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WAEC Mathematics 2014 Objective Past Questions

All 50 questions from the West African Examinations Council (WAEC) Mathematics 2014 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Mathematics 2014 Objective — Question 2

If 23_x = 32_5, find the value of x. (base of first number is x)

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

Explanation

Given 23_x=32_5: 32_5 = 3×5+2 = 17. 23_x = 2x+3=17, so 2x=14, x=7.

Mathematics 2014 Objective — Question 4

The bar chart shows the scores of some students in a test. How many students took the test?

Diagram for question 4
  • A. 18
  • B. 19
  • C. 20Correct
  • D. 22

Explanation

Reading the bar chart, total number of students = 4+2+5+6+3 = 20 students.

Mathematics 2014 Objective — Question 5

Using the same bar chart, if one student is selected at random, find the probability that he/she scored at most 2 marks.

  • A. 11/18
  • B. 11/20Correct
  • C. 7/22
  • D. 5/19

Explanation

Students scoring at most 2 marks = 5+2+4 = 11 (frequencies for scores of 0,1,2 marks). Probability = 11/20.

Mathematics 2014 Objective — Question 7

Which of the following number lines represents the solution to the inequality: −9 ≤ (2/3)x − 7 < 5?

Diagram for question 7
  • A. Number line A
  • B. Number line B
  • C. Number line C
  • D. Number line DCorrect

Explanation

Solving: −9≤(2/3)x−7<5 gives −2≤(2/3)x<12, so −3≤x<18. This is represented by a number line with a closed circle at −3 and an open circle at 18.

Mathematics 2014 Objective — Question 8

In the diagram, the value of x°+y°=220°. Find the value of n°.

Diagram for question 8
  • A. 20°
  • B. 40°Correct
  • C. 60°
  • D. 80°

Explanation

Since x°+y°=220° are the two base exterior-related angles, n° = x°+y°−180° = 220°−180° = 40°.

Mathematics 2014 Objective — Question 9

Given that x>y and 3<y, which of the following is/are true? I. y>3 II. x<3 III. x>y>3

  • A. I only
  • B. I and II only
  • C. I and III onlyCorrect
  • D. I, II and III

Explanation

Since y>3 (statement I is true) and x>y, then x>y>3 (statement III is true). Statement II (x<3) is false since x>y>3.

Mathematics 2014 Objective — Question 10

Three quarter of a number added to two and a half of that number gives 13. Find the number.

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

Explanation

¾n + 2½n = 13 → (3/4+5/2)n=13 → (13/4)n=13 → n=4.

Mathematics 2014 Objective — Question 11

If X={0,2,4,6} and Z={1,3} are subsets of U={x:0≤x≤6}, and Y={1,2,3,4}. Find X∩(Y′∪Z).

  • A. {0,2,6}
  • B. {1,3}
  • C. {0,6}Correct
  • D. { }

Explanation

Y′ (complement of Y in U) = {0,5,6}. Y′∪Z = {0,1,3,5,6}. X∩(Y′∪Z) = {0,2,4,6}∩{0,1,3,5,6} = {0,6}.

Mathematics 2014 Objective — Question 12

Find the truth set of the equation x² = 3(2x+9).

  • A. {x:x=3,x=−9}
  • B. {x:x=−3,x=−9}
  • C. {x:x=3,x=9}
  • D. {x:x=−3,x=9}Correct

Explanation

x²=6x+27 → x²−6x−27=0 → (x−9)(x+3)=0 → x=9 or x=−3.

Mathematics 2014 Objective — Question 13

The coordinates of points P and Q are (4,3) and (2,−1) respectively. Find the shortest distance between P and Q.

  • A. 10√2
  • B. 4√5
  • C. 5√2
  • D. 2√5Correct

Explanation

|PQ| = √((2−4)²+(−1−3)²) = √(4+16) = √20 = 2√5.

Mathematics 2014 Objective — Question 14

Make u the subject of the formula, E = (m/2g)(v²−u²).

  • A. u=√(v²−2Eg/m)Correct
  • B. u=√(v²−2Eg/(4))
  • C. u=√(v−2Eg/m)
  • D. u=√(2v²Eg/m)

Explanation

Multiply through by 2g/m: 2Eg/m = v²−u², so u² = v²−2Eg/m, giving u=√(v²−2Eg/m).

Mathematics 2014 Objective — Question 15

In the diagram, ∠QPT=∠PTS=90°, ∠PQR=110° and ∠TSR=20°. Find the size of the obtuse angle QRS.

Diagram for question 15
  • A. 140°
  • B. 130°Correct
  • C. 120°
  • D. 110°

Explanation

Using angle sum properties of the quadrilateral/triangles formed: ∠PRQ=50°, so ∠QRS=180°−50°=130°.

Mathematics 2014 Objective — Question 16

If x varies inversely as y and y varies directly as z, what is the relationship between x and z?

  • A. x∝z
  • B. x∝1/zCorrect
  • C. x∝z²
  • D. x∝1/z²

Explanation

x∝1/y and y∝z, so substituting gives x∝1/z.

Mathematics 2014 Objective — Question 17

Find the gradient of the line joining the points (2,−3) and (2,5).

  • A. 0
  • B. 1
  • C. 2
  • D. undefinedCorrect

Explanation

Gradient = (5−(−3))/(2−2) = 8/0, which is undefined (a vertical line).

Mathematics 2014 Objective — Question 18

If (x−a) is a factor of bx−ax+x²−ab, find the other factor.

  • A. (x+b)Correct
  • B. (x−b)
  • C. (a+b)
  • D. (a−b)

Explanation

bx−ax+x²−ab = x(x+b)−a(x+b) = (x−a)(x+b), so the other factor is (x+b).

Mathematics 2014 Objective — Question 19

The table shows the distribution of the height of plants in a nursery (Height: 2,3,4,5,6; Frequency: 2,4,5,3,1). Calculate the mean height of the plants.

  • A. 3.8Correct
  • B. 3.0
  • C. 2.8
  • D. 2.3

Explanation

Mean = Σfx/Σf = (2×2+3×4+4×5+5×3+6×1)/15 = 57/15 = 3.8.

Mathematics 2014 Objective — Question 20

In the diagram, PQR is a straight line, (m°+n°)=120° and (n+r)°=100°. Find (m+r)°.

Diagram for question 20
  • A. 110°
  • B. 120°
  • C. 140°Correct
  • D. 160°

Explanation

Since PQR is a straight line, m°+n°+r°=180°. From m°+n°=120°, r°=60°. From n°+r°=100°, m°=80°. So m°+r°=80°+60°=140°.

Mathematics 2014 Objective — Question 21

In the diagram, line |SR| is parallel to line |UW|, ∠WVT=x°, ∠VUT=y°, ∠RSV=45° and ∠VTU=20°. Find the value of x°.

Diagram for question 21
  • A. 20
  • B. 45Correct
  • C. 65
  • D. 135

Explanation

∠SVU=45° (alternate angle to ∠RSV). x°=∠WVT is vertically opposite to ∠SVU, so x°=45°.

Mathematics 2014 Objective — Question 22

Using the same diagram, calculate the value of y°.

  • A. 20°
  • B. 25°
  • C. 45°Correct
  • D. 65°

Explanation

Using the angle relationships in the triangle UVT (∠VUT=y°, ∠VTU=20°, and the vertically opposite/exterior angle relationships), y°=45°.

Mathematics 2014 Objective — Question 23

The area of a sector of a circle with diameter 12cm is 66cm². If the sector is folded to form a cone, calculate the radius of the base of the cone. [Take π=22/7]

  • A. 3.0cm
  • B. 3.5cmCorrect
  • C. 7.0cm
  • D. 7.5cm

Explanation

Using arc length = base circumference: r = Rθ/360°, and solving with R=6cm and sector area 66cm² gives r=3.5cm.

Mathematics 2014 Objective — Question 24

A chord, 7cm long, is drawn in a circle with radius 3.7cm. Calculate the distance of the chord from the centre of the circle.

  • A. 0.7cm
  • B. 1.2cmCorrect
  • C. 2.0cm
  • D. 2.5cm

Explanation

By Pythagoras: 3.7² = 3.5² + h², so h² = 13.69−12.25 = 1.44, h = √1.44 = 1.2cm.

Mathematics 2014 Objective — Question 25

Which of the following is a measure of dispersion?

  • A. RangeCorrect
  • B. Percentile
  • C. Median
  • D. Quartile

Explanation

Range, standard deviation, and variance are measures of dispersion, while median, percentile, and quartile are measures of position/central tendency.

Mathematics 2014 Objective — Question 26

A box contains 13 currency notes, all of which are either #50 or #20 notes. The total value of the currency notes is #530. How many #50 notes are in the box?

  • A. 4
  • B. 6
  • C. 8
  • D. 9Correct

Explanation

Let x=#50 notes, y=#20 notes. x+y=13 and 50x+20y=530. Solving simultaneously gives x=9.

Mathematics 2014 Objective — Question 27

The graph below is for the relation y=2x²+x−1. From the graph, what are the coordinates of the point S?

Diagram for question 27
  • A. (1,0.2)
  • B. (1,0.4)
  • C. (1,2.0)Correct
  • D. (1,4.0)

Explanation

When x=1: y=2(1)²+1−1=2, giving the coordinate (1,2.0), which is point S on the graph.

Mathematics 2014 Objective — Question 28

Using the same graph, find the minimum value of y.

  • A. 0.00
  • B. −0.65
  • C. −1.25Correct
  • D. −2.10

Explanation

From the graph, the minimum value of y (the vertex of the parabola) is −1.25.

Mathematics 2014 Objective — Question 29

A ship sails x km due east to a point E and continues x km due north to a point F. Find the bearing of F from the starting point.

  • A. 045°Correct
  • B. 090°
  • C. 135°
  • D. 225°

Explanation

tan θ = opposite/adjacent = x/x = 1, so θ=45°. The bearing of F from the starting point (north-east direction) is 045°.

Mathematics 2014 Objective — Question 30

If x:y=3:2 and y:z=5:4, find the value of x in the ratio x:y:z.

  • A. 8
  • B. 10
  • C. 15Correct
  • D. 20

Explanation

Combining the ratios: x:y=3:2=15:10 and y:z=5:4=10:8, giving x:y:z=15:10:8. So x=15.

Mathematics 2014 Objective — Question 31

A trader bought sachet water for GH¢55.00 per dozen and sold them at 10 for GH¢50.00. Calculate, correct to 2 decimal places, his percentage gain.

  • A. 8.00%
  • B. 8.30%
  • C. 9.09%Correct
  • D. 10.00%

Explanation

Cost price per unit = 55/12 ≈ 4.583. Selling price per 10 units after adjustment ≈ 45.83 for 10. Percentage gain = (50.00−45.83)/45.83×100% ≈ 9.09%.

Mathematics 2014 Objective — Question 32

In the figure, PQ is a tangent to the circle at R and UT is parallel to PQ. If ∠TRQ=x°, find ∠URT in terms of x°.

Diagram for question 32
  • A. 2x°
  • B. (90−x)°
  • C. (90+x)°
  • D. (180−2x)°Correct

Explanation

∠TRQ=∠UTR=x° (alternate angles, since UT∥PQ). Also ∠TRQ=∠TUR=x° (angles in alternate segment). In triangle TUR: ∠TUR+∠UTR+∠URT=180°, so ∠URT=180°−2x°.

Mathematics 2014 Objective — Question 33

Given that cos x = 12/13, evaluate (1−tan x)/tan x.

  • A. 5/13
  • B. 5/7
  • C. 7/5Correct
  • D. 13/5

Explanation

Since cos x=12/13, sin x=5/13, so tan x=5/12. (1−tanx)/tanx = (1−5/12)/(5/12) = (7/12)/(5/12) = 7/5.

Mathematics 2014 Objective — Question 36

If 2/(x−3) − 3/(x−2) is equal to P/((x−3)(x−2)), find P.

  • A. −x−5
  • B. −(x+3)
  • C. 5x−13
  • D. 5−xCorrect

Explanation

Combining over the common denominator: P = 2(x−2)−3(x−3) = 2x−4−3x+9 = 5−x.

Mathematics 2014 Objective — Question 37

Subtract ½(a−b−c) from the sum of ⅓(a−b+c) and ½(a+b−c).

  • A. ½(a+b+c)Correct
  • B. ½(a−b+c)
  • C. ½(a−b+c)
  • D. ½(a+b−c)

Explanation

Summing ½(a−b+c) and ½(a+b−c) gives 'a'. Subtracting ½(a−b−c) from a gives ½(a+b+c).

Mathematics 2014 Objective — Question 38

A man's eye level is 1.7m above the horizontal ground and 13m from a vertical pole. If the pole is 8.3m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes.

  • A. 33°
  • B. 32°
  • C. 27°Correct
  • D. 26°

Explanation

Height above eye level = 8.3−1.7=6.6m. tanθ=6.6/13=0.5077, θ=tan⁻¹(0.5077)≈27° (to the nearest degree).

Mathematics 2014 Objective — Question 39

A chord subtends an angle of 120° at the centre of a circle of radius 3.5cm. Find the perimeter of the minor sector containing the chord. [Take π=22/7]

  • A. 14⅓cmCorrect
  • B. 12⅚cm
  • C. 8⅓cm
  • D. 7⅓cm

Explanation

Perimeter = arc length + 2r = (120/360)×2×(22/7)×3.5 + 2(3.5) = 7.333+7 = 14⅓cm.

Mathematics 2014 Objective — Question 40

In parallelogram PQRS, line QR is produced to M such that |QR|=|RM|. What fraction of the area of PQMS is the area of PRMS?

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

Explanation

Since PQMS is a trapezium formed with QM=2QR, and PRMS is a parallelogram with base RM=QR, the area of PRMS is 2/3 of the area of the trapezium PQMS.

Mathematics 2014 Objective — Question 41

Determine the value of m° in the diagram above (a cyclic quadrilateral inscribed in a circle with one angle 80°).

Diagram for question 41
  • A. 80°
  • B. 90°
  • C. 110°Correct
  • D. 150°

Explanation

Since the quadrilateral is cyclic, opposite angles are supplementary: m° + 80° = 180°... using the properties of the inscribed figure (isosceles triangles from the diagonals), m°=110°.

Mathematics 2014 Objective — Question 42

In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years?

  • A. 15%
  • B. 35%
  • C. 65%Correct
  • D. 85%

Explanation

At most 18 years = 25% (lower quartile = 25th percentile). Greater than 48 years (60th percentile) = 40%. Total = 25%+40% = 65%.

Mathematics 2014 Objective — Question 43

If a number is selected at random from each of the sets P={1,2,3} and Q={2,3,4,5}, find the probability that the sum of the numbers is prime.

  • A. 5/9
  • B. 4/9Correct
  • C. 1/3
  • D. 2/9

Explanation

Total possible sums form a grid of outcomes; counting the prime sums out of the total equally likely outcomes gives a probability of 4/9.

Mathematics 2014 Objective — Question 44

In the diagram, O is the centre of the circle, line PR is a tangent to the circle at Q and ∠SOQ=86°. Calculate the value of ∠SQR°.

Diagram for question 44
  • A. 43°Correct
  • B. 47°
  • C. 54°
  • D. 86°

Explanation

Since OQ⊥PR (radius to tangent) and triangle OQS is isosceles with ∠SOQ=86°, ∠OQS=(180°−86°)/2=47°. ∠SQR = 90°−47° = 43°.

Mathematics 2014 Objective — Question 46

The probability of an event P happening is 1/5 and that of event Q is 1/4. If the events are independent, what is the probability that neither of them happens?

  • A. 4/5
  • B. 3/4
  • C. 3/5Correct
  • D. 1/20

Explanation

P(neither) = (1−1/5)(1−1/4) = (4/5)(3/4) = 3/5.

Mathematics 2014 Objective — Question 47

Each exterior angle of a polygon is 30°. Calculate the sum of the interior angles.

  • A. 540°
  • B. 720°
  • C. 1080°
  • D. 1800°Correct

Explanation

Number of sides = 360°/30° = 12. Sum of interior angles = (12−2)×180° = 1800°.

Mathematics 2014 Objective — Question 48

Find the number of terms in the Arithmetic Progression (A.P) 2, −9, −20, …, −141.

  • A. 11
  • B. 12
  • C. 13
  • D. 14Correct

Explanation

Common difference d=−11. Using Tn=a+(n−1)d: −141=2+(n−1)(−11), solving gives n=14.

Mathematics 2014 Objective — Question 50

The radii of the bases of two cylindrical tins, P and Q are r and 2r respectively. If the water level in P is 10cm high, what would be the height of the same quantity of water in Q?

  • A. 2.5cmCorrect
  • B. 5.0cm
  • C. 7.5cm
  • D. 20.0cm

Explanation

Equal volumes: πr²(10) = π(2r)²h, so h = 10r²/4r² = 2.5cm.

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