Question

The ages of two persons are in the ratio 5 : 7 Eighteen years ago their ages were in the ratio 8 : 13 . Find their present ages .​

Answer 1

        $\text{It is given that the ages of two people are in the ratio }5:7$

        $\text{Let the common factor be }x$

$\implies \text{Their present ages are }5x \text{ and }7x$

        $\text{18 years ago the ratio was }8:13$

$\implies \: \dfrac{5x-18}{7x-18} = \dfrac{8}{13}$

$\implies \: 13(5x-18) = 8(7x-18)$

$\implies \: 65x-270 = 56x-144$

$\implies \: 65x - 56x = 270 - 144$

$\implies \: 9x = 126$

$\implies \: x = 14$

$\implies \boxed{\text{Their present ages are }5x = 5\times 14 = 70 \text{ years and }7x = 7 \times 14 = 98 \text{ years}}$

Answer 2

Step-by-step explanation:

Let the common ratio be x

Their present age are 5x and 7x

Their ages 18 years ago are 5x-18 and 7x-18

5x-18/7x-18= 8/13

13(5x-18)= 8(7x-18)

65x-234= 56x-144

65x-56x= -144+234

9x=90

x= 10

5*10= 50

7*10= 70

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