Math, asked by Kehkasha8368, 1 year ago

The present ages of a & b are in the ratio of 9:7. 13 years ago their ages were in the ratio 25:18. What is the difference of their ages ?

Answers

Answered by Anonymous
21

Given :-

the present age of a and b are in ratio 9 : 7

let their ages be 9x and 7x respectively.

ATQ,

13 years ago, their ages were in ratio 25 : 18

therefore (9x - 13)/(7x - 13) = 25/18

by cross multiplying, we get

➡ 18(9x - 13) = 25(7x - 13)

➡ 162x - 234 = 175x - 325

➡ 162x - 175x = -325 + 234

➡ -13x = -91

➡ x = -91/-13 = 7 yrs

The present age of

  • a = 9x = 9 × 7 = 63 years

  • b = 7x = 7 × 7 = 49 years

hence,

difference between their ages is = 63 - 49

= 14years final answer!

Answered by BrainlyWriter
19

♊♊YOUR ANSWERS♍♍

✍14 years

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EXPLANATION —

Since the present age of a and b are given in ratio 9:7

Let their present ages are 9x and 7x res.

A/q

The ratio of their age before 13 years ago is 25:18

 \frac{9x - 13}{7x - 13}  =  \frac{25}{18}

Crossmultiplying

⇒18(9x - 13) = 25(7x - 13)

⇒162x - 234 = 175x - 325

⇒13x = 91

On solving

⇒x = 7 years

Therefore there present ages are

⇒a = 9x = 9×7 = 63 years

⇒b = 7x = 7×7 = 49 years

Difference of there current ages = 63 - 49 = 14 years

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