A salesperson earns a basic salary of GH¢550.00 per month plus a commission of 3% on all sales made. In a particular month, the salesperson's total income was GH¢1,225.00. a) Calculate the amount of c

Total income = Salary + Commission
GH¢ 1,225.00 = GH¢ 550.00 + Commission
Commission = GH¢ 1,225.00 - GH¢ 550.00

$$\text{Commission} = \text{GH¢ } 675.00$$

Commission = Commission Rate $\times$ Sales
GH¢ 675.00 = 3% $\times$ Sales
GH¢ 675.00 = 0.03 $\times$ Sales

$$\text{Sales} = \frac{\text{GH¢ } 675.00}{0.03}$$ $$\text{Sales} = \text{GH¢ } 22,500.00$$

The amount of sales for that month was $\boxed{\text{GH¢ } 22,500.00}$.

b)
The given data is: 18, 47, 22, 9, 32, 39, 20, 59, 48, 53, 10, 43, 36, 55, 6, 27, 42, 44, 2, 24, 17, 32, 46, 14, 11, 57, 38, 45, 25, 57.

2, 6, 9, 10, 11, 14, 17, 18, 20, 22, 24, 25, 27, 32, 32, 36, 38, 39, 42, 43, 44, 45, 46, 47, 48, 53, 55, 57, 57, 59.

i. Make a stem and leaf plot of the data.
Key: $0|2$ represents 2

$$\begin{array}{c|l} \text{Stem} & \text{Leaf} \\ \hline 0 & 2, 6, 9 \\ 1 & 0, 1, 4, 7, 8 \\ 2 & 0, 2, 4, 5, 7 \\ 3 & 2, 2, 6, 8, 9 \\ 4 & 2, 3, 4, 5, 6, 7, 8 \\ 5 & 3, 5, 7, 7, 9 \end{array}$$

ii. Find the number of pupils who passed the test, if the pass mark was 45.
Pupils passed if their mark is 45 or greater.
From the ordered data: 45, 46, 47, 48, 53, 55, 57, 57, 59.
There are 9 pupils who passed.
The number of pupils who passed the test is $\boxed{9}$.

iii. Find the modal mark of the distribution.
The mode is the number that appears most frequently in the data set.
From the ordered data, 32 appears twice, and 57 appears twice. All other numbers appear once.
The modal marks are $\boxed{32 \text{ and } 57}$.

c)

Volume = length $\times$ width $\times$ height

$$\text{Volume} = 3.5 \text{ m} \times 4 \text{ m} \times 5 \text{ m}$$ $$\text{Volume} = 70 \text{ m}^3$$

Water left = Initial volume - Water used

$$\text{Water left} = 70 \text{ m}^3 - 38 \text{ m}^3$$ $$\text{Water left} = 32 \text{ m}^3$$

The amount of water left in the tank is $\boxed{32 \text{ m}^3}$.

d)
The relation given is $46t = d + 10t$.

$$d = 46t - 10t$$ $$d = 36t$$

a) Draw a table for the relation $d = 36t$ for values of $t$ in the range $\{1, 2, 3, 4, 5, 6\}$ seconds.

$$\begin{array}{|c|c|} \hline t \text{ (seconds)} & d = 36t \text{ (metres)} \\ \hline 1 & 36 \\ 2 & 72 \\ 3 & 108 \\ 4 & 144 \\ 5 & 180 \\ 6 & 216 \\ \hline \end{array}$$

The table is provided above.
To draw the graph:

• Draw a horizontal axis (t-axis) and a vertical axis (d-axis) intersecting at the origin (0,0).
• On the t-axis, mark units from 0 to 6, with 2cm representing 1 unit.
• On the d-axis, mark units from 0 to 220, with 2cm representing 20 units.

b) On the same graph sheet, mark the t, second axis from 0 to 6 and the d distance axis from 0 to 220.
This is part of the graph setup described in part (a).

c) Plot on the same graph, the ordered pair (t, d) from the relation and join with a straight line.
Plot the points from the table: (1, 36), (2, 72), (3, 108), (4, 144), (5, 180), (6, 216).
Connect these points with a straight line.

d) Using your graph;
i. how long does it take to reach a height of 90metres?
To find this using the graph, locate $d = 90$ on the d-axis, move horizontally to the line, then move vertically down to the t-axis to read the corresponding time.
Algebraically:

$$d = 36t$$ $$90 = 36t$$ $$t = \frac{90}{36}$$ $$t = \frac{5}{2}$$ $$t = 2.5 \text{ seconds}$$

It takes $\boxed{2.5 \text{ seconds}}$ to reach a height of 90 metres.

ii. find the gradient of the line, $46t = d + 10t$.
The equation of the line is $d = 36t$.
This is in the form $y = mx + c$, where $m$ is the gradient.
Here, $m = 36$ and $c = 0$.
The gradient of the line is $\boxed{36}$.