Question

A mother who is 40 years old has a daughter and a son. The son is twice as old as the daugther. In 15 years the sum of all their ages will be 100 years. How old are the siblings now?

Answer 1

Explanation:

Let's call the ages D for daughter and S for son.

We know that the son is twice as old as the daughter, so:

S = 2D

We also know that in 15 years, their ages add up to 100, so:

(40+15) + (S+15) + (D+15) = 100

55 + S + 15 + D + 15 = 100

85 + S + D = 100

S + D = 15

Substituting the first equation:

2D + D = 15

3D = 15

D = 5

Therefore:

S = 2D = 10

The son is 10 and the daughter is 5.

Answer 2

son = 10

daughter = 5

Explanation:

Let the daughter = d

Let the son = s

s = 2*d

there ages in 15 years

Mother = 40 + 15 = 55

Son = s + 15

daughter = d + 15

Total: s + 15 + d+15 + 55 = 100 Combine the like terms.

s + d + 85 = 100 Subtract 85 from both sides.

s + d = 100 - 85

s + d = 15

s = 2*d Substitute for son

2d + d = 15

3d = 15

d = 15/3

d = 5

son = 2*5

son = 10

Check

son = 15 = 25

daughter + 15 = 20

Mother + 15 = 55

Total 100 just as it should be.