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Integers, Fractions and Decimals

7 questions, 24 marks, every one with its mark scheme.

Or open them one at a time, as you finish each question.

Question 1

3 marks
1(a)[3 marks]

Show that 334 ÷ 217 = 134

You must write down all the stages in your working.

Mark scheme
  • Both mixed numbers as improper fractions: 154 and 157M1
  • Multiplies by the reciprocal: 154 × 715M1
  • 10560 = 74 = 134 from correct workingA1*

Do not accept: working in decimals

Question 2

5 marks
2(a)[3 marks]

Write 4116 as a product of powers of its prime factors.

Show your working clearly.

Mark scheme
  • At least two correct stages of splitting 4116 into factors (factor tree or ladder)M1
  • All the prime factors 2, 2, 3, 7, 7, 7 foundM1
  • 22 × 3 × 73A1

Do not accept: 22, 3, 73 (a list); 22 + 3 + 73; 4 × 3 × 343 (composite factors)

2(b)[2 marks]

Hence write 4116 × 104 as a product of powers of its prime factors.

Mark scheme
  • 104 = 24 × 54M1
  • 26 × 3 × 54 × 73A1ft

Do not accept: 22 × 3 × 73 × 104 (10 is not prime)

Question 3

2 marks
3(a)[2 marks]

Find the lowest common multiple (LCM) of 96 and 180.

Show your working clearly.

Mark scheme
  • 96 = 25 × 3 and 180 = 22 × 32 × 5, or lists of at least four multiples of eachM1
  • 1440A1

Do not accept: 12 (the HCF)

Question 4

3 marks
4(a)[3 marks]

Show that 416 − 234 = 1512

You must write down all the stages in your working.

Mark scheme
  • Both mixed numbers as improper fractions: 256 and 114M1
  • Both over a common denominator of 12: 5012 − 3312M1
  • 1712 = 1512 from correct workingA1*

Do not accept: working in decimals

Question 5

2 marks
5(a)[2 marks]

Use algebra to show that 0.218 = 1255

Mark scheme
  • x = 0.21818…, 1000x = 218.1818… and 10x = 2.1818… with 1000x − 10x = 216M1
  • 990x = 216, x = 216990 = 1255A1*

Do not accept: a decimal check with no equations in x

Question 6

2 marks
6(a)[2 marks]

Use algebra to show that 1.207 = 123111

Mark scheme
  • x = 1.207207… and 1000x = 1207.207207… with 1000x − x = 1206M1
  • 999x = 1206, x = 1206999 = 134111 = 123111A1*

Do not accept: a decimal check with no equations in x

Question 7

7 marks

A = 23 × 3 × 52
B = 2 × 34 × 7

7(a)[2 marks]

Find the highest common factor (HCF) of 6A and 5B.

Give your answer as a product of powers of its prime factors.

Mark scheme
  • 6A = 24 × 32 × 52 or 5B = 2 × 34 × 5 × 7M1
  • 2 × 32 × 5A1

Do not accept: 2 × 3 (the HCF of A and B)

7(b)[2 marks]

Find the lowest common multiple (LCM) of 6A and 5B.

Give your answer as a product of powers of its prime factors.

Mark scheme
  • Takes the higher power of every prime in 6A and 5BM1
  • 24 × 34 × 52 × 7A1
7(c)[2 marks]

Work out the value of A2 × B.

Give your answer as a product of powers of its prime factors.

Mark scheme
  • A2 = 26 × 32 × 54M1
  • 27 × 36 × 54 × 7A1
7(d)[1 mark]

C = 2p × 32 × 5

The HCF of A and C is 22 × 3 × 5

Write down the value of p.

Mark scheme
  • p = 2B1
All Edexcel International GCSE Mathematics question setsThe notes for this topic