two years ago rehan was three times as old as his son and two years hence twice his age will be equal to five times that of his son. find the sum of their present ages
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Let Rehan's present age = x
his son's present age = y
two years ago,
rehan's age = x-2
his son's age = y-2
so x-2 = 3(y-2)
⇒ x = 3y - 6 + 2
⇒ x = 3y - 4
two years hence,
rehan's age = x+2
his son's age = y+2
so 2(x+2) = 5(y+2)
⇒ 2x + 4 = 5y +10
⇒ 2(3y-4) = 5y + 10
⇒ 6y - 8 = 5y + 10
⇒ y = 18
x = 3*18 - 4 = 50
Sum of their ages = 50+18 = 68
his son's present age = y
two years ago,
rehan's age = x-2
his son's age = y-2
so x-2 = 3(y-2)
⇒ x = 3y - 6 + 2
⇒ x = 3y - 4
two years hence,
rehan's age = x+2
his son's age = y+2
so 2(x+2) = 5(y+2)
⇒ 2x + 4 = 5y +10
⇒ 2(3y-4) = 5y + 10
⇒ 6y - 8 = 5y + 10
⇒ y = 18
x = 3*18 - 4 = 50
Sum of their ages = 50+18 = 68
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