Mathematics questions, chapter 2

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Factors and multiples

40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

Which of the following is a factor of 54?

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54 ÷ 9 = 6 exactly, so 9 is a factor of 54. 54 ÷ 7, 54 ÷ 8 and 54 ÷ 12 all leave a remainder.

For the teacher: Remembering, low difficulty, standard 2.1.1

Textbook p. 32

Question 2

Diagram 1 shows all the ways to arrange 20 counters in equal rows. Which list gives all the factors of 20?

20 counters arranged as 1 row of 20, 2 rows of 10 and 4 rows of 5 — Diagram 1

1, 2, 4, 5, 10, 20
2, 4, 5, 10
1, 2, 4, 5, 10
1, 2, 3, 4, 5, 10, 20

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Each arrangement gives a factor pair: 1 × 20, 2 × 10, 4 × 5. So the factors are 1, 2, 4, 5, 10 and 20. Do not forget 1 and the number itself. 3 is not a factor because 20 ÷ 3 leaves a remainder of 2.

For the teacher: Understanding, low difficulty, standard 2.1.1

Textbook p. 32

Question 3

Which list contains prime numbers only?

2, 3, 5, 9
1, 3, 5, 7
3, 5, 15, 17
2, 3, 11, 23

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2, 3, 11 and 23 each have exactly two factors. 9 = 3 × 3 and 15 = 3 × 5 are not prime, and 1 is not a prime number.

For the teacher: Remembering, low difficulty, standard 2.1.2

Textbook p. 33

Question 4

Diagram 2 shows a factor tree of 126. Which of the following is the prime factorisation of 126?

Factor tree: 126 splits into 2 and 63; 63 splits into 7 and 9; 9 splits into 3 and 3 — Diagram 2

2 × 63
2 × 3 × 3 × 7
2 × 7 × 9
2 × 3 × 7

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The prime factorisation uses all the numbers at the ends of the branches: 2, 7, 3 and 3. So 126 = 2 × 3 × 3 × 7. 63 and 9 are not prime, and 2 × 3 × 7 = 42 has lost one 3.

For the teacher: Understanding, medium difficulty, standard 2.1.2

Textbook p. 34

Question 5

List all the prime factors of 84.

2, 3, 7
1, 2, 3, 7
2, 3, 4, 7
2, 3, 6, 7

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Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. The prime numbers among them are 2, 3 and 7. 1, 4 and 6 are factors but not prime.

For the teacher: Applying, medium difficulty, standard 2.1.2

Textbook p. 34

Question 6

Which number is a common factor of 24, 40 and 56?

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24 ÷ 8 = 3, 40 ÷ 8 = 5, 56 ÷ 8 = 7, so 8 divides all three completely. 3, 6 and 12 do not divide 40 completely.

For the teacher: Applying, low difficulty, standard 2.1.3

Textbook p. 35

Question 7

Find the highest common factor (HCF) of 42 and 70.

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42 = 2 × 3 × 7 and 70 = 2 × 5 × 7. HCF = 2 × 7 = 14. 2 and 7 are common factors but not the highest; 210 is the LCM.

For the teacher: Applying, low difficulty, standard 2.1.4

Textbook p. 36

Question 8

Diagram 3 shows the repeated division of 48, 72 and 120 to find their HCF. What is the HCF?

Repeated division: 2 into 48, 72, 120 gives 24, 36, 60; 2 gives 12, 18, 30; 2 gives 6, 9, 15; 3 gives 2, 3, 5 — Diagram 3

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For the HCF multiply only the divisors on the left: 2 × 2 × 2 × 3 = 24. 2, 3 and 5 have no common factor, so the division stops. 30 multiplies only the bottom row (2 × 3 × 5); 720 multiplies the divisors and the bottom row (that is the LCM); 12 misses one 2.

For the teacher: Analysing, medium difficulty, standard 2.1.4

Textbook p. 36

Question 9

Which of the following is a common multiple of 4 and 6?

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36 ÷ 4 = 9 and 36 ÷ 6 = 6, so 36 is a multiple of both. 16 is a multiple of 4 only, 18 is a multiple of 6 only, and 10 = 4 + 6 is not a multiple of either.

For the teacher: Understanding, low difficulty, standard 2.2.1

Textbook p. 38

Question 10

Diagram 4 shows jumps of 6 and jumps of 9 on a number line. Which list gives the first four common multiples of 6 and 9?

Number line from 0 to 40. Jumps of 6 land on 6, 12, 18, 24, 30, 36. Jumps of 9 land on 9, 18, 27, 36. 18 and 36 are ringed. — Diagram 4

18, 36, 54, 72
18, 27, 36, 45
6, 9, 12, 18
18, 24, 30, 36

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Both jumps first meet at 18, the first common multiple. Every other common multiple is a multiple of 18: 18, 36, 54, 72. 18, 24, 30, 36 are multiples of 6 only; 18, 27, 36, 45 are multiples of 9 only.

For the teacher: Understanding, medium difficulty, standard 2.2.1

Textbook p. 39

Question 11

Find the lowest common multiple (LCM) of 8, 12 and 18.

2
72
144
1 728

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8 = 2 × 2 × 2, 12 = 2 × 2 × 3, 18 = 2 × 3 × 3. LCM = 2 × 2 × 2 × 3 × 3 = 72. 2 is the HCF, 144 is a common multiple but not the lowest, and 1 728 = 8 × 12 × 18.

For the teacher: Applying, medium difficulty, standard 2.2.2

Textbook p. 40

Question 12

Bus A leaves the station every 15 minutes and bus B every 20 minutes. Both buses leave together at 7:00 a.m. At what time will they next leave together?

7:05 a.m.
7:35 a.m.
8:00 a.m.
12:00 noon

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LCM of 15 and 20 = 60 minutes. 7:00 a.m. + 60 minutes = 8:00 a.m. 7:35 adds 15 + 20; 12:00 noon adds 15 × 20 = 300 minutes; 7:05 uses the HCF, 5.

For the teacher: Applying, high difficulty, standard 2.2.3

Textbook p. 42

Section B: structured questions

28 marks

Answer all questions. Show your working.

Question 13

Answer the questions about the number 45.

[1 mark] (a)List all the factors of 45.

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1, 3, 5, 9, 15, 45 (all six) [1]1

45 = 1 × 45 = 3 × 15 = 5 × 9.

Tick each point you got.

[1 mark] (b)1 is a prime factor of 45.

ATrueBFalse

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False

1 is a factor of 45 but 1 is not a prime number, because it has only one factor.

[1 mark] (c)List all the prime factors of 45.

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3, 5 [1]1

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 2.1.2

Textbook p. 33

Question 14

Diagram 5 shows an incomplete factor tree of 180.

Factor tree: 180 splits into 10 and P; 10 splits into 2 and Q; P splits into 2 and R; R splits into 3 and 3 — Diagram 5

[2 marks] (a)Find the values of P, Q and R.

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P = 180 ÷ 10 = 18 [1]1
Q = 10 ÷ 2 = 5 and R = 18 ÷ 2 = 9 [1]1

Tick each point you got.

[1 mark] (b)Express 180 in the form of prime factorisation.

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180 = 2 × 2 × 3 × 3 × 5 [1]1

Use every prime at the end of a branch: 2, 5, 2, 3, 3. Check: 2 × 2 × 3 × 3 × 5 = 180.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.1.2

Textbook p. 34

Question 15

Diagram 6 shows the factors of 24 and the factors of 36 in a Venn diagram.

Venn diagram. Factors of 24 only: 8, 24. Both: 1, 2, 3, 4, 6, 12. Factors of 36 only: 9, 18, 36. — Diagram 6

[1 mark] (a)List the common factors of 24 and 36.

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1, 2, 3, 4, 6, 12 [1]1

Tick each point you got.

[1 mark] (b)State the HCF of 24 and 36.

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12 [1]1

Tick each point you got.

[1 mark] (c)Explain why 8 is placed only in the circle Factors of 24.

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24 ÷ 8 = 3 but 36 ÷ 8 = 4 remainder 4, so 8 is a factor of 24 only / not a common factor [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 2.1.3

Textbook p. 35

Question 16

Find the HCF of 84, 126 and 210 by using prime factorisation.

[3 marks] (a)Show your working.

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84 = 2 × 2 × 3 × 7, 126 = 2 × 3 × 3 × 7, 210 = 2 × 3 × 5 × 7 [1]1
HCF = 2 × 3 × 7 (common prime factors) [1]1
= 42 [1]1

Check: 84 ÷ 42 = 2, 126 ÷ 42 = 3, 210 ÷ 42 = 5, and 2, 3, 5 have no common factor other than 1.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.1.4

Textbook p. 36

Question 17

Encik Zaki wants to cover the floor in Diagram 7 with square tiles of the same size. No tile may be cut.

Rectangular floor 4.8 m long and 3.6 m wide, with one square tile beside it — Diagram 7

[2 marks] (a)Find the biggest side length, in cm, of the square tiles.

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4.8 m = 480 cm, 3.6 m = 360 cm; find the HCF of 480 and 360 [1]1
HCF = 120, so the side is 120 cm [1]1

480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 and 360 = 2 × 2 × 2 × 3 × 3 × 5, so HCF = 2 × 2 × 2 × 3 × 5 = 120.

Tick each point you got.

[2 marks] (b)How many of these tiles are needed to cover the floor?

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480 ÷ 120 = 4 tiles along the length and 360 ÷ 120 = 3 tiles along the width [1]1
4 × 3 = 12 tiles [1]1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 2.1.5

Textbook p. 44

Question 18

Puan Lim makes hot dogs for a school carnival. Diagram 8 shows how sausages and buns are sold. She wants the same number of sausages and buns, with nothing left over.

A pack of 10 sausages and a pack of 6 hot dog buns — Diagram 8

[2 marks] (a)What is the smallest number of hot dogs she can make?

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Multiples of 10: 10, 20, 30 …; multiples of 6: 6, 12, 18, 24, 30 … / find the LCM of 10 and 6 [1]1
LCM = 30 hot dogs [1]1

Tick each point you got.

[2 marks] (b)How many packs of sausages and how many packs of buns does she buy?

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30 ÷ 10 = 3 packs of sausages [1]1
30 ÷ 6 = 5 packs of buns [1]1

Check: 3 × 10 = 30 sausages and 5 × 6 = 30 buns.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.2.3

Textbook p. 42

Question 19

Hafiz wants the LCM of 12, 18 and 30. Diagram 9 shows his working.

Hafiz's working: 2 into 12, 18, 30 gives 6, 9, 15; 3 gives 2, 3, 5; he writes LCM = 2 × 3 = 6 — Diagram 9

[1 mark] (a)Is Hafiz correct? Explain his mistake.

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No. He stopped dividing when there was no common divisor (that gives the HCF, 6). For the LCM he must keep dividing until all quotients are 1 / an LCM cannot be smaller than 30 [1]1

Tick each point you got.

[2 marks] (b)Find the correct LCM of 12, 18 and 30.

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Continue dividing 2, 3, 5 by 2, 3 and 5 until all are 1 / 2 × 3 × 2 × 3 × 5 [1]1
LCM = 180 [1]1

Prime factorisation: 12 = 2 × 2 × 3, 18 = 2 × 3 × 3, 30 = 2 × 3 × 5. LCM = 2 × 2 × 3 × 3 × 5 = 180. Check: 180 ÷ 12 = 15, 180 ÷ 18 = 10, 180 ÷ 30 = 6.

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 2.2.2

Textbook p. 41

Question 20

Suggest two numbers whose HCF is 6 and whose LCM is 60. Show working to prove that your numbers work.

[3 marks] (a)Write your two numbers and your working.

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12 and 30. 12 = 2 × 2 × 3 and 30 = 2 × 3 × 5. HCF = 2 × 3 = 6. LCM = 2 × 2 × 3 × 5 = 60.

Two correct numbers: 12 and 30 / 6 and 60 [1]1
HCF shown, e.g. 12 = 2 × 2 × 3, 30 = 2 × 3 × 5, HCF = 2 × 3 = 6 [1]1
LCM shown, e.g. 2 × 2 × 3 × 5 = 60 [1]1

Only two pairs work: 12 and 30, or 6 and 60. Check: HCF × LCM = 6 × 60 = 360 = 12 × 30.

Tick each point you got.

For the teacher: Creating, high difficulty, standard 2.2.2

Textbook p. 44

Question 21

The HCF of m and 54 is 6. Given that m is less than 54, find the largest possible value of m.

[2 marks] (a)Show your working.

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54 = 2 × 3 × 3 × 3, so m is a multiple of 6 but not a multiple of 9: 48, 42, 30, 24, 12, 6 … [1]1
m = 48 [1]1

Check: 48 = 2 × 2 × 2 × 2 × 3, so HCF(48, 54) = 2 × 3 = 6. 54 itself is not allowed, and HCF(36, 54) = 18, not 6.

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 2.1.4

Textbook p. 44

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