A father is 30 years older than his son. In 5 years, the father will be three times as old as his son. Find the son's present age.

Here are the solutions to questions 3, 4, and 5:

3. Engineer and Assistant Wages

Step 1: Define variables and set up equations.
Let $E$ be the daily wage of the engineer and $A$ be the daily wage of the assistant.
From the first week's work:

$$6E + 7A = 1,350,000 \quad (1)$$

From the second week's work:

$$5E + 3A = 870,000 \quad (2)$$

Step 2: Solve the system of equations using elimination.
Multiply equation (1) by 3 and equation (2) by 7 to eliminate $A$:

$$3 \times (6E + 7A) = 3 \times 1,350,000 \implies 18E + 21A = 4,050,000 \quad (3)$$ $$7 \times (5E + 3A) = 7 \times 870,000 \implies 35E + 21A = 6,090,000 \quad (4)$$

Step 3: Subtract equation (3) from equation (4) to find $E$.

$$(35E + 21A) - (18E + 21A) = 6,090,000 - 4,050,000$$ $$17E = 2,040,000$$ $$E = \frac{2,040,000}{17}$$ $$E = 120,000$$

Step 4: Substitute the value of $E$ into equation (2) to find $A$.

$$5(120,000) + 3A = 870,000$$ $$600,000 + 3A = 870,000$$ $$3A = 870,000 - 600,000$$ $$3A = 270,000$$ $$A = \frac{270,000}{3}$$ $$A = 90,000$$

The daily wage of the engineer is $\boxed{\text{ugx } 120,000}$ and the daily wage of the assistant is $\boxed{\text{ugx } 90,000}$.

4. Age Problem and School Admissions