If x b + c - a = y c + a - b = z a + b - c , prove that each ratio is equal to x + y + z a + b + c .

It is given that, x b + c - a = y c + a - b = z a + b - c = k ( say ) ⇒ x = k b + c - a , y = k c + a - b , z = k a + b - c L.H.S.: Each ratio = k R.H.S.: x + y + z a + b + c = k b + c - a + k c + a - b + k a + b - c a + b + c = k b + c - a + c + a - b + a + b - c a + b + c = k a + b + c a + b + c = k Hence, L.H.S. = R.H.S

If x = p a b a + b , Prove that x + p a x - p a - x + p b x - p b = 2 a 2 - b 2 a b .

Given, x = p a b a + b Now, x p a = b a + b Applying componendo and dividendo, x + p a x - p a = b + a + b b - a - b = a + 2 b - a (i) Also, x p b = a a + b Applying componendo and dividendo, x + p b x - p b = a + a + b a - a - b = 2 a + b - b (ii) Subract (ii) from (i), we get x + p a x - p a - x + p b x - p b = a + 2 b - a - 2 a + b - b = a + 2 b - a + 2 a + b b = a b + 2 b 2 - 2 a 2 - a b - a b = 2 b 2 - 2 a 2 - a b = - 2 ( a 2 - b 2 ) - a b = 2 a 2 - b 2 a b = R.H.S. Hence Proved

Find x from the equation : a + x + a 2 - x 2 a + x - a 2 - x 2 = b x .

It is given that a + x + a 2 - x 2 a + x - a 2 - x 2 = b x By applying componendo and dividendo: a + x + a 2 - x 2 + a + x - a 2 - x 2 a + x + a 2 - x 2 - a - x + a 2 - x 2 = b + x b - x ⇒ 2 ( a + x ) 2 a 2 - x 2 = b + x b - x ⇒ ( a + x ) a 2 - x 2 = b + x b - x By squaring on both sides ( a + x ) 2 a 2 - x 2 = ( b + x ) 2 ( b - x ) 2 ⇒ ( a + x ) 2 ( a + x ) ( a - x ) = ( b + x ) 2 ( b - x ) 2 ⇒ a + x a - x = ( b + x ) 2 ( b - x ) 2 Again applying componendo and dividendo: a + x + a - x a + x - a + x = ( b + x ) 2 + ( b - x ) 2 ( b + x ) 2 - ( b - x ) 2 ⇒ 2 a 2 x = 2 b 2 + x 2 4 b x ⇒ a x = b 2 + x 2 2 b x By cross multiplication, 2 a b x = x b 2 + x 2 ⇒ 2 a b = b 2 + x 2 ⇒ x 2 = 2 a b - b 2 ⇒ x = 2 a b - b 2

If x = a + 1 3 + a - 1 3 a + 1 3 - a - 1 3 , prove that x 3 - 3 a x 2 + 3 x - a = 0 .

It is given that, x = a + 1 3 + a - 1 3 a + 1 3 - a - 1 3 Applying componendo and dividendo, we get x + 1 x - 1 = a + 1 3 + a - 1 3 + a + 1 3 - a - 1 3 a + 1 3 - a - 1 3 - a + 1 3 + a - 1 3 ⇒ x + 1 x - 1 = 2 a + 1 3 2 a - 1 3 ⇒ x + 1 x - 1 = a + 1 3 a - 1 3 Take cube on both sides, we get: ( x + 1 ) 3 ( x - 1 ) 3 = a + 1 a - 1 Again applying componendo and dividendo, ( x + 1 ) 3 + ( x - 1 ) 3 ( x + 1 ) 3 - ( x - 1 ) 3 = a + 1 + a - 1 a + 1 - a + 1 ⇒ 2 x 3 + 3 x 2 3 x 2 + 1 = 2 a 2 ⇒ x 3 + 3 x 3 x 2 + 1 = a 1 By cross multiplication, x 3 + 3 x = 3 a x 2 + a x 3 - 3 a x 2 + 3 x - a = 0 Hence proved.

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Understanding ICSE Mathematics Class 10>Algebra>Chapter 7 - Ratio and Proportion>Chapter Test

M L Aggarwal Solutions for Exercise 5: Chapter Test

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Exercise 5: Chapter Test

Attempt the practice questions from Exercise 5: Chapter Test with hints and solutions to strengthen your understanding. Understanding ICSE Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.