WAEC Mathematics 2016 Objective Past Questions

All 50 questions from the West African Examinations Council (WAEC) Mathematics 2016 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

Mathematics 2016 Objective — Question 1

1. If 23x + 101x = 130x, find the value of x. A. 7 B. 6 C. 5 D. 4

• A. 7
• B. 6
• C. 5
• D. 4Correct

Explanation

Converting to base 10: 2x²+3+1x²+3x+1=1x²+3x → simplifying gives 4=3x-2x → x=4.

Mathematics 2016 Objective — Question 2

2. Simplify (3/4 - 2/3) x 1(1/5). A. 1/60 B. 1/72 C. 1/10 D. 1(7/10)

• A. 1/60
• B. 1/72
• C. 1/10Correct
• D. 1(7/10)

Explanation

By BODMAS, (3/4-2/3)=1/12; 1/12 × 6/5 = 6/60=1/10.

Mathematics 2016 Objective — Question 3

3. Simplify (10√3/√5 - √15)^2. A. 75.00 B. 15.00 C. 8.66 D. 3.87

• A. 75.00
• B. 15.00Correct
• C. 8.66
• D. 3.87

Explanation

Rationalizing 10√3/√5 = 2√15; (2√15-√15)^2=(√15)^2=15.

Mathematics 2016 Objective — Question 4

4. The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds? A. 90cm B. 135cm C. 180cm D. 225cm

• A. 90cm
• B. 135cm
• C. 180cmCorrect
• D. 225cm

Explanation

d=kt²; k=45/9=5; when t=6, d=5×36=180cm.

Mathematics 2016 Objective — Question 5

5. Which of the following is a valid conclusion from the premise? Nigerian footballers are good footballers. A. Joseph plays football in Nigeria therefore he is a good footballer B. Joseph is a good footballer therefore he is a Nigerian footballer C. Joseph is a Nigerian footballer therefore he is a good footballer D. Joseph plays good football therefore he is a Nigerian footballer

• A. Joseph plays football in Nigeria therefore he is a good footballer
• B. Joseph is a good footballer therefore he is a Nigerian footballer
• C. Joseph is a Nigerian footballer therefore he is a good footballerCorrect
• D. Joseph plays good football therefore he is a Nigerian footballer

Explanation

Joseph is a Nigerian footballer, therefore he is a good footballer — a direct application of the premise.

Mathematics 2016 Objective — Question 6

6. On a map, 1cm represents 5km. Find the area on the map that represents 100km². A. 2cm² B. 4cm² C. 8cm² D. 16cm²

• A. 2cm²
• B. 4cm²Correct
• C. 8cm²
• D. 16cm²

Explanation

1cm²→25km²; x→100km² ⇒ x=100/25×1cm²=4cm².

Mathematics 2016 Objective — Question 7

7. Simplify (3^(n-1) × 27^(n+1)) / 81^n. A. 3^2n B. 9 C. 3^n D. 3^(n+1)

• A. 3^2n
• B. 9Correct
• C. 3^n
• D. 3^(n+1)

Explanation

3^(n-1)×3^(3n+3) / 3^(4n) = 3^((n-1)+(3n+3)-4n) = 3^2 = 9.

Mathematics 2016 Objective — Question 8

8. What sum of money will amount to D10,400.00 in 5 years at 6% simple interest? A. D8,000.00 B. D10,000.00 C. D12,000.00 D. D16,000.00

• A. D8,000.00Correct
• B. D10,000.00
• C. D12,000.00
• D. D16,000.00

Explanation

P = A/(1+RT/100) = 10400/1.3 = D8,000.00.

Mathematics 2016 Objective — Question 9

9. Which of the following number lines illustrates the solution of the inequality 4 < (1/3)(2x-1) < 5? (Options A-D show number lines)

• A. Number line A
• B. Number line B
• C. Number line C
• D. Number line DCorrect

Explanation

Solving 4<(1/3)(2x-1)<5 gives 6.5<x<8, matching the number line shown in option D.

Mathematics 2016 Objective — Question 10

10. The roots of a quadratic equation are 4/3 and -3/7. Find the equation. A. 21x²-19x-12=0 B. 21x²+37x-12=0 C. 21x²-x+12=0 D. 21x²+7x-4=0

• A. 21x²-19x-12=0Correct
• B. 21x²+37x-12=0
• C. 21x²-x+12=0
• D. 21x²+7x-4=0

Explanation

(x-4/3)(x+3/7)=0 → x²+3x/7-4x/3-12/21=0; multiplying through by 21 gives 21x²-19x-12=0.

Mathematics 2016 Objective — Question 11

11. Find the values of y for which the expression (y²-9y+18)/(y²+4y-21) is undefined. A. 6,-7 B. 3,-6 C. 3,-7 D. -3,-7

• A. 6,-7
• B. 3,-6
• C. 3,-7Correct
• D. -3,-7

Explanation

Denominator y²+4y-21=0 → (y-3)(y+7)=0 → y=3, y=-7.

Mathematics 2016 Objective — Question 12

12. Given that 2x+y=7 and 3x-2y=3, by how much is 7x greater than 10? A. 1 B. 3 C. 7 D. 17

• A. 1
• B. 3
• C. 7Correct
• D. 17

Explanation

Solving simultaneously: y=7-2x, substitute into 3x-2(7-2x)=3 → 7x=17; 17 is greater than 10 by 7.

Mathematics 2016 Objective — Question 13

13. Simplify 2/(1-x) - 1/x. A. (x+1)/(x(1-x)) B. (3x-1)/(x(1-x)) C. (3x+1)/(x(1-x)) D. (x-1)/(x(1-x))

• A. (x+1)/(x(1-x))
• B. (3x-1)/(x(1-x))Correct
• C. (3x+1)/(x(1-x))
• D. (x-1)/(x(1-x))

Explanation

2/(1-x) - 1/x = (2x-(1-x))/(x(1-x)) = (3x-1)/(x(1-x)).

Mathematics 2016 Objective — Question 14

14. Make s the subject of the relation p = s + sm²/nr. A. s = mrp/(nr+m²) B. s = (nr+m²)/mrp C. s = nrp/(mr+m²) D. s = nrp/(nr+m²)

• A. s = mrp/(nr+m²)
• B. s = (nr+m²)/mrp
• C. s = nrp/(mr+m²)
• D. s = nrp/(nr+m²)Correct

Explanation

p=s(1+m²/nr)=s(nr+m²)/nr → s = nrp/(nr+m²).

Mathematics 2016 Objective — Question 15

15. Factorize: (2x+3y)²-(x-4y)². A. (3x-y)(x+7y) B. (3x+y)(2x-7) C. (3x+y)(x-7y) D. (3x-y)(2x+7y)

• A. (3x-y)(x+7y)Correct
• B. (3x+y)(2x-7)
• C. (3x+y)(x-7y)
• D. (3x-y)(2x+7y)

Explanation

Difference of two squares: [(2x+3y)-(x-4y)][(2x+3y)+(x-4y)] = (x+7y)(3x-y).

Mathematics 2016 Objective — Question 16

16. The curved surface area of a cylinder, 5cm high, is 110cm². Find the radius of its base. [Take π=22/7] A. 2.6cm B. 3.5cm C. 3.6cm D. 7.0cm

• A. 2.6cm
• B. 3.5cmCorrect
• C. 3.6cm
• D. 7.0cm

Explanation

CSA=2πrh; 110=2×(22/7)×r×5 → r=3.5cm.

Mathematics 2016 Objective — Question 17

17. The volume of a pyramid with height 15cm is 90cm³. If its base is a rectangle with dimensions x cm by 6cm, find the value of x. A. 3 B. 5 C. 6 D. 8

• A. 3Correct
• B. 5
• C. 6
• D. 8

Explanation

Volume=(1/3)×base×height → 90=(1/3)×base×15 → base=18cm²=x×6 → x=3.

Mathematics 2016 Objective — Question 18

18. In the diagram, YW is a tangent to the circle at X, |UV|=|VX| and ∠VXW=50°. Find the value of ∠UXY. A. 70° B. 80° C. 105° D. 110°

• A. 70°
• B. 80°Correct
• C. 105°
• D. 110°

Explanation

∠XUV=∠VXW=50° (angle in alt segment); ∠UXV=∠XUV=50° (base angles of isosceles Δ); ∠UXY=180°-(50°+50°)=80° (angles on a straight line).

Mathematics 2016 Objective — Question 19

19. In the diagram, PF, QT, RG intersect at S and PG//RG. If ∠SPQ=113° and ∠RST=220°, Find ∠PSQ. A. 22° B. 45° C. 67° D. 89°

• A. 22°
• B. 45°Correct
• C. 67°
• D. 89°

Explanation

∠RST=∠QST=22° (vertically opposite angles); ∠SQP=22° (alternate angles); ∠PSQ=180°-22°-113°=45° (sum of angles in a triangle).

Mathematics 2016 Objective — Question 20

20. In the diagram, O is the centre of the circle, ∠XOZ=(10m)° and ∠XWZ=m°. Calculate the value of m°. A. 30° B. 36° C. 40° D. 72°

• A. 30°Correct
• B. 36°
• C. 40°
• D. 72°

Explanation

Reflex ∠XOZ=360-10m=2m° (angle at centre = 2×angle at circumference) → 360=12m → m=30°.

Mathematics 2016 Objective — Question 21

21. Kweku walked 8m up to a slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be? A. 4.5m B. 6.0m C. 7.5m D. 9.0m

• A. 4.5m
• B. 6.0m
• C. 7.5mCorrect
• D. 9.0m

Explanation

By similar triangles, 3/8 = h/20 → h=3×20/8=7.5m.

Mathematics 2016 Objective — Question 22

22. In the diagram, TS is a tangent to the circle at S. |PR| and ∠PQR=117°. Calculate ∠PST. A. 54° B. 44° C. 34° D. 27°

• A. 54°Correct
• B. 44°
• C. 34°
• D. 27°

Explanation

∠PRS=180-117=63° (opposite angles of cyclic quadrilateral); ∠RPS=63°(base angles of isosceles Δ); ∠PRS(final)=180-2(63)=54°; ∠PST=∠PRS=54° (angles in alternate segment).

Mathematics 2016 Objective — Question 23

23. In the diagram above, PR//SV//WY, TX//QY, ∠PQT=48° and ∠TXW=60°. Find ∠TQU. A. 120° B. 108° C. 72° D. 60°

• A. 120°
• B. 108°
• C. 72°Correct
• D. 60°

Explanation

∠TXW=∠STX=60°(alternate angles); ∠QTU=∠PQT=48°(alternate angles); producing XT to P: ∠QTP=180-(48+60)=72°; ∠TQU=∠QTP=72°(alternate angles).

Mathematics 2016 Objective — Question 24

24. [A straight line passes through the points P(1,2) and Q(5,8)] Calculate the gradient of the line PQ. A. 3/5 B. 2/3 C. 3/2 D. 5/3

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

Explanation

Gradient = (y2-y1)/(x2-x1) = (8-2)/(5-1)=6/4=3/2.

Mathematics 2016 Objective — Question 25

25. Calculate the length PQ. A. 4√11 B. 4√10 C. 2√17 D. 2√13

• A. 4√11
• B. 4√10
• C. 2√17
• D. 2√13Correct

Explanation

PQ=√[(8-2)²+(5-1)²]=√(36+16)=√52=2√13.

Mathematics 2016 Objective — Question 26

26. In the diagram, TX is perpendicular to UW, |UX|=1cm and |TX|=√3cm. Find ∠UTW. A. 135° B. 105° C. 75° D. 60°

• A. 135°
• B. 105°Correct
• C. 75°
• D. 60°

Explanation

tanθ1=1/√3 → θ1=30°; using the second triangle segment tanθ2=1 → θ2=45°; per the official key ∠UTW=105°.

Mathematics 2016 Objective — Question 27

27. If cosθ=x and sin60°=x+0.5, 0°<θ<90°, find, correct to the nearest degree, the value of θ°. A. 96° B. 57° C. 60° D. 69°

• A. 96°
• B. 57°
• C. 60°
• D. 69°Correct

Explanation

sin60°=cos(90°-60°)=cos30°=√3/2=x+0.5 → x=(√3-1)/2=0.366=cosθ → θ=cos⁻¹(0.366)≈69°.

Mathematics 2016 Objective — Question 28

28. [Table: Age(years) 13,14,15,16,17; Frequency 10,24,8,5,3] How many students are in the club? A. 50 B. 55 C. 60 D. 65

• A. 50Correct
• B. 55
• C. 60
• D. 65

Explanation

Total = 10+24+8+5+3 = 50.

Mathematics 2016 Objective — Question 29

29. Find the median age. A. 13 B. 14 C. 15 D. 16

• A. 13
• B. 14Correct
• C. 15
• D. 16

Explanation

Median position = (50/2)th = 25th value, which falls in the age-14 group.

Mathematics 2016 Objective — Question 30

30. The figure above is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food? A. Le2,250,000.00 B. Le2,700,000.00 C. Le3,600,000.00 D. Le4,500,000.00

• A. Le2,250,000.00
• B. Le2,700,000.00
• C. Le3,600,000.00Correct
• D. Le4,500,000.00

Explanation

Sector angle for food = 360-(80+70+90)=120°; Amount = (120/360)×10,800,000 = Le3,600,000.00.

Mathematics 2016 Objective — Question 31

31. A fair die is thrown two times. What is the probability that the sum of the scores is at least 10? A. 5/36 B. 1/6 C. 5/18 D. 2/3

• A. 5/36
• B. 1/6Correct
• C. 5/18
• D. 2/3

Explanation

Outcomes with sum≥10 (from a 6x6 table) total 6 out of 36; P=6/36=1/6.

Mathematics 2016 Objective — Question 32

32. The marks of eight students in a test are: 10, 4, 5, 3, 14, 13, 16 and 7. Find the range. A. 16 B. 14 C. 13 D. 11

• A. 16
• B. 14
• C. 13Correct
• D. 11

Explanation

Range = highest-lowest = 16-3=13.

Mathematics 2016 Objective — Question 33

33. If log2^(3x-1)=5, find x. A. 2.00 B. 3.67 C. 6.67 D. 11.00

• A. 2.00
• B. 3.67
• C. 6.67
• D. 11.00Correct

Explanation

3x-1=2^5=32 → 3x=33 → x=11.

Mathematics 2016 Objective — Question 34

34. A sphere of radius r cm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r cm. A. 2/3 B. 2 C. 3 D. 6

• A. 2/3
• B. 2
• C. 3Correct
• D. 6

Explanation

(4/3)πr³=π(3²)(4) → r³=27 → r=3cm.

Mathematics 2016 Objective — Question 35

35. Express 1975 correct to 2 significant figures. A. 20 B. 1,900 C. 1,980 D. 2,000

• A. 20
• B. 1,900
• C. 1,980
• D. 2,000Correct

Explanation

1975 to 2 s.f. = 2,000.

Mathematics 2016 Objective — Question 36

36. In the diagram above, MOPQ is a trapezium with QP//MO, MQ//NP, NQ//OP, |QP|=9cm and the height of ΔQNP=6cm, calculate the area of the trapezium. A. 96cm² B. 90cm² C. 81cm² D. 27cm²

• A. 96cm²
• B. 90cm²
• C. 81cm²Correct
• D. 27cm²

Explanation

Area = area of parallelogram QPNM + area of ΔPNO = (b×h) + (½×b×h) = (9×6) + (½×9×6) = 54+27 = 81cm².

Mathematics 2016 Objective — Question 37

37. The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector. [Take π=22/7] A. 70° B. 60° C. 55° D. 42°

• A. 70°
• B. 60°
• C. 55°
• D. 42°Correct

Explanation

Perimeter = (θ/360)×2πr + 2r; 64 = (θ×2×22×21)/(360×7) + 42 → θ = 22/360 × 7920/(22×... ) = 42° per official key.

Mathematics 2016 Objective — Question 38

38. Examine M'∩N from the Venn diagram above. A. {f,g} B. {e} C. {e,f,g} D. {e,f,g}

• A. {f,g}Correct
• B. {e}
• C. {e,f,g}
• D. {e,f,g}

Explanation

M'={e,f,g}, N={c,f,g}; M'∩N={f,g}.

Mathematics 2016 Objective — Question 39

39. If 20(mod 9) is equivalent to y(mod 6), find y. A. 1 B. 2 C. 3 D. 4

• A. 1
• B. 2Correct
• C. 3
• D. 4

Explanation

20(mod9)=2; comparing to y(mod6), y=2.

Mathematics 2016 Objective — Question 40

40. Simplify (p-r)²-r² / (2p²-4pr). A. 1/2 B. p-2r C. 1/(p-2r) D. 2p/(p-2r)

• A. 1/2Correct
• B. p-2r
• C. 1/(p-2r)
• D. 2p/(p-2r)

Explanation

By difference of two squares, numerator=(p-r+r)(p-r-r)=p(p-2r); denominator=2p(p-2r); simplifies to 1/2.

Mathematics 2016 Objective — Question 41

41. In the diagram above, O is the centre of the circle, ∠QPS=100°, ∠PSQ=60° and ∠QSR=80°. Calculate ∠SQR. A. 20° B. 40° C. 60° D. 80°

• A. 20°Correct
• B. 40°
• C. 60°
• D. 80°

Explanation

∠RPQ=80° (angle in the same segment); ∠SPR=100-80=20°; ∠SQR=∠SPR=20° (angles in the same segment).

Mathematics 2016 Objective — Question 42

42. A bag contains 5 red and 4 blue identical balls. If two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue. A. 2/9 B. 5/18 C. 20/81 D. 5/9

• A. 2/9
• B. 5/18
• C. 20/81Correct
• D. 5/9

Explanation

P(1st red)×P(2nd blue)=5/9×4/9=20/81.

Mathematics 2016 Objective — Question 43

43. The relation y=x²+2x+k passes through the point (2,0). Find the value of k. A. -8 B. -4 C. 4 D. 8

• A. -8Correct
• B. -4
• C. 4
• D. 8

Explanation

0=2²+2(2)+k → 0=4+4+k → k=-8.

Mathematics 2016 Objective — Question 44

44. Find the next three terms of the sequence: 0,1,1,2,3,5,8... A. 13,19,23 B. 9,11,13 C. 11,15,19 D. 13,21,34

• A. 13,19,23
• B. 9,11,13
• C. 11,15,19
• D. 13,21,34Correct

Explanation

This is a Fibonacci sequence; next terms: 8+5=13, 13+8=21, 21+13=34.

Mathematics 2016 Objective — Question 45

45. Find the lower quartile of the distribution illustrated by the cumulative frequency curve. A. 17.5 B. 19.0 C. 27.5 D. 28.0

• A. 17.5
• B. 19.0Correct
• C. 27.5
• D. 28.0

Explanation

First quartile = 25th percentile = 25% of 600 = 150th term, whose corresponding score is 19.

Mathematics 2016 Objective — Question 46

46. The ratio of the exterior angle to the interior angle of a regular polygon is 1:11. How many sides has the polygon? A. 30 B. 24 C. 18 D. 12

• A. 30
• B. 24Correct
• C. 18
• D. 12

Explanation

Exterior:Interior=1:11 → (360/n):((n-2)180/n)=1:11 → 22/(n-2)=1 → n=22+2=24.

Mathematics 2016 Objective — Question 47

47. Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother? A. n/2 + 5/2 B. n/2 - 5 C. 5 - n/2 D. n/2 + 5

• A. n/2 + 5/2
• B. n/2 - 5
• C. 5 - n/2
• D. n/2 + 5Correct

Explanation

Half of Halima's age = n/2; brother's age = n/2 + 5.

Mathematics 2016 Objective — Question 48

48. In the diagram above, MN is a chord of a circle KMN centre O and radius 10cm. If ∠MON=140°, find, correct to the nearest cm, the length of the chord MN. A. 19cm B. 18cm C. 17cm D. 12cm

• A. 19cmCorrect
• B. 18cm
• C. 17cm
• D. 12cm

Explanation

Length of chord = 2r·sin(θ/2) = 2×10×sin70° = 20×0.9397 ≈ 19cm.

Mathematics 2016 Objective — Question 49

49. An object is 6cm away from the base of a mast. If the angle of depression of the object from the top of the mast is 50°, find, correct to 2 decimal places, the height of the mast. A. 8.60m B. 7.83m C. 7.51m D. 7.15m

• A. 8.60m
• B. 7.83m
• C. 7.51m
• D. 7.15mCorrect

Explanation

tan50°=h/6 → h=6tan50°=7.15m (2 d.p.).

Mathematics 2016 Objective — Question 50

50. From the diagram, which of the following is true? A. m°+n°+p°=180° B. m°+n°=180° C. m°=p°+n° D. n°=m°+p°

• A. m°+n°+p°=180°
• B. m°+n°=180°
• C. m°=p°+n°
• D. n°=m°+p°Correct

Explanation

m°+p°+180°-n°=180° (sum of angles in a triangle) → m°+p°=n°.