Chapter 9 — Full Mock Examinations

If $\displaystyle\sum_{k=1}^{n} (2k - 1) = 144$, the value of $n$ is:

A) 12

B) 72

C) 144

D) 24

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In how many ways can 4 books be arranged on a shelf if two particular books must stand next to each other?

A) 48

B) 24

C) 6

D) 12

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$P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cap B) = 0.2$. Then $P(A \cup B)$ equals:

A) 1.1

B) 0.9

C) 0.2

D) 0.7

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In a class, 12 learners scored 60 % and 8 learners scored 85 %. The mean mark for the class is:

A) 75 %

B) 72.5 %

C) 70 %

D) 68 %

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The points $(1;\ 3)$, $(5;\ 11)$ and $(k;\ 19)$ are collinear. The value of $k$ is:

A) 8

B) 7

C) 13

D) 9

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If $m\%$ of $r$ items are defective, the number of items that are not defective is:

A) $\dfrac{100r}{m}$

B) $\dfrac{mr}{100}$

C) $\dfrac{r - m}{100}$

D) $\dfrac{100r - mr}{100}$

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Each interior angle of a regular polygon is $150°$. The polygon has:

A) 15 sides

B) 10 sides

C) 12 sides

D) 6 sides

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A straight line passes through $(-1;\ 4)$ and is perpendicular to $y = \dfrac{1}{3}x + 2$. The equation of this line is:

A) $y = 3x + 7$

B) $y = -3x - 3$

C) $y = -3x + 1$

D) $y = \dfrac{1}{3}x + \dfrac{13}{3}$

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$PQ$ is a tangent to the circle with centre $O$, touching the circle at $P$. If $OP = 5$ cm and $OQ = 13$ cm, the area of $\triangle OPQ$ is:

A) $32.5\ \text{cm}^2$

B) $30\ \text{cm}^2$

C) $60\ \text{cm}^2$

D) $39\ \text{cm}^2$

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Which one of the following is the equation of the graph shown?

A) $y = 2^x - 3$

B) $y = 2^x + 3$

C) $y = 3^x - 2$

D) $y = -2^x - 3$

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$M(1;\ 2)$ is the midpoint of $AB$. If $A$ is the point $(-2;\ 6)$, then $B$ is:

A) $(3;\ -4)$

B) $(-5;\ 10)$

C) $(4;\ -2)$

D) $(0;\ 4)$

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The graph of $f$ is shown above. For which values of $x$ is $f(x) \geq 0$?

A) $x \leq 1$ or $x \geq 5$

B) $1 \leq x \leq 5$

C) $-5 \leq x \leq 4$

D) $x \geq 4$

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A bag contains 3 red and 7 blue marbles. Two marbles are drawn at random without replacement. The probability that both are red is:

A) $\dfrac{3}{10}$

B) $\dfrac{9}{100}$

C) $\dfrac{1}{15}$

D) $\dfrac{1}{10}$

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If $(x + 3)$ is a factor of $2x^2 + bx - 3$, the value of $b$ is:

A) 5

B) $-5$

C) 3

D) 15

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One solution of $\cos(x - 20°) = k$ is $x = 80°$. Which one of the following is also a solution?

A) $260°$

B) $100°$

C) $-40°$

D) $-80°$

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A fair coin is tossed 3 times. The probability of getting at least 2 heads is:

A) $\dfrac{1}{4}$

B) $\dfrac{3}{8}$

C) $\dfrac{7}{8}$

D) $\dfrac{1}{2}$

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The graph of $y = f(x)$ is shifted 2 units to the right and 3 units down. The equation of the new graph is:

A) $y = f(x + 2) + 3$

B) $y = f(x + 2) - 3$

C) $y = f(x - 2) + 3$

D) $y = f(x - 2) - 3$

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In $\triangle ABC$, $AB = 6$, $AC = 8$ and $\hat{A} = 60°$. The length of $BC$ is:

A) $2\sqrt{13}$

B) $10$

C) $2\sqrt{37}$

D) $14$

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The graph of $y = \log_2(x + k)$ has an $x$-intercept at $(3;\ 0)$. The value of $k$ is:

A) 2

B) $-2$

C) $-3$

D) 1

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Simplify $\dfrac{\sqrt{50} - \sqrt{18}}{\sqrt{2}}$.

A) $2\sqrt{2}$

B) 2

C) 4

D) $\sqrt{2}$

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If $\sin\theta = \dfrac{2}{3}$ and $\theta$ is acute, then $\cos^2\theta$ equals:

A) $\dfrac{\sqrt{5}}{3}$

B) $\dfrac{1}{3}$

C) $\dfrac{5}{9}$

D) $\dfrac{4}{9}$

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A tourist changes $500 into rand at R18.50 per dollar, and is charged 2 % commission on the rand amount. He receives:

A) R9 250

B) R9 065

C) R9 435

D) R8 875

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The graph of $y = a \cdot 2^x$ passes through the point $(3;\ 24)$. The value of $a$ is:

A) 3

B) 8

C) 21

D) 12

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The median of a data set is 12 and the interquartile range is 6. If $Q_1 = 9$, then $Q_3$ equals:

A) 21

B) 18

C) 3

D) 15

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$f$ is an even function and $f(3) = 7$. The value of $f(-3) + f(3)$ is:

A) 14

B) 0

C) 7

D) $-14$

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A triangle has sides of 5, 12 and 13 units. Its area is:

A) 32.5 square units

B) 30 square units

C) 60 square units

D) 39 square units

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From a class of 30 students, 12 play sport and 10 study music. 5 do both. How many do neither?

A) 8

B) 12

C) 17

D) 13

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The value of $\dfrac{\tan 45° + \sin 30°}{\cos 60°}$ is:

A) $\dfrac{3}{2}$

B) 2

C) 3

D) 1

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In a geometric sequence, $T_2 = 6$ and $T_5 = 48$. The common ratio is:

A) 2

B) 8

C) 3

D) 4

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For which value of $k$ is the set of ordered pairs $\{(1;\ 2),\ (2;\ 5),\ (k;\ 7)\}$ not a function?

A) 7

B) 2

C) 3

D) 0

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If $\sin 2\theta = \dfrac{3}{5}$, the value of $(\sin\theta + \cos\theta)^2$ is:

A) $\dfrac{3}{5}$

B) 1

C) $\dfrac{2}{5}$

D) $\dfrac{8}{5}$

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The sum of the first $n$ terms of an arithmetic sequence is $S_n = 2n^2 + n$. The 5th term of the sequence is:

A) 21

B) 55

C) 19

D) 15

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The graph of $f$ has a turning point at $(-2;\ 4)$. Which one of the following graphs will cut the $x$-axis at exactly one point?

A) $y = f(x) + 4$

B) $y = f(x) - 4$

C) $y = f(x + 4)$

D) $y = -f(x)$

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The line through $(2;\ k)$ and $(4;\ 7)$ is perpendicular to $y = -2x + 1$. The value of $k$ is:

A) 5

B) 8

C) 6

D) 3

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In the diagram the two horizontal lines are parallel. The value of $x$ is:

A) $62°$

B) $118°$

C) $28°$

D) $128°$

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The circle $x^2 + y^2 = 25$ and the line $y = x + 1$ intersect. One point of intersection is:

A) $(4;\ 3)$

B) $(3;\ 4)$

C) $(5;\ 0)$

D) $(-3;\ -4)$

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If $\log_{10} 2 = a$, then $\log_{10} 5$ equals:

A) $\dfrac{a}{2}$

B) $a - 1$

C) $5a$

D) $1 - a$

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Which one of the following is irrational?

A) $\sqrt{8} \times \sqrt{2}$

B) $\sqrt{8} + \sqrt{2}$

C) $\sqrt{8} \div \sqrt{2}$

D) $(\sqrt{8})^2$

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$O$ is the centre of the circle. The chord $AB = 24$ cm and the radius is 13 cm. The area of $\triangle AOB$ is:

A) $156\ \text{cm}^2$

B) $60\ \text{cm}^2$

C) $120\ \text{cm}^2$

D) $65\ \text{cm}^2$

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Written in standard form, $\dfrac{(3 \times 10^5)(4 \times 10^{-2})}{6 \times 10^{-4}}$ is:

A) $12 \times 10^7$

B) $2 \times 10^{11}$

C) $2 \times 10^7$

D) $2 \times 10^{-1}$

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The mean of the data set $\{3;\ 8;\ x;\ 15;\ 7;\ 1\}$ is 7. The value of $x$ is:

A) 42

B) 7

C) 6

D) 8

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Two triangles are similar. Their perimeters are 18 cm and 30 cm. If the shorter triangle has an area of $27\ \text{cm}^2$, what is the area of the larger triangle?

A) $45\ \text{cm}^2$

B) $54\ \text{cm}^2$

C) $75\ \text{cm}^2$

D) $108\ \text{cm}^2$

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An infinite geometric series has $S_\infty = 12$ and first term $a = 8$. The common ratio is:

A) $\dfrac{1}{3}$

B) $\dfrac{2}{3}$

C) $\dfrac{3}{2}$

D) $\dfrac{1}{2}$

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$AB$ is a diameter of a circle and $C$ is a point on the circle. If $\hat{A} = 35°$, then $\hat{B}$ equals:

A) $35°$

B) $55°$

C) $90°$

D) $45°$

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If $\dfrac{4^{x+1}}{2^{x-1}} = 2^9$, the value of $x$ is:

A) 6

B) 9

C) 3

D) 12

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The interquartile range of $\{2;\ 5;\ 8;\ 10;\ 14;\ 18\}$ is:

A) 6

B) 16

C) 8

D) 9

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After a 15 % increase, the price of an item is R276. The original price was:

A) R261

B) R234.60

C) R240

D) R320

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How many real solutions does $|x - 3| + 4 = 2$ have?

A) 0

B) 1

C) 2

D) Infinitely many

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The graph of $y = \dfrac{a}{x}$ passes through the point $(-2;\ 3)$. Its two branches lie in quadrants:

A) II and IV

B) I and III

C) I and II

D) III and IV

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R8 000 grows to R9 680 in 2 years at compound interest. The annual interest rate is:

A) 21 %

B) 10 %

C) 10.5 %

D) 11 %

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The roots of $2x^2 - 6x + 1 = 0$ are $\alpha$ and $\beta$. The value of $\alpha + \beta$ is:

A) $-3$

B) $\dfrac{1}{2}$

C) 6

D) 3

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If $f(x) = 3x - 6$, the value of $f^{-1}(0)$ is:

A) 2

B) $-6$

C) $\dfrac{1}{6}$

D) $-2$

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If $\sqrt{x} : \sqrt{2} = 3 : 1$, the value of $x$ is:

A) 9

B) 6

C) 18

D) 36

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For the arithmetic sequence $6;\ 10;\ 14;\ \ldots$, which term is equal to 102?

A) The 25th

B) The 24th

C) The 26th

D) The 102nd

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A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm. If its area is $21\ \text{cm}^2$, the value of $x$ is:

A) 3

B) $-6$

C) 6

D) 4

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A sphere has a volume of $36\pi\ \text{cm}^3$. Its radius is:

A) 27 cm

B) 9 cm

C) 3 cm

D) 6 cm

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For which value of $x$ is $(x + 2)(x - 3) - x(x + 3) = 0$?

A) $-\dfrac{3}{2}$

B) $\dfrac{3}{2}$

C) $-6$

D) 0

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The domain of $f(x) = \dfrac{\sqrt{x + 2}}{x - 3}$ is:

A) $x \geq -2,\ x \neq 3$

B) $x > -2$

C) $x \neq 3$

D) $x \geq -2$

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A car depreciates at 20 % p.a. on the reducing balance. After how many complete years is it worth less than half its original value?

A) 3

B) 4

C) 5

D) 2

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Every value in a data set is increased by 5. Which one of the following does not change?

A) The mode

B) The mean

C) The median

D) The standard deviation

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Mock Exam 1 — Answer Key

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	A	2	D	3	D	4	C	5	D
6	D	7	C	8	C	9	B	10	A
11	C	12	B	13	C	14	A	15	C
16	D	17	D	18	A	19	B	20	B
21	C	22	B	23	A	24	D	25	A
26	B	27	D	28	C	29	A	30	B
31	D	32	C	33	B	34	C	35	B
36	B	37	D	38	B	39	B	40	C
41	D	42	C	43	A	44	B	45	A
46	D	47	C	48	A	49	A	50	B
51	D	52	A	53	C	54	A	55	D
56	C	57	A	58	A	59	B	60	D

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Mock Exam 1 — Worked Solutions

Mark your paper against the key above first, then read only the solutions for the questions you missed. Each one shows the route, not a full lesson — the chapters teach the ideas in depth.

Q1 — A. The sum of the first $n$ odd numbers is $n^2$. $n^2 = 144 \Rightarrow n = 12$.

Why the others are wrong: B halves 144 instead of taking its square root. C repeats the sum itself. D doubles the answer.

Q2 — D. Tie the pair together as one block: $3! = 6$ arrangements of the three items, times $2!$ for the order inside the block $= 12$.

Why the others are wrong: A is $4! \times 2$ — the pair was tied but the block never collapsed, so four items were still arranged. B is $4!$: the restriction was ignored altogether. C is $3!$: the block was collapsed but the two orders inside it were forgotten.

Q3 — D. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$.

Why the others are wrong: A adds the overlap instead of subtracting it. B forgets the overlap entirely. C gives $P(A \cap B)$, the overlap on its own.

Q4 — C. Weight by how many learners: $\dfrac{12(60) + 8(85)}{20} = \dfrac{1\,400}{20} = 70\%$. (The plain average of 60 and 85 is 72.5 % — wrong, because the groups differ in size.)

Why the others are wrong: B is the plain average of 60 and 85 — right only if the two groups were the same size, and they are not. A and D sit either side of that average and follow from no consistent weighting.

Q5 — D. Gradient of the first pair $= \dfrac{11 - 3}{5 - 1} = 2$. Collinear means the same gradient throughout: $\dfrac{19 - 3}{k - 1} = 2 \Rightarrow k = 9$.

Why the others are wrong: C uses a gradient of $\tfrac{4}{3}$ instead of 2. A and B come from dividing 16 by something other than 2 — check the gradient on the first pair before using it on the second.

Q6 — D. Defective $= \dfrac{mr}{100}$, so the rest is $r - \dfrac{mr}{100} = \dfrac{100r - mr}{100}$. (Check with $m = 20$, $r = 50$: 40 are not defective. ✓)

Why the others are wrong: B is the number that IS defective. A turns the fraction upside down. C subtracts the percentage as though it were a count of items.

Q7 — C. Each exterior angle is $180° - 150° = 30°$, and exterior angles total $360°$: $n = \dfrac{360}{30} = 12$.

Why the others are wrong: Each wrong option divides 360 by the wrong exterior angle: A by 24, B by 36 and D by 60. Only $180° - 150° = 30°$ is the exterior angle here.

Q8 — C. Perpendicular gradient $= -\dfrac{1}{1/3} = -3$. $y - 4 = -3(x + 1) \Rightarrow y = -3x + 1$. Check: at $x = -1$, $y = 4$. ✓

Why the others are wrong: D keeps the gradient $\tfrac{1}{3}$ — that line is parallel, not perpendicular. A inverts without changing the sign. B has the right gradient but does not pass through $(-1;\ 4)$.

Q9 — B. A tangent meets the radius at $90°$, so $\triangle OPQ$ is right-angled at $P$. $PQ = \sqrt{13^2 - 5^2} = 12$ (the 5-12-13 triple). Area $= \tfrac{1}{2}(5)(12) = 30\ \text{cm}^2$.

Why the others are wrong: A uses $OQ = 13$ as the second leg; it is the hypotenuse. C forgets to halve. D follows from no consistent pair of sides.

Q10 — A. The horizontal asymptote is $y = -3$ and the curve rises, which rules out C and D; the $y$-intercept is $-2$, and $2^0 - 3 = -2$. So $y = 2^x - 3$.

Why the others are wrong: B has its asymptote at $y = +3$. C has its asymptote at $y = -2$ and cuts the $y$-axis at $-1$. D falls; the graph shown rises.

Q11 — C. The midpoint is the average, so $B = (2 \times 1 - (-2);\ 2 \times 2 - 6) = (4;\ -2)$.

Why the others are wrong: B goes from $M$ in the wrong direction, landing as far past $A$ as $B$ is past $M$. D averages $M$ and $A$ again instead of stepping on from $M$. A gets both coordinates wrong by a step in each.

Q12 — B. The parabola opens downward with roots at $x = 1$ and $x = 5$, so it is on or above the $x$-axis between them: $1 \leq x \leq 5$.

Why the others are wrong: A is where the curve lies BELOW the axis — it opens downward, so the inequality holds between the roots, not outside them. C reads the roots off with the wrong signs. D gives only part of the interval, and starts at the turning point rather than at a root.

Q13 — C. Without replacement the second draw changes: $\dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} = \dfrac{1}{15}$. (With replacement it would be $\tfrac{9}{100}$.)

Why the others are wrong: B draws WITH replacement: $\tfrac{3}{10} \times \tfrac{3}{10}$. A gives the first draw only. D neither keeps all ten marbles nor removes one properly.

Q14 — A. $(x+3)$ a factor $\Rightarrow f(-3) = 0$: $2(9) - 3b - 3 = 0 \Rightarrow 15 = 3b \Rightarrow b = 5$.

Why the others are wrong: B loses the sign solving $15 = 3b$. C reads off the constant term. D stops at 15, before dividing by 3.

Q15 — C. $x = 80°$ gives $\cos 60°$. Cosine is even, so $\cos(-60°)$ is the same value: $x - 20° = -60° \Rightarrow x = -40°$.

Why the others are wrong: A and D put the argument where the cosine is NEGATIVE — $\cos 240°$ and $\cos(-100°)$ — while $\cos 60°$ is positive. B gives $\cos 80°$, which is not $\cos 60°$.

Q16 — D. "At least 2" means exactly 2 or exactly 3: $\dfrac{3}{8} + \dfrac{1}{8} = \dfrac{1}{2}$.

Why the others are wrong: B is EXACTLY two heads, leaving out the three-head case. C is at least ONE head. A counts only two of the eight outcomes.

Q17 — D. Right by 2 replaces $x$ with $x - 2$; down by 3 subtracts 3 outside. So $y = f(x - 2) - 3$. (Horizontal shifts act inside and look reversed.)

Why the others are wrong: B and A use $f(x + 2)$, which moves the graph LEFT — a horizontal shift acts inside the bracket and looks reversed. C moves up instead of down, and A gets both directions wrong.

Q18 — A. Cosine rule: $BC^2 = 6^2 + 8^2 - 2(6)(8)\cos 60° = 100 - 48 = 52$, so $BC = \sqrt{52} = 2\sqrt{13}$.

Why the others are wrong: B uses Pythagoras and ignores the $60°$ altogether. C adds the $2ab\cos C$ term instead of subtracting it. D simply adds the two sides.

Q19 — B. $\log_2(3 + k) = 0 \Rightarrow 3 + k = 2^0 = 1 \Rightarrow k = -2$.

Why the others are wrong: A has the sign of $k$ the wrong way round. C solves $3 + k = 0$ — but a logarithm is zero when its argument is 1, not 0. D gives the argument, 1, rather than $k$.

Q20 — B. $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$, so the top is $2\sqrt{2}$. Dividing by $\sqrt{2}$ leaves 2.

Why the others are wrong: A is the numerator, $2\sqrt{2}$, before dividing. C subtracts under the roots, as if $\sqrt{50} - \sqrt{18}$ were $\sqrt{32}$. D divides only one of the two terms.

Q21 — C. $\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{4}{9} = \tfrac{5}{9}$.

Why the others are wrong: A is $\cos\theta$ itself, not its square. B subtracts $\tfrac{2}{3}$ without squaring it first. D is $\sin^2\theta$.

Q22 — B. $500 \times 18.50 = 9\,250$, then keep 98 %: $9\,250 \times 0.98 = \text{R}9\,065$.

Why the others are wrong: A is the rand amount before the commission is taken off. C adds the 2 % instead of subtracting it. D takes off twice as much as the bank charges.

Q23 — A. $24 = a \cdot 2^3 = 8a$, so $a = 3$.

Why the others are wrong: B gives $2^3$, the power, rather than the coefficient in front of it. D divides 24 by 2 instead of by 8. C subtracts 3 from 24 instead of dividing.

Q24 — D. $\text{IQR} = Q_3 - Q_1$, so $Q_3 = 9 + 6 = 15$. The median is not needed.

Why the others are wrong: B adds the IQR to the median instead of to $Q_1$ — the median is not needed at all here. A does the same starting from 12 and 9 together. C subtracts where it should add.

Q25 — A. Even means $f(-x) = f(x)$, so $f(-3) = f(3) = 7$ and the sum is 14.

Why the others are wrong: B treats $f$ as ODD, so the two values cancel; an even function repeats them. C gives $f(3)$ alone. D has the sign of both terms wrong.

Q26 — B. $5^2 + 12^2 = 169 = 13^2$, so the triangle is right-angled with legs 5 and 12. Area $= \tfrac{1}{2}(5)(12) = 30$.

Why the others are wrong: A uses 13, the hypotenuse, as one of the perpendicular sides. C forgets to halve. D follows from no consistent pair of sides.

Q27 — D. $12 + 10 - 5 = 17$ do at least one, so $30 - 17 = 13$ do neither. (Subtract the overlap once, or it is counted twice.)

Why the others are wrong: C is the number who do AT LEAST ONE, not the number who do neither. A subtracts the overlap twice instead of once. B copies a number straight out of the question.

Q28 — C. $\tan 45° = 1$, $\sin 30° = \tfrac{1}{2}$, $\cos 60° = \tfrac{1}{2}$: $\dfrac{1 + \frac{1}{2}}{\frac{1}{2}} = 3$.

Why the others are wrong: A is the numerator, $\tfrac{3}{2}$, before dividing by $\cos 60°$. B leaves out $\sin 30°$. D gives $\tan 45°$ on its own.

Q29 — A. $\dfrac{T_5}{T_2} = r^3 = \dfrac{48}{6} = 8$, so $r = 2$.

Why the others are wrong: B gives $r^3 = 8$ rather than $r$ itself. C counts the three steps from $T_2$ to $T_5$. D takes the cube root of 8 as 4.

Q30 — B. A set of pairs fails to be a function when one input has two outputs. $k = 2$ repeats the input 2 (already paired with 5), so it is not a function.

Why the others are wrong: A is an OUTPUT in this set, not a repeated input. C and D each add a fresh input, which leaves it a function — only a repeated input with a different output breaks it.

Q31 — D. $(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + \sin 2\theta = 1 + \tfrac{3}{5} = \tfrac{8}{5}$. You never need $\theta$.

Why the others are wrong: A gives $\sin 2\theta$ back without adding the 1. B drops the cross term $2\sin\theta\cos\theta$ and keeps only the identity. C subtracts the cross term instead of adding it.

Q32 — C. $T_5 = S_5 - S_4 = (50 + 5) - (32 + 4) = 55 - 36 = 19$. (A term is the difference of two sums — no need to build the sequence.)

Why the others are wrong: B gives $S_5$, the SUM of the first five terms, not the fifth term. A and D come from subtracting $S_4$ from $S_5$ incorrectly — work both sums out fully before taking the difference.

Q33 — B. The turning point is at height 4. Subtracting 4 lowers it onto the $x$-axis, where the graph touches at exactly one point — and this works whether the parabola opens up or down. So $y = f(x) - 4$.

Why the others are wrong: A raises the graph, lifting the turning point clear of the axis. C shifts it sideways, which never changes how often it crosses. D reflects it, which keeps two crossings, just mirrored.

Q34 — C. Perpendicular to gradient $-2$ means gradient $\tfrac{1}{2}$: $\dfrac{7 - k}{4 - 2} = \tfrac{1}{2} \Rightarrow 7 - k = 1 \Rightarrow k = 6$.

Why the others are wrong: D uses the given gradient $-2$ instead of its perpendicular. A uses a gradient of 1. B has the subtraction the wrong way round, giving $k = 7 + 1$.

Q35 — B. The two marked angles are co-interior (same side of the transversal, between the parallels), so they are supplementary: $x = 180° - 62° = 118°$.

Why the others are wrong: A treats the pair as corresponding or alternate angles, which would make them equal; co-interior angles are supplementary. C takes the complement instead. D subtracts 62 from 190.

Q36 — B. Substitute: $x^2 + (x+1)^2 = 25 \Rightarrow 2x^2 + 2x - 24 = 0 \Rightarrow (x+4)(x-3) = 0$. So $x = 3$ gives $(3;\ 4)$. $(4;\ 3)$ is on the circle but not on the line.

Why the others are wrong: A, C and D all lie on the CIRCLE and none of them on the line: $3 \neq 4 + 1$, $0 \neq 5 + 1$ and $-4 \neq -3 + 1$. A point of intersection has to satisfy both equations, so test it against the line as well.

Q37 — D. $\log_{10} 5 = \log_{10}\dfrac{10}{2} = \log_{10}10 - \log_{10}2 = 1 - a$.

Why the others are wrong: B has the subtraction the wrong way round. A halves $a$, as if 5 were half of 10 in the logarithm as well as in the number. C multiplies the logarithm by 5.

Q38 — B. $\sqrt{8}\times\sqrt{2} = \sqrt{16} = 4$, $\sqrt{8}\div\sqrt{2} = \sqrt{4} = 2$ and $(\sqrt{8})^2 = 8$ — all rational. Only the sum resists: $\sqrt{8} + \sqrt{2} = 3\sqrt{2}$.

Why the others are wrong: A is $\sqrt{16} = 4$, C is $\sqrt{4} = 2$ and D is 8 — all rational. Multiplying, dividing and squaring surds can remove the root; adding them cannot.

Q39 — B. The perpendicular from $O$ bisects the chord, so the half-chord is 12. Distance $= \sqrt{13^2 - 12^2} = 5$. Area $= \tfrac{1}{2}(24)(5) = 60\ \text{cm}^2$.

Why the others are wrong: A uses the radius, 13, as the height instead of the perpendicular distance, 5. C forgets to halve. D follows from no consistent base and height.

Q40 — C. Coefficients: $\dfrac{3 \times 4}{6} = 2$. Exponents: $5 + (-2) - (-4) = 7$. So $2 \times 10^7$.

Why the others are wrong: A never divides the coefficients: $3 \times 4$ is left as 12, and 12 is not between 1 and 10, so it is not standard form either. B adds the $-4$ instead of subtracting it. D gets the signs of the exponents wrong throughout.

Q41 — D. The six values must total $6 \times 7 = 42$. The five known ones sum to 34, so $x = 8$.

Why the others are wrong: A is the TOTAL of all six values, not the missing one. B assumes the missing value must equal the mean. C comes from a slip in $42 - 34$.

Q42 — C. Linear scale factor $= \dfrac{30}{18} = \dfrac{5}{3}$, so areas scale by its square: $27 \times \left(\dfrac{5}{3}\right)^2 = 27 \times \dfrac{25}{9} = 75\ \text{cm}^2$.

Why the others are wrong: A scales the area by the LINEAR factor $\tfrac{5}{3}$ instead of its square. B doubles. D squares 2 rather than $\tfrac{5}{3}$.

Q43 — A. $S_\infty = \dfrac{a}{1 - r}$: $12 = \dfrac{8}{1 - r}$, so $1 - r = \tfrac{2}{3}$ and $r = \tfrac{1}{3}$.

Why the others are wrong: B gives $1 - r$ rather than $r$. C is greater than 1, so that series would not converge at all. D would give a sum of 16, not 12.

Q44 — B. $AB$ is a diameter, so $\hat{C} = 90°$ (angle in a semicircle). Then $\hat{B} = 180° - 90° - 35° = 55°$.

Why the others are wrong: C is the angle at $C$, the right angle, not the angle at $B$. A assumes the triangle is isosceles. D halves the remaining $90°$ as though the other two angles were equal.

Q45 — A. $4^{x+1} = 2^{2x+2}$, so $\dfrac{2^{2x+2}}{2^{x-1}} = 2^{x+3}$. Then $x + 3 = 9 \Rightarrow x = 6$.

Why the others are wrong: B reads the 9 straight off the right-hand side. C solves $x + 3 = 6$. D doubles the answer. Write $4^{x+1}$ as $2^{2x+2}$ first — the two sides cannot be compared until they share a base.

Q46 — D. Six values, so split into halves of three. Lower $\{2;5;8\} \Rightarrow Q_1 = 5$; upper $\{10;14;18\} \Rightarrow Q_3 = 14$. $\text{IQR} = 14 - 5 = 9$.

Why the others are wrong: B is the RANGE, $18 - 2$, not the interquartile range. C uses the two middle values instead of the quartiles. A comes from placing $Q_1$ and $Q_3$ wrongly in the two halves.

Q47 — C. An increase of 15 % multiplies by 1.15, so reverse it by dividing: $276 \div 1.15 = \text{R}240$. (Taking 15 % off 276 gives R234.60 — the usual trap.)

Why the others are wrong: B takes 15 % OFF 276, which is the usual trap: a rise of 15 % is undone by dividing, not by subtracting the same percentage. A subtracts 15 rand. D divides by the wrong multiplier.

Q48 — A. $|x - 3| = -2$. An absolute value is a distance and can never be negative, so there are no solutions. (Isolate the absolute value before counting.)

Why the others are wrong: B would need the absolute value to be 0 and C would need it positive; here it must equal $-2$, which no absolute value can. D would need the variable to vanish from the equation.

Q49 — A. $3 = \dfrac{a}{-2} \Rightarrow a = -6$. A negative $a$ puts the branches in quadrants II and IV.

Why the others are wrong: B is where a POSITIVE $a$ puts the branches; here $a = -6$. C and D name the upper and lower halves of the plane, which is not how a hyperbola's branches sit — they are diagonally opposite.

Q50 — B. $\dfrac{9\,680}{8\,000} = 1.21 = (1 + i)^2$, so $1 + i = 1.1$ and $i = 10\%$.

Why the others are wrong: A is the TOTAL growth over both years, not the annual rate. C halves that total, which is not how compounding works. D comes from a slip in the square root of 1.21.

Q51 — D. Vieta: for $ax^2 + bx + c = 0$, $\alpha + \beta = -\dfrac{b}{a} = \dfrac{6}{2} = 3$. No need to solve the quadratic.

Why the others are wrong: A keeps the minus sign from $-\tfrac{b}{a}$ when $b$ is already negative. B is the PRODUCT of the roots, $\tfrac{c}{a}$. C forgets to divide by $a$.

Q52 — A. $f^{-1}(0)$ is the $x$ for which $f(x) = 0$: $3x - 6 = 0 \Rightarrow x = 2$. The inverse formula is never needed.

Why the others are wrong: C treats the inverse as a reciprocal — $f^{-1}$ is a function, not $\tfrac{1}{f}$. B reads off the constant term. D has the sign of the answer wrong.

Q53 — C. $\dfrac{\sqrt{x}}{\sqrt{2}} = 3 \Rightarrow \sqrt{x} = 3\sqrt{2} \Rightarrow x = 18$.

Why the others are wrong: A squares the 3 and stops, forgetting the $\sqrt{2}$. B multiplies 3 by 2. D squares 6 instead of $3\sqrt{2}$.

Q54 — A. $T_n = 6 + 4(n - 1) = 4n + 2$. Set $4n + 2 = 102 \Rightarrow n = 25$.

Why the others are wrong: D takes the VALUE of the term, 102, as its position. B and C are one term short and one too far — check by substituting back: the 24th term is 98 and the 26th is 106.

Q55 — D. $(x+3)(x-1) = 21 \Rightarrow x^2 + 2x - 24 = 0 \Rightarrow (x+6)(x-4) = 0$. $x = -6$ would make the width negative, so $x = 4$.

Why the others are wrong: B is the other root of the quadratic, and it makes the width $-7$ — a rectangle cannot have a negative side, so it must be discarded. A and C come from factorising $x^2 + 2x - 24$ wrongly.

Q56 — C. $\tfrac{4}{3}\pi r^3 = 36\pi \Rightarrow r^3 = 27 \Rightarrow r = 3$ cm.

Why the others are wrong: A gives $r^3 = 27$ rather than $r$. D gives the diameter. B takes the cube root of 27 as 9.

Q57 — A. Expand first: $(x^2 - x - 6) - (x^2 + 3x) = -4x - 6$. Then $-4x - 6 = 0 \Rightarrow x = -\tfrac{3}{2}$.

Why the others are wrong: D assumes the expression is a product of factors, so that setting one to zero would do; expanding first shows the squares cancel. B loses the sign. C gives the constant left after expanding.

Q58 — A. Two conditions: $x + 2 \geq 0 \Rightarrow x \geq -2$ (the root), and $x - 3 \neq 0$ (the denominator). Both must hold: $x \geq -2,\ x \neq 3$.

Why the others are wrong: D forgets that $x = 3$ makes the denominator zero. C forgets that the root needs a non-negative argument. B excludes $-2$, where the root is 0 and perfectly defined, and forgets the denominator too. Both conditions have to hold at once.

Q59 — B. Each year multiplies by 0.8. $0.8^3 = 0.512$ (still above half) and $0.8^4 = 0.4096$ (below), so 4 complete years.

Why the others are wrong: A stops at $0.8^3 = 0.512$, which is still above half. D stops at $0.8^2 = 0.64$. C goes one year further than necessary. The question asks for the first COMPLETE year below half.

Q60 — D. Adding 5 to every value shifts the whole distribution, so the mean, median and mode all rise by 5. Spread is unaffected — the standard deviation does not change.

Why the others are wrong: A, B and C all rise by 5, because adding a constant slides the whole distribution along without changing how spread out it is.

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Mock Exam 2

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The graph of $f(x) = x^2 + bx + 8$ has its axis of symmetry at $x = 3$. The value of $b$ is:

A) 6

B) $-6$

C) $-3$

D) 3

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A cube has a total surface area of $150\ \text{cm}^2$. Its volume is:

A) $100\ \text{cm}^3$

B) $25\ \text{cm}^3$

C) $216\ \text{cm}^3$

D) $125\ \text{cm}^3$

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An exterior angle of a triangle is $110°$ and one of the non-adjacent interior angles is $45°$. The other non-adjacent interior angle is:

A) $55°$

B) $70°$

C) $35°$

D) $65°$

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The roots of $x^2 - 5x + 6 = 0$ are $\alpha$ and $\beta$. The value of $\alpha^2 + \beta^2$ is:

A) 13

B) 25

C) 6

D) 37

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If $2x + y = 10$ and $x - y = 2$, the value of $xy$ is:

A) 8

B) 12

C) 6

D) 20

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A machine worth R60 000 depreciates at 10 % p.a. on the straight-line method. After how many years is its book value R30 000?

A) 3

B) 5

C) 6

D) 10

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For which values of $x$ is $x^2 < 4x$?

A) $0 < x < 4$

B) $x < 4$

C) $x < 0$ or $x > 4$

D) $-2 < x < 2$

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The sum of all the solutions of $2|x| - 3 = 7$ is:

A) 0

B) 10

C) 5

D) $-5$

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If $\log_2 x + \log_2(x - 2) = 3$, the value of $x$ is:

A) 8

B) $-2$

C) 4

D) 2

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In a geometric sequence, $T_2 = 6$ and $T_5 = 162$. The first term is:

A) 2

B) 3

C) 6

D) 18

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For the arithmetic sequence $2;\ 6;\ 10;\ \ldots$ the sum of the first $n$ terms is 200. The value of $n$ is:

A) 10

B) 20

C) 100

D) 14

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The expression $\log_a(xy) =$

A) $\log_a x \cdot \log_a y$

B) $(\log_a x)(\log_a y)$

C) $\log_a x - \log_a y$

D) $\log_a x + \log_a y$

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A line is perpendicular to $y = 4x - 1$ and passes through $(4;\ 2)$. Its equation is:

A) $y = -\dfrac{1}{4}x + 2$

B) $y = -4x + 18$

C) $y = \dfrac{1}{4}x + 1$

D) $y = -\dfrac{1}{4}x + 3$

Q56 — C. $\tfrac{4}{3}\pi r^3 = 36\pi \Rightarrow r^3 = 27 \Rightarrow r = 3$ cm.

Q57 — A. Expand first: $(x^2 - x - 6) - (x^2 + 3x) = -4x - 6$. Then $-4x - 6 = 0 \Rightarrow x = -\tfrac{3}{2}$.

Q58 — A. Two conditions: $x + 2 \geq 0 \Rightarrow x \geq -2$ (the root), and $x - 3 \neq 0$ (the denominator). Both must hold: $x \geq -2,\ x \neq 3$.

Q59 — B. Each year multiplies by 0.8. $0.8^3 = 0.512$ (still above half) and $0.8^4 = 0.4096$ (below), so 4 complete years.

Q60 — D. Adding 5 to every value shifts the whole distribution, so the mean, median and mode all rise by 5. Spread is unaffected — the standard deviation does not change.

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