# WAEC GCE Mathematics Questions, Answers 2020/2021 & Syllabus

I am sure, a lot of WAEC GCE candidates have been expecting free and correct WAEC GCE Mathematics questions and answers for the 2020 expo.

If so, this is a detailed WAEC GCE Mathematics questions 2020 samples and a correctly solved WAEC GCE Mathematics answers 2020 expo for the two papers.

The first paper is paper 1 which is known as the objective or OBJ paper while the second paper is paper 2 which is known as theory or essay.

Instructively, the OBJ section contains 50 multi-choice questions and options to choose from and the theory paper has instructed numbers of questions to be answered.

More so, you will see and download the free WAEC GCE Mathematics syllabus for 2020 and the area of concentration at the bottom of this page.

Page Contents

## Expo WAEC GCE Mathematics OBJ questions 2020 and answers

Here are some Sample WAEC GCE Mathematics answers 2020 to the questions below. Start by practicing with this, while expecting the actual answers.

1. How many liters of oil will a cylindrical drum of 28cm diameter and 50cm deep hold?

- 18.5 liters
- 30.8 liters
- 100 liters
- 25 liters
- 6 liters

**30. 8 liters is the correct answer and B is the correct option**

**Solving for question 1 solution:**

The volume of the drum is πr^{2}h = ^{22}/_{7} x 14^{2 }x 50cm^{3}

Capacity of the drum is^{ 22 x }14^{2 x 50}/_{7 x 1000} = 30.8 litres

Use this question to answer questions 2 & 3; A solid is made up of a cylinder with a hemisphere on top as shown in the image below.

2. What is the surface area of the solid?

- 342cm
^{3} - 542cm
^{2} - 1342cm
^{2} - 42cm
^{2} - 3421cm
^{2}

**1342cm ^{2 }is the correct answer and the correct option is C**

**Sloving for question 2 solution:**

Total area surface = πr^{2} + 2πrh + ½(4πr^{2})

=π x 7^{2} + 2π x 7 x 20 + 2 π x 7^{2} cm^{2}

=7 π (7 + 40 +14) cm^{2}

=7 x 22/7 x 61 cm^{2}

= 1342cm^{2}

3. Calculate the volume of the solid.

- 3800cm
^{3} - 800cm
^{3} - 2800cm
^{3} - 3300cm
^{3} - 3820cm
^{3}

**3800cm ^{3 }is the correct answer and the correct option is A**

**Solving for question 3 solution:**

Volume = πr^{2}h + ½ (^{4}/_{3} πr^{3})

= π x 7^{2} x 20 + ^{2}/_{3}π x 73cm^{3}

= 49π(20 + ^{2}/_{3} x 7)cm^{3}

= 49 x ^{22}/_{7} x ^{74}/_{3} cm^{3}

=3800cm^{3}

4. The Length of a rectangular compound is 5m more than the width. Its area is 500cm^{2}. Find the width and length of the compound.

- 40m x 45m
- 10cm x 15m
- 5m x 10m
- 5m x 15m
- 20m x 25m

**20m x 25m is the correct answer, the correct option is E.**

**Solving for question 4 solution:**

Let the width be XM. Then, from the 1^{st} sentence, the length is (x+5)m. The area is x(x+5)m2. From the second sentence, we have;

x(x+5) =500

X2 +5x – 500 =0

(x – 20)(x+25)=0

x=20 or -25

An answer to +25m is clearly not appropriate in this case. Hence the width is 20m and the length, 5m more, is 25m. Validation: 20m x 25m =500m^{2}

5. A student bought some packets of pens for N2160. If she had paid N24 less for each packet, she could have bought three more packets. How many packets did she buy?

- 25
- 115
- 100
- 15
- 35

**15 is the correct answer and the correct option is D. 15**

**Solving for question 5 Solution;**

Let the number of packets she bought be n; Let Nx be the cost of each packet of pens.

From the first sentence, ^{N2160}/_{Nx} = n ………… equation 1

Then from the second sentence, ^{N2160}/_{N}(x – 24) = n + 3 ……….. equation2

From equation 1, x = ^{2160}/_{n} ……. equation 3

From equation 2, x – 24 = ^{2160}/_{n} + 3 ………equation 4

Subtracting equation 4 from 3, we have 24 = ^{2160}/_{n} – ^{2160}/_{n + 3}

>>> 24(n)(n + 3) = 2160(n + 3) – 2160n

>>> 24(n2 + 3n) = 3 x 2160

>>>n^{2} + 3n = 270 = 270

>>>n^{2} + 3n – 270 = 0

(n – 15) (n + 18) = 0

n = 15 or -18

**Therefore, n = 15, since n cannot be negative.**

### 2020 WAEC GCE Mathematics Theory Answers Rules to follow.

Here are the 2020 WAEC GCE Mathematics Theory questions and answers rules that candidates can benefit from in the course of the exam.

**Addition Rules **

- Any Positive number + another positive number is positive. So you add up. An Example is: 5 + 4 = 9
- A negative number + another negative is negative = So, you add up without changing the negative sign. An Example : -5 + (-4) = -9

**Subtraction Rules**

- Any Negative number + a positive number is equal to subtraction depending on the sign of the number with the highest absolute value. An Example : -4 + 5 = 1
- A positive number + a negative is equal to subtraction depending on the sign of the number with the highest absolute value. An example : 4 + (-5) = -1

**Multiplication Rules**

- A positive number x a positive number is equal to a positive number. An Example is: 5 x 4 = 20
- A negative number x a negative number is equal to a positive number. An example is: (-5) x (-4) = 20. Here, the – cancels – leaving the two numbers positive.
- A positive number x a negative number is equal to a negative number. An example is: 4 x (-5) = -20
- negative number x a positive number is equal to a negative number. An example is: -4 x 5 = -20.

**Division Rules**

- A positive number ÷ a positive number is equal to a positive number. An example is: 30 ÷ 5 = 6
- Negative number ÷ a negative number is equal to a positive number. An example is: -30 ÷ (-5) = 6
- A positive number ÷ a negative number is equal to a negative number. An example is: 30 ÷ (-5) = -6
- Any Negative number ÷ a positive number is equal to a negative number. Example : -30 ÷ 5 = -6.

### Download WAEC GCE Mathematics Syllabus in PDF

The WAEC GCE Mathematics syllabus for 2020 will give candidates a better explanation of the topics that questions will set from.

This 2020 WAEC GCE syllabus for mathematics is free and it contains the area of concentration for WAEC GCE candidates to follow.

I am sure, this will help you to perform better in this exam. More information on this will be updated anytime from now.

So do well to always revisit this page often to see the actual and correct WAEC GCE Mathematics questions and answers for 2020.