Question

a woman's son is two years older than her daughter her age is three times the sum of Ages of children after 5 years she will be 47 years old find the present ages of the children​

Answer 1

let day daughter age = D years

Son age = D + 2 years

woman age = 3 × (son + daughter age)

woman age = 3×(D+2 + D)

= 3 ×(2D +2)

= 6D + 6

after 5 years woman age will be

6D + 6 + 5= 6D + 11 years

6D + 11 = 47

6D = 36

D = 6

Age of daughter = 6 years

Age of Son = 6+2 = 8 Years

Answer 2

Solution:-

• The present age of the woman is 47 - 5 = 42.

• Since the woman's age is 3 times the sum of the ages of the children,

• This sum is 42/3 = 14.

• So for children ages you have two equations:

=> s + d = 14,

=> s - d = 2.

• Every time when you see two equations like these, add them.

• You will get

=> 2s = 14 + 2,

=> 2s = 16,

=> s = 16/2 = 8.

Thus son is 8 years now.

The daughter is 8 - 2 = 6 years.

i hope it helps you.