Question

Five years ago,a man was seven times as old as his son , while five years hence,the man will be four times as old as his son. Find their percent age?

Answer 1

Answer:

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The man's present age is: 75 years.

His son's present age is: 15 years.

Let the man's present age be 'X' years.

Let his son's present age be 'Y' years.

Five years ago,the man's age was:(X-5) years.

Five years ago his son's age was:(Y-5) years.

First equation is:-

(X-5)=7(Y-5)

or, X-5 =7Y-35

or, X=7Y-35+5

So, X =7Y-30 ......................(i)

Five years after the man's age will be:(X+5) years.

Five years after his son's age will be:(Y+5) years.

Second equation is:-

(X+5)=4(Y+5)

or, X+5=4Y+20

or, X =4Y+20-5

So, X =4Y+15................(ii)

By substituting the value of (i) equation here we get,

7Y-30=4y+15

or, 7Y-4Y=15+30

or, 3Y =45

or, Y =45/3

So, Y =15

So, Son's present age is: 15

years

By substituting the value of Y in the (i) equation we get,

X =7×15-30

or, X =105-30

So, X =75

So, the man's present age

is:75 years.

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