PAST QUESTIONS 1992

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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.

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Question 1

1. Solve \(5 - 2x > x + 2\), where \(x\) is a real number.

   Illustrate your result on the number line.

   Show Solution

   Question 1. \(a\)

   \(5 - 2x > x + 2\)

   \(\Rightarrow -2x - x > 2 -5\)

   \(\Rightarrow \hspace{0.6cm} -3x > -3\)

   \(\Rightarrow \hspace{0.6cm} \dfrac{-3x}{-3} < \dfrac{-3}{-3}\)

   \(\Rightarrow \hspace{1cm} x < 1\)

   \(\therefore\) {\(x:x < 1, \ x\) is a real number}

   
   

   Illutrating on the number line:

   
   
   

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2. Find the truth set of the equation:

   \(\frac{2}{3}(3y - 1) - (y + 2) = \frac{1}{3}\)

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   Solution

   \(\frac{2}{3}(3y - 1) - (y + 2) = \frac{1}{3}\)

   Multiplying through by the L.C.M of 3

   \(\Rightarrow 3(\frac{2}{3}(3y - 1)) - 3(y + 2) \)\(= 3(\frac{1}{3})\)

   \(\Rightarrow 2(3y - 1) - 3(y + 2) = 1\)

   \(\Rightarrow 6y - 2 - 3y - 6 = 1\)

   \(\Rightarrow 6y - 3y - 2 - 6 = 1\)

   \(\Rightarrow \hspace{1.3cm} 3y - 8 = 1\)

   \(\Rightarrow \hspace{1.88cm} 3y = 1 + 8\)

   \(\Rightarrow \hspace{1.88cm} 3y = 9 \)

   \(\Rightarrow \hspace{1.76cm} \dfrac{3y}{3} = \dfrac{9}{3}\)

   \(\Rightarrow \hspace{2.1cm} y = 3\)

   \(\therefore y\) is 3

   Hence, the truth set of the equation is \(\{y:y = 3\}\).

   
   
   

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3. Factorise completely: \(mp + np - mt - nt\)

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   Solution

   \(mp + np - mt - nt\)

   \(\Rightarrow (mp + np) - (mt + nt)\)

   \(\Rightarrow p(m + n) - t(m + n)\)

   \(\Rightarrow (m + n)(p - t)\)

   
   
   

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4. Make \(t\) the subject of the relation \(v = u + at\)

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   Solution

   \(v = u + at\)

   Making \(t\) the subject

   \(\Rightarrow v - u = at\)

   \(\Rightarrow \dfrac{v - u}{a} = \dfrac{at}{a}\)

   \(\Rightarrow \dfrac{v - u}{a} = t\)

   Re-arranging the relation

   \(\Rightarrow t = \dfrac{v - u}{a} \)

   
   
   

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Question 2

A landlady rented out her house for ₵ 240,000.00 for one year. During the year, she paid 15% of the rent as income tax. She also paid 25% of the rent as property tax and spent ₵ 10,000.00 on repairs. Calculate

1. The landlady's total expenses.

Show Solution

Question 2\(a\)

Amount gotten \(= ₵ 240,000.00\)

Amount spent on income tax

\(\Rightarrow 15\% \ \ of \ \ ₵ 240,000.00 \)

\(\Rightarrow \dfrac{15}{100} \times ₵ 240,000.00 \)

\(\Rightarrow 15 \times ₵ 2400\)

\(\Rightarrow ₵ 36,000.00\)

Amount spent on property tax

\(\Rightarrow 25\% \ \ of \ \ ₵ 240,000.00 \)

\(\Rightarrow \dfrac{25}{100} \times ₵ 240,000.00 \)

\(\Rightarrow 25 \times ₵ 2400\)

\(\Rightarrow ₵ 60,000.00\)

Amount spent on repairs

\(\Rightarrow ₵ 10,000.00\)

Total expense = income tax + property tax + repairs

Hence, Total expense

\(\Rightarrow\)\(₵ 36,000.00 +\)\( ₵ 60,000.00 + \)\(₵ 10,000.00\)

\(\Rightarrow\)\(₵ 106,000.00\)

\(\therefore\) the landlady's total expense was \(₵ 106,000.00\)





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2. The remainder of the rent after the landlady's expenses.

Show Solution

Question 2\(b\)

Amount gotten \(= ₵ 240,000.00\)

Total expense \(= ₵ 106,000.00\)

Remaining amount \(=\) Amount gotten \(-\) total expense

Hence, Remaining amount \(= ₵ 240,000.00 -\)\(₵ 106,000.00\)

\(\Rightarrow\) Remaining amount \(= ₵ 134,000.00\)

\(\therefore\) she had \(₵ 134,000.00\) remaining after her expenses.





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3. The percentage of the rent she spent on repairs.

Show Solution

Question 2\(c\)

Percentage spent on repairs.

\(\Rightarrow \dfrac{amount \ on \ repairs}{amount \ gotten} \times 100\%\)

\(\Rightarrow \dfrac{₵10,000}{₵240,000} \times 100\%\)

\(\Rightarrow \dfrac{1}{24} \times 100\%\)

\(\Rightarrow \dfrac{100}{24}\%\)

\(\Rightarrow \dfrac{4 \times 25}{4 \times 6}\%\)

\(\Rightarrow \dfrac{25}{6}\%\)

\(\Rightarrow 4.1667\%\)

\(\therefore\) percentage of rent spent on repairs is \(4.1667\%\)





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Question 3

1. Using a scale of 2cm 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 \(A(1, 1)\), \(B(4, 3)\) and \(C(2, 5)\). Join the points \(A, B\) and \(C\) to form a triangle.

4. Using the \(y\)-axis as mirror line, draw the image of the triangle \(ABC\) such that \(A \rightarrow A^\prime\), \(B \rightarrow B^\prime\) and \(C \rightarrow C^\prime\).

5. Using the \(y\)-axis as the mirror line, draw the image of triangle \(ABC\) such that \(A \rightarrow A^{\prime\prime}\), \(B \rightarrow B^{\prime\prime}\) and \(C \rightarrow C^{\prime\prime}\).
   Write down the coordinates of \(A^{\prime\prime}\), \(B^{\prime\prime}\) and \(C^{\prime\prime}\).

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Question 4

The table below gives the frequency distribution of the marks obtained in a class test by a group of 64 pupils.

1. Draw a bar chart for the distribution.

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Solution

Question 4\((a)\)






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2. A pupil is chosen at random from the class, what is the probability that the pupil obtained 7 marks?

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Solution

Question 4\((b)\)


Total number of pupils:
\(\Rightarrow 9 + 14 + 13 + 10 + 5 + 8 + 2 + 3\)
\(\Rightarrow 64\)

Pupils who obtained 7 marks \(= 8\)


Probability \(= \dfrac{successful \ outcomes}{total \ outcomes}\)

Prob.(obtaining 7 marks):

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

\(\Rightarrow \dfrac{1\times 8}{8 \times 8}\)

\(\Rightarrow \dfrac{1}{8}\)

\(\therefore\) the probability of choosing a pupil who obtained 7 marks is \(\frac{1}{8}\).





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Question 5

1. Using a ruler and a pair of compasses only:

2. Draw \(|PQ|=9cm\)

3. Construct a perpendicular to \(PQ\) at \(Q\)

4. Construct \(\angle QPS = 60^\circ\) at the point \(P\) on \(PQ\) such that \(PS = 6.5 cm\)

5. Construct a line parallel to \(PS\) through \(S\). Let the perpendicular through \(Q\) and the parallel through \(S\) meet at \(R\). Measure \(|PR|\)

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