The sum of the ages of two brothers, having a | vedclass

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(D) Let the present age of the elder brother be $A$ and the younger brother be $B$.Given that the difference between their ages is $8 \, 	ext{years}$

Vedclass · Accepted Answer

$3:7$

Vedclass · Answer

(D) Let the present age of the elder brother be $A$ and the younger brother be $B$.Given that the difference between their ages is $8 \, 	ext{years}$, so $A - B = 8$ (Equation $1$).After $10 \, 	ext{years}$, the age of the elder brother will be $(A + 10)$ and the younger brother will be $(B + 10)$.The sum of their ages after $10 \, 	ext{years}$ will be double their present sum $(A + B)$.So, $(A + 10) + (B + 10) = 2(A + B)$.$A + B + 20 = 2A + 2B$.$A + B = 20$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(A - B) + (A + B) = 8 + 20 \Rightarrow 2A = 28 \Rightarrow A = 14$.Substituting $A = 14$ in Equation $2$: $14 + B = 20 \Rightarrow B = 6$.The ratio of the younger brother's age to the elder brother's age is $B : A = 6 : 14 = 3 : 7$.

The sum of the ages of two brothers, having a difference of $8 \, \text{years}$ between them, will double after $10 \, \text{years}$. What is the ratio of the age of the younger brother to that of the elder brother?

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