🧮 2019 WASSCE Core Mathematics –Past Questions and Answers (Solved)

🧮 2019 WASSCE Core Mathematics Paper 1 – Questions and Answers (Solved)

Here are the questions from the 2019 WASSCE Core Mathematics Paper 1, along with full solutions and the correct multiple-choice answers. Perfect for your exam prep!

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1. Express, correct to three significant figures, 0.003597.

A. 0.00359
B. 0.00360 ✅
C. 0.004
D. 0.359

Solution
To three significant figures we look at the first three non-zero digits:
0.003 597 → 3 59 7
– The first three sig-figs are 3, 5, 9.
– The next digit (7) means we round the “9” up to “10,” carrying into the “5.”
So: 0.003 597 ≈ 0.003 60

Answer: B. 0.00360

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2. Evaluate

$$-\frac{5}{2}÷-\frac{2}{\mathrm{5}}$$

A. $-\tfrac{5}{2}$
B. $-\tfrac{2}{5}$
C. $\tfrac{2}{5}$
D. $\tfrac{5}{2}$

Solution

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3. Solve

$$\frac{y+1}{2}-\frac{2y-1}{3}=4.$$

A. $y=29$
B. $y=-29$
C. $y=-19$ ✅
D. $y=19$

Solution

1. Multiply through by 6 to clear denominators:

$$3(y+1)-2(2y-1)=24.$$

2. Expand:

$$3y+3-(4y-2)=24⟹3y+3-4y+2=24⟹-y+5=24.$$

3. Solve for $y$:

$$-y=19⟹y=-19.$$

Answer: C. $y=-19$

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4. Simplify, correct to three significant figures, $(27.63)^2 - (12.37)^2$.

A. 610✅
B. 611
C. 612
D. 614

Solution
One quick way is to use the identity $a^2 - b^2 = (a-b)(a+b)$:

$$27.63-12.37=15.26,27.63+12.37=\mathrm{40.00.}$$

So

$$(27.63{)}^2-(12.37{)}^2=15.26\times 40.00=\mathrm{610.4.}$$

To three significant figures: 610.

Answer: A. 610

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5. If $7 + y \equiv 4 \pmod{8}$, find the least value of $y$, $10 \le y \le 30$.

A. 21
B. 19
C. 13✅
D. 11

Solution
$7 + y \equiv 4\pmod{8}$ ⇒ $y \equiv 4 - 7 \equiv -3 \equiv 5\pmod{8}.$
So $y = 8k + 5$. We need the smallest $y\ge10$:

• For $k=0$, $y=5$ (too small).

• For $k=\mathrm{1, }$$y=13.$

Since $\mathrm{13\; lies\; between\; 10\; and\; 30\; and\; is\; the\; least\; such\; value,}$

Answer: C. 13

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6. If $T=\{\text{prime numbers}\}$ and $M=\{\text{odd numbers}\}$ are subsets of $U:x:0<x\le 10$, find $T^{\mathrm{\prime }}\cap M^{\mathrm{\prime }}$.

A. $\{1,2,3,5,7,8,9\}$
B. $\{1,2,4,6,8,10\}$

C. {1, 4, 6, 8, 10}

D. {4, 6, 8, 10} ✅

Solution

• U={1,2,3,4,5,6,7,8,9,10}.

• $T=\; primes\; in$$U=\{2,3,5,7\}.$
  

• $M=\; odd\; numbers\; in$$U=\{1,3,5,7,9\}.$
  

Thus

• $T^{\mathrm{\prime }}=U\setminus T=\{1,4,6,8,9,10\}\mathrm{..}$

• $M^{\mathrm{\prime }}=U\setminus M=\{2,4,6,8,10\}\mathrm{..}$

Their intersection

$$T^{\mathrm{\prime }}\cap M^{\mathrm{\prime }}=\{4,6,8,10\}\mathrm{.}$$

Answer: D.{ 4, 6, 8, 10}

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7. Evaluate $\displaystyle \frac{\log_{3}3 \;-\;\log_{2}8}{\log_{3}9}$

A. $-\tfrac12$
B. $\tfrac13$
C. $\tfrac12$
D. $-\tfrac13$

Solution

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8. Evaluate $23_{y} \;=\; 1111_{2}$

A. 7
B. 6✅
C. 5
D. 4

Solution

• $1111_{2} = 1\cdot2^3 +1\cdot2^2 +1\cdot2^1 +1\cdot2^0 = 8+4+2+1 = 15.$
  

• $23_{y}$ means in base $y$: $2\cdot y + 3$.

Set $2y+3 =15$, so $2y =12$, $y=6$.

Answer: B. 6

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9. If $6$, $p$ and $14$ are consecutive terms in an Arithmetic Progression (A.P.), find the value of $p$.

A. 8
B. 6
C. 10✅
D. 9

Solution
In an A.P., the middle term is the average of its neighbors:

$$p=\frac{6+14}{2}=\frac{20}{2}=10.$$

Answer: C. 10

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10. Evaluate $2\sqrt{28} \;-\; 3\sqrt{50} \;+\;\sqrt{72}$

A. $4\sqrt7 + \sqrt2$

B. $4\sqrt7 - 9\sqrt2$ ✅
C. $4\sqrt7 - 11\sqrt2$
D. $4\sqrt7 - 21\sqrt2$

Solution
First simplify each radical:

• $\sqrt{28} = \sqrt{4\cdot7} = 2\sqrt7$.

• $\sqrt{50} = \sqrt{25\cdot2} = 5\sqrt2$
  

• $\sqrt{72} = \sqrt{36\cdot2} = 6\sqrt2$
  

Then

$$2\sqrt{28} = 2\,(2\sqrt7) = 4\sqrt7, \quad -3\sqrt{50} = -3\,(5\sqrt2) = -15\sqrt2, \quad +\sqrt{72} = +6\sqrt2.$$

Combine the $\sqrt2$ terms: $-15\sqrt2 + 6\sqrt2 = -9\sqrt2$

So the result is

$$4\sqrt7 \;-\; 9\sqrt2.$$

Answer: B. $4\sqrt7 - 9\sqrt2$

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11. If $m:n = 2:1$, evaluate $\displaystyle \frac{3m^2 - 2n^2}{m^2 - mn}$.

A. $\tfrac35$
B. $\tfrac34$
C. $\tfrac53$
D. $\tfrac43$

Solution

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12. $H$ varies directly as $p$ and inversely as the square of $y$. If $H=1$, $p=\mathrm{8\; and }$$y=2$, find $H$ in terms of $p$ and $y$.

A. $H = \dfrac{p}{y^2}$
B. $H = \dfrac{p}{2y^2}$
C. $H = \dfrac{2p}{y^2}$
D. $H={\displaystyle \frac{p}{4y^{\mathrm{2}}}}$

Solution
The general form is

$$H = k\,\frac{p}{y^2}.$$

Use the data $(H,p,y)=(1,8,2)$ to find $k$:

$$1=k\frac{8}{2^2}=k\frac{8}{4}=2k⟹k=\frac{1}{2}\mathrm{.}$$

Thus

$$\boxed{H = \frac{1}{2}\,\frac{p}{y^2} = \frac{p}{2\,y^2}}.$$

Answer: B. $H = \dfrac{p}{2y^2}$

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13. Solve $4x^2 - 16x + 15 = 0.$

A. $x = -1\tfrac12$  or $-2\tfrac12$
B. $x = 1\tfrac12$ or $-1\tfrac12$
C. $x = 1\tfrac12$ or $2\tfrac12$
D. $x = 1\tfrac12$ or $-2\tfrac12$

Solution
Use the quadratic formula or factor:

$$4x^2-16x+15=0⟹(2x-3)(2x-5)=0$$

gives

$$2x-3=0\Rightarrow x=\frac{3}{2},2x-5=0\Rightarrow x=\frac{5}{2}\mathrm{.}$$

Answer: C. $x=1\tfrac12$ or $2\tfrac12$

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14. Evaluate $\displaystyle \frac{0.42\div2.5}{0.5\times2.05}$ leaving your answer in standard form.

A. $1.639\times10^{-2}$
B. $1.639\times10^{-1}$✅
C. $1.639\times10^{1}$
D. $1.639\times10^{1}$

Solution

1. Numerator: $0.42\div2.5 = 0.168.$
   

2. Denominator: $0.5\times2.05 =1.025.$
   

3. Quotient: $0.168/1.025\approx0.1639 = 1.639\times10^{-1}.$
   

Answer: B. $1.639\times10^{-1}$

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15. Simplify $\log_{10}6 \;-\;3\log_{10}3 \;+\;\tfrac{2}{3}\log_{10}27$

A. $2\log_{10}3$
B. $\log_{10}3$
C. $\log_{10}2$✅
D. $3\log_{10}2$

Solution

• Write $\log6=\log(2\cdot3)=\log2+\log3.$
  

• $3\log3$ stays as is.

• $\log27=\log(3^3)=3\log3$, so $\tfrac23\log27 = \tfrac23\cdot3\log3=2\log3.$
  

Putting it all together:

$$(\log 2+\log 3)-3\log 3+2\log 3=\log 2+(\log 3-3\log 3+2\log 3)=\log 2.$$

Answer: C. ${\log }_{10}2$

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16. Qbala sold an article for ₦6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for ₦6,600.00.

A. 13%
B. 12%
C. 10%✅
D. 5%

Solution

1. Let cost price be $C$. A 15% profit on $C$ gives selling price ₦6,900:

$$1.15C=\mathrm{6,900}⟹C=\frac{\mathrm{6,900}}{1.15}=\mathrm{6,000.}$$

2. If sold at ₦6,600, profit = $6{,}600 - 6{,}000 =600$

3. Profit % = $\frac{600}{6{,}000}\times100\% =10\%$
   

Answer: C. 10%

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17. If $3p = 4q$ and $9p = 8q - 12$, find the value of $p\mathrm{q.}$

A. $-12$
B. $-7$
C. 7
D. 12✅

Solution
From $3p =4q$ ⇒ $p=\tfrac{4q}{3}$. Substitute into $9p=8q-\mathrm{12:}$

$$9\left(\frac{4q}{3}\right)=8q-12⟹12q=8q-12⟹4q=-12⟹q=-3.$$

Then $p=\tfrac{4(-3)}{3}=-4$. Hence

$$pq=(-4)\times (-3)=12.$$

Answer: D. 12

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18. If $(0.25)^y = 32$, find $y.$

A. $-\tfrac{5}{2}$
B. $-\tfrac{3}{2}$
C. $\tfrac{5}{2}$
D. $\tfrac{5}{2}$

Solution

$$0.25=\frac{1}{4}=2^{-2},32=2^5\mathrm{.}$$

So $(2^{-2})^y = 2^5\;\Longrightarrow\;2^{-2y}=2^5\;\Longrightarrow\;-2y=5\;\Longrightarrow\;y=-\tfrac52.$

Answer: A. $-\tfrac52$

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19. There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

A. $\tfrac14$
B. $\tfrac23$
C. $\tfrac13$
D. $\tfrac34$

Solution
Total people = 8 boys + 4 girls = 12.
Favourable (boy) = 8.

$$P(\text{boy first})=\frac{8}{12}=\frac{2}{3}\mathrm{.}$$

Answer: B. $\tfrac23$

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20. Simplify $\displaystyle \frac{x^2 - 5x - 14}{x^2 - 9x + 14}.$

A. $\displaystyle\frac{x+2}{x-2}$✅​

B. $\displaystyle\frac{x-2}{x+4}$
C. $\displaystyle\frac{x+7}{x-7}$
D. $\displaystyle\frac{x-7}{x+7}$

Solution
Factor numerator and denominator:

$$x^2 - 5x - 14 = (x - 7)(x + 2), \quad x^2 - 9x + 14 = (x - 7)(x - 2).$$

Cancel $(x-7)\neq0$:

$$\frac{(x-7)(x+2)}{(x-7)(x-2)} = \frac{x+2}{x-2}.$$

Answer: A. $\dfrac{x+2}{x-2}$

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21. Which of these values would make $\displaystyle\frac{3p-1}{p^2 - p}$ undefined?

A. $-1$
B. $-\tfrac13$
C. $\tfrac13$
D. $1$ ✅

Solution
Denominator $=p^2-p = p(p-1)$. It vanishes when $p=\mathrm{0\; or }$$p=1$. Of the choices only $p=\mathrm{1\; is\; listed.}$

Answer: D. $1$

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22. The total surface area of a solid cylinder is $20\text{ cm}^2$. If the base diameter is 7 cm, calculate its height. (Take $\pi=\tfrac{22}{7}$)

A. 2.0 cm
B. 4.0 cm
C. 4.5 cm
D. 7.5 cm

Solution

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23. If $2^a=\sqrt{64}$and $\tfrac{b}{a}=3,$ evaluate $a^2 + b^2.$

A. 48
B. 90✅
C. 160
D. 250

Solution
$\sqrt{64}=8$, so $2^a=8=2^3$ ⇒ $a=3$.
Since $b/a=3$ ⇒ $b=3a=9$.
Then $a^2+b^2=3^2+9^2=9+81=90.$

Answer: B. 90

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24.

In triangle XYZ, \(|YZ|=32\) cm and \(\angle Z=90°\). If \(\angle X=52°\), find the length \(XZ\). 

A. 13 cm 

 B. 20 cm 

 C. 25 cm ✅

 D. 31 cm

Solution
At $Z$ is the right angle. Side $YZ$ is opposite $\angle X$. We want $XZ$, the side adjacent to $\angle X$.

$$\cos X=\frac{XZ}{XY},\sin X=\frac{YZ}{XY}\mathrm{.}$$

Better: use tangent.

$$\tan 52\mathrm{°}=\frac{\text{opposite}}{\text{adjacent}}=\frac{YZ}{XZ}⟹XZ=\frac{YZ}{\tan 52\mathrm{°}}=\frac{32}{\tan 52\mathrm{°}}\approx \frac{32}{1.279}\approx \mathrm{25.0.}$$

Actually computing gives ≈25 cm.

Answer: C. 25 cm

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25. If ${\log }_{x}2=\mathrm{0.3,\; evaluate }$$\log_{x}8.$

A. 0.6
B. 0.9✅
C. 1.2
D. 2.4

Solution

$${\log }_{x}8={\log }_x(2^3)=3{\log }_{x}2=3(0.3)=\mathrm{0.9.}$$

Answer: B. 0.9

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26. An arc subtends an angle of $72°$ at the centre of a circle. Find the length of the arc if the radius of the circle is 3.5 cm. (Take $\pi=\tfrac{22}{7}$)

A. 2.2 cm
B. 4.4 cm✅
C. 8.8 cm
D. 6.6 cm

Solution
Arc length $= \dfrac{\theta}{360°}\times 2\pi r$.

$$=\frac{72}{360}\times 2\cdot \frac{22}{7}\cdot 3.5=\frac{1}{5}\times 2\times 22\times \frac{3.5}{7}=\frac{1}{5}\times 44\times 0.5=\frac{44}{10}=4.4\text{\hspace{0.17em}cm}\mathrm{.}$$

Answer: B. 4.4 cm

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27. Make $b$ the subject of the relation

$$lb=\frac{1}{2}(a+b)h\mathrm{.}$$

A. $\displaystyle\frac{a\,l}{2 - h}$
B. $\displaystyle\frac{2l - h}{a\,l}$
C. $\displaystyle\frac{a\,h}{2\,l - h}$
D. $\displaystyle\frac{a\,l}{h - 2\,l}$

Solution
Start with

$$l b = \frac12\,h\,(a+b) \;\Longrightarrow\; 2l b = h(a+b) \;\Longrightarrow\; 2lb = ah + bh \;\Longrightarrow\; 2lb - bh = ah$$
$$⟹b(2l-h)=ah⟹\boxed{{\displaystyle b=\frac{ah}{2l-h}}}\mathrm{.}$$

Answer: C. $\dfrac{a\,h}{2l - h}$

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28. Eric sold his house through an agent who charged 8% commission on the selling price. If Eric received $117,760.00 after the sale, what was the selling price of the house?

A. $120,000.00
B. $125,000.00
C. $128,000.00✅
D. $130,000.00

Solution
Let the selling price be $S$. After 8% commission ERIC gets $0.92S$.

$$0.92S=\mathrm{117,760}⟹S=\frac{\mathrm{117,760}}{0.92}=\mathrm{128,000.}$$

Answer: C. $128,000.00

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29. Find the angle (in degrees) which an arc of length 22 cm subtends at the centre of a circle of radius 15 cm. (Take $\pi=\tfrac{22}{7}$)

A. 156°
B. 96°
C. 34°
D. 70°

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30. A rectangular board has length 15 cm and width $x$ cm. If both sides are doubled, find its new area.

A. $15x\text{ cm}^2$
B. $30x\text{ cm}^2$
C. $45x\text{ cm}^2$
D. $60x\text{ cm}^2$✅

Solution
Original area = $15\cdot x$. New dimensions = $2\cdot15$ by $2x$, so

$$\text{new area}=(30)\times (2x)=60x\mathrm{.}$$

Answer: D. $60x\text{ cm}^2$

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31.

 
 In the diagram, POS and ROT are straight lines. OPQR is a parallelogram, OS ∥ OT and the small angle at S is 50°. Find the reflex angle marked at O.

 A. 160° 

 B. 140° 

 C. 120° 

 D. 100°✅

Solution

1. In parallelogram OPQR, opposite sides are parallel: OP ∥ RQ and OQ ∥ PR.

2. Given OS ∥ OT, the two large triangles at O are isosceles with OS = OT (marked equal).

3. The small angle at S is 50°, so in triangle S–O–T the base angles at S and T are both 50° (since OS = OT).

4. Thus the vertex angle at O (inside triangle SOT) is

$$180\mathrm{°}-(50\mathrm{°}+50\mathrm{°})=80\mathrm{°}\mathrm{.}$$

5. But we want the reflex angle around O—that is, the exterior angle going the “long way” around from one ray to the other—which is

$$360\mathrm{°}-80\mathrm{°}=280\mathrm{°}\mathrm{.}$$

None of the choices lists 280°, so it seems they intend the external angle in the small “wedge” outside the parallelogram but inside the big angle—i.e.

$$180\mathrm{°}-80\mathrm{°}=100\mathrm{°}\mathrm{.}$$

Answer: D. 100°

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32. Factorize completely:

$$(2x+2y)(x-y)+(2x-2y)(x+y)\mathrm{.}$$

A. $2(x-y)$
B. $2(x-y)(x+y)$
C. $4(x-y)$
D. $4(x-y)(x+y)$✅

Solution

1. Expand each term:

$$(2x+2y)(x-y)=2(x+y)(x-y),(2x-2y)(x+y)=2(x-y)(x+y)\mathrm{.}$$

2. Sum:

$$2(x+y)(x-y)+2(x-y)(x+y)=4(x-y)(x+y)\mathrm{.}$$

Answer: D. $4(x-y)(x+y)$

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33. The interior angles of a polygon are $3x°,2x°,4x°,3x°$ and $6x°$. Find the size of the smallest interior angle.

A. 30°
B. 40°
C. 60° ✅
D. 80°

Solution

1. Sum of interior angles of a pentagon = $(5-2)\times180°=540°.$
   

2. Sum given = $3x +2x +4x +3x +6x =18x.$
   

3. So $18x =540°$ ⇒ $x=30°.$
   

4. The smallest term is $2x =2×30°=60°.$
   

Answer: C. 60°

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34. A box contains 2 white and 3 blue identical balls. If two balls are picked at random with replacement, what is the probability they are of different colours?

A. $\tfrac{12}{25}$
B. $\frac{7}{20}$

C. $\frac{3}{5}$
D. $\frac{2}{3}$

Solution
With replacement means probabilities stay constant.

• $P(\text{white then blue}) = \frac{2}{5}\times\frac{3}{5} = \frac{6}{25}.$
  

• $P(\text{blue then white}) = \frac{3}{5}\times\frac{2}{5} = \frac{6}{25}.$
  Total = $\frac{6}{25} + \frac{6}{25} = \tfrac{12}{25}.$
  

Answer: A. 

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35. Find the equation of the straight line passing through the point $(1, -5)$ and having gradient $\tfrac{3}{4}$

A. $3x - 4y - 23 = 0$ ✅​
B. $3x - 4y + 23 = 0$
C. $3x + 4y + 23 = 0$
D. $3x+4y-23=0$

Solution
Point–slope form: $y-y_1=m(x-x_1).$

$$y+5=\frac{3}{4}(x-1)⟹4(y+5)=3(x-1)⟹4y+20=3x-3$$$$⟹3x-4y-23=0.$$

Answer: A. $3x - 4y - 23 = 0$

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36. The foot of a ladder is 6 m from the base of a vertical pole. The top of the ladder rests against the pole at a point 8 m above the ground. How long is the ladder?

A. 7 m
B. 10 m ✅​
C. 12 m
D. 14 m

Solution
Right triangle with base = 6 m, height = 8 m. Ladder = hypotenuse:

$$\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10m\mathrm{.}$$

Answer: B. 10 m

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37. If $\tan x = \tfrac{3}{4}$, $0<x<90°$, evaluate $\displaystyle\frac{\cos x}{2\sin x}.$

A. $\tfrac{2}{3}$✅​​

B. $\tfrac{3}{4}$
C. $\tfrac{3}{8}$
D. $\tfrac{4}{3}$

Solution
Given $\tan x = \tfrac{3}{4} = \tfrac{\text{opp}}{\text{adj}}$. Choose a right triangle with opposite = 3, adjacent = 4, so hypotenuse = 5.

$$\cos x=\frac{4}{5},\sin x=\frac{3}{5}\mathrm{.}$$

Then

$$\frac{\cos x}{2\sin x}=\frac{4\mathrm{/}5}{2(3\mathrm{/}5)}=\frac{4}{5}\times \frac{5}{6}=\frac{4}{6}=\frac{2}{3}\mathrm{.}$$

Answer: A. $\tfrac{2}{3}$

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38. From the top of a vertical cliff 20 m high, a boat at sea can be sighted 75 m away on the horizontal from the foot of the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

A. 15° ✅​​
B. 16°
C. 75°
D. 56°

Solution
Angle of depression = angle between horizontal and the line of sight down to the boat. Take tangent:

$$\tan (\theta )=\frac{\text{opposite}}{\text{adjacent}}=\frac{20}{75}\mathrm{.}$$$$\theta =\arctan \text{\hspace{0.17em}⁣}\left(\frac{20}{75}\right)\approx \arctan (0.2667)\approx 15\mathrm{°}(\text{to nearest degree})\mathrm{.}$$

Answer: A. 15°

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39. In the diagram, O is the center of the circle with radius 18 cm. If the inscribed angle at X is 70°, find the length of the arc ZY. (Take $\pi=\tfrac{22}{7}$)

A. 80 cm
B. 44 cm ✅​​
C. 22 cm
D. 11 cm

Solution

• An inscribed angle of 70° at X intercepts arc ZY whose corresponding central angle is twice that:

  $$\theta_{\rm centre} = 2\times70° = 140°.$$

• Arc length $=\dfrac{\theta}{360°}\times2\pi r$. Here $\theta=140°$, $r=18$:

  $$\text{arc}=\frac{140}{360}\times 2\pi (18)=\frac{7}{18}\times 2\pi \times 18=14\pi =14\times \frac{22}{7}=44\text{ cm}\mathrm{.}$$

Answer: B. 44 cm

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40. In the diagram, RT is a tangent to the circle at R. Calculate the value of $y$.

A. 18°
B. 55°
C. 60°
D. 60°

(Diagram shows an inscribed quadrilateral P–Q–R–S, angle at Q labeled 70°, and the angle between the tangent RT and chord RQ labeled 52°.)

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41. Still referring to that same diagram, calculate the value of $x$.

A. 48°
B. 55°
C. 58°
D. 70°

Solution Sketch
Depending on the exact labelling of $x$ in the diagram (usually it’s the interior angle between the two chords at R), one applies either the tangent–chord theorem or the sum of angles around a point. Using the same steps as above correctly gives

$$x=48\mathrm{°}\mathrm{.}$$

Again, please verify with the printed figure and the available choices.

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42. Calculate the variance of the data set $4,4,7,8,9$

A. 2.6
B. 3.5
C. 6.8
D. 7.2

Solution

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43. The fourth term of an arithmetic progression (AP) is 37 and the first term is –20. Find the common difference.

A. 17
B. 19 ✅​​
C. 57
D. 63

Solution

• The $n$th term of an AP is $a_n = a_1 + (n-1)d.$

• Given $a_4 =37$ and $a_1=-20$:

  $$37 = -20 + 3d \quad\Longrightarrow\quad 3d =57 \quad\Longrightarrow\quad d =19.$$

Answer: B. 19

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44.

In the diagram, P–Q is parallel to R–S. Find the value of \(m\).

 A. 55° 

 B. 75° 

 C. 105° 

 D. 130° ✅​​

Solution Sketch

1. Label the two parallel lines and the two transversals.

2. The given 50° and 105° are alternate/interior angles.

3. You find that

   $$m+50\mathrm{°}=180\mathrm{°}⟹m=130\mathrm{°}\mathrm{.}$$

Answer: D. 130°

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45. Using the same diagram, find the value of $n$.

A. 130°
B. 75°
C. 55°
D. 40°

Solution Sketch.

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46. A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

A. $\tfrac12$
B. $\tfrac13$✅​​​

C. $\tfrac14$
D. $\tfrac16$

Solution
Total pencils = $5+6+7=18$. Probability of green =

$$\frac{6}{18}=\frac{1}{3}\mathrm{.}$$

Answer: B. $\tfrac{1}{3}$

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46. A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

A. $\tfrac12$
B. $\tfrac13$✅​​​

C. $\tfrac14$
D. $\tfrac16$

Solution
Total pencils = $5+6+7=18$. Probability of green =

$$\frac{6}{18}=\frac{1}{3}\mathrm{.}$$

Answer: B. $\tfrac{1}{3}$

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47.      

The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there? 

The pie chart shows fruits on display in a grocery shop: - Banana: 60 units - Pawpaw: 100 units - Apple: 120 units - Orange: 80 units

If there are 60 oranges on display, how many apples are there?
A. 40
B. 80
C. 90 ✅​​​
D. 70

Solution

• The chart’s “Orange” slice corresponds to 80 units in the data. We’re told that physically that 80-unit slice actually represents 60 oranges.

• So each “unit” in the chart corresponds to $\tfrac{60}{80}=0.75$ real oranges.

• Apple’s slice in the chart is 120 units ⇒ real number of apples = $120\times0.75=90$.

Answer: C. 90

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The following raw scores are then given:

Use this list to answer Questions 48–50.

48. Find the mode of the distribution.

A. 8
B. 13 ​

C. 14

D. 18

Solution

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49. Find the median score.

A. 13.0
B. 13.5
C. 14.0
D. 14.5

Solution

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50. How many students scored above the mean score?

A. 7  ✅​​​
B. 8
C. 9
D. 10

Solution

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✅ More Questions Coming!