# NECO Mathematics Questions and Answers 2021/2022 (OBJ & Theory)

Are you searching for NECO mathematics answers for the 2021/2022 examination session?

This is material for NECO Mathematics questions 2021 for both theory and OBJ answers 2021. The material covers NECO 2021 Mathematics answers for paper 1 (objectives) and paper 2 (essay) and is meant for practice because they are only likely questions and their answers.

NECO candidates are advised to use this material to pass their NECO Mathematics exam for free. With this, you will know the possible 2021 NECO questions that you will see in this Mathematics 2021 exam.

If you wish to get the correct NECO Mathematics answers for 2021 before the exam, then this is for you. I will be sharing NECO Mathematics answers 2021, including questions and answers for Theory and Objectives.

## NECO Mathematics Objective (OBJ) Answers to questions for 2021

The NECO Mathematics OBJ questions are those sets of questions that are usually accompanied by 4 or 5 different options (A, B, C, D, or E), and each of these questions has just one correct option. They are called multiple-choice questions.

Please, these questions are not the actual questions for this year’s exam but there are NECO 2021 possible questions and answers for Mathematics.

1. Evaluate log927

1. x =11/3
2. x = 1½
3. x =2½
4. x =21/3
5. x =12/3

The correct answer is B. x =1½; scroll to the bottom to see working…

2. Solve log50.04

1. r =-1
2. r =+1
3. r =-2
4. r= +2
5. r =-1/25

The correct answer is C. r =-2; scroll to the bottom to see working…

3. Solve 2x+ 9x =5.

1. x = ½ or x = -5
2. x = -½ or x = -5
3. x = -½ or x = 5
4. x = ½ or x = 5
5. x = 1 or x = 5

The correct answer is A. x = ½ or x = -5; scroll to the bottom to see working…

4. Solve 3x2 –XY = 0 and 2y – 5x = 1 simultaneously

1. (0, ½) and (1, 2).
2. (2, ½) and (2, 3).
3. (1, ½) and (1, 2).
4. (0, ½) and (1, 3).
5. (0, ½) and (2, 4).

The correct answer is D. (0, ½) and (1, 3); scroll to the bottom to see working…

5. An item is marked ‘cost N7000 VAT inclusive’. If the rate of VAT is 5%, how much VAT does the Government receive?

1. N3331/3
2. N3231/2
3. N3311/3
4. N3321/2
5. N3131/3

The correct answer is A. N333 1/3; scroll to the bottom to see working…

6. Use the chain rule to differentiate (3x + 8)6

1. 18 (3x + 8)5
2. 8 (3x + 8)5
3. 18 (3x + 8)6
4. 16 (3x + 5)8
5. 18 (3x + 8)5

The correct answer is = B. 8 (3x + 8)5; scroll to the bottom to see working…

7. Find the equation of the tangent to the curve y = x2 – 4x + 3 at the point (3, 1).

1. y = 5x – 2
2. y = 2x + 5
3. y = 3x + 4
4. y = 4x – 3
5. y = 2x – 5

The correct is E. y = 2x – 5; scroll to the bottom to see working…

8. Factorize this 5y2 + 2y – 3a2?

1. (y + a) (5y – 3a)
2. (5y + a) (y – 3a)
3. 5y2 + a (2y – 3a)
4. (y – a) (5y + 3a)
5. (5y – a) (y + 3a)

Solution Solving: 5y 2 + 5ay – 3ay – 3a 2
5y(y + a) – 3a(y + a)
(y + a) (5y – 3a)

Therefore the Correct Answer is A

9. The radius of a circle of latitude is 6370km and Abidjan is 40 west of Accra but on the same circle of latitude. Calculate how far Abidjan is west of Accra.

The answer should be corrected to the nearest kilometer. [π = 22/7]?

1. 222km
2. 5005km
3. 890km
4. 445km
5. 10010km

Solution Solving to question 9: θ/360 * 2πr = Rcos4o = 4/360 * 40041/1 = 445km

Therefore, the answer is D

10. Find the area of the curved surface of the cone if the cone is made from a sector of a circle that has a radius of 14cm and an angle of 90. (x = 22/7)?

1. 308cm2
2. 22cm2
3. 154cm2
4. 38.5cm2
5. 77cm2

Solution Solving: Length of arc = θ/360 * 2πr => r = 90/360 * 14/4 = 3.5
slant height t = 14cm
the curved surface area of cone = πrl
=> 3.142 x 3.5 x 14 = 153.95

Therefore, the answer is C

11. Find the size of the angle of the sector if a rope of length 18m is used to form a sector of a circle of radius 3.5m on a school playing field.

Correct your answer to the nearest degree.

1. 330
2. 1800
3. 900
4. 400
5. 2700

Solution Solving: Circumference = 2πr

2 x 22/7 x 3.5 ≅ 22m, 22m = 360

11m = 360/22 x 11 = 1800

The correct answer is B.

12. Find Q ∩ P if P = (all odd numbers from 1 to 10) and Q = (all perfect squares less than 30)?

1. (1, 9)
2. (1, 4, 9, 16, 25)
3. (1, 3, 4, 5, 7, 9, 16, 25)
4. (1, 3, 5, 7, 9)
5. E. Φ

Solution Solving: The only numbers that intersect between all odd numbers from 1 to 10 and all perfect squares that are less than 30 are [1 & 9]

Therefore, the answer = A

## Sample NECO Mathematics Theory Questions 2021 (Paper 2)

Below are the sample NECO Mathematics theory questions 2021. They are essay questions gotten from last year’s exam. Study them effectively because they might be repeated in this year’s NECO Mathematics exam.

1. Evaluate log927

Solution; let’s assume x = log927

Then 9= 27

Therefore, (32)x = 33

32x = 33

Divide it through by the coefficient of 2x

i.e. 32x /3 = 33/3

= 2x = 3 (equating indices) and x =1½

2. Solve log50.04

Solution: Let r be = log50.04, then 5r = 0.04 = 1/25 = 5-2 (equating indices)

Hence r = -2 Ans.

3. Solve 2x+ 9x =5.

Solving solution: 2x+ 9x =5

2x+ 9x -5 =0

(2x – 1) (x + 5) = 0

Either 2x – 1 = 0 or x + 5 = 0

Therefore 2x = 1 or x = -5 so, x = ½ or x = -5 Ans.

4. Solve 3x2 –xy = 0 and 2y – 5x = 1 simultaneously

Solution solving

3x2 – xy = 0 ———————- equation 1

2y – 5x = 1 ———————-equation 2

From equation 2, we have 2y = 1 + 5x

Y = ½ (1 + 5x) ——————-equation 3

Substitute ½(1 + 5x) for y in equation 1

3x2 – x [½(1 + 5x)] = 0

3x2 – ½x – 2½x2 = 0

½x2 – ½x = 0

½x (x – 1) = 0

Therefore, x =0 or x =1

Substituting for x in equation 3

When x =0, y = ½(1 + 5 X 0) = ½

And, when x =1, y = ½(1 + 5 X 0) =3

Therefore, the solutions are x =0 and y =½

x =1 and y =3

Or, in ordered pairs (0, ½) and (1, 3) Ans.

5. An item is marked ‘cost N7000 VAT inclusive’. If the rate of VAT is 5%, how much VAT does the Government receive?

Solution: The price tag means that a 5% VAT surcharge is included in the total price. Thus N7000 represents 105% of the actual cost of the item.

105% of the cost price =N7000

1% of the cost of the item = N7000/105

5% of the cost (i.e. the VAT) =N7000/105 X 5

=N333 1/Ans.

6. Use the chain rule to differentiate (3x + 8)6

Solution: Let y = (3x + 8)and t = 3x + 8 then,

y = t6, dt/dx =3 and dy/dt =6t5

By the chain rule, we have; dy/dx = dy/dt X dt/dx

= 6t5 X 3

= 6 (3x + 8)5 X 3

= 18 (3x + 8)Ans.

7. Find the equation of the tangent to the curve y = x2 – 4x + 3 at the point (3, 1).

Solution: y = x2 – 4x + 3

= dy/dx = 2x – 4, the gradient function.

If x = 3, the gradient of the curve at this point is 2 X 3 – 4 =2

Hence, the gradient of the tangent at x=3 is 2, i.e. the same as the gradient of the curve. The equation of the tangent is, y-1/x-3 = 2

• → y – 1 = 2(x-3)
• → y = 2x – 5 Ans.

### 2021 NECO Exam Scheme and Format for Mathematics.

The NECO Mathematics exam will be in two papers (paper 1 and paper 2).

• PAPER 1: Consist of compulsory questions which would last for fifty (50) minutes and carry fifty (50) marks. Each question has multiple-choice options ranging from A to D or E and only one of the options is to be selected for each question.
• PAPER 2: Paper 2 known as theory paper is a “show working paper”. Candidates are to answer the questions by solving them on the essay solving sheet. All solving steps are to be properly shown on the working sheet.

### 1 thought on “NECO Mathematics Questions and Answers 2021/2022 (OBJ & Theory)”

1. CM_Flashschoolgist.com_Education_WC3
2. Very good and thanks