Q2. A trader bought goods for GH₵500 and sold them for GH₵650. Find the percentage profit. b) Find the simple interest on GH₵1000 at 5% per annum for 3 years. Q3. a) The marks obtained by 10 students are: 12, 15, 10, 18, 20, 15, 17, 19, 14, 10 i. Find the mean ii. Find the mode b) Find the range of the data. Q4. a) A point B is on a bearing of 060° from A. Find the bearing of A from B. b) A man walks 5 km east and then 12 km north. Find the distance from the starting point.

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Q2.

  • a) Step 1: Calculate the profit.

    Profit=Selling Price−Cost Price\text{Profit} = \text{Selling Price} - \text{Cost Price} Profit=GH₵650−GH₵500=GH₵150\text{Profit} = \text{GH₵}650 - \text{GH₵}500 = \text{GH₵}150

  • Step 2: Calculate the percentage profit.

    Percentage Profit=ProfitCost Price×100%\text{Percentage Profit} = \frac{\text{Profit}}{\text{Cost Price}} \times 100\% Percentage Profit=GH₵150GH₵500×100%\text{Percentage Profit} = \frac{\text{GH₵}150}{\text{GH₵}500} \times 100\% Percentage Profit=0.3×100%=30%\text{Percentage Profit} = 0.3 \times 100\% = 30\%

    The percentage profit is 30%\boxed{\text{30\%}}.

  • b) Step 1: Identify the principal, rate, and time. Principal (PP) = GH₵1000 Rate (RR) = 5% per annum Time (TT) = 3 years

  • Step 2: Use the simple interest formula.

    Simple Interest (SI)=P×R×T100\text{Simple Interest (SI)} = \frac{P \times R \times T}{100} SI=1000×5×3100\text{SI} = \frac{1000 \times 5 \times 3}{100}

  • Step 3: Calculate the simple interest.

    SI=15000100=GH₵150\text{SI} = \frac{15000}{100} = \text{GH₵}150

    The simple interest is GH₵150\boxed{\text{GH₵}150}.

Q3.

  • a) The marks obtained by 10 students are: 12, 15, 10, 18, 20, 15, 17, 19, 14, 10.

Part i· 1 step

Step 1: Sum all the marks.

    $$
    \text{Sum} = 12 + 15 + 10 + 18 + 20 + 15 + 17 + 19 + 14 + 10 = 150
    $$

  1. Calculate the mean.

        $$
    \text{Mean} = \frac{\text{Sum of marks}}{\text{Number of students}} = \frac{150}{10} = 15
    $$
    The mean mark is $\boxed{15}$.

Part ii· 5 steps

Step 1: Identify the frequency of each mark.

    10 appears 2 times.  
    12 appears 1 time.  
    14 appears 1 time.  
    15 appears 2 times.  
    17 appears 1 time.  
    18 appears 1 time.  
    19 appears 1 time.  
    20 appears 1 time.

  1. Determine the mark(s) that appear most frequently.

        Both 10 and 15 appear 2 times, which is the highest frequency.  
    The mode is $\boxed{10 \text{ and } 15}$.
    • b) Step 1: Identify the highest and lowest marks from the data. Highest mark = 20 Lowest mark = 10

  2. Calculate the range.

    $$
\text{Range} = \text{Highest mark} - \text{Lowest mark}
$$
$$
\text{Range} = 20 - 10 = 10
$$
The range of the data is $\boxed{10}$.

    Q4.

    • a) Step 1: Understand the relationship between bearings. If the bearing of B from A is θ\theta, then the bearing of A from B is θ+180∘\theta + 180^\circ if θ<180∘\theta < 180^\circ, or θ−180∘\theta - 180^\circ if θ>180∘\theta > 180^\circ.

  3. Apply the rule for the given bearing.

    Given bearing of B from A = $060^\circ$. Since $060^\circ < 180^\circ$, we add $180^\circ$.
$$
\text{Bearing of A from B} = 060^\circ + 180^\circ = 240^\circ
$$
The bearing of A from B is $\boxed{240^\circ}$.
    • b) Step 1: Visualize the movement as a right-angled triangle. The man walks 5 km East (one leg of the triangle) and 12 km North (the other leg of the triangle). The distance from the starting point is the hypotenuse.

  4. Use the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2).

    Let $a = 5$ km and $b = 12$ km. Let $c$ be the distance from the starting point.
$$
c^2 = (5 \text{ km})^2 + (12 \text{ km})^2
$$
$$
c^2 = 25 \text{ km}^2 + 144 \text{ km}^2
$$
$$
c^2 = 169 \text{ km}^2
$$

  5. Solve for cc.

    $$
c = \sqrt{169 \text{ km}^2}
$$
$$
c = 13 \text{ km}
$$
The distance from the starting point is $\boxed{13 \text{ km}}$.
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