# Expo: WAEC Mathematics Questions and Answers 2022/2023 – theory / Objective

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## WAEC 2022/2023 Essay and Objectives – General Mathematics

2022 NECO Chemistry Expo: Questions and Answers Obj Theory

2022 NECO Physics Questions and Answers is Out

WAEC Further Mathematics Questions and Answers theory / Objective

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This post contains both the mathematics Objectives and the Theory.

OBJECTIVE TEST

Symbols used:

^ means raise to power
/ division

SQRT means square root

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### 2022 WAEC Mathematics Questions – Expo

(1a)

Given A={2,4,6,8,…}
B={3,6,9,12,…}
C={1,2,3,6}
U= {1,2,3,4,5,6,7,8,9,10}

A’ = {1,3,5,7,9}
B’ = {1,2,4,5,7,8,10}
C’ = {4,5,7,8,9,10}
A’nB’nC’ = {5, 7}

(1b)
Cost of each premiere ticket = \$18.50
At bulk purchase, cost of each = \$80.00/50 = \$16.00

Amount saved = \$18.50 – \$16.00
=\$2.50
=======================================

(2ai)
P = (rk/Q – ms)⅔
P^3/2 = rk/Q – ms
rk/Q = P^3/2 + ms
Q= rk/P^3/2 + ms

(2aii)
When P =3, m=15, s=0.2, k=4 and r=10
Q = rk/p^3/2 + ms = 10(4)/(3)^3/2 + (15)(0.2)
= 40/8.196 = 4.88(1dp)

(2b)
x + 2y/5 = x – 2y
Divide both sides by y
X/y + 2/5 = x/y – 2
Cross multiply
5(x/y) – 10 = x/y + 2
5(x/y) – x/y = 2 + 10
4x/y = 12
X/y = 3
X : y = 3 : 1
=======================================

(3a)
Diagram
CBD = CDB (base angles an scales D)
BCD+CBD+CDB=180° (Sum of < in a D)
2CDB+BCD=180°
2CDB+108°=180°
2CDB=180°-108°=72°
CDB=72/2=36°
BDE=90°(Angle in semi circle)
CDE=CDB+BDE
=36°+90
=126

(3b)
(Cosx)² – Sinx given
(Sinx)² + Cosx
Using Pythagoras theory thrid side of triangle
y²= 1²+√3
y²= 1+ 3=4
y=√4=2
(Cosx)² – sinx/(sinx)² + cosx

(1/2)² – √3/2/
(√3/2)² + 1/2 = 1/4 – √3/2 = 1-2√3/4
3/4+1/2 = 3+2/4
=1-2√3/4 * 4/5
=1-2√3/5
=======================================

(4a)
Total Surface Area = 224πcm²
r:l = 2:5
r/l = 2/5
Cross multiply
2l/2 = 5r/2
L = 5r / 2
Total surface = πrl + πr²
= πr (l + r)
24π/π = πr (5r/2 + r )/ π
224 = 5r²/2 + r²/1
L.c.m = 2
448 = 5r² + 2r²
448 / 7= 7r²/7
r² = 64
r = √64 = 8cm
L = 5*8/2 = 20cm

(4b)
Volume = 1/2πr²h
= 1/3 * 22/7 * 8 * 8 * 18.33
= 1228.98cm³
L² = h² + r ²
20² = h² + 8²
400 – 64 = h²
h² = 336
h = √ 336
h = 18.33cm
=======================================

(5a)
Total income = 32+m+25+40+28+45
=170+m
PR(²)=m/170+m = 0.15/1
M=0.15(170+m)
M=25.5+0.15m
0.85m/0.85=25.5/0.85
M=30

(5b)
Total outcome = 170 + 30 = 200

(5c)
PR(even numbers) = 30+40+50/200
=115/200 = 23/40
=======================================

(7a)
Diagram

Using Pythagoras theorem, l²=48² + 14²
l²=2304 + 196
l²=2500
l=√2500
l=50m
Area of Cone(Curved) =πrl
Area of hemisphere=2πr²
Total area of structure =πrl + 2πr²
=πr(l + 2r)
=22/7 * 14 [50 + 2(14)]
=22/7 * 14 * 78
=3432cm²
~3430cm² (3 S.F)

(7b)
let the percentage of Musa be x
Let the percentage of sesay be y
x + y=100 ——————-1
(x – 5)=2(y – 5)
x – 5=2y – 10
x – 2y=-5 ——————-2
Equ (1) minus equ (2)
y – (-2y)=100 – (-5)
3y=105
y=105/3
y=35
Sesay’s present age is 35years
=======================================

(8a)
Let Ms Maureen’s Income = Nx
1/4x = shopping mall
1/3x = at an open market

Hence shopping mall and open market = 1/4x + 1/3x
= 3x + 4x/12 = 7/12x

Hence the remaining amount
= X-7/12x = 12x-7x/12 =5x/12

Then 2/5(5x/12) = mechanic workshop
= 2x/12 = x/6
Amount left = N225,000
Total expenses
= 7/12x + X/6 + 225000
= Nx

7x+2x+2,700,000/12 =Nx
9x + 2,700,000 = 12x
2,700,000 = 12x – 9x
2,700,000/3 = 3x/3
X = N900,000

(ii) Amount spent on open market = 1/3X
= 1/3 × 900,000
= N300,000

(8b)
T3 = a + 2d = 4m – 2n
T9 = a + 8d = 2m – 8n
-6d = 4m – 2m – 2n + 8n
-6d = 2m + 6n
-6d/-6 = 2m+6n/-6
d = -m/3 – n
d = -1/3m – n
=======================================

(9a)
Draw the triangle

(9b)
(i)Using cosine formulae
q² = x² + y² – 2xycosQ
q² = 9² + 5² – 2×9×5cos90°
q² = 81 + 25 – 90 × 0
q² = 106
q = square root 106
q = 10.30 = 10km/h
Distance = 10 × 2 = 20km

(ii)
Using sine formula
y/sin Y = q/sin Q
5/sin Y = 10.30/sin 90°
Sin Y = 5 × sin90°/10.30
Sin Y = 5 × 1/10.30
Sin Y = 0.4854
Y = sin‐¹(0.4854), Y = 29.04

Bearing of cyclist X from y
= 90° + 19.96°
= 109.96° = 110°

(9c)
Speed = 20/4, average speed = 5km/h

(12a)
BCD=ABC=40°(alternate D)

DDE=2*BCD(<at centre = twice < at circle)

DDE = 2*40 = 80°
OD3=OED(base < of I sealed D ODE)
ODE + OED + DOE= 180°(sum of < is in D)
2ODE+DOE=180°
2ODE+80°=180
2ODE+180=180
2ODE+100°
ODE+100/2=50°

(12bi)
Digram

(12bii)
Area of parallelogram = absin
=5*7*sin125°
=35*sin55°
=35*0.8192
=28.67
=28.7cm²(1dp)

(12c)
Given x=1/2(1-√2)
2x²-2x=2[1/2(1-√2]²-2(1/2(1-√2)}
=2[1-2√2+2/4]-(1-√2)
=(3-2√2/2)-(1-√2)
=3-2√2-2+2√2/2=1/2

### WAEC Maths OBJ Questions

1. Express, correct to three significant figures, 0.003597.

A. 0.359

B. 0.004

C. 0.00360

D. 0.00359

2. Evaluate: (0.064)^-1/3

A. 5/2

B. 2/5

C. -2/5

D. -5/2

3. Solve: y+1/2 – 2y-1/3 = 4

A. y = 19

B. y = -19

C. y = -29

D. y = 29

4. Simplify, correct to three significant figures, (27.63)^2 – (12.37)^2

A. 614

B. 612

C. 611

D. 610

5. If 7 + y = 4 (mod 8), find the least value of y, 10<=y<=30

A. 11

B. 13

C. 19

D. 21

6. If T = (prime numbers) and

M = (odd numbers) are subsets of

U = (x: 0<x<=10, and x is an integer), find (T’ n M’).

A. (4, 6, 8, 10)

B. (1, 4, 6, 8, 10)

C. (1, 2, 4, 6, 8, 10)

D. (1, 2, 3, 5, 7, 8, 9)

7. Evaluate: log9 base 3 – log8 base 2 /log9 base 3

A. -1/3

B. 1/2

C. 1/3

D. -1/2

8. If 23y = 1111two, find the value of y.

A. 4

B. 5

C. 6

D. 7

9. If 6, P and 14 are consecutive terms in an Arithmetic Progression (A.P), find the value of P.

A. 9

B. 10

C. 6

D. 8

10. Evaluate: 2 (SQRT 28) – 3 (SQRT 50) + (SQRT 50)

A. 4 (SQRT 7) – 21 (SQRT 2)

B. 4 (SQRT 7) – 11 (SQRT 2)

C. 4 (SQRT 7) – 9 (SQRT 2)

D. 4 (SQRT 7) + (SQRT 2)

11. If m : n = 2 : 1, evaluate 3m^2 – 2n^2 /m^2 + mn

A. 4/3

B. 5/3

C. 3/4

D. 3/5

12. H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y

A. H = p /4y^2

B. H = 2p / y^2

C. H = p / 2 y^2

D. H = p / y^2

13. Solve 4x^2 – 16x + 15

A. X = 1 (1/2) or X = -2 (1/2)

B. X = 1 (1/2) or X = 2 (1/2)

C. X = 1 (1/2) or X = -1 (1/2)

D. X = -1 (1/2) or X = -2 (1/2)

14. Evaluate 0.42 divided by 2.5 /0.5 x 2.05, leaving the answer in standard form.

A. 1.639 x 10^2

B. 1.639 x 10^1

C. 1.639 x 10^-1

D. 1.639 x 10^-2

15. Simplify: log6 – 3log3 + 2/3log27.

A. 3log 2

B. Log2

C. Log3

D. 2log3

16. Bala sold an article for 6,900.00 naira and made a profit of 15 percent. Calculate his percentage profit if he had sold for 6600.00.

A. 5 percent

B. 10 percent

C. 12 percent

D. 13 percent

17. If 3p = 4q and 9p = 8q-12, find the value of pq.

A. 12

B. 7

C. -7

D. -12

18. If (0.25)^y = 32, find the value of y.

A. y = -5/2

B. y = -3/2

B. y = 3/2

D. y = 5/2

19. There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

A. 3/4

B. 1/3

C. 2/3

D. 1/4

20. Simplify: x^2 – 5x – 14 / x^2 – 9x + 14

A. X – 7 /x + 7

B. X + 7 /x – 7

C. X – 2 /x + 4

D. X + 2 /x – 2

21. Which of these values would make 3p – 1 /p^2 – p undefined?

A. 1

B. 1/3

C. -1/3

D. -1

22. The total surface area of a solid cylinder is 165 cm2. If the base diameter is 7 cm, calculate its height. (Take pi = 22/7)

A. 7.5 cm

B. 4.5 cm

C. 4.0 cm

D. 2.0 cm

23. If 2^a = SQRT(64) and b/a = 3, evaluate a^2 + b^2.

A. 250

B. 160

C. 90

D. 48

24. In triangle XYZ, line XZ = 32 cm, angle YXZ = 52 degrees and XZY = 90 degrees. Find, correct to the nearest centimeter, line XZ.

A. 31 cm

B. 25 cm

C. 20 cm

D. 13 cm

25. If log 2 base x = 0.3, evaluate log 8 base x.

A. 2.4

B. 1.2

C. 0.9

D. 0.6

26. An arc subtends an angle of 72 degrees at the centre of a circle. Find the length of the arc if the radius of the circle is 3.5 cm. (Take pi = 22/7)

A. 6.6 cm

B. 8.8 cm

C. 4.4 cm

D. 2.2 cm

27. Make b the subject of the relation

lb = 1/2(a+b)h

A. ah /2l – h

B. 2l – h/al

C. al/2l – h

D. al/2 – h

29. Eric sold his house through an agent who charged 8 percent commission on the selling price. If Eric received 117,760.00 dollars after the sale, what was the selling price of the house?

A. 130,000.00 dollars

B. 128,000.00 dollars

C. 125,000.000 dollars

D. 120,000.00 dollars

29. Find the angle which an arc of length 22 cm subtends at the centre of a circle of radius 15 cm. (take pi = 22/7)

A. 70 degrees

B. 84 degrees

C. 96 degrees

D. 156 degrees

30. A rectangular board has a length of 15 cm and width x cm. If its sides are doubled, find its new area.

A. 60x cm squared

B. 45x cm squared

C. 30x cm squared

D. 15x cm squared

31. In the diagram, POS and ROT are straight lines. OPQR is a parallelogram, line OS = line OT and angle OST = 50 degrees. Calculate the value of angle OPQ.

A. 100 degrees

B. 120 degrees

C.140 degrees

D. 160 degrees

32. Factorize completely: (2x + 2y)(x-y) + (2x – 2y)(x + y)

A. 4(x – y)(x + y)

B. 4(x – y)

C. 2(x – y) (x + y)

D. 2(x – y)

33. The interior angles of a polygon are 3x, 2x, 4x, 3x and 6x. Find the size of the smallest angle of the polygon.

A. 80 degrees

B. 60 degrees

C. 40 degrees

D. 30 degrees

34. A box contains 2 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours?

A. 2/3

B. 3/5

C. 7/20

D. 12/25

35. Find the equation of a straight line passing through the points (1, -5) and having gradient of ¾.

A. 3x + 4y – 23 = 0

B. 3x + 4y + 23 = 0

C. 3x – 4y + 23 = 0

D. 3x – 4y – 23 = 0

36. The foot of a ladder is 6 m from the base of an electric pole. The top of the ladder rest against the pole at a point 8 m above the ground. How long is the ladder?

A. 14 m

B. 12 m

C. 10 m

D. 7 m

37. If tan x = 3/4, 0<x<90,

evaluate cos x/2sin x

A. 8/3

B. 3/2

C. 4/3

D. 2/3

38. From the top of a vertical cliff 20 m high, a boat at sea can be sighted 75 m away and on the same horizontal position as the foot of the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

A. 56 degrees

B. 75 degrees

C. 16 degrees

D. 15 degrees

39. In the diagram, O is the centre of the circle of radius 18 cm. If angle ZXY = 70 degrees, calculate the length of arc ZY. (Take pi = 22/7)

A. 11 cm

B. 22 cm

C. 44 cm

D. 80 cm

In the diagram, RT is tangent to the circle at R, angle PQR = 70 degrees, angle QRT = 52 degrees, angle QSR = y and angle PRQ = x. Use the diagram to answer questions 40 and 41

40. Find the value of y.

A. 70 degrees

B. 60 degree

C. 52 degree

D. 18 degree

41. Calculate the value of x

A. 70 degrees

B. 58 degrees

C. 55 degrees

D. 48 degrees

42. Calculate the variance of 2, 4, 7, 8, 9

A. 7.2

B. 6.8

C. 3.5

D. 2.6

43. The fourth term of an Arithmetic Progression (A.P.) is 37 and the first term is -20. Find the common difference.

A. 63

B. 57

C. 19

D. 17

In the diagram, PQ is parallel to RS, angle QFG = 105 degrees and angle FEG = 50 degrees. Use the diagram to answer questions 44 and 45.

44. Find the value of m

A. 130 degrees

B. 105 degrees

C. 75 degrees

D. 55 degrees

45. Find the values of n

A. 40 degrees

B. 55 degrees

C. 75 degrees

D. 130 degrees

46. A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

A. 1/6

B. 1/4

C. 1/3

D. 1/2

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

A. 90

B. 80

C. 70

D. 40

The following are scores obtained by some students in a test.

8 18 10 14 18 11 13 14 17 15 8 16 13

Use this information to answer questions 48 to 50

48. Find the mode of the distribution

A. 18

B. 14

C. 13

D. 8

49. Find the median score.

A. 14.5

B. 14.0

C. 13.5

D. 13.0

50. How many students scored above the mean score?

A. 10

B. 9

C. 8

D. 7

1-10 CBCDACDCCD

31-40 CDACCCCCDA

41-50 BBBCDCACDB

Are the WAEC objective questions difficult? Let’s quickly move to the Solutions of 2022/2023 WAEC Mathematics Questions

## 2022 WAEC Possible Mathematics Questions and Answers

1.A

The two zeros before 3 are not counted since they are not between whole numbers

2. A

0.064^(-1/3) = 5/2

3. y+1/2 – 2y-1/3 = 4

Multiply through by 6

3y + 3 – 4y + 2 = 24

Collecting like terms

Implies:

y =-19

4. D

5. B

6. U = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

T = (2, 3, 5, 7)

M = (1, 3, 5, 7, 5)

(T’ n M’) = the numbers in the union set U but not in T and M

(T’ n M’) = (4, 6, 8, 10)

7. Log9 base 3 – log 8 base 2 / log 9 base 3

Log3^2 base 3 – log2^3 base 2 / log 3^2

2log3 base 3 – 3log2 base 2 / 2log3 base 3

Log3 base 3 = 1

Log 2 base = 1

Reason: Log to its own base is 1

This implies:

2 -3 /2 = -1/2

8. Convert both sides to base ten

2xy^1 + 3Xy^0 = 1×2^3 + 1×2^2 + 1x 2^1 + 1×2^0

2y +3 = 8 + 4 + 2 + 2 + 1

2y + 3 = 15

2y = 15-3

Y = 12/2 = 6

9. 6, p, 14

First term a = 6, third term = 14

Number of terms of A.P = a + (n-1)d

T3 =14 = a + 2d

6 + 2d = 14

Common difference d = 14

P = second term

T2 = a + d = 6 + 4 = 10

10. 4 SQRT(7) – 15 SQRT(2) + 6 SQRT(2)

4 SRT(7) – 9 SRT(2)

11. m/n = 2/1

m = 2n

3(2n)^2 – 2n^2 / 2n^2 +2n^2

12n^2 – 2n^2 / 4n^2 + 2n^2 = 10n^2/6n^2

= 5/3

12. H = kp / y^2 where k = constant, H = 1, p = 8, y = 2

k = Hy^2 /p = 4×1/8 = ½

H = p/2y^2

13. Factorizing:

(4x^2 -6x)(-10x +15)

2x(2x – 3)-5(2x – 3)

2x-5 = 0 or 2x-3 = 0

X = 5/2 or x = 3/2

14. C

15. B

16. 15 percent profit = 115 = 6,900 naira

Let the percent Bala sold the article at 6600 naira = x

115 = 6,900

X = 6600

Cross multiplying

6900x = 759000

X = 110 percent

The percentage profit if he had sold it at 6600 naira = 110 – 100

= 10 percent

17. 3p = 4q

9p = 8q – 12

p = 4q/3

8q – 12 -(36q/3)

24q – 36 – 36q

-12q = 36

q = -3

3p = -12

p = -4, pq = -3x-4 = 12

18. (0.25)^y = 32

(1/4)^y = 32

2^-2 = 2^5

-2y = 5

y = -5/2

19. C

20. (x-7)(x+2)/ (x+7)(x+2) = (x-7) / (x+7)

21. 3p – 1/p^2 -p

It will be undefined when the denominator p^2 – p = 0

therefore, p = 1 would make it zero

22. Given:

T.S.A = 165 cm^2

D = 7 cm

h = ?

pi = 22/7

T.S.A = 2pirh + 2pir^2

165 -2x(22/7)x3.5^2 = 2x(22/7) x3.5h

165 – 77 = 22h,

h = 88/22

h = 4.0 cm

Are you enjoying the Solutions to WAEC Mathematics Questions? Let’s continue.

23. 2^a = SQRT(64)

2^a = 8

2^a = 2^3

a = 3

but b/a = 3

therefore, b = 3a = 3×3 = 9

a^2 + b^2 = 3^2 + 9^2 = 90

24. Angle XYZ = 180 – (90+52)

180-142 = 38 degrees

From sin rule,

XZ/sin38 = 32/sin52

XZ = 25 cm

26. tita = 72

r = 3.5

pi = 22/7

arc length = 2pir(tita/360)

arc length = 2x( 22/7 )x3.5x(72/360) = 4.4 cm

27. 2lb = (a+b)h

2lb = ah + bh

2lb- bh = ah

b(2l-h) = ah

b = ah/2l-h

28. agent commission = 8 percent

percentage received by eric = 100 – 8 = 92

let the selling price be x

92x/100 = 117760

x = 11776000/92

x = 128000.00 dollars

29. arc length = 2pir(tita/360)

22 = 2×22/7x15xtita/360

tita = 55440/660

tita = 84 degrees

30. l = 15

width = x

if its sides are doubled,

l = 30

width = 2x

area = 30* 2x = 60x cm^2

31. A 32. A 33. B 34. D 35. D 36. C 37.D 38. D 39. C 40. C 41. B42. C 43. C 44. D 45. C46 A 47.A 48. C 49. C 50. A

These questions and solutions are WAEC practice questions to get you prepared for your 2022/2023 WAEC Mathematics Examination.

### WAEC Maths Essay and Objective 2022 Expo

The above questions are not exactly 2022 WAEC Mathematics questions and answers but likely WAEC Maths repeated questions and answers.

These questions are for practice. The 2022 WAEC correct Mathematics expo will be posted on this page 2 hours before the WAEC Mathematics examination starts.

WAEC English Language Questions and Answers

## How to pass WAEC Mathematics Examination 2022

To pass your WAEC successfully there are major things you have to do and they include the items below:

### Solve mathematical problems everyday

There is this informal saying that ‘when you leave mathematics, mathematics will also leave you‘. Ensure you solve at least one mathematical problem each day in order to get well acquainted with it.

However, you may not have all it takes to solve mathematics as frequent as that. Ensure you solve at least twice or thrice a week. Each time you want to solve, give it at least 1 hour 30 minutes and you will see yourself doing well.