Sum of items = Mean x Number of items.

Mean = 17.6

Number of girls in the class = t

Sum of the ages of girls = 17.6 x t = 17.6t

When the girls where dismissed, the sum of the ages reduced to 17.6t - (16 + 19 + 20 + 17) and the number of girls reduced to t - 4

Note: 4 girls were dismissed.

New sum of ages = 17.6t - (16 + 19 + 20 + 17)

New sum of ages = 17.6t - 72

New number of girls = t - 4

New mean = $\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$

The new mean age of girls in the class became 0.2 less than the original mean.

Original mean = 17.6

New mean = 17.6 - 0.2 = 17.4

$\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$ = 17.4

Get rid of the fraction by multiplying both sides by the denominator t - 4

$\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$ x (t - 4) = 17.4 x (t - 4)

17.6t - 72 = 17.4t - 69.6

17.6t - 17.4t = 72 - 69.6

0.2t = 2.4

Divide both sides by 0.2

t = $\frac{2.4}{0.2}$ = 12

Mean = $\frac{\mathrm{Sum\; of\; numbers}}{\mathrm{Number\; of\; numbers}}$

Let x = sum of the ages of the second year class before the promotion examination
Let n = number of students in the second year class before the promotion examination

Mean = $\frac{\mathrm{x}}{\mathrm{n}}$

But the mean before the promotion examination = 18$\frac{2}{5}$

18$\frac{2}{5}$ = $\frac{\mathrm{5\; x\; 18\; +\; 2}}{5}$ = $\frac{92}{5}$

$\frac{92}{5}$ = $\frac{\mathrm{x}}{\mathrm{n}}$

Cross multiply

5 x x = 92 x n

Divide both sides by 5

x = $\frac{\mathrm{92n}}{5}$

3 students were repeated.

Sum of ages of the repeated students = 20 + 19 + 19
Sum of ages of the repeated students = 58

Sum of ages of the promoted students = Sum of ages before the promotion - Sum of ages of the repeated students
Sum of ages of the promoted students = x - 58

Number of promoted students = Number of students before the promotion - Number of students repeated
Number of promoted students = n - 3

Mean age of the promoted students = $\frac{\mathrm{x\; -\; 58}}{\mathrm{n\; -\; 3}}$

The new mean age of the class = 18$\frac{1}{3}$

18$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 18\; +\; 1}}{3}$ = $\frac{55}{3}$

$\frac{\mathrm{x\; -\; 58}}{\mathrm{n\; -\; 3}}$ = $\frac{55}{3}$

Cross multiply

3 x (x - 58) = 55 x (n - 3)

3x - 174 = 55n - 165

3x - 174 + 165 = 55n

3x - 9 = 55n

But x = $\frac{\mathrm{92n}}{5}$

3 x $\frac{\mathrm{92n}}{5}$ - 9 = 55n

Get rid of the fraction by multiplying both sides by 5

3 x 92n - 9 x 5 = 5 x 55n

276n - 45 = 275n

276n - 275n = 45

n = 45

∴ the number of students who were in the class before the promotion examination = 45

Mean/Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{(x\; –\; 2)\; +\; 4\; +\; (x\; +\; 2)\; +\; (2x\; +\; 1)\; +\; 9}}{5}$

Mean = $\frac{\mathrm{x\; –\; 2\; +\; 4\; +\; x\; +\; 2\; +\; 2x\; +\; 1\; +\; 9}}{5}$

Mean = $\frac{\mathrm{4x\; +\; 14}}{5}$

Median is the middle item. The middle item is x + 2

But median is equal to the mean

$\frac{\mathrm{4x\; +\; 14}}{5}$ = x + 2

Get rid of the fraction by multiplying both sides by the denominator.

4x + 14 = 5 x (x + 2)

4x + 14 = 5x + 10

14 - 10 = 5x - 4x

4 = x

x = 4

Angle of sector = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 360°

Total percentage = 100%

Angle of sector for plastic = $\frac{\mathrm{35\%}}{\mathrm{100\%}}$ x 360° = 126°

Angle of sector for liquid = $\frac{\mathrm{27.5\%}}{\mathrm{100\%}}$ x 360° = 99°

Angle of sector for metal = $\frac{\mathrm{17.5\%}}{\mathrm{100\%}}$ x 360° = 63°

Angle of sector for vegetable = $\frac{\mathrm{12.5\%}}{\mathrm{100\%}}$ x 360° = 45°

Angle of sector for others = $\frac{\mathrm{7.5\%}}{\mathrm{100\%}}$ x 360° = 27°

Note: the sum of the angle of sectors should be 360°