A Math Problem: Family’s Age

This year, sum of the family ‘s age is 100. There are Dad, Mum, Brother, and Sister in the family. And we know:

• Brother is 8 years younger than Sister.
• Mum is 2 years younger than Dad.
• 10 Years ago, the sum of the family’s age was 65.

So, today, how old is everybody in the family?

Solution

Let’s represent the ages of everybody using the following letters:

B – brother
D – dad
M – mum
S – sister

So, B + D + M + S = 100
B + 8 = S
M + 2 = D

Ten years ago, the age difference should be -40 (four people each 10 years younger), however, the difference is only 35 (100 – 65), that means little brother was not born yet 10 years ago.

The actual difference is (65 – 60) = 5, which mean that little brother was born 5 years ago – therefore, the sum was 5 years less than expected 10 years ago.

Therefore,
B = 5
S = 13
B + S = 5 + 13 = 18
Then, D + M = 100 – (B + S) = 82
We know D – M = 2

Adding both sides:
2 * D + (M – M) = 82 + 2
D = 42
M = 40

Let’s verify the answer:
This year, D + M + B + S = 42 + 40 + 5 + 13 = 100
D – M = 42 – 40 = 2
S – B = 13 – 5 = 8

10 years ago, B = 0 (brother was not born), and
(D – 10) + (M – 10) + (S – 10) + 0 = 32 + 30 + 3 = 65

Bingo!

–EOF (The Ultimate Computing & Technology Blog) —

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