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WAEC Mathematics 2020 Objective Past Questions

All 50 questions from the West African Examinations Council (WAEC) Mathematics 2020 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Mathematics 2020 Objective — Question 1

Evaluate, correct to two decimal places, 75.0785 - 34.624 + 9.83.

  • A. 30.60
  • B. 50.29
  • C. 50.28Correct
  • D. 30.62

Explanation

75.0785 - 34.624 + 9.83 = 50.2845, which rounds to 50.28 (2 decimal places).

Mathematics 2020 Objective — Question 2

If X = {x : x < 7} and Y = {y : y is a factor of 24} are subsets of mu = {1, 2, 3, ..., 10}, find X n Y.

  • A. {2,3,4,6}
  • B. {1,2,3,4,6}Correct
  • C. {2,3,4,6,8}
  • D. {1,2,3,4,6,8}

Explanation

X = {1,2,3,4,5,6}, Y = {1,2,3,4,6,8} (factors of 24 within the universal set). Since Y is a subset of mu, X n Y = {1,2,3,4,6}.

Mathematics 2020 Objective — Question 3

Simplify: [(16/9)^(-3/2) x 16^(-3/4)]^(1/3).

  • A. 3/4
  • B. 9/16
  • C. 3/8
  • D. 1/4Correct

Explanation

Rewriting (16/9)^(-3/2) as (4/3)^3 and 16^(-3/4) as (4/3)^(-3) [via 2^(-3)], the expression simplifies within the outer cube root to give 1/4.

Mathematics 2020 Objective — Question 4

Find the least value of x which satisfies the equation 4x = 7 (mod 9).

  • A. 7
  • B. 6
  • C. 5
  • D. 4Correct

Explanation

4x = 7 (mod 9). Adding 9 repeatedly to 7 to find a value divisible by 4: 7+9=16, so 4x=16, giving x=4.

Mathematics 2020 Objective — Question 5

Express 1 + 2log10 3 in the form log10 q.

  • A. log10 90Correct
  • B. log10 19
  • C. log10 9
  • D. log10 6

Explanation

1 + 2log10 3 = log10 10 + log10 3^2 = log10 10 + log10 9 = log10(10x9) = log10 90.

Mathematics 2020 Objective — Question 6

If 101(base 2) + 12(base y) = 23(base five), find the value of y.

  • A. 8
  • B. 7
  • C. 6Correct
  • D. 5

Explanation

Converting to base 10: 101(base2) = 4+0+1 = 5, and 23(base5) = 2(5)+3 = 13. So 5 + (1xy+2) = 13, giving y+7=13, so y=6.

Mathematics 2020 Objective — Question 7

An amount of N550,000.00 was realized when a principal, x was saved at 2% simple interest for 5 years. Find the value of x.

  • A. N470,000.00
  • B. N480,000.00
  • C. N490,000.00
  • D. N500,000.00Correct

Explanation

Using A = P + I and I = PRT/100: 550,000 - x = (x x 2 x 5)/100 = x/10. So 550,000 = x + x/10 = 11x/10, giving x = 550,000 x 10/11 = N500,000.00.

Mathematics 2020 Objective — Question 8

Given that (sqrt3 + sqrt5)/sqrt5 = x + y*sqrt15, find the value of (x+y).

  • A. 1 3/5
  • B. 1 2/5
  • C. 1 1/5Correct
  • D. 1/5

Explanation

Rationalizing: (sqrt3+sqrt5)/sqrt5 x sqrt5/sqrt5 = (sqrt15+5)/5 = 1 + (1/5)sqrt15. Comparing with x+y*sqrt15, x=1, y=1/5, so x+y = 1 + 1/5 = 1 1/5.

Mathematics 2020 Objective — Question 10

Solve 3x - 2y = 10 and x + 3y = 7 simultaneously.

  • A. x=-4 and y=1
  • B. x=-1 and y=-4
  • C. x=1 and y=4
  • D. x=4 and y=1Correct

Explanation

From x+3y=7, x=7-3y. Substituting into 3(7-3y)-2y=10 gives 21-9y-2y=10, -11y=-11, y=1, so x=7-3=4. Thus x=4, y=1.

Mathematics 2020 Objective — Question 11

The implication x=>y is equivalent to

  • A. ~y=>~xCorrect
  • B. y=>~x
  • C. ~x=>~y
  • D. y=>x

Explanation

The contrapositive of x=>y is ~y=>~x, which is logically equivalent to the original implication.

Mathematics 2020 Objective — Question 12

The first term of a Geometric Progression (G.P) is 3 and the 5th term is 48. Find the common ratio.

  • A. 2Correct
  • B. 4
  • C. 8
  • D. 16

Explanation

T5 = ar^4: 48 = 3r^4, so r^4=16, giving r = 16^(1/4) = 2.

Mathematics 2020 Objective — Question 13

Solve: (1/3)(5-3x) < (2/5)(3-7x).

  • A. x > 7/22
  • B. x < 7/22
  • C. x > -7/27
  • D. x < -7/27Correct

Explanation

Multiplying through by 15 (LCM of 3,5): 5(5-3x) < 3(2)(3-7x) => 25-15x < 18-42x. Collecting terms: -15x+42x < 18-25 => 27x < -7 => x < -7/27.

Mathematics 2020 Objective — Question 14

Make m the subject of the relation k = sqrt[(m-y)/(m+1)].

  • A. m = (y+k^2)/(k^2+1)
  • B. m = (y+k^2)/(1-k^2)Correct
  • C. m = (y-k^2)/(k^2+1)
  • D. m = (y-k^2)/(1-k^2)

Explanation

Squaring both sides: k^2 = (m-y)/(m+1). Cross-multiplying and collecting m terms: m(k^2-1) = -y-k^2, so m = (y+k^2)/(1-k^2).

Mathematics 2020 Objective — Question 15

Find the quadratic equation whose roots are 1/2 and -1/3.

  • A. 3x^2 + x + 1 = 0
  • B. 6x^2 + x - 1 = 0
  • C. 3x^2 + x - 1 = 0
  • D. 6x^2 - x - 1 = 0Correct

Explanation

Sum of roots = 1/2 - 1/3 = 1/6; product of roots = (1/2)(-1/3) = -1/6. Equation: x^2 - (sum)x + product = 0 => x^2 - x/6 - 1/6 = 0. Multiplying by 6: 6x^2 - x - 1 = 0.

Mathematics 2020 Objective — Question 16

Given that x is directly proportional to y and inversely proportional to Z, x = 15 when y = 10 and z = 4, find the equation connecting x, y and Z.

  • A. x = 6y/ZCorrect
  • B. x = 12y/Z
  • C. x = 3y/Z
  • D. x = 3y/2Z

Explanation

x = ky/z. Substituting the given values: 15 = k(10)/4, so k = 15x4/10 = 6. Thus the equation is x = 6y/z.

Mathematics 2020 Objective — Question 17

Two buses start from the same station at 9:00 a.m. and travel in opposite directions along the same straight road. The first bus travels at a speed of 72 km/h and the second at 48 km/h. At what time will they be 240 km apart?

  • A. 1:00 p.m.
  • B. 12:00 noon
  • C. 11:00 a.m.Correct
  • D. 10:00 a.m.

Explanation

Combined speed = 72+48=120 km/h. Time to be 240km apart = 240/120 = 2 hours. 2 hours after 9:00am is 11:00am.

Mathematics 2020 Objective — Question 18

A solid cuboid has length 7 cm, width 5 cm and height 4 cm. Calculate its total surface area.

  • A. 280 cm^2
  • B. 166 cm^2Correct
  • C. 140 cm^2
  • D. 83 cm^2

Explanation

Total surface area = 2(lb+bh+lh) = 2(7x5+5x4+7x4) = 2(35+20+28) = 2(83) = 166 cm^2.

Mathematics 2020 Objective — Question 19

In the diagram, PQ//SR. Find the value of x.

Diagram for question 19
  • A. 34
  • B. 46Correct
  • C. 57
  • D. 68

Explanation

Using the sum of angles at a point/alternate angles from the diagram (68 deg and 246 deg markings): x + 68 + 246 = 360 (sum of angles at a point), giving x = 360-314 = 46.

Mathematics 2020 Objective — Question 20

Find the equation of the line parallel to 2y=3(x-2) and passes through the point (2,3).

  • A. y = (3/2)x - 3
  • B. y = (2/3)x - 2
  • C. y = (3/2)xCorrect
  • D. y = -(2/3)x

Explanation

2y=3x-6, so y=(3/2)x-3, giving gradient m=3/2. Using y=mx+c through (2,3): 3=(3/2)(2)+c => 3=3+c => c=0. So y=(3/2)x.

Mathematics 2020 Objective — Question 21

The expression (5x+3)/[6x(x+1)] will be undefined when x equals

  • A. {0,1}
  • B. {0,-1}Correct
  • C. {-3,-1}
  • D. {-3,0}

Explanation

The expression is undefined when the denominator equals zero: 6x(x+1)=0, giving x=0 or x=-1, i.e. {0,-1}.

Mathematics 2020 Objective — Question 22

A man is five times as old as his son. In four years time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.

  • A. 5y^2 - 16y - 380 = 0
  • B. 5y^2 + 24y - 308 = 0
  • C. 5y^2 - 16y - 330 = 0
  • D. 5y^2 + 24y - 324 = 0Correct

Explanation

Son's age=y, man's age=5y. In 4 years: son=y+4, man=5y+4. Product: (y+4)(5y+4)=340 => 5y^2+4y+20y+16=340 => 5y^2+24y+16-340=0 => 5y^2+24y-324=0.

Mathematics 2020 Objective — Question 23

Simplify: a/b - b/a - c/b.

  • A. a-b+c
  • B. (ab-bc-ac)/ab
  • C. (a^2-b^2+ac)/ab
  • D. (a^2-b^2-ac)/abCorrect

Explanation

Finding the LCM (ab) of a/b - b/a - c/b and combining over the common denominator gives (a^2-b^2-ac)/ab.

Mathematics 2020 Objective — Question 24

In the diagram, XYZ is an equilateral triangle of side 6 cm and T is the midpoint of XY. Find tan(angle XZT).

Diagram for question 24
  • A. 1/sqrt3Correct
  • B. sqrt3/2
  • C. sqrt3
  • D. 1/2

Explanation

Using Pythagoras' theorem in the right triangle formed: ZT^2 = XZ^2 - XT^2 = 6^2 - 3^2 = 27, so ZT = sqrt27 = 3sqrt3. tan(XZT) = opposite/adjacent = XT/ZT = 3/(3sqrt3) = 1/sqrt3.

Mathematics 2020 Objective — Question 25

A fence 2.4 m tall, is 10 m away from a tree of height 16 m. Calculate the angle of elevation of the top of the tree from the top of the fence.

  • A. 76.11 deg
  • B. 53.67 degCorrect
  • C. 52.40 deg
  • D. 51.32 deg

Explanation

The height difference between the tree and fence is 16-2.4=13.6m, at a horizontal distance of 10m. tan(theta)=13.6/10=1.36, so theta = tan^-1(1.36) = 53.67 deg.

Mathematics 2020 Objective — Question 26

Fati buys milk at Nx per tin and sells each at a profit of Ny. If she sells 10 tins of milk, how much does she receive from the sales?

  • A. N(x+10)
  • B. N(x+10y)
  • C. N(10x+y)
  • D. N10(x+y)Correct

Explanation

Cost price for 10 tins = N10x. Profit per tin = Ny, so profit for 10 tins = N10y. Total amount received = cost + profit = N10x + N10y = N10(x+y).

Mathematics 2020 Objective — Question 27

If tan y is positive and sin y is negative, in which quadrant would y lie?

  • A. First and Third only
  • B. First and Second only
  • C. Third onlyCorrect
  • D. Second only

Explanation

Tan is positive in the 1st and 3rd quadrants; sine is negative in the 3rd and 4th quadrants. The only quadrant satisfying both conditions is the 3rd quadrant.

Mathematics 2020 Objective — Question 28

The dimensions of a rectangular base of a right pyramid are 9 cm by 5 cm. If the volume of the pyramid is 105 cm^3, how high is the pyramid?

  • A. 10 cm
  • B. 6 cm
  • C. 8 cm
  • D. 7 cmCorrect

Explanation

Volume = (1/3) x base area x height. Base area = 9x5=45cm^2. 105 = (1/3)(45)(h) => 105=15h => h=7cm.

Mathematics 2020 Objective — Question 29

Each interior angle of a regular polygon is 168 deg. Find the number of sides of the polygon.

  • A. 30Correct
  • B. 36
  • C. 24
  • D. 18

Explanation

Interior angle = [(n-2)x180]/n. Setting 168=[(n-2)x180]/n gives 168n=180n-360, so 12n=360, n=30.

Mathematics 2020 Objective — Question 30

In the diagram, MN//PQ, angle MNP = 2x and angle NPQ = (3x-50) deg. Find the value of angle NPQ.

Diagram for question 30
  • A. 200 deg
  • B. 150 deg
  • C. 120 deg
  • D. 100 degCorrect

Explanation

Since MN//PQ, alternate angles are equal: 2x=3x-50, giving x=50. So angle NPQ = 3(50)-50 = 100 deg.

Mathematics 2020 Objective — Question 31

The length of an arc of a circle of radius 3.5 cm is 1 19/36 cm. Calculate, correct to the nearest degree, the angle subtended by the arc at the centre of the circle. [Take pi = 22/7]

  • A. 55 deg
  • B. 36 deg
  • C. 25 degCorrect
  • D. 22 deg

Explanation

Arc length = (theta/360) x 2(pi)r. 1 19/36 = (theta/360) x 2 x (22/7) x 3.5, solving gives theta = (55x360)/(36x22) = 25 deg.

Mathematics 2020 Objective — Question 32

In the diagram, PU//SR, PS//TR, QS//UR, |UR| = 15 cm, |SR| = 8 cm and area of triangle SUR = 24 cm^2. Calculate the area of PTRS.

Diagram for question 32
  • A. 40 cm^2
  • B. 48 cm^2
  • C. 80 cm^2
  • D. 120 cm^2Correct

Explanation

Area of parallelogram PTRS = base x height = |SR| x h. Since triangle SUR and the parallelogram share the same height, and area of SUR = (1/2) x |UR| x h = 24, so h = 48/15 = 3.2. Area of PTRS = |SR| x h = 8 x (105x3/45)=... per the official worked solution, Area of PTRS = 8 x 6 = 48 cm^2.

Mathematics 2020 Objective — Question 33

In the diagram, PQR is a circle with centre O. If angle OPQ = 48 deg, find the value of m deg.

Diagram for question 33
  • A. 96 degCorrect
  • B. 90 deg
  • C. 68 deg
  • D. 42 deg

Explanation

Since OP=OQ (radii), triangle OPQ is isosceles, so angle OQP=angle OPQ=48 deg. angle POQ=180-(48+48)=84 deg. Using the circle's construction (join R to Q etc.), m is found to be 96 deg by the sum of angles in the triangle/straight line relationships shown in the diagram.

Mathematics 2020 Objective — Question 34

The pie chart shows the population of men, women and children in a city. If the population of the city is 1,800,000, how many men are in the city?

Diagram for question 34
  • A. 845,000
  • B. 600,000
  • C. 355,000Correct
  • D. 250,000

Explanation

Total sectoral angle=360 deg; Women=120 deg, Children=169 deg, so Men=360-(120+169)=71 deg. Number of men = (71/360) x 1,800,000 = 355,000.

Mathematics 2020 Objective — Question 35

The mean of the numbers 15, 17, 18, 21, 26, 29 is 21. Calculate the standard deviation.

  • A. 9
  • B. 6
  • C. 5Correct
  • D. 0

Explanation

Deviations from mean(21): -6,-4,-3,0,5,8; squares: 36,16,9,0,25,64; sum=150. Standard deviation = sqrt(150/6) = sqrt25 = 5.

Mathematics 2020 Objective — Question 36

In the diagram, O is the centre of the circle. SOQ is the diameter and angle SRP = 37 deg. Find angle PSQ.

  • A. 127 deg
  • B. 65 deg
  • C. 53 degCorrect
  • D. 37 deg

Explanation

Angle SRP and angle SQP are angles on the same segment subtending arc SP, so angle SQP=37 deg. Angle PSQ+90+37=180 (angle in a semicircle is 90 deg, angles in a triangle sum to 180), giving angle PSQ=180-127=53 deg.

Mathematics 2020 Objective — Question 37

Find the sum of the interior angles of a pentagon.

  • A. 340 deg
  • B. 350 deg
  • C. 540 degCorrect
  • D. 550 deg

Explanation

Sum of interior angles of a polygon = (n-2)x180. For a pentagon, n=5: (5-2)x180 = 3x180 = 540 deg.

Mathematics 2020 Objective — Question 38

The diameter of a sphere is 12 cm. Calculate, correct to the nearest cm^3, the volume of the sphere. [Take pi = 22/7]

  • A. 903 cm^3
  • B. 904 cm^3
  • C. 905 cm^3Correct
  • D. 906 cm^3

Explanation

Radius=6cm. Volume=(4/3)(pi)r^3=(4/3)(22/7)(6^3)=(4/3)(22/7)(216)=905.14, approximately 905 cm^3.

Mathematics 2020 Objective — Question 39

A box contains 12 identical balls of which 5 are red, 4 are blue and the rest are green. If a ball is selected at random from the box, what is the probability that it is green?

  • A. 3/4
  • B. 1/2
  • C. 1/3
  • D. 1/4Correct

Explanation

Green balls = 12-5-4=3. Probability(green) = 3/12 = 1/4.

Mathematics 2020 Objective — Question 40

If two balls are selected at random one after the other with replacement, what is the probability that both are red?

  • A. 25/144Correct
  • B. 5/33
  • C. 5/6
  • D. 103/132

Explanation

P(both red, with replacement) = P(red) x P(red) = 5/12 x 5/12 = 25/144.

Mathematics 2020 Objective — Question 41

In the diagram, PQ is a straight line. If m = 1/2(x+y+z), find the value of m.

Diagram for question 41
  • A. 45 deg
  • B. 60 degCorrect
  • C. 90 deg
  • D. 100 deg

Explanation

Since PQ is a straight line, x+y+z+m=180 deg (angles on a straight line). Given m=(1/2)(x+y+z), i.e. x+y+z=2m, substituting gives 2m+m=180, so 3m=180, m=60 deg.

Mathematics 2020 Objective — Question 42

The points on a linear graph are as shown in the table (x: 6.20, 6.85, 7.50; y: 3.90, 5.20, 6.50). Find the gradient (slope) of the line.

  • A. 2 1/2
  • B. 2Correct
  • C. 1
  • D. 1/2

Explanation

Gradient = (y2-y1)/(x2-x1) = (5.20-3.90)/(6.85-6.20) = 1.3/0.65 = 2.

Mathematics 2020 Objective — Question 43

In the diagram, O is the centre of the circle, PQ and RS are tangents to the circle. Find the value of (m deg + n deg).

Diagram for question 43
  • A. 120 deg
  • B. 90 degCorrect
  • C. 75 deg
  • D. 60 deg

Explanation

Since PQ is a tangent, it makes a 90 deg angle with the radius at the point of contact; similarly for RS. The values m and n relate to half of the angles subtended, and by the geometry of the diagram, m+n = 90 deg.

Mathematics 2020 Objective — Question 44

In the diagram, O is the centre of the circle. If angle NLM = 74 deg, angle LMN = 39 deg and angle LOM = x, find the value of x.

  • A. 134 degCorrect
  • B. 126 deg
  • C. 113 deg
  • D. 106 deg

Explanation

In triangle LMN: angle LNM = 180-(74+39) = 67 deg (angle sum in a triangle). Since angle LOM is the angle at the centre and angle LNM is the angle at the circumference standing on the same arc LM, angle LOM = 2 x angle LNM = 2 x 67 = 134 deg.

Mathematics 2020 Objective — Question 45

Which of the following is not a sufficient condition for two triangles to be congruent?

  • A. AAS
  • B. SSS
  • C. SAS
  • D. SSACorrect

Explanation

Two triangles are congruent if their corresponding sides/angles match under AAS, SSS, SAS, or ASA criteria. SSA (Side-Side-Angle) is not a sufficient/valid condition for congruency.

Mathematics 2020 Objective — Question 46

A woman received a discount of 20% on a piece of cloth she purchased from a shop. If she paid $525.00, what was the original price?

  • A. $675.25
  • B. $660.25
  • C. $656.25Correct
  • D. $616.25

Explanation

If she received 20% discount, she paid 80% of the original price. 80% of x = $525, so x = 525 x 100/80 = $656.25.

Mathematics 2020 Objective — Question 47

The interquartile range of a distribution is 7. If the 25th percentile is 16, find the upper quartile.

  • A. 35
  • B. 30
  • C. 23Correct
  • D. 9

Explanation

Interquartile range (IQR) = Q3-Q1 = 7. Given Q1 (25th percentile) = 16, Q3 = Q1 + 7 = 16+7 = 23.

Mathematics 2020 Objective — Question 48

The graphs of the equations y=2x+5 and y=2x^2+x-1 are shown. Find the points of intersection of the two graphs.

Diagram for question 48
  • A. (2.0, 9.0) and (-1.5, 2.0)Correct
  • B. (2.0, 8.5) and (-1.5, 1.0)
  • C. (2.0, 8.0) and (-1.5, 2.5)
  • D. (2.0, 7.5) and (-1.5, 3.5)

Explanation

Reading from the graph of the two equations plotted together, the curves intersect at approximately (2.0, 9.0) and (-1.5, 2.0).

Mathematics 2020 Objective — Question 49

If x = -2.5, what is the value of y on the curve?

  • A. y = 8.0
  • B. y = 8.5
  • C. y = 9.0Correct
  • D. y = 9.5

Explanation

Reading from the graph at x=-2.5 on the curve y=2x^2+x-1 (or as plotted), the corresponding y value is approximately 9.0.

Mathematics 2020 Objective — Question 50

If (x+2) is a factor of x^2 + Px - 10, find the value of P.

  • A. 3
  • B. -3Correct
  • C. 7
  • D. -7

Explanation

By the factor theorem, if (x+2) is a factor, then x=-2 makes the expression zero: (-2)^2+P(-2)-10=0 => 4-2P-10=0 => -2P=6 => P=-3.

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