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.

Here's the solution for question 8:

a) The value of the angle marked $x$.

Step 1: Identify the relevant triangle and trigonometric ratio. 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. $\cos(x) = \frac{Adjacent}{Hypotenuse} = \frac{MN}{MR}$

Step 2: Substitute the values and solve for $x$. $\cos(x) = \frac{15}{10\sqrt{3}}$ 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 $\boxed{30^\circ}$.

b) The length of $QR$.

Step 1: Calculate the length of $NR$ using the Pythagorean theorem in $\triangle MNR$. $MR^2 = MN^2 + NR^2$ $(10\sqrt{3})^2 = 15^2 + NR^2$ $100 \times 3 = 225 + NR^2$ $300 = 225 + NR^2$ $NR^2 = 300 - 225$ $NR^2 = 75$ $NR = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} m$

Step 2: Calculate the length of $NQ$ using trigonometry in $\triangle MNQ$. In the right-angled triangle $\triangle MNQ$, we have $MN = 15$ m and $\angle MQN = 72^\circ$. We use the tangent function. $\tan(\angle MQN) = \frac{Opposite}{Adjacent} = \frac{MN}{NQ}$ $\tan(72^\circ) = \frac{15}{NQ}$ $NQ = \frac{15}{\tan(72^\circ)}$ Using a calculator, $\tan(72^\circ) \approx 3.07768$. $NQ \approx \frac{15}{3.07768} \approx 4.8739 m$

Step 3: Calculate $QR$. $QR = NR - NQ$ $QR = 5\sqrt{3} - \frac{15}{\tan(72^\circ)}$ $QR \approx 5(1.73205) - 4.8739$ $QR \approx 8.66025 - 4.8739$ $QR \approx 3.78635 m$ Rounding to the nearest whole number: The length of $QR$ is $\boxed{4 m}$.

c) The area of $\triangle MQR$.

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

Step 2: Substitute the values and calculate the area. Using the more precise value for $QR \approx 3.78635$ m and $MN = 15$ m: $Area = \frac{1}{2} \times 3.78635 \times 15$ $Area \approx 28.3976 m^2$ Rounding to the nearest whole number: The area of $\triangle MQR$ is $\boxed{28 m^2}$.

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