Question

Two years ago father was three times as old as his son two years hence twice his age will be equal to five times that of his son find their ages?i want to take father's age be x please

Answer 1

Let age of son be x
father's age two yeas back was 3(x-2) and now is 3(x-2)+2
Fathers age two years hence = 5(x+2)/2 and now is 5(x+2) -2
Hence 3x-6+2=((5x+10)/2) -2
⇒ 6x - 8=5x+10-4
⇒ x=14
son's age is 14 and fathers age is 38

Answer 2

Hii....☺

Here is your answer.....☺

===============================

→ let father's age = x
And age of his son be "y"

Two years ago,

Age of father = ( x - 2 ) yrs
Age of son = (y - 2) yrs

According to question we have,

→ x - 2 = 3( y - 2 )
→ x - 2 = 3y - 6
→ x - 3y = - 6 + 2
→ x - 3y = -4
→ x = 3y -4 ----------------(1)

Also we have,

Two years hence,
Age of father = ( x + 2 ) yrs
Age of son = ( y + 2) yrs

According to question we have,

→ 2(x+2) = 5(y+2)
→ 2x + 4 = 5y + 10
→ 2x - 5y = 10 - 4
→ 2x - 5y = 6 --------------(2)

Putting value of x in equation (2) we get,

→ 2(3y - 4) - 5y =6
→ 6y -8 - 5y = 6
→ y - 8 = 6
→ y = 8 + 6 = 14

Putting value of y in equation (1) we get,

→ x = 3y - 4
→ x = 3 × 14 - 4
→ x = 42 - 4
→ x = 38

===============================

Hence, age of father = x = 38 yrs.
and,
Age of his son = y = 14 yrs.

================================

Hope this helps you....☺

Thanks....☺