfrac{x}{a + b - c} = 5 \\[1em] \Rightarrow \dfrac{x}{7 - c - c} = 5 \\[1em] \Rightarrow x = 5(7 - 2c). \\[2em] \phantom{\Rightarrow} \dfrac{y}{b + c - a} = 5 \\[1em] \Rightarrow \dfrac{y}{7 - a - a} = 5 \\[1em] \Rightarrow y = 5(7 - 2a). \\[2em] \phantom{\Rightarrow} \dfrac{z}{c + a - b} = 5 \\[1em] \Rightarrow \dfrac{z}{7 - b - b} = 5 \\[1em] \Rightarrow z = 5(7 - 2b). \\[1em] ⇒ a + b − c x = 5 ⇒ 7 − c − c x = 5 ⇒ x = 5 ( 7 − 2 c ) . ⇒ b + c − a y = 5 ⇒ 7 − a − a y = 5 ⇒ y = 5 ( 7 − 2 a ) . ⇒ c + a − b z = 5 ⇒ 7 − b − b z = 5 ⇒ z = 5 ( 7 − 2 b ) .
x + y + z = 5(7 - 2c) + 5(7 - 2a) + 5(7 - 2b)
= 35 - 10c + 35 - 10a + 35 - 10b
= 105 - 10(a + b + c)
= 105 - 10 × 7
= 105 - 70
= 35.
Hence, Option 1 is the correct option.
If a : b = c : d, prove that :
5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d
Answer
Given,
⇒ a : b = c : d
∴ a b = c d \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] ∴ b a = d c
Multiplying both sides by 5 7 \dfrac{5}{7} 7 5
⇒ 5 a 7 b = 5 c 7 d \Rightarrow \dfrac{5a}{7b} = \dfrac{5c}{7d} ⇒ 7 b 5 a = 7 d 5 c
Applying componendo and dividendo:
⇒ 5 a + 7 b 5 a − 7 b = 5 c + 7 d 5 c − 7 d \Rightarrow \dfrac{5a + 7b}{5a - 7b} = \dfrac{5c + 7d}{5c - 7d} ⇒ 5 a − 7 b 5 a + 7 b = 5 c − 7 d 5 c + 7 d
Hence, proved that 5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d.
If a : b = c : d, prove that :
xa + yb : xc + yd = b : d
Answer
Given,
a : b = c : d
∴ a b = c d \therefore \dfrac{a}{b} = \dfrac{c}{d} ∴ b a = d c
Multiplying both sides by x y \dfrac{x}{y} y x
⇒ x a y b = x c y d \Rightarrow \dfrac{xa}{yb} = \dfrac{xc}{yd} ⇒ y b x a = y d x c
Applying componendo: ⇒ x a + y b y b = x c + y d y d \Rightarrow \dfrac{xa + yb}{yb} = \dfrac{xc + yd}{yd} ⇒ y b x a + y b = y d x c + y d
On cross-multiplication:
⇒ x a + y b x c + y d = y b y d ⇒ x a + y b x c + y d = b d . \Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{yb}{yd} \\[1em] \Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{b}{d}. ⇒ x c + y d x a + y b = y d y b ⇒ x c + y d x a + y b = d b .
Hence, proved that xa + yb : xc + yd = b : d.
If (7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d);
prove that a : b = c : d.
Answer
Given,
(7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d)
∴ 7 a + 8 b 7 a − 8 b = 7 c + 8 d 7 c − 8 d \therefore \dfrac{7a + 8b}{7a - 8b} = \dfrac{7c + 8d}{7c - 8d} ∴ 7 a − 8 b 7 a + 8 b = 7 c − 8 d 7 c + 8 d
Applying componendo and dividendo: ⇒ 7 a + 8 b + 7 a − 8 b 7 a + 8 b − ( 7 a − 8 b ) = 7 c + 8 d + 7 c − 8 d 7 c + 8 d − ( 7 c − 8 d ) ⇒ 14 a 16 b = 14 c 16 d ⇒ a b = c d . \Rightarrow \dfrac{7a + 8b + 7a - 8b}{7a + 8b - (7a - 8b)} = \dfrac{7c + 8d + 7c - 8d}{7c + 8d - (7c - 8d)} \\[1em] \Rightarrow \dfrac{14a}{16b} = \dfrac{14c}{16d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}. ⇒ 7 a + 8 b − ( 7 a − 8 b ) 7 a + 8 b + 7 a − 8 b = 7 c + 8 d − ( 7 c − 8 d ) 7 c + 8 d + 7 c − 8 d ⇒ 16 b 14 a = 16 d 14 c ⇒ b a = d c .
Hence, proved that a : b = c : d.
If x = 6 a b a + b \dfrac{6ab}{a + b} a + b 6 ab
x + 3 a x − 3 a + x + 3 b x − 3 b \dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b} x − 3 a x + 3 a + x − 3 b x + 3 b
Answer
(i) Given,
⇒ x = 6 a b a + b ⇒ x 3 a = 2 b a + b \Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3a} = \dfrac{2b}{a + b} ⇒ x = a + b 6 ab ⇒ 3 a x = a + b 2 b
Applying componendo and dividendo:
⇒ x + 3 a x − 3 a = 2 b + ( a + b ) 2 b − ( a + b ) ⇒ x + 3 a x − 3 a = 3 b + a b − a . . . . . . . ( i ) \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{2b + (a + b)}{2b - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{3b + a}{b - a} .......(i) ⇒ x − 3 a x + 3 a = 2 b − ( a + b ) 2 b + ( a + b ) ⇒ x − 3 a x + 3 a = b − a 3 b + a ....... ( i )
Again,
⇒ x = 6 a b a + b ⇒ x 3 b = 2 a a + b \Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3b} = \dfrac{2a}{a + b} ⇒ x = a + b 6 ab ⇒ 3 b x = a + b 2 a
Applying componendo and dividendo:
⇒ x + 3 b x − 3 b = 2 a + ( a + b ) 2 a − ( a + b ) ⇒ x + 3 b x − 3 b = 3 a + b a − b . . . . . . . ( i i ) \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{2a + (a + b)}{2a - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{3a + b}{a - b} .......(ii) ⇒ x − 3 b x + 3 b = 2 a − ( a + b ) 2 a + ( a + b ) ⇒ x − 3 b x + 3 b = a − b 3 a + b ....... ( ii )
Adding (i) and (ii) we get,
⇒ x + 3 a x − 3 a + x + 3 b x − 3 b = 3 b + a b − a + 3 a + b a − b = 3 b + a b − a + ( − 3 a + b b − a ) = 3 b + a b − a − 3 a + b b − a = 3 b + a − 3 a − b b − a = 2 b − 2 a b − a = 2 ( b − a ) b − a = 2. \Rightarrow \dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b} = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] = \dfrac{3b + a}{b - a} + \Big(-\dfrac{3a + b}{b - a}\Big) \\[1em] = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] = \dfrac{3b + a - 3a - b}{b - a} \\[1em] = \dfrac{2b - 2a}{b - a} \\[1em] = \dfrac{2(b - a)}{b - a} \\[1em] = 2. ⇒ x − 3 a x + 3 a + x − 3 b x + 3 b = b − a 3 b + a + a − b 3 a + b = b − a 3 b + a + ( − b − a 3 a + b ) = b − a 3 b + a − b − a 3 a + b = b − a 3 b + a − 3 a − b = b − a 2 b − 2 a = b − a 2 ( b − a ) = 2.
Hence, x + 3 a x − 3 a + x + 3 b x − 3 b \dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b} x − 3 a x + 3 a + x − 3 b x + 3 b
If a = 4 6 2 + 3 \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} 2 + 3 4 6
a + 2 2 a − 2 2 + a + 2 3 a − 2 3 \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} a − 2 2 a + 2 2 + a − 2 3 a + 2 3
Answer
Given,
⇒ a = 4 6 2 + 3 ⇒ a 2 2 = 2 3 2 + 3 \Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{2}} = \dfrac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}} ⇒ a = 2 + 3 4 6 ⇒ 2 2 a = 2 + 3 2 3
Applying componendo and dividendo:
⇒ a + 2 2 a − 2 2 = 2 3 + ( 2 + 3 ) 2 3 − ( 2 + 3 ) ⇒ a + 2 2 a − 2 2 = 3 3 + 2 3 − 2 . . . . . . . ( i ) \Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} = \dfrac{2\sqrt{3} + (\sqrt{2} + \sqrt{3})}{2\sqrt{3} - (\sqrt{2} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} .......(i) ⇒ a − 2 2 a + 2 2 = 2 3 − ( 2 + 3 ) 2 3 + ( 2 + 3 ) ⇒ a − 2 2 a + 2 2 = 3 − 2 3 3 + 2 ....... ( i )
Again,
⇒ a = 4 6 2 + 3 ⇒ a 2 3 = 2 2 2 + 3 \Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{3}} = \dfrac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}} ⇒ a = 2 + 3 4 6 ⇒ 2 3 a = 2 + 3 2 2
Applying componendo and dividendo:
⇒ a + 2 3 a − 2 3 = 2 2 + ( 2 + 3 ) 2 2 − ( 2 + 3 ) ⇒ a + 2 3 a − 2 3 = 3 2 + 3 2 − 3 . . . . . . . ( i i ) \Rightarrow \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{2\sqrt{2} + (\sqrt{2} + \sqrt{3})}{2\sqrt{2} - (\sqrt{2} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} .......(ii) ⇒ a − 2 3 a + 2 3 = 2 2 − ( 2 + 3 ) 2 2 + ( 2 + 3 ) ⇒ a − 2 3 a + 2 3 = 2 − 3 3 2 + 3 ....... ( ii )
Adding (i) and (ii) we get,
⇒ a + 2 2 a − 2 2 + a + 2 3 a − 2 3 = 3 3 + 2 3 − 2 + 3 2 + 3 2 − 3 = 3 3 + 2 3 − 2 + ( − 3 2 + 3 3 − 2 ) = 3 3 + 2 3 − 2 − 3 2 + 3 3 − 2 = 3 3 + 2 − 3 2 − 3 3 − 2 = 2 3 − 2 2 3 − 2 = 2 ( 3 − 2 ) 3 − 2 = 2. \Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \Big(-\dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}}\Big) \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2} - 3\sqrt{2} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{2(\sqrt{3} - \sqrt{2})}{\sqrt{3} - \sqrt{2}} \\[1em] = 2. ⇒ a − 2 2 a + 2 2 + a − 2 3 a + 2 3 = 3 − 2 3 3 + 2 + 2 − 3 3 2 + 3 = 3 − 2 3 3 + 2 + ( − 3 − 2 3 2 + 3 ) = 3 − 2 3 3 + 2 − 3 − 2 3 2 + 3 = 3 − 2 3 3 + 2 − 3 2 − 3 = 3 − 2 2 3 − 2 2 = 3 − 2 2 ( 3 − 2 ) = 2.
Hence, a + 2 2 a − 2 2 + a + 2 3 a − 2 3 \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} a − 2 2 a + 2 2 + a − 2 3 a + 2 3
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)
⇒ 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:
⇒ 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 ) \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)} ⇒ 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 )
Applying alternendo:
⇒ ( 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 )
Applying componendo and dividendo:
⇒ 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{2a}{2b} = \dfrac{2c}{2d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[1em] ⇒ 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.
If a − 2 b − 3 c + 4 d a + 2 b − 3 c − 4 d = a − 2 b + 3 c − 4 d a + 2 b + 3 c + 4 d \dfrac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \dfrac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d} a + 2 b − 3 c − 4 d a − 2 b − 3 c + 4 d = a + 2 b + 3 c + 4 d a − 2 b + 3 c − 4 d
show that : 2ad = 3bc.
Answer
Given,
⇒ a − 2 b − 3 c + 4 d a + 2 b − 3 c − 4 d = a − 2 b + 3 c − 4 d a + 2 b + 3 c + 4 d \Rightarrow \dfrac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \dfrac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d} ⇒ a + 2 b − 3 c − 4 d a − 2 b − 3 c + 4 d = a + 2 b + 3 c + 4 d a − 2 b + 3 c − 4 d
Applying componendo and dividendo:
⇒ a − 2 b − 3 c + 4 d + a + 2 b − 3 c − 4 d a − 2 b − 3 c + 4 d − ( a + 2 b − 3 c − 4 d ) = a − 2 b + 3 c − 4 d + a + 2 b + 3 c + 4 d a − 2 b + 3 c − 4 d − ( a + 2 b + 3 c + 4 d ) ⇒ 2 ( a − 3 c ) 2 ( 4 d − 2 b ) = 2 ( a + 3 c ) 2 ( − 4 d − 2 b ) ⇒ ( a − 3 c ) ( 4 d − 2 b ) = ( a + 3 c ) ( − 4 d − 2 b ) \Rightarrow \dfrac{a - 2b - 3c + 4d + a + 2b - 3c - 4d}{a - 2b - 3c + 4d - (a + 2b - 3c - 4d)} = \dfrac{a - 2b + 3c - 4d + a + 2b + 3c + 4d}{a - 2b + 3c - 4d - (a + 2b + 3c + 4d)} \\[1em] \Rightarrow \dfrac{2(a - 3c)}{2(4d - 2b)} = \dfrac{2(a + 3c)}{2(-4d - 2b)} \\[1em] \Rightarrow \dfrac{(a - 3c)}{(4d - 2b)} = \dfrac{(a + 3c)}{(-4d - 2b)} ⇒ a − 2 b − 3 c + 4 d − ( a + 2 b − 3 c − 4 d ) a − 2 b − 3 c + 4 d + a + 2 b − 3 c − 4 d = a − 2 b + 3 c − 4 d − ( a + 2 b + 3 c + 4 d ) a − 2 b + 3 c − 4 d + a + 2 b + 3 c + 4 d ⇒ 2 ( 4 d − 2 b ) 2 ( a − 3 c ) = 2 ( − 4 d − 2 b ) 2 ( a + 3 c ) ⇒ ( 4 d − 2 b ) ( a − 3 c ) = ( − 4 d − 2 b ) ( a + 3 c )
Applying alternendo:
⇒ ( a − 3 c ) ( a + 3 c ) = ( 4 d − 2 b ) ( − 4 d − 2 b ) ⇒ a − 3 c + a + 3 c a − 3 c − ( a + 3 c ) = 4 d − 2 b + ( − 4 d − 2 b ) 4 d − 2 b − ( − 4 d − 2 b ) ⇒ 2 a − 6 c = − 4 b 8 d ⇒ a − 3 c = − b 2 d ⇒ − 2 a d = − 3 b c ⇒ 2 a d = 3 b c . \Rightarrow \dfrac{(a - 3c)}{(a + 3c)} = \dfrac{(4d - 2b)}{(-4d - 2b)} \\[1em] \Rightarrow \dfrac{a - 3c + a + 3c}{a - 3c - (a + 3c)} = \dfrac{4d - 2b + (-4d - 2b)}{4d - 2b - (-4d - 2b)} \\[1em] \Rightarrow \dfrac{2a}{-6c} = \dfrac{-4b}{8d} \\[1em] \Rightarrow \dfrac{a}{-3c} = \dfrac{-b}{2d} \\[1em] \Rightarrow -2ad = -3bc \\[1em] \Rightarrow 2ad = 3bc. ⇒ ( a + 3 c ) ( a − 3 c ) = ( − 4 d − 2 b ) ( 4 d − 2 b ) ⇒ a − 3 c − ( a + 3 c ) a − 3 c + a + 3 c = 4 d − 2 b − ( − 4 d − 2 b ) 4 d − 2 b + ( − 4 d − 2 b ) ⇒ − 6 c 2 a = 8 d − 4 b ⇒ − 3 c a = 2 d − b ⇒ − 2 a d = − 3 b c ⇒ 2 a d = 3 b c .
Hence, proved that 2ad = 3bc.
If (a2 + b2 )(x2 + y2 ) = (ax + by)2 ; prove that a x = b y \dfrac{a}{x} = \dfrac{b}{y} x a = y b
Answer
Given,
⇒ (a2 + b2 )(x2 + y2 ) = (ax + by)2
⇒ a2 x2 + a2 y2 + b2 x2 + b2 y2 = a2 x2 + b2 y2 + 2abxy
⇒ a2 y2 + b2 x2 - 2abxy = 0
⇒ (ay - bx)2 = 0
⇒ ay - bx = 0
⇒ ay = bx
⇒ a x = b y \dfrac{a}{x} = \dfrac{b}{y} x a = y b
Hence, proved that a x = b y \dfrac{a}{x} = \dfrac{b}{y} x a = y b
If a, b and c are in continued proportion, prove 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
Answer
Given,
a, b and c are in continued proportion
∴ a b = b c Let a b = b c = k ⇒ a = b k , b = c k ⇒ a = ( c k ) k = c k 2 \therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \text{Let } \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow a = bk, b = ck \\[1em] \Rightarrow a = (ck)k = ck^2 ∴ b a = c b Let b a = c b = k ⇒ a = bk , b = c k ⇒ a = ( c k ) k = c k 2
To prove :
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
Substituting value of a and b in L.H.S. of above equation,
⇒ ( 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 2 ( k 2 + k + 1 ) c 2 ( k 2 + k + 1 ) ⇒ k 2 . \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^2k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow k^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 2 ( k 2 + k + 1 ) ⇒ k 2 .
Substituting value of a and b in R.H.S. of above equation,
⇒ c k 2 c ⇒ k 2 . \Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2. ⇒ c c k 2 ⇒ k 2 .
Since, L.H.S. = R.H.S. = k2 .
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
Using properties of proportion, solve for x :
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 \Rightarrow \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:
⇒ x + 5 + x − 16 + x + 5 − x − 16 x + 5 + x − 16 − ( x + 5 − x − 16 ) = 7 + 3 7 − 3 ⇒ 2 x + 5 2 x − 16 = 10 4 ⇒ x + 5 x − 16 = 10 4 ⇒ x + 5 x − 16 = 100 16 ⇒ 16 ( x + 5 ) = 100 ( x − 16 ) ⇒ 16 x + 80 = 100 x − 1600 ⇒ 100 x − 16 x = 1680 ⇒ 84 x = 1680 ⇒ x = 20. \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{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \dfrac{10}{4} \\[1em] \Rightarrow \dfrac{\sqrt{x + 5}}{\sqrt{x - 16}} = \dfrac{10}{4} \\[1em] \Rightarrow \dfrac{x + 5}{x - 16} = \dfrac{100}{16} \\[1em] \Rightarrow 16(x + 5) = 100(x - 16) \\[1em] \Rightarrow 16x + 80 = 100x - 1600 \\[1em] \Rightarrow 100x - 16x = 1680 \\[1em] \Rightarrow 84x = 1680 \\[1em] \Rightarrow x = 20. ⇒ x + 5 + x − 16 − ( x + 5 − x − 16 ) x + 5 + x − 16 + x + 5 − x − 16 = 7 − 3 7 + 3 ⇒ 2 x − 16 2 x + 5 = 4 10 ⇒ x − 16 x + 5 = 4 10 ⇒ x − 16 x + 5 = 16 100 ⇒ 16 ( x + 5 ) = 100 ( x − 16 ) ⇒ 16 x + 80 = 100 x − 1600 ⇒ 100 x − 16 x = 1680 ⇒ 84 x = 1680 ⇒ x = 20.
Hence, x = 20.
Using properties of proportion, solve for x :
3 x + 9 x 2 − 5 3 x − 9 x 2 − 5 = 5 \dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 5 3 x − 9 x 2 − 5 3 x + 9 x 2 − 5 = 5
Answer
Applying componendo and dividendo,
⇒ 3 x + 9 x 2 − 5 + 3 x − 9 x 2 − 5 3 x + 9 x 2 − 5 − ( 3 x − 9 x 2 − 5 ) = 5 ⇒ 6 x 2 9 x 2 − 5 = 5 + 1 5 − 1 ⇒ 6 x 2 9 x 2 − 5 = 6 4 ⇒ 6 x 2 9 x 2 − 5 = 3 2 ⇒ 2 x 9 x 2 − 5 = 1 ⇒ 9 x 2 − 5 = 2 x \Rightarrow \dfrac{3x + \sqrt{9x^2 - 5} + 3x - \sqrt{9x^2 - 5}}{3x + \sqrt{9x^2 - 5} - (3x - \sqrt{9x^2 - 5})} = 5 \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{9x^2 - 5}} = 1 \\[1em] \Rightarrow \sqrt{9x^2 - 5} = 2x \\[1em] ⇒ 3 x + 9 x 2 − 5 − ( 3 x − 9 x 2 − 5 ) 3 x + 9 x 2 − 5 + 3 x − 9 x 2 − 5 = 5 ⇒ 2 9 x 2 − 5 6 x = 5 − 1 5 + 1 ⇒ 2 9 x 2 − 5 6 x = 4 6 ⇒ 2 9 x 2 − 5 6 x = 2 3 ⇒ 9 x 2 − 5 2 x = 1 ⇒ 9 x 2 − 5 = 2 x
Squaring both sides of the equation,
⇒ 9 x 2 − 5 = 4 x 2 ⇒ 9 x 2 − 4 x 2 = 5 ⇒ 5 x 2 = 5 ⇒ x 2 = 1 ⇒ x = ± 1. \Rightarrow 9x^2 - 5 = 4x^2 \\[1em] \Rightarrow 9x^2 - 4x^2 = 5 \\[1em] \Rightarrow 5x^2 = 5 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \pm1. ⇒ 9 x 2 − 5 = 4 x 2 ⇒ 9 x 2 − 4 x 2 = 5 ⇒ 5 x 2 = 5 ⇒ x 2 = 1 ⇒ x = ± 1.
Hence, x = ±1.
If x = a + 3 b + a − 3 b a + 3 b − a − 3 b \dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}} a + 3 b − a − 3 b a + 3 b + a − 3 b
3bx2 - 2ax + 3b = 0
Answer
Given,
⇒ x = a + 3 b + a − 3 b a + 3 b − a − 3 b \Rightarrow x = \dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}} ⇒ x = a + 3 b − a − 3 b a + 3 b + a − 3 b
Applying componendo and dividendo,
⇒ x + 1 x − 1 = a + 3 b + a − 3 b + a + 3 b − a − 3 b a + 3 b + a − 3 b − ( a + 3 b − a − 3 b ) ⇒ x + 1 x − 1 = 2 a + 3 b 2 a − 3 b ⇒ x + 1 x − 1 = a + 3 b a − 3 b \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a + 3b} + \sqrt{a - 3b} + \sqrt{a + 3b} - \sqrt{a - 3b}}{\sqrt{a + 3b} + \sqrt{a - 3b} - (\sqrt{a + 3b} - \sqrt{a - 3b})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{a + 3b}}{2\sqrt{a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a + 3b}}{\sqrt{a - 3b}} ⇒ x − 1 x + 1 = a + 3 b + a − 3 b − ( a + 3 b − a − 3 b ) a + 3 b + a − 3 b + a + 3 b − a − 3 b ⇒ x − 1 x + 1 = 2 a − 3 b 2 a + 3 b ⇒ x − 1 x + 1 = a − 3 b a + 3 b
Squaring both sides:
⇒ ( x + 1 ) 2 ( x − 1 ) 2 = a + 3 b a − 3 b ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = a + 3 b a − 3 b ⇒ ( x 2 + 1 + 2 x ) ( a − 3 b ) = ( x 2 + 1 − 2 x ) ( a + 3 b ) ⇒ a x 2 + a + 2 a x − 3 b x 2 − 3 b − 6 b x = a x 2 + a − 2 a x + 3 b x 2 + 3 b − 6 b x ⇒ a x 2 − a x 2 + a − a + 2 a x + 2 a x − 3 b x 2 − 3 b x 2 − 3 b − 3 b − 6 b x + 6 b x = 0 ⇒ 4 a x − 6 b x 2 − 6 b = 0 ⇒ 6 b x 2 − 4 a x + 6 b = 0 ⇒ 2 ( 3 b x 2 − 2 a x + 3 b ) = 0 ⇒ 3 b x 2 − 2 a x + 3 b = 0. \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{a + 3b}{a - 3b} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{a + 3b}{a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(a - 3b) = (x^2 + 1 - 2x)(a + 3b) \\[1em] \Rightarrow ax^2 + a + 2ax - 3bx^2 - 3b - 6bx = ax^2 + a - 2ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow ax^2 - ax^2 + a - a + 2ax + 2ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 4ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 4ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 2ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 2ax + 3b = 0. ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = a − 3 b a + 3 b ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = a − 3 b a + 3 b ⇒ ( x 2 + 1 + 2 x ) ( a − 3 b ) = ( x 2 + 1 − 2 x ) ( a + 3 b ) ⇒ a x 2 + a + 2 a x − 3 b x 2 − 3 b − 6 b x = a x 2 + a − 2 a x + 3 b x 2 + 3 b − 6 b x ⇒ a x 2 − a x 2 + a − a + 2 a x + 2 a x − 3 b x 2 − 3 b x 2 − 3 b − 3 b − 6 b x + 6 b x = 0 ⇒ 4 a x − 6 b x 2 − 6 b = 0 ⇒ 6 b x 2 − 4 a x + 6 b = 0 ⇒ 2 ( 3 b x 2 − 2 a x + 3 b ) = 0 ⇒ 3 b x 2 − 2 a x + 3 b = 0.
Hence, proved that 3bx2 - 2ax + 3b = 0.
Using the properties of proportion, solve for x,
given x 4 + 1 2 x 2 = 17 8 \dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8} 2 x 2 x 4 + 1 = 8 17
Answer
Given,
⇒ x 4 + 1 2 x 2 = 17 8 \Rightarrow \dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8} ⇒ 2 x 2 x 4 + 1 = 8 17
Applying componendo and dividendo, we get
⇒ x 4 + 1 + 2 x 2 x 4 + 1 − 2 x 2 = 17 + 8 17 − 8 ⇒ ( x 2 + 1 ) 2 ( x 2 − 1 ) 2 = 25 9 ⇒ x 2 + 1 x 2 − 1 = 5 3 \Rightarrow \dfrac{x^4 + 1 + 2x^2}{x^4 + 1 - 2x^2} = \dfrac{17 + 8}{17 - 8} \\[1em] \Rightarrow \dfrac{(x^2 + 1)^2}{(x^2 - 1)^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x^2 - 1} = \dfrac{5}{3} \\[1em] ⇒ x 4 + 1 − 2 x 2 x 4 + 1 + 2 x 2 = 17 − 8 17 + 8 ⇒ ( x 2 − 1 ) 2 ( x 2 + 1 ) 2 = 9 25 ⇒ x 2 − 1 x 2 + 1 = 3 5
Applying componendo and dividendo, we get
⇒ x 2 + 1 + x 2 − 1 x 2 + 1 − ( x 2 − 1 ) = 5 + 3 5 − 3 ⇒ 2 x 2 2 = 8 2 ⇒ x 2 = 4 ⇒ x = ± 2. \Rightarrow \dfrac{x^2 + 1 + x^2 - 1}{x^2 + 1 - (x^2 - 1)} = \dfrac{5 + 3}{5 - 3} \\[1em] \Rightarrow \dfrac{2x^2}{2} = \dfrac{8}{2} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \pm 2. ⇒ x 2 + 1 − ( x 2 − 1 ) x 2 + 1 + x 2 − 1 = 5 − 3 5 + 3 ⇒ 2 2 x 2 = 2 8 ⇒ x 2 = 4 ⇒ x = ± 2.
Hence, x = ±2.
If x = m + n + m − n m + n − m − n x = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} x = m + n − m − n m + n + m − n
Answer
Given,
x = m + n + m − n m + n − m − n x = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} x = m + n − m − n m + n + m − n
Applying componendo and dividendo,
⇒ x + 1 x − 1 = m + n + m − n + m + n − m − n m + n + m − n − ( m + n − m − n ) ⇒ x + 1 x − 1 = 2 m + n 2 m − n ⇒ x + 1 x − 1 = m + n m − n ⇒ m + n m − n = ( x + 1 ) 2 ( x − 1 ) 2 ⇒ m + n m − n = x 2 + 1 + 2 x x 2 + 1 − 2 x \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{m + n} + \sqrt{m - n} + \sqrt{m + n} - \sqrt{m - n}}{\sqrt{m + n} + \sqrt{m - n} - (\sqrt{m + n} - \sqrt{m - n})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{m + n}}{2\sqrt{m - n}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{m + n}}{\sqrt{m - n}} \\[1em] \Rightarrow \dfrac{m + n}{m - n} = \dfrac{(x + 1)^2}{(x - 1)^2} \\[1em] \Rightarrow \dfrac{m + n}{m - n} = \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} \\[1em] ⇒ x − 1 x + 1 = m + n + m − n − ( m + n − m − n ) m + n + m − n + m + n − m − n ⇒ x − 1 x + 1 = 2 m − n 2 m + n ⇒ x − 1 x + 1 = m − n m + n ⇒ m − n m + n = ( x − 1 ) 2 ( x + 1 ) 2 ⇒ m − n m + n = x 2 + 1 − 2 x x 2 + 1 + 2 x
Applying componendo and dividendo,
⇒ m + n + m − n m + n − ( m − n ) = x 2 + 1 + 2 x + x 2 + 1 − 2 x x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) ⇒ 2 m 2 n = 2 ( x 2 + 1 ) 4 x ⇒ m n = x 2 + 1 2 x ⇒ n = 2 m x x 2 + 1 . \Rightarrow \dfrac{m + n + m - n}{m + n - (m - n)} = \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow \dfrac{2m}{2n} = \dfrac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{x^2 + 1}{2x} \\[1em] \Rightarrow n = \dfrac{2mx}{x^2 + 1}. ⇒ m + n − ( m − n ) m + n + m − n = x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) x 2 + 1 + 2 x + x 2 + 1 − 2 x ⇒ 2 n 2 m = 4 x 2 ( x 2 + 1 ) ⇒ n m = 2 x x 2 + 1 ⇒ n = x 2 + 1 2 m x .
Hence, n = 2 m x x 2 + 1 . \dfrac{2mx}{x^2 + 1}. x 2 + 1 2 m x .
If x 3 + 3 x y 2 3 x 2 y + y 3 = m 3 + 3 m n 2 3 m 2 n + n 3 \dfrac{x^3 + 3xy^2}{3x^2y + y^3} = \dfrac{m^3 + 3mn^2}{3m^2n + n^3} 3 x 2 y + y 3 x 3 + 3 x y 2 = 3 m 2 n + n 3 m 3 + 3 m n 2
show that : nx = my.
Answer
Given,
x 3 + 3 x y 2 3 x 2 y + y 3 = m 3 + 3 m n 2 3 m 2 n + n 3 \dfrac{x^3 + 3xy^2}{3x^2y + y^3} = \dfrac{m^3 + 3mn^2}{3m^2n + n^3} 3 x 2 y + y 3 x 3 + 3 x y 2 = 3 m 2 n + n 3 m 3 + 3 m n 2
Applying componendo and dividendo,
⇒ x 3 + 3 x y 2 + 3 x 2 y + y 3 x 3 + 3 x y 2 − ( 3 x 2 y + y 3 ) = m 3 + 3 m n 2 + 3 m 2 n + n 3 m 3 + 3 m n 2 − ( 3 m 2 n + n 3 ) ⇒ ( x + y ) 3 ( x − y ) 3 = ( m + n ) 3 ( m − n ) 3 ⇒ ( x + y ) ( x − y ) = ( m + n ) ( m − n ) \Rightarrow \dfrac{x^3 + 3xy^2 + 3x^2y + y^3}{x^3 + 3xy^2 - (3x^2y + y^3)} = \dfrac{m^3 + 3mn^2 + 3m^2n + n^3}{m^3 + 3mn^2 - (3m^2n + n^3)} \\[1em] \Rightarrow \dfrac{(x + y)^3}{(x - y)^3} = \dfrac{(m + n)^3}{(m - n)^3} \\[1em] \Rightarrow \dfrac{(x + y)}{(x - y)} = \dfrac{(m + n)}{(m - n)} ⇒ x 3 + 3 x y 2 − ( 3 x 2 y + y 3 ) x 3 + 3 x y 2 + 3 x 2 y + y 3 = m 3 + 3 m n 2 − ( 3 m 2 n + n 3 ) m 3 + 3 m n 2 + 3 m 2 n + n 3 ⇒ ( x − y ) 3 ( x + y ) 3 = ( m − n ) 3 ( m + n ) 3 ⇒ ( x − y ) ( x + y ) = ( m − n ) ( m + n )
Applying componendo and dividendo,
⇒ x + y + x − y x + y − ( x − y ) = m + n + m − n m + n − ( m − n ) ⇒ 2 x 2 y = 2 m 2 n ⇒ x y = m n ⇒ n x = m y . \Rightarrow \dfrac{x + y + x - y}{x + y - (x - y)} = \dfrac{m + n + m - n}{m + n - (m - n)} \\[1em] \Rightarrow \dfrac{2x}{2y} = \dfrac{2m}{2n} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{m}{n} \\[1em] \Rightarrow nx = my. ⇒ x + y − ( x − y ) x + y + x − y = m + n − ( m − n ) m + n + m − n ⇒ 2 y 2 x = 2 n 2 m ⇒ y x = n m ⇒ n x = m y .
Hence, proved that nx = my.
From the following table, find the values of a, b and c:
Length of cloth bought (in m) 10 a 40 c Cost of cloth (in ₹) 40 100 b 180
Answer
It is a case of direct variation:
∴ 10 40 = a 100 = 40 b = c 180 \therefore \dfrac{10}{40} = \dfrac{a}{100} = \dfrac{40}{b} = \dfrac{c}{180} ∴ 40 10 = 100 a = b 40 = 180 c
Taking,
10 40 = a 100 \dfrac{10}{40} = \dfrac{a}{100} 40 10 = 100 a
⇒ 10 x 100 = 40 x a
⇒ a = 1000 40 \dfrac{1000}{40} 40 1000
⇒ a = 25 m
Taking,
10 40 = 40 b \dfrac{10}{40} = \dfrac{40}{b} 40 10 = b 40
⇒ 10 x b = 40 x 40
⇒ b = 1600 10 \dfrac{1600}{10} 10 1600
⇒ b = ₹ 160
Taking,
10 40 = c 180 \dfrac{10}{40} = \dfrac{c}{180} 40 10 = 180 c
⇒ 10 x 180 = 40 x c
⇒ c = 1800 40 \dfrac{1800}{40} 40 1800
⇒ c = 45 m
Hence, the value of a = 25 m, b = ₹ 160 and c = 45 m.
Height of vertical pole (in cm) 56 672 448 2016 Length of its shadow on a horizontal ground (in cm) 32 a b c
Answer
It is a case of direct variation:
∴ 56 32 = 672 a = 448 b = 2016 c \therefore \dfrac{56}{32} = \dfrac{672}{a} = \dfrac{448}{b} = \dfrac{2016}{c} ∴ 32 56 = a 672 = b 448 = c 2016
Taking,
56 32 = 672 a \dfrac{56}{32} = \dfrac{672}{a} 32 56 = a 672
⇒ 56 x a = 672 x 32
⇒ a = 672 × 32 56 \dfrac{672 \times 32}{56} 56 672 × 32
⇒ a = 12 x 32
⇒ a = 384 cm
Taking,
56 32 = 448 b \dfrac{56}{32} = \dfrac{448}{b} 32 56 = b 448
⇒ 56 x b = 32 x 448
⇒ b = 32 × 448 56 \dfrac{32 \times 448}{56} 56 32 × 448
⇒ b = 32 x 8
⇒ b = 256 cm
Taking,
56 32 = 2016 c \dfrac{56}{32} = \dfrac{2016}{c} 32 56 = c 2016
⇒ 56 x c = 2016 x 32
⇒ c = 2016 × 32 56 \dfrac{2016 \times 32}{56} 56 2016 × 32
⇒ c = 36 x 32
⇒ c = 1,152 cm
Hence, the value of a = 384 cm, b = 256 cm and c = 1,152 cm.
From the following table, find the values of a, b and c:
Number of men 130 70 b 120 Number of days a 39 30 c
Answer
It is a case of inverse variation:
∴ 130 x a = 70 x 39 = b x 30 = 120 x c
Taking,
130 x a = 70 x 39
⇒ a = 70 × 39 130 \dfrac{70 \times 39}{130} 130 70 × 39
⇒ a = 7 x 3
⇒ a = 21
Taking,
70 x 39 = b x 30
⇒ b = 70 × 39 30 \dfrac{70 \times 39}{30} 30 70 × 39
⇒ b = 7 x 13
⇒ b = 91
Taking,
70 x 39 = 120 x c
⇒ c = 70 × 39 120 \dfrac{70 \times 39}{120} 120 70 × 39
⇒ c = 22.75
Hence, the value of a = 21, b = 91 and c = 22.75
A car covers a certain distance with uniform speed (y km/h) in x hrs. Use the following table to find the values of a, b and c.
Speed (y km/h) a 60 150 c Time taken (x hrs) 8 b 2 6
Answer
It is a case of inverse variation:
∴ a x 8 = 60 x b = 150 x 2 = c x 6
Taking,
a x 8 = 150 x 2
⇒ a = 150 × 2 8 \dfrac{150 \times 2}{8} 8 150 × 2
⇒ a = 300 8 \dfrac{300}{8} 8 300
⇒ a = 37.5
Taking,
60 x b = 150 x 2
⇒ b = 150 × 2 60 \dfrac{150 \times 2}{60} 60 150 × 2
⇒ b = 300 60 \dfrac{300}{60} 60 300
⇒ b = 5
Taking,
150 x 2 = c x 6
⇒ c = 150 × 2 6 \dfrac{150 \times 2}{6} 6 150 × 2
⇒ c = 300 6 \dfrac{300}{6} 6 300
⇒ c = 50
Hence, the value of a = 37.5, b = 5 and c = 50
The mean proportional of 3 + 2 \sqrt{3} + \sqrt{2} 3 + 2 3 − 2 \sqrt{3} - \sqrt{2} 3 − 2
5 \sqrt{5} 5
5
1
0
Answer
Let x be the mean proportion of 3 + 2 and 3 − 2 \sqrt{3} + \sqrt{2} \text{ and } \sqrt{3} - \sqrt{2} 3 + 2 and 3 − 2
∴ 3 + 2 x = x 3 − 2 ⇒ ( 3 + 2 ) ( 3 − 2 ) = x 2 ⇒ x 2 = ( 3 ) 2 − ( 2 ) 2 [ ∵ ( a + b ) ( a − b ) = a 2 − b 2 ] ⇒ x 2 = 3 − 2 ⇒ x 2 = 1 ⇒ x = 1 = ± 1. \therefore \dfrac{\sqrt{3} + \sqrt{2}}{x} = \dfrac{x}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = x^2 \\[1em] \Rightarrow x^2 = (\sqrt{3})^2 - (\sqrt{2})^2 \quad [\because (a + b)(a - b) = a^2 - b^2] \\[1em] \Rightarrow x^2 = 3 - 2 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1. ∴ x 3 + 2 = 3 − 2 x ⇒ ( 3 + 2 ) ( 3 − 2 ) = x 2 ⇒ x 2 = ( 3 ) 2 − ( 2 ) 2 [ ∵ ( a + b ) ( a − b ) = a 2 − b 2 ] ⇒ x 2 = 3 − 2 ⇒ x 2 = 1 ⇒ x = 1 = ± 1.
Since, geometrical mean is always positive.
∴ x = 1.
Hence, Option 3 is the correct option.
If (a + b) : (a - b) = 13 : 3, a : b is :
13 3 \dfrac{13}{3} 3 13
3 13 \dfrac{3}{13} 13 3
5 8 \dfrac{5}{8} 8 5
8 5 \dfrac{8}{5} 5 8
Answer
Given,
⇒ a + b a − b = 13 3 ⇒ 3 ( a + b ) = 13 ( a − b ) ⇒ 3 a + 3 b = 13 a − 13 b ⇒ 13 a − 3 a = 3 b + 13 b ⇒ 10 a = 16 b ⇒ a b = 16 10 ⇒ a b = 8 5 . \Rightarrow \dfrac{a + b}{a - b} = \dfrac{13}{3} \\[1em] \Rightarrow 3(a + b) = 13(a - b) \\[1em] \Rightarrow 3a + 3b = 13a - 13b \\[1em] \Rightarrow 13a - 3a = 3b + 13b \\[1em] \Rightarrow 10a = 16b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{16}{10} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{8}{5}. ⇒ a − b a + b = 3 13 ⇒ 3 ( a + b ) = 13 ( a − b ) ⇒ 3 a + 3 b = 13 a − 13 b ⇒ 13 a − 3 a = 3 b + 13 b ⇒ 10 a = 16 b ⇒ b a = 10 16 ⇒ b a = 5 8 .
Hence, Option 4 is the correct option.
The table, given below, shows the values of x and y, where x is proportional (directly proportional) to y.
The values of A and B are :
A = 16 and B = 18
A = 32 and B = 9
A = 9 and B = 32
A = 18 and B = 16
Answer
Given,
x is proportional to y.
∴ A 24 = 12 B \therefore \dfrac{A}{24} = \dfrac{12}{B} ∴ 24 A = B 12
and
∴ 24 15 = B 20 \therefore \dfrac{24}{15} = \dfrac{B}{20} ∴ 15 24 = 20 B
Solving equation (2), we get :
⇒ B = 24 × 20 15 ⇒ B = 32. \Rightarrow B = \dfrac{24 \times 20}{15} \\[1em] \Rightarrow B = 32. ⇒ B = 15 24 × 20 ⇒ B = 32.
Substituting value of B in equation 1 :
⇒ A 24 = 12 32 ⇒ A = 12 × 24 32 ⇒ A = 9. \Rightarrow \dfrac{A}{24} = \dfrac{12}{32} \\[1em] \Rightarrow A = \dfrac{12 \times 24}{32} \\[1em] \Rightarrow A = 9. ⇒ 24 A = 32 12 ⇒ A = 32 12 × 24 ⇒ A = 9.
Hence, Option 3 is the correct option.
If (m + n) : (n - m) = 5 : 2; m : n is :
3 : 7
7 : 3
5 : 3
3 : 5
Answer
Given,
(m + n) : (n - m) = 5 : 2
∴ m + n n − m = 5 2 ⇒ 2 ( m + n ) = 5 ( n − m ) ⇒ 2 m + 2 n = 5 n − 5 m ⇒ 2 m + 5 m = 5 n − 2 n ⇒ 7 m = 3 n ⇒ m n = 3 7 ⇒ m : n = 3 : 7. \therefore \dfrac{m + n}{n - m} = \dfrac{5}{2} \\[1em] \Rightarrow 2(m + n) = 5(n - m) \\[1em] \Rightarrow 2m + 2n = 5n - 5m \\[1em] \Rightarrow 2m + 5m = 5n - 2n \\[1em] \Rightarrow 7m = 3n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{3}{7} \\[1em] \Rightarrow m : n = 3 : 7. ∴ n − m m + n = 2 5 ⇒ 2 ( m + n ) = 5 ( n − m ) ⇒ 2 m + 2 n = 5 n − 5 m ⇒ 2 m + 5 m = 5 n − 2 n ⇒ 7 m = 3 n ⇒ n m = 7 3 ⇒ m : n = 3 : 7.
Hence, Option 1 is the correct option.
If x = y, the value of (3x + y) : (5x - 3y) is :
1 : 2
2 : 1
3 : 2
2 : 3
Answer
Substituting x = y in (3x + y) : (5x - 3y), we get :
⇒ 3 x + x 5 x − 3 x ⇒ 4 x 2 x ⇒ 2 1 ⇒ 2 : 1. \Rightarrow \dfrac{3x + x}{5x - 3x} \\[1em] \Rightarrow \dfrac{4x}{2x} \\[1em] \Rightarrow \dfrac{2}{1} \\[1em] \Rightarrow 2 : 1. ⇒ 5 x − 3 x 3 x + x ⇒ 2 x 4 x ⇒ 1 2 ⇒ 2 : 1.
Hence, Option 2 is the correct option.
x : y = 3 : 2 and (x2 + y2 ) : (x2 - y2 )
Assertion (A) : The value of (x2 + y2 ) : (x2 - y2 ) = 13 : 5
Reason (R) : x : y = 3 : 2
⇒ x 2 + y 2 x 2 − y 2 = ( 3 k ) 2 + ( 2 k ) 2 ( 3 k ) 2 − ( 2 k ) 2 = 13 5 \dfrac{x^2 + y^2}{x^2 - y^2} = \dfrac{(3k)^2 + (2k)^2}{(3k)^2 - (2k)^2} = \dfrac{13}{5} x 2 − y 2 x 2 + y 2 = ( 3 k ) 2 − ( 2 k ) 2 ( 3 k ) 2 + ( 2 k ) 2 = 5 13
A is true, R is false.
A is false, R is true.
Both A and R are true and R is the correct reason for R.
Both A and R are true and R is the incorrect reason for R.
Answer
Both A and R are true and R is the correct reason for R.
Reason
Given,
x : y = 3 : 2
Let the value of x be 3k and y be 2k.
The value of
⇒ x 2 + y 2 x 2 − y 2 = ( 3 k ) 2 + ( 2 k ) 2 ( 3 k ) 2 − ( 2 k ) 2 = 9 k 2 + 4 k 2 9 k 2 − 4 k 2 = 13 k 2 5 k 2 = 13 5 \Rightarrow\dfrac{x^2 + y^2}{x^2 - y^2}\\[1em] = \dfrac{(3k)^2 + (2k)^2}{(3k)^2 - (2k)^2}\\[1em] = \dfrac{9k^2 + 4k^2}{9k^2 - 4k^2}\\[1em] = \dfrac{13k^2}{5k^2}\\[1em] = \dfrac{13}{5} ⇒ x 2 − y 2 x 2 + y 2 = ( 3 k ) 2 − ( 2 k ) 2 ( 3 k ) 2 + ( 2 k ) 2 = 9 k 2 − 4 k 2 9 k 2 + 4 k 2 = 5 k 2 13 k 2 = 5 13
According to Assertion; the value of (x2 + y2 ) : (x2 - y2 ) = 13 : 5, which is true.
According to Reason; x : y = 3 : 2
⇒ x 2 + y 2 x 2 − y 2 = ( 3 k ) 2 + ( 2 k ) 2 ( 3 k ) 2 − ( 2 k ) 2 = 13 5 \dfrac{x^2 + y^2}{x^2 - y^2} = \dfrac{(3k)^2 + (2k)^2}{(3k)^2 - (2k)^2} = \dfrac{13}{5} x 2 − y 2 x 2 + y 2 = ( 3 k ) 2 − ( 2 k ) 2 ( 3 k ) 2 + ( 2 k ) 2 = 5 13
Hence, option 3 is the correct option.
( x + y ) 4 ( x − y ) 4 = 16 1 \dfrac{(x + y)^4}{(x - y)^4} = \dfrac{16}{1} ( x − y ) 4 ( x + y ) 4 = 1 16
Assertion (A) : x : y = 3 : 1
Reason (R) : x + y x − y = 2 1 \dfrac{x + y}{x - y} = \dfrac{2}{1} x − y x + y = 1 2 x + y + x − y x + y − x + y = 2 + 1 2 − 1 \dfrac{x + y + x - y}{x + y - x + y} = \dfrac{2 + 1}{2 - 1} x + y − x + y x + y + x − y = 2 − 1 2 + 1
A is true, R is false.
A is false, R is true.
Both A and R are true and R is the correct reason for R.
Both A and R are true and R is the incorrect reason for R.
Answer
Both A and R are true and R is the correct reason for R.
Reason
Given,
⇒ ( x + y ) 4 ( x − y ) 4 = 16 1 ⇒ ( x + y ) ( x − y ) = 16 1 4 ⇒ ( x + y ) ( x − y ) = 2 1 ⇒ ( x + y ) + ( x − y ) ( x + y ) − ( x − y ) = 2 + 1 2 − 1 ⇒ x + y + x − y x + y − x + y = 3 1 ⇒ 2 x 2 y = 3 1 ⇒ x y = 3 1 \Rightarrow \dfrac{(x + y)^4}{(x - y)^4} = \dfrac{16}{1}\\[1em] \Rightarrow \dfrac{(x + y)}{(x - y)} = \sqrt[4]{\dfrac{16}{1}}\\[1em] \Rightarrow \dfrac{(x + y)}{(x - y)} = \dfrac{2}{1}\\[1em] \Rightarrow \dfrac{(x + y) + (x - y)}{(x + y) - (x - y)} = \dfrac{2 + 1}{2 - 1}\\[1em] \Rightarrow \dfrac{x + y + x - y}{x + y - x + y} = \dfrac{3}{1}\\[1em] \Rightarrow \dfrac{2x}{2y} = \dfrac{3}{1}\\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{3}{1} ⇒ ( x − y ) 4 ( x + y ) 4 = 1 16 ⇒ ( x − y ) ( x + y ) = 4 1 16 ⇒ ( x − y ) ( x + y ) = 1 2 ⇒ ( x + y ) − ( x − y ) ( x + y ) + ( x − y ) = 2 − 1 2 + 1 ⇒ x + y − x + y x + y + x − y = 1 3 ⇒ 2 y 2 x = 1 3 ⇒ y x = 1 3
According to Assertion; x : y = 3 : 1 , which is true.
According to Reason; x + y x − y = 2 1 \dfrac{x + y}{x - y} = \dfrac{2}{1} x − y x + y = 1 2 x + y + x − y x + y − x + y = 2 + 1 2 − 1 \dfrac{x + y + x - y}{x + y - x + y} = \dfrac{2 + 1}{2 - 1} x + y − x + y x + y + x − y = 2 − 1 2 + 1
Hence, option 3 is the correct option.
Two irrational numbers 6 \sqrt{6} 6 5 \sqrt{5} 5
Statement 1: The mean proportion of 6 \sqrt{6} 6 5 \sqrt{5} 5 6 + 5 2 \dfrac{\sqrt{6} + \sqrt{5}}{2} 2 6 + 5
Statement 2: The mean proportion of two positive real numbers x and y is x × y \sqrt{x \times y} x × y
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
Statement 1 is false, and statement 2 is true.
Reason
The mean proportion of two positive real numbers x and y is x × y \sqrt{x \times y} x × y
The mean proportion of 6 \sqrt{6} 6 5 \sqrt{5} 5 6 × 5 \sqrt{\sqrt{6} \times \sqrt{5}} 6 × 5
= 30 = 30 4 \sqrt{\sqrt{30}} = \sqrt[4]{30} 30 = 4 30
So, statement 1 is false but statement 2 is true.
Hence, option 4 is the correct option.
Numbers a, b and c are in continued proportion.
Statement 1: (a + b + c)(a - b + c) = a2 + b2 + c2 .
Statement 2: b2 = ac and (a + b + c)(a - b + c) = (a + c)2 - b2
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
Both the statements are true.
Reason
Numbers a, b, c are in continued proportion.
Now, (a + b + c)(a - b + c)
= [(a + c) + b][(a + c) - b]
= (a + c)2 - b2
= a2 + c2 + 2ac - b2
Given, a, b and c are in continued proportion,
∴ a b = b c ⇒ b 2 = a c \therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow b^2 = ac ∴ b a = c b ⇒ b 2 = a c
Substituting the value b2 = ac in above equation,
= a2 + c2 + 2b2 - b2
= a2 + c2 + b2
So, Statement 1 correctly states that: (a + b + c)(a - b + c) = a2 + b2 + c2 .
and
Statement 2 correctly states that: b2 = ac
Hence, option 1 is the correct option.
If a : b = 3 : 5, find :
(10a + 3b) : (5a + 2b)
Answer
Given, a : b = 3 : 5,
If a = 3k, b = 5k.
Substituting value of a and b in (10a + 3b) : (5a + 2b),
⇒ 10 ( 3 k ) + 3 ( 5 k ) 5 ( 3 k ) + 2 ( 5 k ) ⇒ 30 k + 15 k 15 k + 10 k ⇒ 45 k 25 k ⇒ 9 5 . \Rightarrow \dfrac{10(3k) + 3(5k)}{5(3k) + 2(5k)} \\[1em] \Rightarrow \dfrac{30k + 15k}{15k + 10k} \\[1em] \Rightarrow \dfrac{45k}{25k} \\[1em] \Rightarrow \dfrac{9}{5}. ⇒ 5 ( 3 k ) + 2 ( 5 k ) 10 ( 3 k ) + 3 ( 5 k ) ⇒ 15 k + 10 k 30 k + 15 k ⇒ 25 k 45 k ⇒ 5 9 .
Hence, (10a + 3b) : (5a + 2b) = 9 : 5.
If 5x + 6y : 8x + 5y = 8 : 9, find : x : y.
Answer
Given,
5x + 6y : 8x + 5y = 8 : 9
⇒ 5 x + 6 y 8 x + 5 y = 8 9 ⇒ 9 ( 5 x + 6 y ) = 8 ( 8 x + 5 y ) ⇒ 45 x + 54 y = 64 x + 40 y ⇒ 64 x − 45 x = 54 y − 40 y ⇒ 19 x = 14 y ⇒ x y = 14 19 \Rightarrow \dfrac{5x + 6y}{8x + 5y} = \dfrac{8}{9} \\[1em] \Rightarrow 9(5x + 6y) = 8(8x + 5y) \\[1em] \Rightarrow 45x + 54y = 64x + 40y \\[1em] \Rightarrow 64x - 45x = 54y - 40y \\[1em] \Rightarrow 19x = 14y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{14}{19} ⇒ 8 x + 5 y 5 x + 6 y = 9 8 ⇒ 9 ( 5 x + 6 y ) = 8 ( 8 x + 5 y ) ⇒ 45 x + 54 y = 64 x + 40 y ⇒ 64 x − 45 x = 54 y − 40 y ⇒ 19 x = 14 y ⇒ y x = 19 14
Hence, x : y = 14 : 19.
Find the duplicate ratio of 2 2 : 3 5 2\sqrt{2} : 3\sqrt{5} 2 2 : 3 5
Answer
Duplicate ratio of 2 2 : 3 5 2\sqrt{2} : 3\sqrt{5} 2 2 : 3 5
= ( 2 2 ) 2 : ( 3 5 ) 2 = 8 : 45. = (2\sqrt{2})^2 : (3\sqrt{5})^2 \\[1em] = 8 : 45. = ( 2 2 ) 2 : ( 3 5 ) 2 = 8 : 45.
Hence, duplicate ratio of 2 2 : 3 5 2\sqrt{2} : 3\sqrt{5} 2 2 : 3 5
Find the triplicate ratio of 2a : 3b
Answer
Triplicate ratio of 2a : 3b,
= (2a)3 : (3b)3
= 8a3 : 27b3
Hence, triplicate ratio of 2a : 3b = 8a3 : 27b3 .
Find the sub-duplicate ratio of 9x2 a4 : 25y6 b2
Answer
Sub-duplicate ratio of 9x2 a4 : 25y6 b2
= 9 x 2 a 4 : 25 y 6 b 2 = 3 x a 2 : 5 y 3 b = \sqrt{9x^2a^4} : \sqrt{25y^6b^2} \\[1em] = 3xa^2 : 5y^3b = 9 x 2 a 4 : 25 y 6 b 2 = 3 x a 2 : 5 y 3 b
Hence, sub-duplicate ratio of 9x2 a4 : 25y6 b2 = 3xa2 : 5y3 b.
Find the sub-triplicate ratio of 216 : 343
Answer
Sub-triplicate ratio of 216 : 343
= 216 3 : 343 3 \sqrt[3]{216} : \sqrt[3]{343} 3 216 : 3 343
= 6 : 7.
Hence, sub-triplicate ratio of 216 : 343 = 6 : 7.
Find the reciprocal ratio of 3 : 5
Answer
Reciprocal ratio of 3 : 5,
⇒ 1 3 : 1 5 ⇒ 5 : 3. \Rightarrow \dfrac{1}{3} : \dfrac{1}{5} \\[1em] \Rightarrow 5 : 3. ⇒ 3 1 : 5 1 ⇒ 5 : 3.
Hence, reciprocal ratio of 3 : 5 = 5 : 3.
Find the ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Answer
Ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49,
⇒ 5 2 6 2 × 1 25 1 42 × 36 49 ⇒ 25 36 × 42 25 × 6 7 ⇒ 1 1 . \Rightarrow \dfrac{5^2}{6^2} \times \dfrac{\dfrac{1}{25}}{\dfrac{1}{42}} \times \dfrac{\sqrt{36}}{\sqrt{49}} \\[1em] \Rightarrow \dfrac{25}{36} \times \dfrac{42}{25} \times \dfrac{6}{7} \\[1em] \Rightarrow \dfrac{1}{1}. ⇒ 6 2 5 2 × 42 1 25 1 × 49 36 ⇒ 36 25 × 25 42 × 7 6 ⇒ 1 1 .
Hence, resultant ratio = 1 : 1.
Find the value of x, if (2x + 3) : (5x - 38) is the duplicate ratio of 5 : 6 \sqrt{5} : \sqrt{6} 5 : 6
Answer
According to question,
⇒ 2 x + 3 5 x − 38 = ( 5 ) 2 ( 6 ) 2 ⇒ 2 x + 3 5 x − 38 = 5 6 ⇒ 6 ( 2 x + 3 ) = 5 ( 5 x − 38 ) ⇒ 12 x + 18 = 25 x − 190 ⇒ 25 x − 12 x = 190 + 18 ⇒ 13 x = 208 ⇒ x = 16. \Rightarrow \dfrac{2x + 3}{5x - 38} = \dfrac{(\sqrt{5})^2}{(\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{2x + 3}{5x - 38} = \dfrac{5}{6} \\[1em] \Rightarrow 6(2x + 3) = 5(5x - 38) \\[1em] \Rightarrow 12x + 18 = 25x - 190 \\[1em] \Rightarrow 25x - 12x = 190 + 18 \\[1em] \Rightarrow 13x = 208 \\[1em] \Rightarrow x = 16. ⇒ 5 x − 38 2 x + 3 = ( 6 ) 2 ( 5 ) 2 ⇒ 5 x − 38 2 x + 3 = 6 5 ⇒ 6 ( 2 x + 3 ) = 5 ( 5 x − 38 ) ⇒ 12 x + 18 = 25 x − 190 ⇒ 25 x − 12 x = 190 + 18 ⇒ 13 x = 208 ⇒ x = 16.
Hence, x = 16.
Find the value of x, if (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25
Answer
According to question,
⇒ 2 x + 1 3 x + 13 = 9 25 ⇒ 2 x + 1 3 x + 13 = 3 5 ⇒ 5 ( 2 x + 1 ) = 3 ( 3 x + 13 ) ⇒ 10 x + 5 = 9 x + 39 ⇒ 10 x − 9 x = 39 − 5 ⇒ x = 34. \Rightarrow \dfrac{2x + 1}{3x + 13} = \dfrac{\sqrt{9}}{\sqrt{25}} \\[1em] \Rightarrow \dfrac{2x + 1}{3x + 13} = \dfrac{3}{5} \\[1em] \Rightarrow 5(2x + 1) = 3(3x + 13) \\[1em] \Rightarrow 10x + 5 = 9x + 39 \\[1em] \Rightarrow 10x - 9x = 39 - 5 \\[1em] \Rightarrow x = 34. ⇒ 3 x + 13 2 x + 1 = 25 9 ⇒ 3 x + 13 2 x + 1 = 5 3 ⇒ 5 ( 2 x + 1 ) = 3 ( 3 x + 13 ) ⇒ 10 x + 5 = 9 x + 39 ⇒ 10 x − 9 x = 39 − 5 ⇒ x = 34.
Hence, x = 34.
Find the value of x, if (3x - 7) : (4x + 3) is the sub-triplicate ratio of 8 : 27
Answer
According to question,
⇒ 3 x − 7 4 x + 3 = 8 3 27 3 ⇒ 3 x − 7 4 x + 3 = 2 3 ⇒ 3 ( 3 x − 7 ) = 2 ( 4 x + 3 ) ⇒ 9 x − 21 = 8 x + 6 ⇒ 9 x − 8 x = 6 + 21 ⇒ x = 27. \Rightarrow \dfrac{3x - 7}{4x + 3} = \dfrac{\sqrt[3]{8}}{\sqrt[3]{27}} \\[1em] \Rightarrow \dfrac{3x - 7}{4x + 3} = \dfrac{2}{3} \\[1em] \Rightarrow 3(3x - 7) = 2(4x + 3) \\[1em] \Rightarrow 9x - 21 = 8x + 6 \\[1em] \Rightarrow 9x - 8x = 6 + 21 \\[1em] \Rightarrow x = 27. ⇒ 4 x + 3 3 x − 7 = 3 27 3 8 ⇒ 4 x + 3 3 x − 7 = 3 2 ⇒ 3 ( 3 x − 7 ) = 2 ( 4 x + 3 ) ⇒ 9 x − 21 = 8 x + 6 ⇒ 9 x − 8 x = 6 + 21 ⇒ x = 27.
Hence, x = 27.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Answer
Let the number to be added be a
⇒ x + a y + a = c d ⇒ d ( x + a ) = c ( y + a ) ⇒ d x + d a = c y + c a ⇒ d a − c a = c y − d x ⇒ a ( d − c ) = c y − d x ⇒ a = c y − d x d − c . \Rightarrow \dfrac{x + a}{y + a} = \dfrac{c}{d} \\[1em] \Rightarrow d(x + a) = c(y + a) \\[1em] \Rightarrow dx + da = cy + ca \\[1em] \Rightarrow da - ca = cy - dx \\[1em] \Rightarrow a(d - c) = cy - dx \\[1em] \Rightarrow a = \dfrac{cy - dx}{d - c}. ⇒ y + a x + a = d c ⇒ d ( x + a ) = c ( y + a ) ⇒ d x + d a = cy + c a ⇒ d a − c a = cy − d x ⇒ a ( d − c ) = cy − d x ⇒ a = d − c cy − d x .
Hence, number to be added = c y − d x d − c . \dfrac{cy - dx}{d - c}. d − c cy − d x .
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg ?
Answer
Let woman's reduced weight become x kg.
Since, weight is reduced in the ratio 7 : 5.
∴ 84 x = 7 5 ⇒ x = 84 × 5 7 ⇒ x = 60. \therefore \dfrac{84}{x} = \dfrac{7}{5} \\[1em] \Rightarrow x = \dfrac{84 \times 5}{7} \\[1em] \Rightarrow x = 60. ∴ x 84 = 5 7 ⇒ x = 7 84 × 5 ⇒ x = 60.
Hence, reduced weight = 60 kg.
If 15(2x2 - y2 ) = 7xy, find x : y; if x and y both are positive.
Answer
Given,
⇒ 15 ( 2 x 2 − y 2 ) = 7 x y ⇒ 30 x 2 − 15 y 2 = 7 x y ⇒ 30 x 2 − 15 y 2 x y = 7 x y x y ⇒ 30 x y − 15 y x = 7 \Rightarrow 15(2x^2 - y^2) = 7xy \\[1em] \Rightarrow 30x^2 - 15y^2 = 7xy \\[1em] \Rightarrow \dfrac{30x^2 - 15y^2}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow 30\dfrac{x}{y} - 15\dfrac{y}{x} = 7 ⇒ 15 ( 2 x 2 − y 2 ) = 7 x y ⇒ 30 x 2 − 15 y 2 = 7 x y ⇒ x y 30 x 2 − 15 y 2 = x y 7 x y ⇒ 30 y x − 15 x y = 7
Let x y \dfrac{x}{y} y x
⇒ 30 t − 15 1 t = 7 ⇒ 30 t 2 − 15 t = 7 ⇒ 30 t 2 − 15 = 7 t ⇒ 30 t 2 − 7 t − 15 = 0 ⇒ 30 t 2 − 25 t + 18 t − 15 = 0 ⇒ 5 t ( 6 t − 5 ) + 3 ( 6 t − 5 ) = 0 ⇒ ( 5 t + 3 ) ( 6 t − 5 ) = 0 ⇒ 5 t + 3 = 0 or 6 t − 5 = 0 ⇒ t = − 3 5 or t = 5 6 . \Rightarrow 30t - 15\dfrac{1}{t} = 7 \\[1em] \Rightarrow \dfrac{30t^2 - 15}{t} = 7 \\[1em] \Rightarrow 30t^2 - 15 = 7t \\[1em] \Rightarrow 30t^2 - 7t - 15 = 0 \\[1em] \Rightarrow 30t^2 - 25t + 18t - 15 = 0 \\[1em] \Rightarrow 5t(6t - 5) + 3(6t - 5) = 0 \\[1em] \Rightarrow (5t + 3)(6t - 5) = 0 \\[1em] \Rightarrow 5t + 3 = 0 \text{ or } 6t - 5 = 0 \\[1em] \Rightarrow t = -\dfrac{3}{5} \text{ or } t = \dfrac{5}{6}. ⇒ 30 t − 15 t 1 = 7 ⇒ t 30 t 2 − 15 = 7 ⇒ 30 t 2 − 15 = 7 t ⇒ 30 t 2 − 7 t − 15 = 0 ⇒ 30 t 2 − 25 t + 18 t − 15 = 0 ⇒ 5 t ( 6 t − 5 ) + 3 ( 6 t − 5 ) = 0 ⇒ ( 5 t + 3 ) ( 6 t − 5 ) = 0 ⇒ 5 t + 3 = 0 or 6 t − 5 = 0 ⇒ t = − 5 3 or t = 6 5 .
Since, x and y both are positive,
∴ t ≠ − 3 5 -\dfrac{3}{5} − 5 3
Hence, x : y = 5 : 6.
Find the fourth proportional to 2xy, x2 and y2
Answer
Let fourth proportional to 2xy, x2 and y2 be n.
∴ 2 x y x 2 = y 2 n ⇒ n = x 2 y 2 2 x y ⇒ n = x y 2 . \therefore \dfrac{2xy}{x^2} = \dfrac{y^2}{n} \\[1em] \Rightarrow n = \dfrac{x^2y^2}{2xy}\\[1em] \Rightarrow n = \dfrac{xy}{2}. ∴ x 2 2 x y = n y 2 ⇒ n = 2 x y x 2 y 2 ⇒ n = 2 x y .
Hence, fourth proportional = x y 2 \dfrac{xy}{2} 2 x y
Find the third proportional to a2 - b2 and a + b.
Answer
Let third proportional to a2 - b2 and a + b be x,
⇒ a 2 − b 2 a + b = a + b x ⇒ x = ( a + b ) ( a + b ) a 2 − b 2 ⇒ x = ( a + b ) ( a + b ) ( a + b ) ( a − b ) ⇒ x = ( a + b ) ( a − b ) . \Rightarrow \dfrac{a^2 -b^2}{a + b} = \dfrac{a + b}{x} \\[1em] \Rightarrow x = \dfrac{(a + b)(a + b)}{a^2- b^2} \\[1em] \Rightarrow x = \dfrac{(a + b)(a + b)}{(a + b)(a - b)} \\[1em] \Rightarrow x = \dfrac{(a + b)}{(a - b)}. ⇒ a + b a 2 − b 2 = x a + b ⇒ x = a 2 − b 2 ( a + b ) ( a + b ) ⇒ x = ( a + b ) ( a − b ) ( a + b ) ( a + b ) ⇒ x = ( a − b ) ( a + b ) .
Hence, third proportional to a2 - b2 and a + b = ( a + b ) ( a − b ) . \dfrac{(a + b)}{(a - b)}. ( a − b ) ( a + b ) .
Find the mean proportion to (x - y) and (x3 - x2 y)
Answer
Let mean proportion to (x - y) and (x3 - x2 y) be a.
∴ x − y a = a x 3 − x 2 y ⇒ a 2 = ( x − y ) ( x 3 − x 2 y ) ⇒ a 2 = ( x − y ) . x 2 . ( x − y ) ⇒ a 2 = x 2 . ( x − y ) 2 ⇒ a = x ( x − y ) . \therefore \dfrac{x - y}{a} = \dfrac{a}{x^3 - x^2y} \\[1em] \Rightarrow a^2 = (x - y)(x^3 - x^2y) \\[1em] \Rightarrow a^2 = (x - y).x^2.(x - y) \\[1em] \Rightarrow a^2 = x^2.(x - y)^2 \\[1em] \Rightarrow a = x(x - y). ∴ a x − y = x 3 − x 2 y a ⇒ a 2 = ( x − y ) ( x 3 − x 2 y ) ⇒ a 2 = ( x − y ) . x 2 . ( x − y ) ⇒ a 2 = x 2 . ( x − y ) 2 ⇒ a = x ( x − y ) .
Hence, mean proportion to (x - y) and (x3 - x2 y) = x(x - y).
Find two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.
Answer
Let two numbers be x and y.
Given, 14 is mean proportional to x and y,
∴ x 14 = 14 y ⇒ x y = 196 . . . . . ( i ) \therefore \dfrac{x}{14} = \dfrac{14}{y} \\[1em] \Rightarrow xy = 196 \space .....(i) ∴ 14 x = y 14 ⇒ x y = 196 ..... ( i )
Given, 112 is third proportional to x and y,
∴ x y = y 112 ⇒ y 2 = 112 x ⇒ x = y 2 112 . . . . . ( i i ) \therefore \dfrac{x}{y} = \dfrac{y}{112} \\[1em] \Rightarrow y^2 = 112x \\[1em] \Rightarrow x = \dfrac{y^2}{112} \space .....(ii) ∴ y x = 112 y ⇒ y 2 = 112 x ⇒ x = 112 y 2 ..... ( ii )
Substituting value of x from (ii) in (i) we get,
⇒ y 2 112 . y = 196 ⇒ y 3 = 196 × 112 ⇒ y 3 = 21952 ⇒ y = 21952 3 ⇒ y = 28. \Rightarrow \dfrac{y^2}{112}.y = 196 \\[1em] \Rightarrow y^3 = 196 \times 112 \\[1em] \Rightarrow y^3 = 21952 \\[1em] \Rightarrow y = \sqrt[3]{21952} \\[1em] \Rightarrow y = 28. ⇒ 112 y 2 . y = 196 ⇒ y 3 = 196 × 112 ⇒ y 3 = 21952 ⇒ y = 3 21952 ⇒ y = 28.
x = 28 2 112 = 784 112 x = \dfrac{28^2}{112} = \dfrac{784}{112} x = 112 2 8 2 = 112 784
Hence, numbers are 7 and 28.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Answer
According to question,
⇒ x y = ( x + z ) 2 ( y + z ) 2 ⇒ x y = x 2 + z 2 + 2 x z y 2 + z 2 + 2 y z ⇒ x ( y 2 + z 2 + 2 y z ) = y ( x 2 + z 2 + 2 x z ) ⇒ x y 2 + x z 2 + 2 x y z = y x 2 + y z 2 + 2 x y z ⇒ x z 2 − y z 2 = 2 x y z − 2 x y z + y x 2 − x y 2 ⇒ z 2 ( x − y ) = x y ( x − y ) ⇒ z 2 = x y . \Rightarrow \dfrac{x}{y} = \dfrac{(x + z)^2}{(y + z)^2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{x^2 + z^2 + 2xz}{y^2 + z^2 + 2yz} \\[1em] \Rightarrow x(y^2 + z^2 + 2yz) = y(x^2 + z^2 + 2xz) \\[1em] \Rightarrow xy^2 + xz^2 + 2xyz = yx^2 + yz^2 + 2xyz \\[1em] \Rightarrow xz^2 - yz^2 = 2xyz - 2xyz + yx^2- xy^2 \\[1em] \Rightarrow z^2(x - y) = xy(x - y) \\[1em] \Rightarrow z^2 = xy. ⇒ y x = ( y + z ) 2 ( x + z ) 2 ⇒ y x = y 2 + z 2 + 2 yz x 2 + z 2 + 2 x z ⇒ x ( y 2 + z 2 + 2 yz ) = y ( x 2 + z 2 + 2 x z ) ⇒ x y 2 + x z 2 + 2 x yz = y x 2 + y z 2 + 2 x yz ⇒ x z 2 − y z 2 = 2 x yz − 2 x yz + y x 2 − x y 2 ⇒ z 2 ( x − y ) = x y ( x − y ) ⇒ z 2 = x y .
Since, z2 = xy hence, proved that z is mean proportional between x and y.
If a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
(a + b) : (c + d) = a 2 + b 2 : c 2 + d 2 \sqrt{a^2 + b^2} : \sqrt{c^2 + d^2} a 2 + b 2 : c 2 + d 2
Answer
Let a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
a = bk, c = dk.
Substituting a = bk, c = dk in L.H.S. of (a + b) : (c + d) = a 2 + b 2 : c 2 + d 2 \sqrt{a^2 + b^2} : \sqrt{c^2 + d^2} a 2 + b 2 : c 2 + d 2
L.H.S. = a + b c + d = b k + b d k + d = b ( k + 1 ) d ( k + 1 ) = b d . \text{L.H.S.} = \dfrac{a + b}{c + d} \\[1em] = \dfrac{bk + b}{dk + d} \\[1em] = \dfrac{b(k + 1)}{d(k + 1)} \\[1em] = \dfrac{b}{d}. L.H.S. = c + d a + b = d k + d bk + b = d ( k + 1 ) b ( k + 1 ) = d b .
Substituting a = bk, c = dk in R.H.S. of (a + b) : (c + d) = a 2 + b 2 : c 2 + d 2 \sqrt{a^2 + b^2} : \sqrt{c^2 + d^2} a 2 + b 2 : c 2 + d 2
R.H.S. = a 2 + b 2 c 2 + d 2 = ( b k ) 2 + b 2 ( d k ) 2 + d 2 = b 2 ( k 2 + 1 ) d 2 ( k 2 + 1 ) = b 2 d 2 = b d . \text{R.H.S.} = \dfrac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \\[1em] = \dfrac{\sqrt{(bk)^2 + b^2}}{\sqrt{(dk)^2 + d^2}} \\[1em] = \dfrac{\sqrt{b^2(k^2 + 1)}}{\sqrt{d^2(k^2 + 1)}} \\[1em] = \dfrac{\sqrt{b^2}}{\sqrt{d^2}} \\[1em] = \dfrac{b}{d}. R.H.S. = c 2 + d 2 a 2 + b 2 = ( d k ) 2 + d 2 ( bk ) 2 + b 2 = d 2 ( k 2 + 1 ) b 2 ( k 2 + 1 ) = d 2 b 2 = d b .
Since, L.H.S. = R.H.S. = b d \dfrac{b}{d} d b
Hence, proved that (a + b) : (c + d) = a 2 + b 2 : c 2 + d 2 \sqrt{a^2 + b^2} : \sqrt{c^2 + d^2} a 2 + b 2 : c 2 + d 2
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so ratio of number of boys to number of girls may be 9 : 5 ?
Answer
Ratio of the number of boys to the number of girls is 3 : 1.
Total members = 36
No. of boys = 3 3 + 1 × 36 = 3 4 × 36 = 27. \dfrac{3}{3 + 1} \times 36 = \dfrac{3}{4} \times 36 = 27. 3 + 1 3 × 36 = 4 3 × 36 = 27.
No. of girls = 36 - 27 = 9.
Let no. of girls to be added be x.
∴ 27 9 + x = 9 5 ⇒ 135 = 9 ( 9 + x ) ⇒ 135 = 81 + 9 x ⇒ 54 = 9 x ⇒ x = 6. \therefore \dfrac{27}{9 + x} = \dfrac{9}{5} \\[1em] \Rightarrow 135 = 9(9 + x) \\[1em] \Rightarrow 135 = 81 + 9x \\[1em] \Rightarrow 54 = 9x \\[1em] \Rightarrow x = 6. ∴ 9 + x 27 = 5 9 ⇒ 135 = 9 ( 9 + x ) ⇒ 135 = 81 + 9 x ⇒ 54 = 9 x ⇒ x = 6.
Hence, 6 girls must be added to council so ratio of number of boys to number of girls becomes 9 : 5.
If 7x - 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of :
(i) 9 x + 5 y 9 x − 5 y \dfrac{9x + 5y}{9x - 5y} 9 x − 5 y 9 x + 5 y
(ii) 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
Answer
7x - 15y = 4x + y
⇒ 7x - 4x = y + 15y
⇒ 3x = 16y
⇒ x y = 16 3 \dfrac{x}{y} = \dfrac{16}{3} y x = 3 16
(i) 9 x + 5 y 9 x − 5 y \dfrac{9x + 5y}{9x - 5y} 9 x − 5 y 9 x + 5 y
⇒ x y = 16 3 9 x 5 y = 9 × 16 5 × 3 9 x 5 y = 144 15 \phantom{\Rightarrow} \dfrac{x}{y} = \dfrac{16}{3} \\[1em] \dfrac{9x}{5y} = \dfrac{9 \times 16}{5 \times 3} \\[1em] \dfrac{9x}{5y} = \dfrac{144}{15} ⇒ y x = 3 16 5 y 9 x = 5 × 3 9 × 16 5 y 9 x = 15 144
Applying componendo and dividendo:
9 x + 5 y 9 x − 5 y = 144 + 15 144 − 15 9 x + 5 y 9 x − 5 y = 159 129 = 53 43 . \dfrac{9x + 5y}{9x - 5y} = \dfrac{144 + 15}{144 - 15} \\[1em] \dfrac{9x + 5y}{9x - 5y} = \dfrac{159}{129} = \dfrac{53}{43}. 9 x − 5 y 9 x + 5 y = 144 − 15 144 + 15 9 x − 5 y 9 x + 5 y = 129 159 = 43 53 .
Hence, 9 x + 5 y 9 x − 5 y = 53 43 \dfrac{9x + 5y}{9x - 5y} = \dfrac{53}{43} 9 x − 5 y 9 x + 5 y = 43 53
(ii) 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
⇒ x y = 16 3 ⇒ x 2 y 2 = 256 9 ⇒ 3 x 2 2 y 2 = 3 × 256 2 × 9 ⇒ 3 x 2 2 y 2 = 768 18 \Rightarrow \dfrac{x}{y} = \dfrac{16}{3} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{256}{9} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = \dfrac{3 \times 256}{2 \times 9} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = \dfrac{768}{18} ⇒ y x = 3 16 ⇒ y 2 x 2 = 9 256 ⇒ 2 y 2 3 x 2 = 2 × 9 3 × 256 ⇒ 2 y 2 3 x 2 = 18 768
Applying componendo and dividendo:
⇒ 3 x 2 + 2 y 2 3 x 2 − 2 y 2 = 768 + 18 768 − 18 ⇒ 3 x 2 + 2 y 2 3 x 2 − 2 y 2 = 786 750 ⇒ 3 x 2 + 2 y 2 3 x 2 − 2 y 2 = 131 125 . \Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{768 + 18}{768 - 18} \\[1em] \Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{786}{750} \\[1em] \Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{131}{125}. ⇒ 3 x 2 − 2 y 2 3 x 2 + 2 y 2 = 768 − 18 768 + 18 ⇒ 3 x 2 − 2 y 2 3 x 2 + 2 y 2 = 750 786 ⇒ 3 x 2 − 2 y 2 3 x 2 + 2 y 2 = 125 131 .
Hence, 3 x 2 + 2 y 2 3 x 2 − 2 y 2 = 131 125 \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{131}{125} 3 x 2 − 2 y 2 3 x 2 + 2 y 2 = 125 131 .
If 4 m + 3 n 4 m − 3 n = 7 4 \dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4} 4 m − 3 n 4 m + 3 n = 4 7
(i) m : n
(ii) 2 m 2 − 11 n 2 2 m 2 + 11 n 2 \dfrac{2m^2 - 11n^2}{2m^2 + 11n^2} 2 m 2 + 11 n 2 2 m 2 − 11 n 2
Answer
(i) Given,
4 m + 3 n 4 m − 3 n = 7 4 \dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4} 4 m − 3 n 4 m + 3 n = 4 7
Applying componendo and dividendo:
4 m + 3 n + 4 m − 3 n 4 m + 3 n − ( 4 m − 3 n ) = 7 + 4 7 − 4 ⇒ 8 m 6 n = 11 3 ⇒ m n = 11 × 6 3 × 8 ⇒ m n = 11 4 . \dfrac{4m + 3n + 4m - 3n}{4m + 3n - (4m - 3n)} = \dfrac{7 + 4}{7 - 4} \\[1em] \Rightarrow \dfrac{8m}{6n} = \dfrac{11}{3} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{11 \times 6}{3 \times 8} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{11}{4}. 4 m + 3 n − ( 4 m − 3 n ) 4 m + 3 n + 4 m − 3 n = 7 − 4 7 + 4 ⇒ 6 n 8 m = 3 11 ⇒ n m = 3 × 8 11 × 6 ⇒ n m = 4 11 .
Hence, m : n = 11 : 4.
(ii) We know,
⇒ m n = 11 4 ⇒ m 2 n 2 = 121 16 ⇒ 2 m 2 11 n 2 = 2 × 121 11 × 16 ⇒ 2 m 2 11 n 2 = 242 176 \phantom{\Rightarrow} \dfrac{m}{n} = \dfrac{11}{4} \\[1em] \Rightarrow \dfrac{m^2}{n^2} = \dfrac{121}{16} \\[1em] \Rightarrow \dfrac{2m^2}{11n^2} = \dfrac{2 \times 121}{11 \times 16} \\[1em] \Rightarrow \dfrac{2m^2}{11n^2} = \dfrac{242}{176} ⇒ n m = 4 11 ⇒ n 2 m 2 = 16 121 ⇒ 11 n 2 2 m 2 = 11 × 16 2 × 121 ⇒ 11 n 2 2 m 2 = 176 242
Applying componendo and dividendo:
⇒ 2 m 2 + 11 n 2 2 m 2 − 11 n 2 = 242 + 176 242 − 176 ⇒ 2 m 2 + 11 n 2 2 m 2 − 11 n 2 = 418 66 = 19 3 \Rightarrow \dfrac{2m^2 + 11n^2}{2m^2 - 11n^2} = \dfrac{242 + 176}{242 - 176} \\[1em] \Rightarrow \dfrac{2m^2 + 11n^2}{2m^2 - 11n^2} = \dfrac{418}{66} = \dfrac{19}{3} ⇒ 2 m 2 − 11 n 2 2 m 2 + 11 n 2 = 242 − 176 242 + 176 ⇒ 2 m 2 − 11 n 2 2 m 2 + 11 n 2 = 66 418 = 3 19
Applying invertendo:
⇒ 2 m 2 − 11 n 2 2 m 2 + 11 n 2 = 3 19 \Rightarrow \dfrac{2m^2 - 11n^2}{2m^2 + 11n^2} = \dfrac{3}{19} ⇒ 2 m 2 + 11 n 2 2 m 2 − 11 n 2 = 19 3
Hence, 2 m 2 − 11 n 2 2 m 2 + 11 n 2 = 3 19 . \dfrac{2m^2 - 11n^2}{2m^2 + 11n^2} = \dfrac{3}{19}. 2 m 2 + 11 n 2 2 m 2 − 11 n 2 = 19 3 .
If x, y and z are in continued proportion, prove that :
( x + y ) 2 ( y + z ) 2 = x z \dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z} ( y + z ) 2 ( x + y ) 2 = z x
Answer
Given, x, y and z are in continued proportion
∴ x y = y z \therefore \dfrac{x}{y} = \dfrac{y}{z} ∴ y x = z y
⇒ y2 = xz
Taking LHS,
( x + y ) 2 ( y + z ) 2 = x 2 + y 2 + 2 x y y 2 + z 2 + 2 y z \dfrac{(x + y)^2}{(y + z)^2} \\[1em] = \dfrac{x^2 + y^2 + 2xy}{y^2 + z^2 + 2yz} ( y + z ) 2 ( x + y ) 2 = y 2 + z 2 + 2 yz x 2 + y 2 + 2 x y
Substituting y2 = xz in above we get,
⇒ x 2 + x z + 2 x y x z + z 2 + 2 y z ⇒ x ( x + z + 2 y ) z ( x + z + 2 y ) ⇒ x z = RHS \Rightarrow \dfrac{x^2 + xz + 2xy}{xz + z^2 + 2yz} \\[1em] \Rightarrow \dfrac{x(x + z + 2y)}{z(x + z + 2y)} \\[1em] \Rightarrow \dfrac{x}{z} = \text{ RHS } ⇒ x z + z 2 + 2 yz x 2 + x z + 2 x y ⇒ z ( x + z + 2 y ) x ( x + z + 2 y ) ⇒ z x = RHS
Hence, proved that ( x + y ) 2 ( y + z ) 2 = x z \dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z} ( y + z ) 2 ( x + y ) 2 = z x
Given, x = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − a 2 − b 2 \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}} a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2
Use componendo and dividendo to prove that :
b2 = 2 a 2 x x 2 + 1 \dfrac{2a^2x}{x^2 + 1} x 2 + 1 2 a 2 x
Answer
Given,
⇒ x = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − a 2 − b 2 \Rightarrow x = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}} ⇒ x = a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2
Applying componendo and dividendo:
⇒ x + 1 x − 1 = a 2 + b 2 + a 2 − b 2 + a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2 − ( a 2 + b 2 − a 2 − b 2 ) ⇒ x + 1 x − 1 = 2 a 2 + b 2 2 a 2 − b 2 ⇒ x + 1 x − 1 = a 2 + b 2 a 2 − b 2 ⇒ ( x + 1 ) 2 ( x − 1 ) 2 = a 2 + b 2 a 2 − b 2 ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = a 2 + b 2 a 2 − b 2 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} + \sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} - (\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{a^2 + b^2}}{2\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2}}{\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{a^2 + b^2}{a^2 - b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{a^2 + b^2}{a^2 - b^2} ⇒ x − 1 x + 1 = a 2 + b 2 + a 2 − b 2 − ( a 2 + b 2 − a 2 − b 2 ) a 2 + b 2 + a 2 − b 2 + a 2 + b 2 − a 2 − b 2 ⇒ x − 1 x + 1 = 2 a 2 − b 2 2 a 2 + b 2 ⇒ x − 1 x + 1 = a 2 − b 2 a 2 + b 2 ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = a 2 − b 2 a 2 + b 2 ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = a 2 − b 2 a 2 + b 2
Applying componendo and dividendo:
⇒ x 2 + 1 + 2 x + x 2 + 1 − 2 x x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − ( a 2 − b 2 ) ⇒ 2 ( x 2 + 1 ) 4 x = 2 a 2 2 b 2 ⇒ x 2 + 1 2 x = a 2 b 2 ⇒ b 2 = 2 a 2 x x 2 + 1 . \Rightarrow \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} = \dfrac{a^2 + b^2 + a^2 - b^2}{a^2 + b^2 - (a^2 - b^2)} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{4x} = \dfrac{2a^2}{2b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = \dfrac{a^2}{b^2} \\[1em] \Rightarrow b^2 = \dfrac{2a^2x}{x^2 + 1}. ⇒ x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) x 2 + 1 + 2 x + x 2 + 1 − 2 x = a 2 + b 2 − ( a 2 − b 2 ) a 2 + b 2 + a 2 − b 2 ⇒ 4 x 2 ( x 2 + 1 ) = 2 b 2 2 a 2 ⇒ 2 x x 2 + 1 = b 2 a 2 ⇒ b 2 = x 2 + 1 2 a 2 x .
Hence, proved that b2 = 2 a 2 x x 2 + 1 \dfrac{2a^2x}{x^2 + 1} x 2 + 1 2 a 2 x
If x 2 + y 2 x 2 − y 2 = 2 1 8 \dfrac{x^2 + y^2}{x^2 - y^2} = 2\dfrac{1}{8} x 2 − y 2 x 2 + y 2 = 2 8 1
(i) x y \dfrac{x}{y} y x
(ii) x 3 + y 3 x 3 − y 3 \dfrac{x^3 + y^3}{x^3 - y^3} x 3 − y 3 x 3 + y 3
Answer
Given,
⇒ x 2 + y 2 x 2 − y 2 = 2 1 8 ⇒ x 2 + y 2 x 2 − y 2 = 17 8 \Rightarrow \dfrac{x^2 + y^2}{x^2 - y^2} = 2\dfrac{1}{8} \\[1em] \Rightarrow \dfrac{x^2 + y^2}{x^2 - y^2} = \dfrac{17}{8} ⇒ x 2 − y 2 x 2 + y 2 = 2 8 1 ⇒ x 2 − y 2 x 2 + y 2 = 8 17
Applying componendo and dividendo:
⇒ x 2 + y 2 + x 2 − y 2 x 2 + y 2 − ( x 2 − y 2 ) = 17 + 8 17 − 8 ⇒ 2 x 2 2 y 2 = 25 9 ⇒ x 2 y 2 = 25 9 ⇒ x y = 5 3 . \Rightarrow \dfrac{x^2 + y^2 + x^2 - y^2}{x^2 + y^2 - (x^2 - y^2)} = \dfrac{17 + 8}{17 - 8} \\[1em] \Rightarrow \dfrac{2x^2}{2y^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{5}{3}. ⇒ x 2 + y 2 − ( x 2 − y 2 ) x 2 + y 2 + x 2 − y 2 = 17 − 8 17 + 8 ⇒ 2 y 2 2 x 2 = 9 25 ⇒ y 2 x 2 = 9 25 ⇒ y x = 3 5 .
Hence, x y = 5 3 = 1 2 3 \dfrac{x}{y} = \dfrac{5}{3} = 1\dfrac{2}{3} y x = 3 5 = 1 3 2
(ii) We know that,
⇒ x y = 5 3 ⇒ x 3 y 3 = 125 27 ⇒ x 3 + y 3 x 3 − y 3 = 125 + 27 125 − 27 ⇒ x 3 + y 3 x 3 − y 3 = 152 98 ⇒ x 3 + y 3 x 3 − y 3 = 76 49 ⇒ x 3 + y 3 x 3 − y 3 = 1 27 49 . \phantom{\Rightarrow} \dfrac{x}{y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x^3}{y^3} = \dfrac{125}{27} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{125 + 27}{125 - 27} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{152}{98} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{76}{49} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = 1\dfrac{27}{49}. ⇒ y x = 3 5 ⇒ y 3 x 3 = 27 125 ⇒ x 3 − y 3 x 3 + y 3 = 125 − 27 125 + 27 ⇒ x 3 − y 3 x 3 + y 3 = 98 152 ⇒ x 3 − y 3 x 3 + y 3 = 49 76 ⇒ x 3 − y 3 x 3 + y 3 = 1 49 27 .
Hence, x 3 + y 3 x 3 − y 3 = 1 27 49 \dfrac{x^3 + y^3}{x^3 - y^3} = 1\dfrac{27}{49} x 3 − y 3 x 3 + y 3 = 1 49 27
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 + 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 + 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 − 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 ) ⇒ 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{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 ⇒ 4 2 x = 6 2 y ⇒ 2 x = 3 y ⇒ y x = 3 2 ⇒ x : y = 2 : 3.
Hence, x : y = 2 : 3.
If x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
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
Answer
Let x a = y b = z c = k \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k a x = b y = c z = k
∴ x = ak, y = bk and z = ck.
Substituting x = ak, y = bk and z = ck in L.H.S. of 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
⇒ ( a k ) 3 a 3 + ( b k ) 3 b 3 + ( c k ) 3 c 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 . \Rightarrow \dfrac{(ak)^3}{a^3} + \dfrac{(bk)^3}{b^3} + \dfrac{(ck)^3}{c^3} \\[1em] \Rightarrow \dfrac{a^3k^3}{a^3} + \dfrac{b^3k^3}{b^3} + \dfrac{c^3k^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3. ⇒ a 3 ( ak ) 3 + b 3 ( bk ) 3 + c 3 ( c k ) 3 ⇒ a 3 a 3 k 3 + b 3 b 3 k 3 + c 3 c 3 k 3 ⇒ k 3 + k 3 + k 3 ⇒ 3 k 3 .
Substituting x = ak, y = bk and z = ck in R.H.S. of 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 ( a k ) ( b k ) ( c k ) a b c ⇒ 3 a b c k 3 a b c ⇒ 3 k 3 . \Rightarrow \dfrac{3(ak)(bk)(ck)}{abc} \\[1em] \Rightarrow \dfrac{3abck^3}{abc} \\[1em] \Rightarrow 3k^3. ⇒ ab c 3 ( ak ) ( bk ) ( c k ) ⇒ ab c 3 ab c k 3 ⇒ 3 k 3 .
Since, L.H.S. = R.H.S. = 3k3
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
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^2b^2 + b^4}{b^4 + b^2c^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,
b is the mean proportion between a and c
∴ a b = b c \therefore \dfrac{a}{b} = \dfrac{b}{c} ∴ b a = c b
⇒ b2 = ac
Substituting b2 = ac in L.H.S. of the equation a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 \dfrac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^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
⇒ 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.(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}. ⇒ ( 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^2b^2 + b^4}{b^4 + b^2c^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 x and y both are positive and (2x2 - 5y2 ) : xy = 1 : 3, find x : y.
Answer
Given,
⇒ 2 x 2 − 5 y 2 x y = 1 3 ⇒ 3 ( 2 x 2 − 5 y 2 ) x y = 1 ⇒ 6 x 2 − 15 y 2 x y = 1 ⇒ 6 x y − 15 y x = 1 \Rightarrow \dfrac{2x^2 - 5y^2}{xy} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3(2x^2 - 5y^2)}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x^2 - 15y^2}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x}{y} - \dfrac{15y}{x} = 1 ⇒ x y 2 x 2 − 5 y 2 = 3 1 ⇒ x y 3 ( 2 x 2 − 5 y 2 ) = 1 ⇒ x y 6 x 2 − 15 y 2 = 1 ⇒ y 6 x − x 15 y = 1
Let x y \dfrac{x}{y} y x
⇒ 6 t − 15 t = 1 ⇒ 6 t 2 − 15 t = 1 ⇒ 6 t 2 − 15 = t ⇒ 6 t 2 − t − 15 = 0 ⇒ 6 t 2 − 10 t + 9 t − 15 = 0 ⇒ 2 t ( 3 t − 5 ) + 3 ( 3 t − 5 ) = 0 ⇒ ( 2 t + 3 ) ( 3 t − 5 ) = 0 ⇒ 2 t + 3 = 0 or 3 t − 5 = 0 ⇒ t = − 3 2 or t = 5 3 . \Rightarrow 6t - \dfrac{15}{t} = 1 \\[1em] \Rightarrow \dfrac{6t^2 - 15}{t} = 1 \\[1em] \Rightarrow 6t^2 - 15 = t \\[1em] \Rightarrow 6t^2 - t - 15 = 0 \\[1em] \Rightarrow 6t^2 - 10t + 9t - 15 = 0 \\[1em] \Rightarrow 2t(3t - 5) + 3(3t - 5) = 0 \\[1em] \Rightarrow (2t + 3)(3t - 5) = 0 \\[1em] \Rightarrow 2t + 3 = 0 \text{ or } 3t - 5 = 0 \\[1em] \Rightarrow t = -\dfrac{3}{2} \text{ or } t = \dfrac{5}{3}. ⇒ 6 t − t 15 = 1 ⇒ t 6 t 2 − 15 = 1 ⇒ 6 t 2 − 15 = t ⇒ 6 t 2 − t − 15 = 0 ⇒ 6 t 2 − 10 t + 9 t − 15 = 0 ⇒ 2 t ( 3 t − 5 ) + 3 ( 3 t − 5 ) = 0 ⇒ ( 2 t + 3 ) ( 3 t − 5 ) = 0 ⇒ 2 t + 3 = 0 or 3 t − 5 = 0 ⇒ t = − 2 3 or t = 3 5 .
Since, x and y are positive.
∴ t ≠ − 3 2 -\dfrac{3}{2} − 2 3
∴ t = x y = 5 3 \dfrac{x}{y} = \dfrac{5}{3} y x = 3 5
⇒ x : y = 5 : 3.
Hence, x : y = 5 : 3.
Find x, if 16 ( a − x a + x ) 3 = a + x a − x 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{a + x}{a - x} 16 ( a + x a − x ) 3 = a − x a + x
Answer
Given,
⇒ 16 ( a − x a + x ) 3 = a + x a − x ⇒ 16 = ( a + x ) 3 . ( a + x ) ( a − x ) 3 . ( a − x ) ⇒ ( a + x ) 4 ( a − x ) 4 = 16 ⇒ ( a + x ) 4 ( a − x ) 4 = 2 4 ⇒ a + x a − x = ± 2 \Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{a + x}{a - x} \\[1em] \Rightarrow 16 = \dfrac{(a + x)^3.(a + x)}{(a - x)^3.(a - x)} \\[1em] \Rightarrow \dfrac{(a + x)^4}{(a - x)^4} = 16 \\[1em] \Rightarrow \dfrac{(a + x)^4}{(a - x)^4} = 2^4 \\[1em] \Rightarrow \dfrac{a + x}{a - x} = \pm 2 ⇒ 16 ( a + x a − x ) 3 = a − x a + x ⇒ 16 = ( a − x ) 3 . ( a − x ) ( a + x ) 3 . ( a + x ) ⇒ ( a − x ) 4 ( a + x ) 4 = 16 ⇒ ( a − x ) 4 ( a + x ) 4 = 2 4 ⇒ a − x a + x = ± 2
In first case, let a + x a − x = 2 \dfrac{a + x}{a - x} = 2 a − x a + x = 2
Applying componendo and dividendo we get,
⇒ a + x + a − x a + x − ( a − x ) = 2 + 1 2 − 1 ⇒ 2 a 2 x = 3 1 ⇒ a x = 3 ⇒ x = a 3 . \Rightarrow \dfrac{a + x + a - x}{a + x - (a - x)} = \dfrac{2 + 1}{2 - 1} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{3}{1} \\[1em] \Rightarrow \dfrac{a}{x} = 3 \\[1em] \Rightarrow x = \dfrac{a}{3}. ⇒ a + x − ( a − x ) a + x + a − x = 2 − 1 2 + 1 ⇒ 2 x 2 a = 1 3 ⇒ x a = 3 ⇒ x = 3 a .
In second case, let a + x a − x = − 2 \dfrac{a + x}{a - x} = -2 a − x a + x = − 2
Applying componendo and dividendo we get,
⇒ a + x + a − x a + x − ( a − x ) = − 2 + 1 − 2 − 1 ⇒ 2 a 2 x = − 1 − 3 ⇒ a x = 1 3 ⇒ x = 3 a . \Rightarrow \dfrac{a + x + a - x}{a + x - (a - x)} = \dfrac{-2 + 1}{-2 - 1} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{-1}{-3} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3a. ⇒ a + x − ( a − x ) a + x + a − x = − 2 − 1 − 2 + 1 ⇒ 2 x 2 a = − 3 − 1 ⇒ x a = 3 1 ⇒ x = 3 a .
Hence, x = 3a or a 3 \dfrac{a}{3} 3 a
