Latest Book Aptitude Question SOLUTION: An old man distributed all the coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference be

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(#M40009748) BOOK question Keep an EYE

An old man distributed all the coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference between the two numbers. How many coins did the old man have?

Asked In Book Devendra Marghade (11 years ago)

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Sanjana Patnaik  (11 years ago)

Given a^2-b^2=36(a-b)
Solving this we get a+b=36
Thus, the old man had 36 coins

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Aman  (11 years ago)

36 coins

If coins are divided as x and y, then
x^2-y^2 = 36(x-y)
(x+y)*(x-y)=36*(x-y)
x+y=36

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chaitanya  (11 years ago)

Let x and y
be the number of gold coins the two sons received. Since we are given that the differencebetween the squares of the two numbers is 36 times the difference between the two numbers, we have the equation


(x)2-(y)2= 36(x–y)(x–y)(x+y)
= 36(x–y)
by the Difference of Squares formula
a2–b2=(a–b)(a+b)

x+y = 36
by canceling (x–y) from both sidesHence,
the total number of gold coins the old man had is

x+ y which equals 36.

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