The mean age of a second year class of a school is 18 2/5. At the end of the promotion examination, 3 students aged 20, 19 and 19 years were repeated. The new mean age of the class became 18 1/3. Calculate the number of students who were in the class before the promotion examination.

In the right-angled triangle $\triangle MNR$, we are given the adjacent side $MN = 15$ m and the hypotenuse $MR = 10\sqrt{3}$ m with respect to angle $x$ (which is $\angle RMN$). We use the cosine function.

$$\cos(x) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{MN}{MR}$$

$$\cos(x) = \frac{15}{10\sqrt{3}}$$

Simplify the fraction:

$$\cos(x) = \frac{3}{2\sqrt{3}}$$

Rationalize the denominator:

$$\cos(x) = \frac{3\sqrt{3}}{2\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2}$$

Therefore, $x$ is the angle whose cosine is $\frac{\sqrt{3}}{2}$.

$$x = \arccos\left(\frac{\sqrt{3}}{2}\right)$$ $$x = 30^\circ$$

The value of $x$ to the nearest whole number is $\boxed{\text{30}^\circ}$.

$$MR^2 = MN^2 + NR^2$$ $$(10\sqrt{3})^2 = 15^2 + NR^2$$ $$100 \times 3 = 225 + NR^2$$ $$300 = 225 + NR^2$$ $$NR^2 = 300 - 225$$ $$NR^2 = 75$$ $$NR = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \text{ m}$$

In the right-angled triangle $\triangle MNQ$, we have $MN = 15$ m and $\angle MQN = 72^\circ$. We use the tangent function.

$$\tan(\angle MQN) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{MN}{NQ}$$ $$\tan(72^\circ) = \frac{15}{NQ}$$ $$NQ = \frac{15}{\tan(72^\circ)}$$

Using a calculator, $\tan(72^\circ) \approx 3.07768$.

$$NQ \approx \frac{15}{3.07768} \approx 4.8739 \text{ m}$$

$$QR = NR - NQ$$ $$QR = 5\sqrt{3} - \frac{15}{\tan(72^\circ)}$$ $$QR \approx 5(1.73205) - 4.8739$$ $$QR \approx 8.66025 - 4.8739$$ $$QR \approx 3.78635 \text{ m}$$

Rounding to the nearest whole number:
The length of $QR$ is $\boxed{\text{4 m}}$.

The area of $\triangle MQR$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height}$. Here, $QR$ can be considered the base and $MN$ the height, as $MN$ is perpendicular to $NR$ (which contains $QR$).

$$\text{Area} = \frac{1}{2} \times QR \times MN$$

Using the more precise value for $QR \approx 3.78635$ m and $MN = 15$ m:

$$\text{Area} = \frac{1}{2} \times 3.78635 \times 15$$ $$\text{Area} \approx 28.3976 \text{ m}^2$$

Rounding to the nearest whole number:
The area of $\triangle MQR$ is $\boxed{\text{28 m}^2}$.