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WAEC GCE Mathematics Questions 2019 and How to Check OBJ and Theory Answers

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WAEC GCE Mathematics Questions 2019 and How to Check OBJ and Theory Answers.

WAEC GCE Mathematics Questions 2019 | In this article, I will be showing you past Mathematics objective and theory random repeated questions for free. You will also understand how WAEC Mathematics questions are set and many other examination guides. Stay focus and read through.

WAEC GCE Mathematics Questions

WAEC GCE Mathematics Questions 2018

The West African Examinations Council (WAEC) is an examination board that conducts the West African Senior School Certificate Examination, for University and Jamb entry examination in West Africa countries. … In a year, over three million candidates registered for the exams coordinated by WAEC.

WAEC GCE Mathematics Questions.

Keep accessing this page as update will be made in a jiffy

Question 1

A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universal-set

U = {2, 3, 4, 5, 6, 7, 8, 9}. Find

(a) A n(B’nC’);

(b) (AuB) n(BuC).

Mathematics waec gce 2019 Observation Please ensure to use curly brackets to enclose the elements of the sets.

Furthermore always list the elements of set C hence, this is for you to be able to find its complement.

Do it this way C i.e. C = {4, 5,6, 7, 8}, obtain the compliments of the sets Band C thus B’ = {4,5,6, 8}, C’ = {2, 3, 9,}.

Using these sets, the following procedures were to be followed:

(a)(B’ nC’) = { } Hence An (B’ n C’) = { }. Don’t be like Some candidates know to write like this { 0 } instead of { } or 0.

(b) (A u B) = { 2, 3,4,6, 7, 8, 9 }, (BuC) = {2, 3, 4,5,6, 7, 8, 9 } Therefore { Au B } n (BuC) = { 2, 3, 4, 6, 7, 8, 9 }.

Question 2

(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120 m from the foot of the cliff, calculate the height of the cliff.

(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.

WAEC GCE 2017 Observation.Furthermore, the report stated that in part (a), majority of the candidates could not draw the diagram correctly and this affected their performance significantly. A few others were unable to apply the trigonometric ratios correctly.

Candidates were expected to draw the diagram.

From the diagram, I FMI :: 120 tan 35° = 84.02m. Therefore the height of the cliff:: 2 x 84.02 = 168.04.

In part (b), the most observed weakness was their inability to convert from minutes to hours. They were expected to recall that time :: distance (i.e. t:: ~) and apply this to the problem.

speed v

Time taken by faster motorist « ~ while that of the other motorist = ~ where x is the distance

60 80

X x

from P to Q. – – – = Y2 (30 rnlnutes « Y2 hr). Simplifying this expression gave x = 120km.

60 80

Candidates’ responses to this question were reported to be generally below average.

Question 3.

(a) In the diagram, L PQR = 125°, LQRS = r, LRST = 800 and LSTU = 44°. Calculate the value of r

b) .In the diagram, TS is a tangent to the circle at A. ABI ICE, LAEC = sx”, LADB = 60° and LTAE = xo. Find the value of x”,

WAEC GCE Observation

Part (a) of this question required that candidates drew straight lines, one each passing through points Sand R and parallel to PQ and UT as shown in the diagram below – . LMQR = LQRN = 180 -125 = 55° ( alternate angles), LVST = 44° (alternate angle to LSTU). Therefore LNRS = LVSR = 80 – 44° = 360• Hence, r = 55 + 36 = 91°.

In part (b), candidates’ performance was reported to be worse than part (a). Candidates were reported to have exhibited poor understanding of circle theorems. Teachers were encouraged to do a lot of work in this area.

From the diagram, LBDA = LBAS = 600 (angles in the alternate segment). LBAE = 180° – 5x (adjacent angles on a transversal). Therefore LBAS + LBAE + LEAT = 60 + 180 – 5x + x = 180 (angles on a straight line). Solving this simple equation gave x = 150.

Question 4

The diagram shows a cone with slant height 10.5 cm. If the curved surface area of the cone is 115.5 cm²;, calculate, correct to 3 significant figures, the:

(a)base radius, r;

(b)height, h;

(c)volume of the cone. [Taken= 22]

Candidates are expected to show good understanding of mensuration of right circular cones.

The curved surface area,

iu! = 115.5cm², slant height,l= 10.scm, n = ll, hence r = 115.5

7nl = 11.5 x 7 = 3.S0cm. r2 + h2 = (10.5}2 => h = .J(10.5)2 – (3.5)2 = 9.90cm

22 x 10.5

Volume of cone = .1 x 22 x 3.5 x 9.899 == 127 cm²

3 7 1.

Question 5.

Two fair dice are thrown.

M is the event described by “the sum of the scores is 10” and

N is the event described by “the difference between the scores is 3”.

(a) Write out the elements of M and N.

(b) Find the probability of M or N.

(c) Are M and N mutually exclusive? Give reasons.

Observation.Always remember to list the elements of M and N as required.

Furthermore make sure you state correctly the condition for which events are mutually exclusive.

You are also to show that M = { (4,6) , (5,5), (6,4}. N = { (1,4), (2,5), (3,6), (4,1),

(5,2), (6,3) }. Probability of M = 3/36 = 1/12. Probability of N = 6/36 = 1/6.

Probability of Nor M:1/12 + 1/6 = 1/4. M and N are mutually exclusive because noelement is common to both sets. Thus the two events cannot happen at the same time i.e. n(M n N) = O.

Question 6.

(a) The scale of a map is 1:20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85 km²,

(b) A rectangular playing field is 18 m wide. It is surrounded by a path 6m wide such that its area is equal to the area of the path. Calculate the length of the field.

(c)

lkm = 100,000 ern, therefore 1km2 = 1km x 1km = 100,000 x 100,000 = 10,000 000,OOOcm2•

85km2 = 850,000,000,000 cm². 20,000 em on the ground = 1 cm on the map, hence 400,000,000 cm² on the ground = 1 cm2850,000,000,000 = 2125 cm2 on the map. 400,000,000.

Candidates should draw the diagram correctly.

From the diagram, area of path = 2(30 x 6) + 12a = 360 + 12a. Area of field = 18a. Since they are equal, 18a = 12a + 360. This gave a = 60 m.

see the reflex angle as 360 – x, hence, did not subtract their final answer from 360° when they had calculated the value of the reflex angle. Here, 360 – x x 22x Z x Z = 27.5 cm2

360 7 2 2

This meant that 360 – x = 360 x 2 x 27.5 from where we obtain x = 103° to the nearest degree

77.

Question 8

Using ruler and a pair of compasses only,

(a) construct

(i) a quadrilateral PQRS with IPSI :: 6 cm, LRSP:: 90°,

IRSI = 9 ern, IQRI = 8.4 cm and IPQI :: 5.4 cm;

(ii) the bisectors of LRSP and LSPQ to meet at X;

(iii) The perpendicular XTto meet PS at T.

(b) Measure IXT/’

Math waec gce 2017 Observation.Please construct the angle and not measure them.

construct a perpendicular from a given point to a given line segment.

Question 9.

In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,

(a) the distance AC;

(b) the bearing of C from A;

(c) how far east of B, C is

MATHEMATICS GCE | NECO GCE | NABTEB GCE – Observation.

Always remember to calculate LABC correctly.

Apply the cosine rule correctly.

LASC = 100°. Therefore /AC/ = 82 + 132 – 2 (8)(13)cos100 which gave / AC/ = 16.4 km.

sin (LCAS) = sin100. Hence, sin (LCAB) = 13Sin100

13 16.4 16.4

Simplifying gave LCAB = 51.32°. Bearing of C from A = 180 – (50 + 51.32) = 079°.

If the distance of C east of B = BD, then BO = BC cas 40° = 13 x cas 40° = 9.96 km.

Question 10.

(a) Copy and complete the table of values for the relation V = -X² + X + 2 for -3 ≤ x ≥ 3.

(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the v-axis, draw a graph of the relation

y = -X² + X + 2.

(c) From the graph, find the:

(i)Minimum value of y;

(ii)Roots of the equation X² – x -2 = 0;

(iii)Gradient of the curve at x = -0.5.

Math gce , waec gce neco gce Observation.

x = -0.5.

Question 11.

In the diagram, L.PTQ = L.PSR = 900, /PQ/ = 10 ern, /PS/ = 14.4 cm and /TQ/ = 6 cm.

Calculate the area of quadrilateral QRST.

(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.

Math gce Observation.

We advise that student’s should always remember to apply the concept of similar triangles correctly.

Furthermore student should try and recognize the quadrilateral as a trapezium, this will help in finding its area.

Part (b) Candidates were expected to show that:

/PT/ = …./102 – 62 = 8 cm. m: =!J2L i.e ~ = 14.4.Hence, /SR/ = 10.8 cm.

/TO/ /SR/ 6 /SR/

Area of quadrilateral QRST = Yz (6 + 10.8) x 6.4 = 53.76 cm2. Don’t subtract the area of triangle PQT from triangle PRS.

This was also in order. In part (b) if the side of the square was y, then new breadth = 90 x y = 0.9y.

100

New length = 115 x Y = 1.15y. New area = 1.15y x 0.9y = 1.035/.

100

Hence, ratio = 1.035y² : y² = 1.035 : 1 or 207:200 .

Question 12.

The frequency distribution of the weight of 100 participants in a high jump competition is as

shown below:

Weight (kg) 20-29 30 – 39 40 – 49 SO – 59 60 – 69 70-79

Number of

participants 10 18 22 25 16 9

(a) Construct the cumulative frequency table.

(b) Draw the cumulative frequency curve.

(c) From the curve, estimate the:

(i) median;

(ii) semi-interquartile range;

(iii) probability that a participant chosen at random weighs at least 60 kg.

Further mathematics gce Observation.

Make sure to draw the Ogive using class boundaries.

The median was 49.5 while the first quartile(Ql) was 37.8 and the third Quartile (~)59.5.

Hence, the semi-interquartile range Q1-Q1 = 59.5 – 32.8 = 10.85 ± 1.

2 2

Number of participants who weighed at least 60kg = 25. Therefore probability of choosing a

participant who weighed at least 60kg = 25= -.1 .

100 4.

Question 13.

(a)The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4(20/27) .Find Its first term.

(b)Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.

OBJECTIVES.

1. If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term
A. -45
B. -15
C. 15
D. 33
E. 45

2. If sinθθ = K find tanθθ, 0o ≤≤ θθ ≤≤ 90o
A. 1-K
B. kk−1kk−1
C. k1−k2√k1−k2
D. k1−kk1−k
E. kk2−1√kk2−1

3. Evaluate (101.5)2 – (100.5)2
A. 1
B. 2.02
C. 20.02
D. 202
E. 2020

4. Express the product of 0.06 and 0.09 in standard form
A. 5.4 * 10-1
B. 5.4*10-2
C. 5.4*10-3
D. 5.4*102
E. 5.4*103

5. Simplify 361/2 x 64-1/3 x 50
A. o
B. 1\24
C. 2/3
D. 11/3
E. 71/2

6. Find the quadratic equation whose roots are x = -2 or x = 7
A. x2 + 2x – 7 = 0
B. x2 – 2x + 7 = 0
C. x2 + 5 +14 = 0
D. x2 – 5x – 14 = 0
E. x2 + 5x – 14 = 0

7. A sales girl gave a change of N1.15 to a customer instead of N1.25. Calculate her percentage error
A. 10%
B. 7%
C. 8.0%
D. 2.4%
E. 10%

8. What is the probability of having an odd number in a single toss of a fair die?
A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6

9. If the total surface area of a solid hemisphere is equal to its volume, find the radius
A. 3.0cm
B. 4.5cm
C. 5.0cm
D. 9.0cm

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

11. Simplify: (34−2334−23) x 11515
A. 160160
B. 572572
C. 110110
D. 1710710

12. Simplify:(103√5√−15‾‾‾√1035−15)2
A. 75.00
B. 15.00
C. 8.66
D. 3.87

13. 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

14. Which of following is a valid conclusion from the premise. “Nigeria 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

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

16. Simplify; 3n−1×27n+181n3n−1×27n+181n
A. 32n
B. 9
C. 3n
D. 3n + 1

17. What sum of money will amount to D10,400 in 5 years at 6% simple interest?
A. D8,000.00
B. D10,000.00
C. D12,000.00
D. D16,000.00

18. The roots of a quadratic equation are 4343 and -3737. Find the equation
A. 21×2 – 19x – 12 = 0
B. 21×2 + 37x – 12 = 0
C. 21×2 – x + 12 = 0
D. 21×2 + 7x – 4 = 0

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

20. 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

21. Simplify; 21−x−1×21−x−1x
A. x+1x(1−x)x+1x(1−x)
B. 3x−1x(1−x)3x−1x(1−x)
C. 3x+1x(1−x)3x+1x(1−x)
D. x+1x(1−x)x+1x(1−x)

22. Make s the subject of the relation: P = S + sm2nrsm2nr
A. s = mrpnr+m2mrpnr+m2
B. s = nr+m2mrpnr+m2mrp
C. s = nrpmr+m2nrpmr+m2
D. s = nrpnr+m2nrpnr+m2

23. Factorize; (2x + 3y)2 – (x – 4y)2
A. (3x – y)(x + 7y)
B. (3x + y)(2x – 7y)
C. (3x + y)(x – 7y)
D. (3x – y)(2x + 7y)

24. The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take π=227π=227]
A. 2.6cm
B. 3.5cm
C. 3.6cm
D. 7.0cm

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

26. Calculate the gradient of the line PQ
A. 3535
B. 2323
C. 3232
D. 5353

27. A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
A. 411‾‾‾√411
B. 410‾‾‾√410
C. 217‾‾‾√217
D. 213‾‾‾√213

28. If cos θθ = x and sin 60o = x + 0.5 0o < θθ < 90o, find, correct to the nearest degree, the value of θθ
A. 32o
B. 40o
C. 60o
D. 69o

29. Age(years)Frequency13101424158165173Age(years)1314151617Frequency1024853

The table shows the ages of students in a club. How many students are in the club?
A. 50
B. 55
C. 60
D. 65

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

31. If log2(3x – 1) = 5, find x.
A. 2.00
B. 3.67
C. 8.67
D. 11

32. A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r
A. 2323
B. 2
C. 3
D. 6

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

34. A bag contains 5 red and 4 blue identical balls. Id 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. 2929
B. 518518
C. 20812081
D. 5959

35. 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

36. 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

37. If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x
A. 2
B. 5
C. 15
D. 18

38. The sum 110112, 11112 and 10m10n02. Find the value of m and n.
A. m = 0, n = 0
B. m = 1, n = 0
C. m = 0, n = 1
D. m = 1, n = 1

39. A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.070 to N1.00, how much did the engine cost in Niara?
A. N250,000.00
B. N200,000.00
C. N150,000.00
D. N100,000.00

40. If 27x×31−x92x=127x×31−x92x=1, find the value of x.
A. 1
B. 1212
C. -1212
D. -1

41. Find the 7th term of the sequence: 2, 5, 10, 17, 6,…
A. 37
B. 48
C. 50
D. 63

42. Given that logx 64 = 3, evaluate x log8
A. 6
B. 9
C. 12
D. 24

43. If 2n = y, Find 2(2+n3)(2+n3)
A. 4y1313
B. 4y−3−3
C. 2y1313
D. 2y−3−3

44. Factorize completely: 6ax – 12by – 9ay + 8bx
A. (2a – 3b)(4x + 3y)
B. (3a + 4b)(2x – 3y)
C. (3a – 4b)(2x + 3y)
D. (2a + 3b)(4x -3y)

45. Find the equation whose roots are 3434 and -4
A. 4×2 – 13x + 12 = 0
B. 4×2 – 13x – 12 = 0
C. 4×2 + 13x – 12 = 0
D. 4×2 + 13x + 12 = 0

46. If m = 4, n = 9 and r = 16., evaluate mnmn – 17979 + nrnr
A. 1516516
B. 1116116
C. 516516
D. – 137483748

47. Adding 42 to a given positive number gives the same result as squaring the number. Find the number
A. 14
B. 13
C. 7
D. 6

48. Ada draws the graph of y = x2 – x – 2 and y = 2x – 1 on the same axes. Which of these equations is she solving?
A. x2 – x – 3 = 0
B. x2 – 3x – 1 = 0
C. x2 – 3x – 3 = 0
D. x2 + 3x – 1 = 0

49. The volume of a cone of height 3cm is 381212cm3. Find the radius of its base. [Take π=227π=227]
A. 3.0cm
B. 3.5cm
C. 4.0cm
D. 4.5cm

50. The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take π=227π=227]
A. 14cm
B. 7m
C. 31212m
D. 13434m

51. PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS
A. 53o
B. 55o
C. 63o
D. 65o

52. The angle of elevation of an aircraft from a point K on the horizontal ground 30αα. If the aircraft is 800m above the ground, how far is it from K?
A. 400.00m
B. 692.82m
C. 923.76m
D. 1,600.99m

53. The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 7 students
A. 32o
B. 45o
C. 60o
D. 75o

54. The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.
A. 62
B. 63
C. 64
D. 65

55. Which of the following is used to determine the mode of a grouped data?
A. Bar chart
B. Frequency polygon
C. Ogive
D. Histogram

56. The area of a rhombus is 110cm
A. 5.0
B. 4.0
C. 3.0
D. 2.5

57. Simplify: 3x−yxy+2x+3y2xy+123x−yxy+2x+3y2xy+12
A. 4x+5y−xy2xy4x+5y−xy2xy
B. 5y−4x+xy2xy5y−4x+xy2xy
C. 5x+4y−xy2xy5x+4y−xy2xy
D. 4x−5y+xy2xy4x−5y+xy2xy

58. A farmer uses 2525 of his land to grow cassava, 1313 of the remaining for yam and the rest for maize. Find the part of the land used for maize
A. 215215
B. 2525
C. 2323
D. 45

59. The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?
A. 221212km
B. 30km
C. 331212km
D. 45km

60. A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent
A. 20% profit
B. 20% loss
C. 25% profit
D. 25% loss

Waec GCE 2018 Examination Instructions.

  1. Do not open your question paper until you are told to do so
  2. USE HB pencil throughout in the OBJ Section
  3. You are free to use Biro in the theory part
  4. You are allowed to use calculator to solve
  5. Make sure that you fill your name, Subject, paper, paper code and other examination details where necessary.
  6. Ensure that the texts in your question papers are boldly printed
  7. Behave yourself.
  8. Don’t let invigilators catch you using expo

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Comments (17)

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