## WAEC Mathematics & OBJ Online Questions and Answers Expo for 20th September 2021 – WAEC Maths Expo for today, Monday **20th September 2021 is out**

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**Mathematics – Essay & OBJ Answers for 20th September 2021**

WAEC questions are set and compiled by the West African Senior School Certificate Examination Board (WASSCE). Make sure you follow the instructions as provided by WAEC

### GENERAL MATHEMATICS/MATHEMATICS (CORE) 1

OBJECTIVE TEST

**2021/2022 WAEC Mathematics Questions and Answers**

Symbols used:

^ means raise to power

/ division

SQRT means square root

* Scroll to the bottom to get the 2021 WAEC answers* and refresh the page

The 2021 WAEC Mathematics Answers will be posted here on 18th September 2021 during the examination

WAEC Maths 2021 Answers Loading….

**Solutions to WAEC Mathematics 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 23_{y }= 1111_{two}, 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 cm^{2}. 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

Are the WAEC objective questions difficult? Let’s quickly move to the **Solutions of** **2020/2021 WAEC Mathematics Questions**

**Solutions to WAEC Mathematics Questions**

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

**Answer: A**

**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)

**Answer: A**

**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

**Answer: D**

**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

**Answer: C**

**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

**Answer: B**

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

4 SRT(7) – 9 SRT(2)

**Answer: C**

**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

**Answer: B**

**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

**Answer: C**

**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

**Answer: B**

**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

**Answer: B**

**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

**Answer: A**

**18.** (0.25)^y = 32

(1/4)^y = 32

2^-2 = 2^5

-2y = 5

y = -5/2

**Answer: A**

**19.** C

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

**Answer: A**

**21.** 3p – 1/p^2 -p

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

therefore, p = 1 would make it zero

**Answer: A**

**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

**Answer: C**

**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

**Answer: C**

**24.** Angle XYZ = 180 – (90+52)

180-142 = 38 degrees

From sin rule,

XZ/sin38 = 32/sin52

XZ = 25 cm

**Answer: B**

**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

**Answer: C**

**27.** 2lb = (a+b)h

2lb = ah + bh

2lb- bh = ah

b(2l-h) = ah

b = ah/2l-h

**Answer: A**

**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

**Answer: B**

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

22 = 2×22/7x15xtita/360

tita = 55440/660

tita = 84 degrees

**Answer: B**

**30.** l = 15

width = x

if its sides are doubled,

l = 30

width = 2x

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

**Answer: A**

31. **A** 32. **A** 33. **B** 34. **D** 35. **D** 36. **C** 37.**D** 38. **D** 39. **C** 40. **C** 41. **B****4**2. **C** 43. **C** 44. **D** 45. **C****4**6 **A** 47.**A** 48. **C** 49. **C** 50. **A**

These questions and solutions are WAEC practice questions to get you prepared for your **2021/2022 WAEC Mathematics Questions**.

## WAEC Maths Essay and Objective 2021 (EXPO)

The above questions are not exactly 2021 WAEC Mathematics questions and answers but likely WAEC Maths repeated questions and answers.

These questions are for practice. The 2021 WAEC Mathematics expo will be posted on this page 30 minutes before the WAEC Mathematicsexamination starts. Keep checking and refreshing this page for the answers.

This post will be updated as soon as the **2021/2022 WAEC Mathematics Questions** are out.

What you will see won’t be far from the ones above. I had taken quality time to bring down the solution to a layman’s understanding.

**WAEC Mathematics Questions and Answers 2021 Loading…**

**Today’s OBJ**

(2020 Answers)

**MATHS-OBJ**

**1CBCDACDCCD**

**11AADBDACBBC**

**21BDDABDADAD**

**31CDACCCCCDA**

**41BBBCDCACDB**

**Today’s Mathematics Essay**

**(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

*Objectives Loading….*

What you will see won’t be far from the ones above. I had taken quality time to bring down the solution to a layman’s understanding.