Section B: Answer Any five (5) Questions. 6. Make u the subject of the relation

The mean of $x_1, x_2, x_3, x_4$ is 7. The formula for the mean is $\frac{\text{sum of numbers}}{\text{count of numbers}}$.

$$\frac{x_1 + x_2 + x_3 + x_4}{4} = 7$$

Multiply both sides by 4 to find the sum:

$$x_1 + x_2 + x_3 + x_4 = 7 \times 4 = 28$$

The new numbers are $5+x_1, 5+x_2, 5+x_3, 5+x_4$.
Their sum is:

$$(5+x_1) + (5+x_2) + (5+x_3) + (5+x_4)$$ $$= (5+5+5+5) + (x_1+x_2+x_3+x_4)$$ $$= 20 + 28 = 48$$

There are 4 numbers in the new set.

$$\text{Mean} = \frac{\text{Sum of new numbers}}{\text{Count of new numbers}} = \frac{48}{4} = 12$$

The mean of the numbers $5+x_1, 5+x_2, 5+x_3, 5+x_4$ is $\boxed{12}$.

Multiply each age by its corresponding frequency:

• For age 12: $12 \times 8 = 96$
• For age 13: $13 \times 10 = 130$
• For age 14: $14 \times 4 = 56$
• For age 15: $15 \times 2 = 30$ Sum of (age × frequency) $= 96 + 130 + 56 + 30 = 312$

Add all the frequencies:

$$\text{Total frequency} = 8 + 10 + 4 + 2 = 24$$

The mean age is the sum of (age × frequency) divided by the total frequency:

$$\text{Mean age} = \frac{\text{Sum of (age } \times \text{ frequency)}}{\text{Total frequency}} = \frac{312}{24} = 13$$

The mean age of the class is $\boxed{13 \text{ years}}$.