Question

12. A man's age is three times the sum of the ages of his
two sons, one of whom is twice as old as the other.
In 4 years, the sum of the sons' ages will be half of
their father's age. Calculate their present ages.​

Answer 1

Step-by-step explanation:

let age of man be x and his son's ages be y

and z

according to the question

x = 3(y+z) .......... ......(1)

and

y = 2z................... ....(2)

again

(x+4)/2 = y+z+8........(3)

from (1) and (2)

x= 9z..........................(4)

from (2) and (3)

(x+4)/2 = 3z+8

=> x+4 = 6z+16

=> x = 6z+12.............(5)

from (4) and (5)

9z = 6z+12

=> 3z = 12

=> z = 4 years

substituting the value of z in (2)

y = 2×4 = 8

=> y = 8 years

from (1)

=> x = 3(4+8)

=> x = 36 years

therefore age of man = 36 years

age of younger son = 4 years

age of elder son = 8 years

Hope this helps you