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
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
If a : b :: c : d then prove that:
a : (a - b) :: c : (c - d).
Answer
Given,
a : b :: c : d
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} ⇒ b a = d c
If the original ratios are equal, their reciprocals are also equal (this is known as Invertendo) :
⇒ b a = d c \Rightarrow \dfrac{b}{a} = \dfrac{d}{c} ⇒ a b = c d
Applying dividendo, we get:
⇒ b − a a = d − c c ⇒ − a − b a = − c − d c ⇒ a − b a = c − d c Now, take the reciprocal ∴ a a − b = c c − d \Rightarrow \dfrac{b - a}{a} = \dfrac{d - c}{c} \\[1em] \Rightarrow -\dfrac{a - b}{a} = -\dfrac{c - d}{c} \\[1em] \Rightarrow \dfrac{a - b}{a} = \dfrac{c - d}{c} \\[1em] \text{ Now, take the reciprocal} \\[1em] \therefore \dfrac{a}{a - b} = \dfrac{c}{c - d} ⇒ a b − a = c d − c ⇒ − a a − b = − c c − d ⇒ a a − b = c c − d Now, take the reciprocal ∴ a − b a = c − d c
Hence, proved that a : (a - b) :: c : (c - d).
Using properties of proportion solve:
k 5 + x 5 x 5 − k 5 = 122 121 \dfrac{k^5 + x^5}{x^5 - k^5} = \dfrac{122}{121} x 5 − k 5 k 5 + x 5 = 121 122
Answer
Given,
k 5 + x 5 x 5 − k 5 = 122 121 \dfrac{k^5 + x^5}{x^5 - k^5} = \dfrac{122}{121} x 5 − k 5 k 5 + x 5 = 121 122
Applying componendo and dividendo we get,
⇒ ( k 5 + x 5 ) + ( x 5 − k 5 ) ( k 5 + x 5 ) − ( x 5 − k 5 ) = 122 + 121 122 − 121 ⇒ k 5 + x 5 + x 5 − k 5 k 5 + x 5 − x 5 + k 5 = 243 1 ⇒ 2 x 5 2 k 5 = 243 1 ⇒ ( x k ) 5 = ( 3 ) 5 ⇒ x k = 3 ⇒ x = 3 k . \Rightarrow \dfrac{(k^5 + x^5) + (x^5 - k^5)}{(k^5 + x^5) - (x^5 - k^5)} = \dfrac{122 + 121}{122 - 121} \\[1em] \Rightarrow \dfrac{k^5 + x^5 + x^5 - k^5}{k^5 + x^5 - x^5 + k^5} = \dfrac{243}{1} \\[1em] \Rightarrow \dfrac{2x^5}{2k^5} = \dfrac{243}{1} \\[1em] \Rightarrow \Big(\dfrac{x}{k}\Big)^5 = (3)^5 \\[1em] \Rightarrow \dfrac{x}{k} = 3 \\[1em] \Rightarrow x = 3k. ⇒ ( k 5 + x 5 ) − ( x 5 − k 5 ) ( k 5 + x 5 ) + ( x 5 − k 5 ) = 122 − 121 122 + 121 ⇒ k 5 + x 5 − x 5 + k 5 k 5 + x 5 + x 5 − k 5 = 1 243 ⇒ 2 k 5 2 x 5 = 1 243 ⇒ ( k x ) 5 = ( 3 ) 5 ⇒ k x = 3 ⇒ x = 3 k .
Hence, x = 3k.
Using properties of proportion solve:
x 2 − x + 1 x 2 + x + 1 = 112 ( 1 − x ) 104 ( 1 + x ) \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{112(1 - x)}{104(1 + x)} x 2 + x + 1 x 2 − x + 1 = 104 ( 1 + x ) 112 ( 1 − x )
Answer
Given,
⇒ x 2 − x + 1 x 2 + x + 1 = 112 ( 1 − x ) 104 ( 1 + x ) ⇒ x 2 − x + 1 x 2 + x + 1 = 14 ( 1 − x ) 13 ( 1 + x ) . \Rightarrow \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{112(1 - x)}{104(1 + x)} \\[1em] \Rightarrow \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{14(1 - x)}{13(1 + x)}. ⇒ x 2 + x + 1 x 2 − x + 1 = 104 ( 1 + x ) 112 ( 1 − x ) ⇒ x 2 + x + 1 x 2 − x + 1 = 13 ( 1 + x ) 14 ( 1 − x ) .
Applying componendo and dividendo we get,
⇒ ( x 2 − x + 1 ) + ( x 2 + x + 1 ) ( x 2 − x + 1 ) − ( x 2 + x + 1 ) = 14 ( 1 − x ) + 13 ( 1 + x ) 14 ( 1 − x ) − 13 ( 1 + x ) ⇒ x 2 − x + 1 + x 2 + x + 1 x 2 − x + 1 − x 2 − x − 1 = 14 − 14 x + 13 + 13 x 14 − 14 x − 13 − 13 x ⇒ 2 x 2 + 2 − 2 x = 27 − x 1 − 27 x ⇒ 2 ( x 2 + 1 ) − 2 x = 27 − x 1 − 27 x ⇒ x 2 + 1 − x = 27 − x 1 − 27 x \Rightarrow \dfrac{(x^2 - x + 1) + (x^2 + x + 1)}{(x^2 - x + 1) - (x^2 + x + 1)} = \dfrac{14(1 - x) + 13(1 + x)}{14(1 - x) - 13(1 + x)} \\[1em] \Rightarrow \dfrac{x^2 - x + 1 + x^2 + x + 1}{x^2 - x + 1 - x^2 - x - 1} = \dfrac{14 − 14x + 13 + 13x}{14 − 14x − 13 − 13x} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{-2x} = \dfrac{27 − x}{1 − 27x} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{-2x} = \dfrac{27 - x}{1 - 27x} \\[1em] \Rightarrow \dfrac{x^2 + 1}{-x} = \dfrac{27 - x}{1 - 27x} ⇒ ( x 2 − x + 1 ) − ( x 2 + x + 1 ) ( x 2 − x + 1 ) + ( x 2 + x + 1 ) = 14 ( 1 − x ) − 13 ( 1 + x ) 14 ( 1 − x ) + 13 ( 1 + x ) ⇒ x 2 − x + 1 − x 2 − x − 1 x 2 − x + 1 + x 2 + x + 1 = 14 − 14 x − 13 − 13 x 14 − 14 x + 13 + 13 x ⇒ − 2 x 2 x 2 + 2 = 1 − 27 x 27 − x ⇒ − 2 x 2 ( x 2 + 1 ) = 1 − 27 x 27 − x ⇒ − x x 2 + 1 = 1 − 27 x 27 − x
⇒ (x2 + 1)(1 − 27x) = −x(27 − x)
⇒ x2 + 1 − 27x3 − 27x = −27x + x2
⇒ −27x3 + 1 = 0
⇒ 27x3 = 1
⇒ x3 = 1 27 \dfrac{1}{27} 27 1
⇒ x = 1 27 3 = 1 3 \sqrt[3]{\dfrac{1}{27}} = \dfrac{1}{3} 3 27 1 = 3 1
Hence, x = 1 3 \dfrac{1}{3} 3 1
If x ≠ y and x 2 − x + 1 y 2 − y + 1 = x 2 + x + 1 y 2 + y + 1 \dfrac{x^2 - x + 1}{y^2 - y + 1} = \dfrac{x^2 + x + 1}{y^2 + y + 1} y 2 − y + 1 x 2 − x + 1 = y 2 + y + 1 x 2 + x + 1
Answer
Given,
x 2 − x + 1 y 2 − y + 1 = x 2 + x + 1 y 2 + y + 1 \dfrac{x^2 - x + 1}{y^2 - y + 1} = \dfrac{x^2 + x + 1}{y^2 + y + 1} y 2 − y + 1 x 2 − x + 1 = y 2 + y + 1 x 2 + x + 1
Applying alternendo we get,
x 2 − x + 1 x 2 + x + 1 = y 2 − y + 1 y 2 + y + 1 \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{y^2 - y + 1}{y^2 + y + 1} x 2 + x + 1 x 2 − x + 1 = y 2 + y + 1 y 2 − y + 1
Applying componendo and dividendo we get,
⇒ ( x 2 − x + 1 ) + ( x 2 + x + 1 ) ( x 2 − x + 1 ) − ( x 2 + x + 1 ) = ( y 2 − y + 1 ) + ( y 2 + y + 1 ) ( y 2 − y + 1 ) − ( y 2 + y + 1 ) ⇒ x 2 − x + 1 + x 2 + x + 1 x 2 − x + 1 − x 2 − x − 1 = y 2 − y + 1 + y 2 + y + 1 y 2 − y + 1 − y 2 − y − 1 ⇒ 2 x 2 + 2 − 2 x = 2 y 2 + 2 − 2 y ⇒ 2 ( x 2 + 1 ) − 2 x = 2 ( y 2 + 1 ) − 2 y ⇒ x 2 + 1 x = y 2 + 1 y \Rightarrow \dfrac{(x^2 − x + 1) + (x^2 + x + 1)}{(x^2 − x + 1) − (x^2 + x + 1)} = \dfrac{(y^2 - y + 1)+ (y^2 + y + 1)}{(y^2 - y + 1) - (y^2 + y + 1)} \\[1em] \Rightarrow \dfrac{x^2 − x + 1 + x^2 + x + 1}{x^2 − x + 1 − x^2 - x - 1} = \dfrac{y^2 - y + 1 + y^2 + y + 1}{y^2 - y + 1 - y^2 - y - 1} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{-2x} = \dfrac{2y^2 + 2}{-2y} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{-2x} = \dfrac{2(y^2 + 1)}{-2y} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{y^2 + 1}{y} \\[1em] ⇒ ( x 2 − x + 1 ) − ( x 2 + x + 1 ) ( x 2 − x + 1 ) + ( x 2 + x + 1 ) = ( y 2 − y + 1 ) − ( y 2 + y + 1 ) ( y 2 − y + 1 ) + ( y 2 + y + 1 ) ⇒ x 2 − x + 1 − x 2 − x − 1 x 2 − x + 1 + x 2 + x + 1 = y 2 − y + 1 − y 2 − y − 1 y 2 − y + 1 + y 2 + y + 1 ⇒ − 2 x 2 x 2 + 2 = − 2 y 2 y 2 + 2 ⇒ − 2 x 2 ( x 2 + 1 ) = − 2 y 2 ( y 2 + 1 ) ⇒ x x 2 + 1 = y y 2 + 1
⇒ xy2 + x = x2 y + y
⇒ xy(y - x) - (y - x) = 0
⇒ (y − x)(xy − 1) = 0
⇒ xy = 1
Hence, proved that xy = 1.
