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
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!
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✍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
⇒
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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