Question

paul's ages four times the sum of the ages of his two daughters. four years from now, the difference between his age and his daughter's ages taken together will be thirty two years. and his present age.

Answer 1

Daughters' age  = x + y

Paul's age = 4(x+y)

After 4 years daughters' age = x + 4 + y + 4 = x + y + 8

after 4 years Pauls age = 4(x+y) + 4

(4x + 4y + 4) - (x + y +8) = 32

3x +3y - 4 = 32

3x + 3y = 36

x + y = 12

Paul's age  = 4(x+y) = 4 x 12 = 48years