Question

The ages of A and B are in the ratio 5 : 7. Eight years ago, their ages were in the ratio 7 : 13 . Find their present ages.​

Answer 1

Step-by-step explanation:

Let the Present ages of A and B be 5x and 7x respectively.

$\large\implies{\sf }$ A's age 8 years ago = ( 5x- 8) years .

$\large\implies{\sf }$ B's age 8 years ago = (7x- 8) years.

A/q

$\large{\sf {\frac{5x - 8}{7x -8}}}$ $\large\implies{\sf }$ 13(5x-8) = 7(7x-8)

$\large\implies{\sf }$ 65x - 104 =49x -6

$\large\implies{\sf }$ 65x - 49x = 104-56

$\large\implies{\sf }$16x=48

$\large\implies{\sf }$ X= $\large{\sf {\frac{48}{16}}}$ =3

So, A's Present age = (5x3) =15 Years.

B's Present age = (7x3) =21 Years.

Answer 2

$\bold {\pink{\green {\sf\underline{Step\:by\:step\:explanation:-{\boxed {}}}}}}$

Present ages of A and B be 5x and 7x respectively.

$⟹ \blue{A's\:age\:8\:years\:ago} = \blue{(5x-8)\:Years}$

$⟹ \blue{B's\:age\:8\:years\:ago} = \blue{(7x-8)\:Years}$

According to question,

$\blue{} \frac{5x-8}{7x-8}⟹ {13(5x-8)=7(6x-8) }$

$⟹ \blue{65x - 104} = {49x-6}$

$⟹ \blue{65x - 49x} = {104-56}$

$⟹ \blue{16x} = {48}$

$⟹ \blue{X} = \frac{48}{16} =\red {3}$

So, A's Present age = (5×3) = $\bold{15\:Years.}$

B's Present age = (7×3) = $\bold{21\:Years.}$