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Objective paper · 49 questions · partial

WAEC · 2021 · May/June · General Maths · Paper 1

Topics include Approximation & error, Number bases, Indices & standard form, Logarithms, Sequences & series (AP, GP), Sets & Venn diagrams.

Our copy of this paper is missing question 50.

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Question 1

Correct 0.007985 to three significant figures.

Worked solution (try it first)
  1. The zeros after the point are not significant, so the first three significant figures are 7, 9 and 8.
  2. The next figure is 5, so round the 8 up to 9: 0.00799, option C.

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Question 2

Simplify (11two)2(11_{\text{two}})^2.

Worked solution (try it first)
  1. Change to base ten: 11two=2+1=311_{\text{two}} = 2 + 1 = 3.
  2. Square it: 32=93^2 = 9.
  3. Change back: 9=8+19 = 8 + 1, so the answer is 100121001_2, option A.

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Question 3

Solve 22x+1=322^{\sqrt{2x + 1}} = 32.

Worked solution (try it first)
  1. 32=2532 = 2^5, so the powers are equal: 2x+1=5\sqrt{2x + 1} = 5.
  2. Square both sides: 2x+1=252x + 1 = 25.
  3. Subtract 1 and halve: 2x=242x = 24, so x=12x = 12, option C.

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Question 4

If log⁡102=m\log_{10} 2 = m and log⁡103=n\log_{10} 3 = n, find log⁡1024\log_{10} 24 in terms of mm and nn.

Worked solution (try it first)
  1. Write 24 using 2 and 3: 24=8×3=23×324 = 8 \times 3 = 2^3 \times 3.
  2. The log of a product is the sum of the logs, and a power comes down in front: log⁡24=3log⁡2+log⁡3\log 24 = 3\log 2 + \log 3.
  3. So log⁡24=3m+n\log 24 = 3m + n, option A.

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Question 5

Find the 5th term of the sequence 2,5,10,17,…2, 5, 10, 17, \ldots

Worked solution (try it first)
  1. The gaps between terms are 3, 5 and 7, going up by 2 each time.
  2. So the next gap is 9, and the 5th term is 17+9=2617 + 9 = 26, option D.
  3. Check with the rule n2+1n^2 + 1: 52+1=265^2 + 1 = 26.

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Question 6

If P={−3<x<1}P = \{-3 < x < 1\} and Q={−1<x<3}Q = \{-1 < x < 3\}, where xx is an integer, find P∩QP \cap Q.

Worked solution (try it first)
  1. The inequalities are strict, so the end numbers are left out.
  2. P={−2,−1,0}P = \{-2, -1, 0\}.
  3. Q={0,1,2}Q = \{0, 1, 2\}.
  4. Only 0 is in both, so P∩Q={0}P \cap Q = \{0\}, option A.

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Question 7

Factorize 6pq−3rs−3ps+6qr6pq - 3rs - 3ps + 6qr.

Worked solution (try it first)
  1. Rearrange so the pairs share a factor: 6pq+6qr−3ps−3rs6pq + 6qr - 3ps - 3rs.
  2. Factorise each pair: 6q(p+r)−3s(p+r)6q(p + r) - 3s(p + r).
  3. Take out p+rp + r and then 3: (p+r)(6q−3s)=3(2q−s)(p+r)(p + r)(6q - 3s) = 3(2q - s)(p + r), option C.

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Question 8

What number should be subtracted from the sum of 2162\frac16 and 27122\frac{7}{12} to give 3143\frac14?

Worked solution (try it first)
  1. Add the two numbers.
  2. The LCD of 6 and 12 is 12: 2212+2712=49122\frac{2}{12} + 2\frac{7}{12} = 4\frac{9}{12}
    =434= 4\frac34.
  3. The number is what you take from 4344\frac34 to leave 3143\frac14: 434−314=124=1124\frac34 - 3\frac14 = 1\frac24 = 1\frac12, option B.

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Question 9

Mensah is 5 years old and Joyce is thrice as old as Mensah. In how many years will Joyce be twice as old as Mensah?

Worked solution (try it first)
  1. Joyce is three times Mensah's age: 3×5=153 \times 5 = 15 years old.
  2. In nn years they will be 5+n5 + n and 15+n15 + n.
  3. Joyce is then twice Mensah's age: 15+n=2(5+n)15 + n = 2(5 + n).
  4. Expand: 15+n=10+2n15 + n = 10 + 2n.
  5. Take nn and 10 from both sides: n=5n = 5.
  6. So in 5 years (ages 10 and 20), option C.

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Question 10

If 16×2(x+1)=4x×8(1−x)16 \times 2^{(x + 1)} = 4^x \times 8^{(1 - x)}, find the value of xx.

Worked solution (try it first)
  1. Write everything as a power of 2.
  2. Left: 16×2x+1=24×2x+116 \times 2^{x + 1} = 2^4 \times 2^{x + 1}
    =2x+5= 2^{x + 5}.
  3. Right: 4x=22x4^x = 2^{2x} and 81−x=23−3x8^{1 - x} = 2^{3 - 3x}, so the right side is 22x+3−3x=23−x2^{2x + 3 - 3x} = 2^{3 - x}.
  4. Equate the powers: x+5=3−xx + 5 = 3 - x, so 2x=−22x = -2 and x=−1x = -1, option D.

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Question 11

The circumference of a circular track is 9 km9\text{ km}. A cyclist rides round it a number of times and stops after covering a distance of 302 km302\text{ km}. How far is the cyclist from the starting point (along the track)?

Worked solution (try it first)
  1. Each lap is 9 km.
  2. Divide 302 by 9: 9×33=2979 \times 33 = 297, with 302−297=5302 - 297 = 5 km left over.
  3. After 33 full laps the cyclist is back at the start, then rides 5 km more.
  4. So the cyclist is 5 km from the start, option A.

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Question 12

Simplify 27−147+7212\sqrt7 - \dfrac{14}{\sqrt7} + \dfrac{7}{\sqrt{21}}.

Worked solution (try it first)
  1. Rationalise: 147=1477\dfrac{14}{\sqrt7} = \dfrac{14\sqrt7}{7}, which is 272\sqrt7.
  2. So the first two terms cancel.
  3. Rationalise the last term: 721=72121\dfrac{7}{\sqrt{21}} = \dfrac{7\sqrt{21}}{21}, which is 213\dfrac{\sqrt{21}}{3}.
  4. So the value is 213\dfrac{\sqrt{21}}{3}, option C.

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Question 13

If 4x+2y=164x + 2y = 16 and 6x−2y=46x - 2y = 4, find the value of (y−x)(y - x).

Worked solution (try it first)
  1. Add the two equations so the yy terms cancel: 10x=2010x = 20, so x=2x = 2.
  2. Put x=2x = 2 into 4x+2y=164x + 2y = 16: 8+2y=168 + 2y = 16, so 2y=82y = 8 and y=4y = 4.
  3. So y−x=4−2=2y - x = 4 - 2 = 2, option B.

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Question 14

In the diagram, ∠ABC\angle ABC and ∠BCD\angle BCD are right angles, ∠BAD=t\angle BAD = t and ∠EDF=70∘\angle EDF = 70^\circ. Find the value of tt.

t70°ABCDEF
Worked solution (try it first)
  1. ADFADF and CDECDE are straight lines, so ∠ADC=∠EDF=70∘\angle ADC = \angle EDF = 70^\circ (vertically opposite).
  2. ABAB and CDCD are both perpendicular to BCBC, so AB∥CDAB \parallel CD.
  3. So tt and ∠ADC\angle ADC are co-interior: t=180∘−70∘=110∘t = 180^\circ - 70^\circ = 110^\circ, option D.

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Question 15

The sum of the interior angles of a regular polygon with kk sides is (3k−10)(3k - 10) right angles. Find the size of each exterior angle.

Worked solution (try it first)
  1. A right angle is 90∘90^\circ, and the angle sum is (k−2)×180∘(k - 2) \times 180^\circ.
  2. So (k−2)×180=(3k−10)×90(k - 2) \times 180 = (3k - 10) \times 90.
  3. Divide by 90: 2k−4=3k−102k - 4 = 3k - 10, so k=6k = 6.
  4. Each exterior angle is 360∘÷6=60∘360^\circ \div 6 = 60^\circ, option A.

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Question 16

Make uu the subject in x=2u−33u+2x = \dfrac{2u - 3}{3u + 2}.

Worked solution (try it first)
  1. Multiply both sides by 3u+23u + 2: 3ux+2x=2u−33ux + 2x = 2u - 3.
  2. Collect the uu terms on the right: 2x+3=2u−3ux2x + 3 = 2u - 3ux, so 2x+3=u(2−3x)2x + 3 = u(2 - 3x).
  3. Divide by 2−3x2 - 3x: u=2x+32−3xu = \dfrac{2x + 3}{2 - 3x}, option C.

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Question 17

A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of ₦22,800.00 was paid as import duty, calculate the cost of the engine.

Worked solution (try it first)
  1. 100 kobo make ₦1, so 38 kobo in the naira means a duty of ₦0.38 on every ₦1 of cost: duty =0.38C= 0.38C.
  2. So 0.38C=22 8000.38C = 22\,800.
  3. Divide both sides by 0.38: C=60 000C = 60\,000.
  4. The engine cost ₦60,000.00, option A.

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Question 18

The height of an equilateral triangle is 103 cm10\sqrt3\text{ cm}. Calculate its perimeter.

Worked solution (try it first)
  1. The height of an equilateral triangle of side ss is ssin⁡60∘=32ss\sin60^\circ = \frac{\sqrt3}{2}s.
  2. Set 32s=103\frac{\sqrt3}{2}s = 10\sqrt3 and multiply both sides by 23\frac{2}{\sqrt3}: s=20s = 20 cm.
  3. The perimeter is three sides: 3×20=603 \times 20 = 60 cm, option B.

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Question 19

In △LMN\triangle LMN, ∣LM∣=6 cm|LM| = 6\text{ cm}, ∠LNM=x\angle LNM = x, angle LL is a right angle and sin⁡x=35\sin x = \frac35. Find the area of △LMN\triangle LMN.

Worked solution (try it first)
  1. The right angle is at LL, so MNMN is the hypotenuse and LM=6LM = 6 is the side opposite the angle xx at NN: sin⁡x=6MN=35\sin x = \dfrac{6}{MN} = \dfrac35.
  2. So MN=10MN = 10 cm, and Pythagoras gives LN=102−62=8LN = \sqrt{10^2 - 6^2} = 8 cm.
  3. The two sides at the right angle are the base and height: area =12×6×8= \frac12 \times 6 \times 8
    =24 cm2= 24\text{ cm}^2, option C.

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Question 20

Consider the statements PP: all students offering Literature (LL) also offer History (HH); QQ: students offering History do not offer Geography (GG). Which of the Venn diagrams correctly illustrates the two statements?

A.GHLB.GLHC.GHLD.GLH
Worked solution (try it first)
  1. Statement PP says every Literature student also offers History, so the circle LL sits inside the circle HH.
  2. Statement QQ says History students do not offer Geography, so HH and GG do not overlap.
  3. Diagram C shows LL inside HH with GG separate, option C.

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Question 21

Find the quadratic equation whose roots are −2q-2q and 5q5q.

Worked solution (try it first)
  1. The sum of the roots is −2q+5q=3q-2q + 5q = 3q.
  2. The product is (−2q)(5q)=−10q2(-2q)(5q) = -10q^2.
  3. The equation is x2−(sum)x+(product)=0x^2 - (\text{sum})x + (\text{product}) = 0, which is x2−3qx−10q2=0x^2 - 3qx - 10q^2 = 0, option D.

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Question 22

If tan⁡θ=34\tan\theta = \frac34 and 180∘<θ<270∘180^\circ < \theta < 270^\circ, find the value of cos⁡θ\cos\theta.

Worked solution (try it first)
  1. tan⁡θ=34\tan\theta = \frac34 gives a reference triangle with opposite 3, adjacent 4 and hypotenuse 9+16=5\sqrt{9 + 16} = 5, so the size of cos⁡θ\cos\theta is 45\frac45.
  2. 180∘<θ<270∘180^\circ < \theta < 270^\circ is the third quadrant, where cosine is negative.
  3. So cos⁡θ=−45\cos\theta = -\frac45, option C.

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Question 23

If 2x−3−3x−2=p(x−3)(x−2)\dfrac{2}{x - 3} - \dfrac{3}{x - 2} = \dfrac{p}{(x - 3)(x - 2)}, find pp.

Worked solution (try it first)
  1. Over the LCD (x−3)(x−2)(x - 3)(x - 2), multiply each top by the other bracket: p=2(x−2)−3(x−3)p = 2(x - 2) - 3(x - 3).
  2. Expand: 2x−4−3x+92x - 4 - 3x + 9.
  3. Collect like terms: p=5−xp = 5 - x, option A.

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Question 24

The diagonals of a rhombus are 12 cm12\text{ cm} and 5 cm5\text{ cm}. Calculate its perimeter.

Worked solution (try it first)
  1. The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with legs 6 cm and 2.5 cm.
  2. By Pythagoras, the side is 62+2.52=42.25=6.5\sqrt{6^2 + 2.5^2} = \sqrt{42.25} = 6.5 cm.
  3. The perimeter is 4×6.5=264 \times 6.5 = 26 cm, option A.

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Question 25

In the diagram, △XYZ\triangle XYZ has YZYZ produced to TT. If ∣XY∣=∣ZY∣|XY| = |ZY| and ∠XYT=40∘\angle XYT = 40^\circ, find ∠XZT\angle XZT.

40°XYZT
Worked solution (try it first)
  1. ∣XY∣=∣ZY∣|XY| = |ZY|, so the base angles at XX and ZZ are equal: each is 180∘−40∘2=70∘\frac{180^\circ - 40^\circ}{2} = 70^\circ.
  2. Angles on the straight line YZTYZT: ∠XZT=180∘−70∘\angle XZT = 180^\circ - 70^\circ
    =110∘= 110^\circ, option A.

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Question 26

A solid brass cube is melted and recast as a solid cone of height hh and base radius rr. If the height of the cube is hh, find rr in terms of hh.

Worked solution (try it first)
  1. Melting keeps the volume the same: cube h3h^3 = cone 13πr2h\frac13\pi r^2h.
  2. Divide both sides by hh and multiply by 3: 3h2=πr23h^2 = \pi r^2, so r2=3h2πr^2 = \dfrac{3h^2}{\pi}.
  3. Take the square root: r=3h2πr = \sqrt{\dfrac{3h^2}{\pi}}, option B.

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Question 27

Which of the following is not an exterior angle of a regular polygon?

Worked solution (try it first)
  1. The exterior angles of a regular polygon are equal and add up to 360∘360^\circ, so an exterior angle must divide 360∘360^\circ a whole number of times.
  2. 360÷72=5360 \div 72 = 5, 360÷24=15360 \div 24 = 15 and 360÷15=24360 \div 15 = 24: all whole numbers.
  3. 360÷66≈5.45360 \div 66 \approx 5.45 is not whole, so 66∘66^\circ is not an exterior angle, option A.

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Question 28

From a point TT, a man moves 12 km12\text{ km} due west and then 12 km12\text{ km} due south to another point QQ. Calculate the bearing of TT from QQ.

Worked solution (try it first)
  1. QQ is 12 km west and 12 km south of TT.
  2. So from QQ, TT is 12 km east and 12 km north.
  3. Equal amounts north and east means TT is exactly north-east of QQ.
  4. North-east is a bearing of 045∘045^\circ, option C.

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Question 29

In the diagram, OO is the centre of the circle, ∠PQR=72∘\angle PQR = 72^\circ and OROR is parallel to PSPS. Find ∠OPS\angle OPS.

72°OQRPS
Worked solution (try it first)
  1. ∠POR\angle POR is at the centre on the same arc PRPR as ∠PQR\angle PQR at the circumference.
  2. So ∠POR=2×72∘\angle POR = 2 \times 72^\circ
    =144∘= 144^\circ.
  3. OR∥PSOR \parallel PS, so ∠OPS\angle OPS and ∠POR\angle POR are co-interior angles and add up to 180∘180^\circ.
  4. So ∠OPS=180∘−144∘\angle OPS = 180^\circ - 144^\circ
    =36∘= 36^\circ, option D.

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Question 30✱✱

A trapezium with parallel sides 10 cm10\text{ cm} and 21 cm21\text{ cm} and height 8 cm8\text{ cm} is inscribed in a circle of radius 7 cm7\text{ cm}. Calculate the area of the region not covered by the trapezium. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

Worked solution (try it first)
  1. Area of the circle: 227×72=154 cm2\frac{22}{7} \times 7^2 = 154\text{ cm}^2.
  2. Area of the trapezium: 12(10+21)×8=124 cm2\frac12(10 + 21) \times 8 = 124\text{ cm}^2.
  3. The uncovered region is 154−124=30 cm2154 - 124 = 30\text{ cm}^2, option C.

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Question 31

Find, correct to two decimal places, the mean of 1121\frac12, 2232\frac23, 3343\frac34, 4454\frac45 and 5565\frac56.

Worked solution (try it first)
  1. Write each as an improper fraction over 60: 9060,16060,22560,28860,35060\frac{90}{60}, \frac{160}{60}, \frac{225}{60}, \frac{288}{60}, \frac{350}{60}.
  2. They add up to 111360=18.55\frac{1113}{60} = 18.55.
  3. Divide by 5: 18.555=3.71\frac{18.55}{5} = 3.71, option A.

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Question 32

A cyclist moved at a speed of x km/hx\text{ km/h} for 2 hours. He then increased his speed by 2 km/h2\text{ km/h} for the next 3 hours. If the total distance covered is 36 km36\text{ km}, calculate his initial speed.

Worked solution (try it first)
  1. Distance is speed times time.
  2. The first stretch is 2x2x km.
  3. The new speed is x+2x + 2 for 3 hours, so the second stretch is 3(x+2)=3x+63(x + 2) = 3x + 6 km.
  4. The total is 36 km: 2x+3x+6=362x + 3x + 6 = 36, so 5x=305x = 30.
  5. Divide by 5: x=6x = 6.
  6. His initial speed was 6 km/h6\text{ km/h}, option D.

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Question 33

Find the value of (x+y)(x + y) in the diagram, where the two horizontal lines are parallel.

y35°x110°
Worked solution (try it first)
  1. The transversal marked 110∘110^\circ makes 110∘110^\circ with the rightward direction at both parallel lines (corresponding angles).
  2. So y=180∘−110∘=70∘y = 180^\circ - 110^\circ = 70^\circ (angles on a straight line).
  3. The other transversal makes 35∘35^\circ with the rightward direction at both lines, so at the lower line x=180∘−35∘=145∘x = 180^\circ - 35^\circ = 145^\circ.
  4. So x+y=145∘+70∘=215∘x + y = 145^\circ + 70^\circ = 215^\circ, option A.

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Question 34

In the diagram, MPMP is a tangent to the circle at NN, ∠PNQ=64∘\angle PNQ = 64^\circ and ∣RQ∣=∣RN∣|RQ| = |RN|. Find the angle tt.

64°tNQRMP
Worked solution (try it first)
  1. The angle between tangent NPNP and chord NQNQ equals the angle in the alternate segment, so ∠NRQ=∠PNQ=64∘\angle NRQ = \angle PNQ = 64^\circ.
  2. RQ=RNRQ = RN, so triangle RNQRNQ is isosceles with apex 64∘64^\circ at RR: ∠RQN=180∘−64∘2\angle RQN = \dfrac{180^\circ - 64^\circ}{2}
    =58∘= 58^\circ.
  3. tt is between tangent NMNM and chord NRNR, so it equals the angle opposite NRNR, at QQ.
  4. So t=58∘t = 58^\circ, option C.

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Question 35

Find the first quartile of 7, 8, 7, 9, 11, 8, 7, 9, 6 and 8.

Worked solution (try it first)
  1. Put the 10 numbers in order: 6, 7, 7, 7, 8, 8, 8, 9, 9, 11.
  2. Q1Q_1 is at position 104=2.5\frac{10}{4} = 2.5, halfway between the 2nd and 3rd values.
  3. Both are 7, so Q1=7.0Q_1 = 7.0, option B.

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Question 36

In the cyclic quadrilateral PQRSPQRS, ∠P=∠S=x∘\angle P = \angle S = x^\circ, ∠Q=(2y−30)∘\angle Q = (2y - 30)^\circ and ∠R=(x+y)∘\angle R = (x + y)^\circ. Find the value of xx.

x°x°(2y − 30)°(x + y)°PQRS
Worked solution (try it first)
  1. Opposite angles of a cyclic quadrilateral add up to 180∘180^\circ.
  2. PP and RR: x+(x+y)=180x + (x + y) = 180, so 2x+y=1802x + y = 180.
  3. QQ and SS: (2y−30)+x=180(2y - 30) + x = 180, so x+2y=210x + 2y = 210.
  4. Double the first equation, 4x+2y=3604x + 2y = 360, and take away the second: 3x=1503x = 150, so x=50x = 50 (and y=80y = 80).
  5. That is option A.

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Question 37

A cone has a base radius of 8 cm8\text{ cm} and height 11 cm11\text{ cm}. Calculate, correct to two decimal places, the curved surface area. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

Worked solution (try it first)
  1. The curved surface needs the slant height: l=82+112l = \sqrt{8^2 + 11^2}
    =185= \sqrt{185}
    ≈13.60\approx 13.60 cm.
  2. Curved surface area: πrl=227×8×185\pi rl = \frac{22}{7} \times 8 \times \sqrt{185}.
  3. That is 341.98 cm2341.98\text{ cm}^2 to 2 decimal places, option A.

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Question 38

Given that sin⁡x=35\sin x = \frac35, 0∘≤x≤90∘0^\circ \le x \le 90^\circ, evaluate (tan⁡x+2cos⁡x)(\tan x + 2\cos x).

Worked solution (try it first)
  1. sin⁡x=35\sin x = \frac35 gives a triangle with opposite 3 and hypotenuse 5.
  2. Pythagoras gives the adjacent side 25−9=4\sqrt{25 - 9} = 4.
  3. So cos⁡x=45\cos x = \frac45 and tan⁡x=34\tan x = \frac34.
  4. Then tan⁡x+2cos⁡x=34+85\tan x + 2\cos x = \frac34 + \frac85.
  5. Over 20 this is 1520+3220=4720\frac{15}{20} + \frac{32}{20} = \frac{47}{20}.
  6. As a mixed number, 4720=2720\frac{47}{20} = 2\frac{7}{20}, option C.

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Question 39

In the diagram, ECEC is a diameter of the circle. If ∠ABC=158∘\angle ABC = 158^\circ, find ∠ADE\angle ADE.

158°OABCDE
Worked solution (try it first)
  1. ABCEABCE is a cyclic quadrilateral, so ∠AEC=180∘−158∘\angle AEC = 180^\circ - 158^\circ
    =22∘= 22^\circ.
  2. ECEC is a diameter, so ∠EAC=90∘\angle EAC = 90^\circ.
  3. In triangle EACEAC, ∠ACE=180∘−90∘−22∘\angle ACE = 180^\circ - 90^\circ - 22^\circ
    =68∘= 68^\circ.
  4. Angles in the same segment: ∠ADE\angle ADE and ∠ACE\angle ACE both stand on arc AEAE, so ∠ADE=68∘\angle ADE = 68^\circ, option C.

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Question 40

Height (cm) 160 161 162 163 164 165
No. of players 4 6 3 7 8 9

The table shows the heights of 37 players of a basketball team. Calculate, correct to one decimal place, the mean height of the players.

Worked solution (try it first)
  1. Multiply each height by its number of players: 640+966+486+1141+1312+1485=6030640 + 966 + 486 + 1141 + 1312 + 1485 = 6030.
  2. Divide by the 37 players: 603037=162.97\frac{6030}{37} = 162.97.
  3. To one decimal place the mean is 163.0 cm, option A.

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Question 41

Let XY‾\overline{XY} be a line segment with X(−8,−12)X(-8, -12) and Y(p,q)Y(p, q). If the midpoint of XY‾\overline{XY} is (−4,−2)(-4, -2), find the coordinates of YY.

Worked solution (try it first)
  1. The midpoint is the average of the ends, so each coordinate of YY is twice the midpoint's minus XX's.
  2. p=2(−4)−(−8)=−8+8=0p = 2(-4) - (-8) = -8 + 8 = 0.
  3. q=2(−2)−(−12)=−4+12=8q = 2(-2) - (-12) = -4 + 12 = 8.
  4. So Y=(0,8)Y = (0, 8), option B.

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Question 42

500 tickets for a concert were sold to adults and children at $4.50 and $3.00 respectively. If the total receipts were $1,987.50, how many adult tickets were sold?

Worked solution (try it first)
  1. Let aa adult tickets be sold, so 500−a500 - a children's tickets were sold.
  2. Receipts: 4.50a+3.00(500−a)=1987.504.50a + 3.00(500 - a) = 1987.50, which is 1.5a+1500=1987.51.5a + 1500 = 1987.5.
  3. Take 1500 from both sides: 1.5a=487.51.5a = 487.5.
  4. Divide by 1.5: a=325a = 325 adult tickets, option A.

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Question 43

The distance dd between two villages is more than 18 km18\text{ km} but not more than 23 km23\text{ km}. Which inequality represents the statement?

Worked solution (try it first)
  1. "More than 18 km" means d>18d > 18.
  2. 18 itself is not allowed.
  3. "Not more than 23 km" means d≤23d \le 23.
  4. 23 is allowed.
  5. Together: 18<d≤2318 < d \le 23, option D.

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Question 44

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

Banana 60°Apple 80°Pawpaw 100°Orange
Worked solution (try it first)
  1. The angles add up to 360∘360^\circ, so Orange is 360∘−(80∘+60∘+100∘)=120∘360^\circ - (80^\circ + 60^\circ + 100^\circ) = 120^\circ.
  2. 80∘80^\circ stands for 60 apples, so each degree is 6080=34\frac{60}{80} = \frac34 of a fruit.
  3. Orange: 120×34=90120 \times \frac34 = 90 oranges, option D.

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Question 45

A box contains 40 identical balls of which 10 are red and 12 are blue. A ball is selected at random. What is the probability that it is neither red nor blue?

Worked solution (try it first)
  1. Red and blue together are 10+12=2210 + 12 = 22 balls.
  2. Neither red nor blue leaves 40−22=1840 - 22 = 18 balls.
  3. So the probability is 1840=920\frac{18}{40} = \frac{9}{20}, option A.

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Question 46

A fair die is tossed twice. What is the probability of getting a sum of at least 10?

Worked solution (try it first)
  1. Tossing a die twice gives 36 equally likely outcomes.
  2. A sum of at least 10 comes from (4,6),(5,5),(6,4),(5,6),(6,5),(6,6)(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6), which is 6 outcomes.
  3. So the probability is 636=16\frac{6}{36} = \frac16, option D.

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Question 47

A man will be x+10x + 10 years old in 8 years' time. If 2 years ago he was 63 years old, find the value of xx.

Worked solution (try it first)
  1. Two years ago he was 63, so now he is 63+2=6563 + 2 = 65.
  2. In 8 years' time he will be 65+8=7365 + 8 = 73.
  3. So x+10=73x + 10 = 73.
  4. Take 10 from both sides: x=63x = 63, option B.

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Question 48

The equation of a line is 3x−5y=73x - 5y = 7. Find its gradient.

Worked solution (try it first)
  1. Make yy the subject: subtract 3x3x from both sides to get −5y=−3x+7-5y = -3x + 7.
  2. Divide every term by −5-5: y=35x−75y = \frac35x - \frac75.
  3. The gradient is the coefficient of xx: 35\frac35, option B.

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Question 49

For what value of xx is 4−2xx+1\dfrac{4 - 2x}{x + 1} undefined?

Worked solution (try it first)
  1. A fraction is undefined when its bottom is zero.
  2. x+1=0x + 1 = 0 when x=−1x = -1, option B.

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