Question

Sum of ages of two brothers is 43 years. After 11 years the age of elder brother will be 7/6
times the age of the younger brother then. Find their present ages. (Ans.: 24yrs, 19yrs.)​

Answer 1

Step-by-step explanation:

answer is in the attachment

Answer 2

Step-by-step explanation:

Answer:

Let the Age of Elder Brother be n and of Younger Brother (43 - n).

$\underline{\bigstar\:\textsf{According to the given Question :}}$

After 11 years the age of elder brother will be 7/6 times the age of the younger brother.

$:\implies\sf Elder\:Brother=\dfrac{7}{6} \times Younger\:Brother\\\\\\:\implies\sf (n + 11) = \dfrac{7}{6} \times (43 - n + 11)\\\\\\:\implies\sf 6(n + 11) = 7(54 - n)\\\\\\:\implies\sf 6n + 66 = 378 - 7n\\\\\\:\implies\sf 6n + 7n = 378 - 66\\\\\\:\implies\sf 13n = 312\\\\\\:\implies\sf n = \dfrac{312}{13}\\\\\\:\implies\sf n = 24$

⠀

$\underline{\bigstar\:\textsf{Required Present Age :}}$

$\bullet\:\:\textsf{Elder Brother = n = \textbf{24 years}}\\\bullet\:\:\textsf{Younger Brother = (43 - n) = \textbf{19 years}}$