BECE

Year to Practice:
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PAST QUESTIONS 1993


Section A

Time yourself to improve on your speed. You are to use not more than 60 minutes for this section.

Click on the link below when you are ready.






Section B

Try the questions first, using not more than 15 minutes for each question, and watch the accompanying videos to see how the questions are solved.





Question 1


  1. Simplify \(\dfrac{2a + 4b}{3} - \dfrac{3(a - b)}{2}\)


    2. Question 1a

    \(\dfrac{2a + 4b}{3} - \dfrac{3(a - b)}{2}\)

    \(\Rightarrow \dfrac{2a + 4b}{3} - \dfrac{3a - 3b}{2}\)

    \(\Rightarrow \dfrac{2(2a + 4b) - 3(3a - 3b)}{6}\)

    \(\Rightarrow \dfrac{4a + 8b - 9a + 9b}{6}\)

    Grouping like-terms

    \(\Rightarrow \dfrac{4a - 9a + 8b + 9b}{6}\)

    \(\Rightarrow \dfrac{-5a + 17b}{6}\)

    Re-arranging the expression

    \(\Rightarrow \dfrac{17b - 5a}{6}\)







  2. Solve \(5(a - 5) - \frac{1}{2}(2a + 6) = 4\)


    3. Question 1b

    \(5(a - 5) - \frac{1}{2}(2a + 6) = 4\)

    Multiplying through by an L.C.M of \(2\)

    \(\Rightarrow 2 \times 5(a - 5) - \)\(2 \times \frac{1}{2}(2a + 6)\)\(\ = 2 \times 4\)

    \(\Rightarrow 10(a - 5) - \)\((2a + 6)\)\(\ = 8\)

    Expanding the brackets

    \(\Rightarrow 10a - 50 - 2a - 6 = 8\)

    \(\Rightarrow 10a - 2a - 50 - 6 = 8\)

    \(\Rightarrow 8a - 56 = 8\)

    \(\Rightarrow 8a = 8 + 56\)

    \(\Rightarrow 8a = 64\)

    Dividing both sides by the coefficient of \(a\)

    \(\Rightarrow \dfrac{8a}{8} = \dfrac{64}{8}\)

    \(\Rightarrow a = 8\)

    \(\therefore a \) is \(8\).







  3. If \(\mathbf{r}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix} -2 \\ 1 \end{pmatrix}\),
    calculate \(6(\mathbf{r} + 2\mathbf{q})\)


    4. Question 1c

    \(\mathbf{r}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}\)

    \(\mathbf{q}=\begin{pmatrix} -2 \\ 1 \end{pmatrix}\)

    Hence, \(6(\mathbf{r} + 2\mathbf{q})\)

    \(\Rightarrow 6\begin{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} + 2\begin{pmatrix} -2 \\ 1 \end{pmatrix}\end{pmatrix}\)

    \(\Rightarrow 6\begin{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} -4 \\ 2 \end{pmatrix}\end{pmatrix}\)

    \(\Rightarrow 6\begin{pmatrix} 3 + -4 \\ 1 + 2 \end{pmatrix}\)

    \(\Rightarrow 6\begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

    \(\Rightarrow \begin{pmatrix} -6 \\ 18 \end{pmatrix}\)

    \(\therefore 6(\mathbf{r} + 2\mathbf{q}) = \begin{pmatrix} -6 \\ 18 \end{pmatrix}\)









Question 2


Using a ruler and a pair of compasses only,

  1. Construct a triangle \(ABC\) such that \(|BA|=10 cm\), \(\angle ABC = 90^\circ\) and \(\angle BAC = 30^\circ\). Measure the length \(BC\).

  2. \((i)\) Bisect the angle \(ACB\) to meet \(|BA|\) at \(D\).
    \((ii)\) What type of triangle is \(CDA\)?

  3. Calculate the area of triangle \(ABC\).

Solution









Question 3


Olu bought a radio for ₵65,000.00. After one year, the radio was valued at 75% of the cost price.

  1. What was the value of the radio after one year?

    2. Question 3a

    Cost Price \(= ₵65,000.00\)

    Value after 1 year

    \(\Rightarrow 75\% \ of\) Cost Price

    \(\Rightarrow \dfrac{75}{100} \times ₵65,000.00\)

    \(\Rightarrow 75 \times ₵650\)

    \(\Rightarrow ₵48,750.00\)

    \(\therefore\) the value of the radio after 1 year was \(₵48,750.00\)





  2. If he sold the radio for ₵55,700.00, calculate his profit or loss over the cost price.

    3. Question 3b

    New Cost Price \(= ₵48,750.00\)

    Selling Price \(= ₵55,700.00\)

    Profit \(=\) Selling Price \(-\) Cost Price

    Profit \(=\) \(₵55,700.00 - ₵48,750.00\)

    Profit \(=\) \(₵6,950.00\)

    \(\therefore\) his profit made was \(₵6,950.00\).









Question 4


  1. Using a scale of 2 cm to 1 unit on both axes, draw two perpendicular lines \(Ox\) and \(Oy\) on a graph sheet.

  2. On this graph sheet, mark the \(x-\)axis from \(-5\) to \(5\) and the \(y-\)axis from \(-6\) to \(6\)

  3. Plot on the same graph sheet the points \(A(-2, 4)\) and \(B(4, -5)\). Join the points \(A\) and \(B\) with the help of a ruler.

  4. Using the graph, find:

    \(\hspace{0.5cm}i)\) the gradient (or slope) of the line \(AB\)

    \(\hspace{0.5cm}ii)\) the value of \(x\) when \(y=0\)

    \(\hspace{0.5cm}iii)\) the value of \(y\) when \(x=2\)

  5. Plot on the same graph sheet, the points \(C(-3, -1)\) and \(D(3, 3)\). Join the points \(C\) and \(D\). With the help of a protractor, measure the angle between the lines \(AB\) and \(CD\).
    What is the gradient of the line CD?

    6. Solution









Question 5


The ages of 20 school children are recorded as follows:

13 \(\hspace{0.5cm}\) 9 \(\hspace{0.5cm}\) 15 \(\hspace{0.5cm}\) 17\(\hspace{0.5cm}\) 13
9 \(\hspace{0.7cm}\) 11 \(\hspace{0.5cm}\) 9 \(\hspace{0.5cm}\) 11 \(\hspace{0.5cm}\) 15
17 \(\hspace{0.45cm}\) 15 \(\hspace{0.45cm}\) 11 \(\hspace{0.5cm}\) 9 \(\hspace{0.5cm}\) 9
11 \(\hspace{0.45cm}\) 15 \(\hspace{0.45cm}\) 11 \(\hspace{0.45cm}\) 11 \(\hspace{0.3cm}\) 11

  1. Make a frequency table for the data using the ages 9, 11, 13, ...

  2. Use your table to calculate the mean age (correct to the nearest whole number).

    3. Solution











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