Mathematics Exam Questions for JSS1 Second Term

You’re welcome to our school exams series where we provide you with termly examination questions in different subjects. In today’s post, we will focus on Mathematics exam questions. We will cover Mathematics exam questions for JSS1 second term with answers. This means that we’ll be providing you with answers to the questions at the end. Also, you will get a few success tips on how to pass Mathematics examinations with flying colors. Remember to use the comments sections if you have questions, and don’t forget to join our Free Online Tutorial Classes on Facebook. (Like and Follow Page)

Introduction to Mathematics as a School Subject

Mathematics is a subject that deals with numbers, shapes, and how things work together. It helps us solve everyday problems, whether it’s figuring out how much money we need or measuring the size of something. Math teaches us to think clearly and solve problems step by step. It’s a subject that sharpens our minds, making us better at understanding patterns and making smart decisions. From simple addition to complex calculations, math is everywhere and plays a key role in our daily lives. It’s not just about numbers; it’s about thinking logically and seeing the world in a different way.

The subject is offered by pupils and students in Nursery, Primary and Secondary Schools.

Mathematics Exam Questions for JSS1 Second Term

Mathematics Exam Questions for JSS1 Second Term are divided into two sections:

• Section A
• Section B

The first section, namely, Section A is the objective test, and students are expected to attempt all questions in the section. Section B is the theory part and students are expected to answer three (3) out of five (5) questions.

Note that what you have below are JSS1 Mathematics Second Term Exam Past Questions made freely available to assist students in their revision for 2nd term examinations and also teachers in structuring standard examinations.

SECTION A: Objectives

Instruction: Answer all questions in this section by choosing from the options lettered A—D. Each question carries equal marks.

1. Write two thousand, seven hundred and nine in Roman numerals.
A. MMDCICX         B. MMCDCIX
C. MMDCCIX         D. MMCCDIX

2. What is the sum of the binary numbers 1101, 1110 and 1001?
A. 100000           B. 100010
C. 100100           D. 100110

3. Find the value of (101)² in base two.
A. 10001            B. 10111
C. 11001            D. 11011

4. Express 32% as a fraction.
A. 2/25           B. 4/25
C. 6/25           D. 8/25

5. Write in figure; fifty three million, eight hundred thousand and thirteen.
A. 53 800 013
B. 5 380 013
C. 53 80 13
D. 530 813

6. Approximate 5364 to 3 significant figures
A. 536             B. 5300
D. 5360           D. 5370

7. 24350 rounded to the nearest thousand is ________
A. 24000          B. 24350
C. 20000          D. 25000

8. 178.4673 rounded to 2 decimal place is ________
A. 178.00          B. 178.47
C. 178.46          D. 170.47

9. Round up 0.007081 correct to 2 significant figure is ___________
A. 0.01            B. 0.0070
C. 0.0071        D. 0.00708

10. What is the value of the digit 7 in the number 624.97
A. 7 hundreds         B. 7 tens
C. 7 hundredth       D. 7 tenths

11. What is the place value of 5 in figure 35296?
A. 500000           B. 50000
C. 5000               D. 500

12. Evaluate 1001_two – 110_two
A. 11_two           B. 101_two
C. 111_two         D. 10_two

13. Calculate 1011_two × 11_two
A. 10011_two        B. 100001_two
C. 11101_two        D. 10001_two

14. Convert the denary number 45 to a number in binary.
A. 101101_two             B. 11010 _two
C. 11011011 _two         D. 10010 _two

15. Add 1110_two + 111_two
A. 11010_two         B. 10101_two
C. 101001_two       D. 10011_two

16. Simplify 11_two   + 111_two   + 1101_two
A. 10101_two          B. 10111_two
C. 11011_two          D. 11101_two
E. 11111_two

17. Convert  78₁₀ to a number in base two.
A. 11100110_two         B. 11000111_two
C. 1001111_two           D. 1001110_two

18. Simplify 6 – (-12)
A. -8        B. -18
C. 8          D. 18

19. 10 more than x is _____________.
A. 10x              B. 10x + x
C. x + 10          D. 10 – x

20. Multiply y by itself, the answer is __________
A. 2y              B. y²
C. 2y²                  D. 2y³

21. If x + 5 = 50, what is the value of x?
A. 45           B. 52
C. 55           D. -55

22. If 5 is added to a number, the result is 17, what is the number?
A. 12          B. 22
C. 17         D. 5x

23. If x + 15 = 25, then x = __________
A. 15         B. 25
C. 5           D. 10

24. Evaluate x – 18 + x when x = 9
A. 9          B. 18
C. 0          D. -9

25. Express 20% as a decimal number.
A. 2.00       B. 0.5
C. 0.2         D. 0.02

26. Convert 77₁₀ to a number in base two.
A. 1111111_two
B. 1111101_two
C. 1111011_two
D. 1001101_two

27. Express 6561 as a product of primes in index form.
A. 3⁶             B. 3⁷
C. 3⁸             D. 3⁹

28. Yemi saves N150 daily. How much will he save in two weeks?
A. N300           B. N1050
C. N1500         D. N2100

29. The value of the 1 in 12345 is _____________.
A. one unit
B. one hundred
C. one thousand
D. ten thousand

30. Olu saves N700 daily, how much will he save in 15 days?
A. ₦950          B. ₦10 500
C. ₦1 200       D. ₦1 050

31. Convert 4_ten to base two.
A. 10_two            B. 100_two
C. 1000_two        D. 10000_two

32. Evaluate 101_two + 101_two.
A. 1010_two         B. 1100_two
C. 111_two           D. 1111_two

33. Which of these is an improper fraction?
A. 31/7          B. 6/7
C. 4/5            D. 2/3

34. Express 40% as a decimal number.
A. 0.02          B. 0.20
C. 0.40          D. 4.00

35. List the prime factors of 150.
A. 2, 5, 15
B. 2, 3, 10
C. 3, 5, 10
D. 2, 3, 5

36. Correct 45775 to 1 significant figure.
A. 50 000        B. 5 000
C. 500             D. 50

37. ________ divides a circle into equal parts.
A. Arc               B. Chord
C. Diameter      D. Radius

38. A prime has only two factors which are ___________ and ___________.
A. 1 and any other number
B. 2 and 3
C. 1 and 2
D. itself and 1

39. The result of the expansion of 5(2x – 3) is __________________

40. The coefficient of x in the expression 7 – 9x is _______________

SECTION B: Essay

Instructions: Answer three questions only from this section

1a. Calculate the value of 5.766 + 81.34 and give your answer correct to 3 significant figure (3s.f)
1b. Round off 450 789 to the nearest hundreds (nearest 100’s)
1c. Give 16.8695 correct to 2 d.p.
1d. Subtract 79 P from £3.25

2a. Write 356_eight in expanded form
2b. Evaluate 11001_two – 111_two leaving your answer in binary.
2c. What is the sum of 1011_two and 10011_two
2b. Multiply 1011_two by 101_two

3a. Convert binary number 1110011 to a base ten number
3b. Convert the decimal number 59 to a binary number
3c. Round of 258.909 to the nearest whole number
3d. Calculate the value of 425 x 24 and give your answer correct to 3s.f

4a. Solve for x in the equation 5x – 4 = 16 .
4b. Solve the equation  10 – 2x = 3x + 25
4c. If x = 3, y = 2 and z = 2, then Evaluate z(5x – y)
4d. Simplify 5x + 2y + 7x – y

5a. If  p = -2, q = 5 and  r = 10, evaluate 2p + 5q + 10r
5b. Solve the equation 7 – 2y = 3
5c. Samuel thinks of a number, he multiplied it by 5 and added 3 to the answer. If the result is 13, what is the number he thinks of?
5d. Solve the equation 9x – 20 = 8 – 5x

Answers to Mathematics Exam Questions for JSS1 Second Term

Answers to Section A (Objective Test)

The following table gives the correct answers to the objective section of Mathematics exam questions for JSS1 second term. If you are using a mobile device, hold the table and scroll to the right or left for a complete view.

So here you have the answers to the objective section of Mathematics Exam Questions for JSS1 second term. Use the comments section to let me know if you have any questions you would want me to clarify or discuss further.

Answers to Section B (Theory)

1a. Calculate the value of 5.766 + 81.34 and give your answer correct to 3 significant figures (3s.f):

5.766 + 81.34 = 87.106

Rounded to 3 significant figures: 87.1

1b. Round off 450 789 to the nearest hundreds (nearest 100’s):

450789 rounded to the nearest hundred is 450800

1c. Give 16.8695 correct to 2 decimal places (2 d.p.):

16.8695 rounded to 2 decimal places is 16.87

1d. Subtract 79 P from £3.25:

Since 79 P = £0.79, we subtract: 3.25 − 0.79 = 2.46
So, the result is £2.46.

2a. Write 356₈ in expanded form:

To convert 356₈ (base 8) to expanded form:

356₈ = 3×8²+5×8¹+6×8⁰
= 3×64+5×8+6×1
=192+40+6=238
So, 356₈ in expanded form is: 3×64+5×8+6×1

2b. Evaluate 11001₂ – 111₂ leaving your answer in binary:

First, convert the binary numbers to decimal:

11001₂ = 1 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰ = 16 + 8 + 1 = 25
111₂ = 1 × 2² + 1 × 2¹ + 1 × 2⁰ = 4 + 2 + 1 = 7

Now subtract:

25 − 7 = 18

Convert 18 back to binary:

18 = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ = 10010₂

​So, 11001₂ − 111₂ = 10010₂

2c. What is the sum of 1011₂ and 10011₂:

First, convert the binary numbers to decimal:

1011₂ =1 × 2³ + 0 × 2²+ 1 × 2¹ + 1 × 2⁰ = 8 + 2 + 1 = 11
10011₂ = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰ = 16 + 2 + 1 = 19

Now add:

11 + 19 = 30

Convert 30 back to binary:

30 = 1 × 2⁴ + 1 × 2³ + 1 × 2¹ + 0 × 2⁰ = 11110₂

So, the sum is: 1011₂ + 10011₂ = 11110₂

​2d. Multiply 1011₂ by 101₂:

First, convert the binary numbers to decimal:

1011₂ = 11 (as calculated earlier)

101₂ = 5

Now multiply:

11 × 5 = 55

Convert 55 back to binary:

55 = 1 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 1 × 2¹ + 1 × 2⁰ = 110111₂

​So, 1011₂ × 101₂ = 110111₂

3a. Convert binary number 1110011 to a base ten number:

1110011₂ = 1 × 2⁶ + 1 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
= 64 + 32 + 16 + 2 + 1
= 115 in decimal.

3b. Convert the decimal number 59 to a binary number:

59 in decimal can be converted by repeatedly dividing by 2 and recording the remainders:
59÷2=29 remainder 1
29÷2=14 remainder 1
14÷2=7 remainder 0
7÷2=3 remainder 1
3÷2=1 remainder 1
1÷2=0 remainder 1

Reading the remainders from bottom to top: 5910 = 111011₂

3c. Round off 258.909 to the nearest whole number: 258.909 rounded to the nearest whole number is 259

3d. Calculate the value of 425 × 24 and give your answer correct to 3 significant figures: 425×24=10200
So, the result to 3 significant figures is: 10200

4a. Solve for x in the equation 5x – 4 = 16:
5x−4=16
Add 4 to both sides: 5x=16+4=20
Now divide both sides by 5: x=20/5=4

4b. Solve the equation 10 – 2x = 3x + 25
10−2x=3x+25
Add 2x to both sides: 10=5x+25
Subtract 25 from both sides: −15=5x
Now divide by 5: x=−15/5=−3

4c. If x = 3, y = 2, and z = 2, then evaluate z(5x – y):
z(5x−y)=2(5(3)−2)
=2(15−2)=2×13=26

4d. Simplify 5x + 2y + 7x – y:
5x+7x+2y−y=12x+y

5a. If p = -2, q = 5, and r = 10, evaluate 2p + 5q + 10r:
2p+5q+10r=2(−2)+5(5)+10(10)
=−4+25+100=121

5b. Solve the equation 7 – 2y = 3
Subtract 7 from both sides: −2y=3−7=−4
Now divide by -2: y=−4−2=2

5c. Samuel thinks of a number, he multiplied it by 5 and added 3 to the answer. If the result is 13, what is the number he thinks of?

Let the number be x. The equation is: 5x+3=13
Subtract 3 from both sides: 5x=13−3=10
Now divide by 5: x=10/5=2
So the number Samuel thinks of is 2.

5d. Solve the equation 9x – 20 = 8 – 5x:
9x−20=8−5x
Add 5x to both sides: 14x−20=8
Add 20 to both sides: 14x=28
Now divide by 14:
x=2814=2
x=1428​=2