Question

The sum of the ages of three children is 32. The age of the oldest is twice the age of the youngest. The two older children differ by three years. What is the age of the youngest child?

Answer 1

Answer:-

The youngest child is 7 years old.

Given:-

• The sum of the ages of three children is 32. The age of the oldest is twice the age of the youngest. The two older children differ by three years. What is the age of the youngest child?

Find:-

• What is the age of the youngest child?

Solution:-

Let x represent the age of the youngest child.

Let y represent the age of the middle child.

Let z represent the age of the oldest child.

The sum of the ages of three children is 32. It means that

x + y + z = 32 - - - - - - - - - - - 1

The ages of the oldest is twice the age of the youngest. It means that

x = 2z

z = x/2

The two older children differ by 3 years. It means that

y = x - 3

Substituting z = x/2 and y = x - 3 into equation 1, it becomes

x + x - 3 + x/2 = 32

2x - 3 + x/2 = 32

Crossmultiplying,

4x - 6 + x = 64

5x = 64 + 6 = 70

x = 70/5 = 14

z = x/2 = 14/2

z = 7

Answer 2

Answer:

The youngest child is x years, and the oldest is 2x. The difference between the two older children is three years, thus 2x-3.

Now put it together:

x (youngest) + 2x (oldest) + 2x-3 (differene between the two oldest) hence:

x + 2x + 2x - 3 = 32 (next simplify)

5 x - 3 = 32 (next add 3 to each side)

5x = 35 (next divide both sides by 5)

x = 7

The youngest child is 7 (x), the oldest is 14 (2x) and the middle one is 11 (2x - 3).

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