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​

Answer 1

GIVEN :–

• 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.

TO FIND :–

• Their present ages = ?

SOLUTION :–

• Let the age of younger brother = x

• And the age of another brother = y

• According to the first condition –

$\\ \implies\bf x + y = 43$

$\\ \implies\bf x = 43-y \:\:\:\: ------eq.(1)$

• According to the second condition –

$\\ \implies\bf (y+ 11)= \dfrac{7}{6} (x+ 11)$

$\\ \implies\bf 6(y + 11)=7(x+ 11)$

$\\ \implies\bf 6y+66=7x + 77$

$\\ \implies\bf 6y=7x + 77 - 66$

$\\ \implies\bf 6y=7x +11$

• Using eq.(1) –

$\\ \implies\bf 6y=7(43-y)+11$

$\\ \implies\bf 6y=301-7y +11$

$\\ \implies\bf 6y+7y=312$

$\\ \implies\bf 13y=312$

$\\ \implies\bf y = \cancel\dfrac{312}{13}$

$\\ \large\implies{ \boxed{\bf y =24}}$

• Using eq.(1) –

$\\ \implies\bf x = 43-24$

$\\ \large\implies{ \boxed{\bf x=19}}$

• Hence, The Age of younger brother is 19.

• And The Age of another brother is 24.

Answer 2

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.

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

⠀

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

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