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:

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

   Show Solution

   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 $v = u + at$

   Show Solution

   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 $C^{\prime\prime}$.

Show Solution

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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?

Show Solution

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 $|PR|$

Show Solution

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