Question

Two years ago Bunty was three times as old as his son and twoyear hence twice his age will be equalfive times that of his son . Find their present ages

Answer 1

Let the age of bunty be x
And his son's age be y
ATQ-
2 years ago-
Bunty's age= x-2
His son's age=y-2
Therefore, x-2 = 3(y-2)
= x-2-3y+6
x-3y+4
x-3y= -4 ( equation 1)

Two years later,
Bunty's age= x+2
His son's age= y+2
Now, 2(x+2)= 5(y+2)
= 2x+4-5y-10
2x-5y-6
2x-5y=6 (equation 2)
Multiplying equation 1 by 2,
2x-6y=-8 (equation 3)
Eliminating 1 and 3,
2x-6y=-8
2x-5y=6
(-) (+) (-)
-y=-14
(y=14)
Putting y=14 in equation 2,
2x-5y=6
2x-5(14)=6
2x-70=6
2x=6+70
2x=76
x=76/2
(x=38)
Hence, bunty's age =38 years
And his son's age= 14 years
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Answer 2

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Answer :son's present age=14yrs

Father's present age=38yrs

Explanation in pic..