## Problem 4

The father, mother, son, and daughter together have 85 years. In 15 years, the mother will have the sum of the children's ages. Knowing that the mother is 4 years younger than the father, and the son is 3 years older than his sister, find the age of each.

Note

- Effective working time 2 hours.
- All subjects are mandatory.

7 points are awarded for each correctly solved problem.

## National Mathematics Olympiad  Local stage -20.02.2016  4th Grade  Solutions and grading criteria
## Problem 4

Present
$\mathrm{T}$ $\qquad$ I 4 I
M I $\qquad$ I
Son I $\qquad$ | 3 |
Daughter $\mathrm{I}$ $\qquad$ I

In 15 years:

```
Son + daughter 1 \(\qquad\) \(\qquad\)
I I
I 3 I \(\qquad\)
M I \(\qquad\)
\(1 \quad 15\)
\(\mathrm{T}\) I \(\qquad\)
15
I 4 I ...............1p
\(85+4\) x 15 = 145 years (sum of ages in 15 years)
\(145-4\) = 141 years
\(141: 3\) = 47 years (mother's age in 15 years / sum of the ages of the 2 children in 15
years)
\(1 \mathrm{p}\)
\(47-3=44\) years
\(44: 2\) = 22 years (daughter's age in 15 years)
\(22+3\) = 25 years (son's age in 15 years)
In the present:

![](https://cdn.mathpix.com/cropped/2024_06_07_96faebe7bcd5e4245112g-4.jpg?height=514&width=1674&top_left_y=214&top_left_x=260)

![](https://cdn.mathpix.com/cropped/2024_06_07_96faebe7bcd5e4245112g-4.jpg?height=46&width=1567&top_left_y=233&top_left_x=282)

\(25-15=10\) years - son \(\quad\).......................................................................................1p

\(47-15\) = 32 years - mother..................................................................................1p

32 + 4 = 36 years - father.........................................................................................1p
```