Question

The difference between age of two brothers is 3 years. 5 years ago, the younger was 2/5 times the age of elder brother. Find their present age.

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Answer 1

Given :

• Difference between age of two brothers = 3 years.
• 5 years ago, the younger was 2/5 times the age of elder brother.

To find :

• Their present ages.

Solution :

Consider,

• Age of elder brother = x years
• Age of younger brother = y years

According to the 1st condition :-

• Difference between age of two brothers is 3 years

$\implies\sf{x-y=3}$

$\implies\sf{x=3+y............(1)}$

According to the 2nd condition :-

• 5 years ago, the younger was 2/5 times the age of elder brother.

5 years ago,

• Age of elder brother = (x-5) years
• Age of younger brother = (y-5) years

$\implies\sf{\dfrac{2}{5}\times\:(x-5)=(y-5)}$

$\implies\sf{2x-10=5y-25}$

$\implies\sf{2(3+y)-10=5y-25\:[put\:x=(3+y)\: from\:eq(1)]}$

$\implies\sf{6+2y-10=5y-25}$

$\implies\sf{2y-5y=-25-6+10}$

$\implies\sf{-3y=-21}$

$\implies\sf{y=7}$

Now put y = 7 in eq (1) for getting the value of x .

$\implies\sf{x=3+y}$

$\implies\sf{x=3+7}$

$\implies\sf{x=10}$

Therefore, the present age of elder brother is 10 years and the present age of younger brother is 7 years.