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Exam questions · Maths · Standard Form and Accuracy

Recurring Decimals and Rational Numbers

  • 6 exam questions
  • 15 marks
  • 9 quick checks
  1. 1 Write [2 marks]

    (a) Write \(\dfrac{7}{20}\) as a decimal. [1 mark] (b) Write \(\dfrac{2}{3}\) as a recurring decimal, using dot notation. [1 mark]

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    Model answer

    (a) \(7 \div 20 = 0.35\). (b) \(0.\dot{6}\).

    Mark scheme

    • (a) 0.35 — B1
    • (b) \(0.\dot{6}\) — B1
  2. 2 Write [2 marks]

    Write \(\dfrac{5}{9}\) as a recurring decimal, using dot notation. [2 marks]

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    Model answer

    \(5 \div 9 = 0.5555\ldots = 0.\dot{5}\).

    Mark scheme

    • 0.555 seen — M1
    • \(0.\dot{5}\) — A1
  3. 3 Show that [3 marks]

    Write \(0.\dot{1}\dot{8}\) as a fraction in its simplest form. [3 marks]

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    Model answer

    Let \(x = 0.1818\ldots\). Then \(100x = 18.1818\ldots\), so \(99x = 18\) and \(x = \dfrac{18}{99} = \dfrac{2}{11}\).

    Mark scheme

    • \(100x = 18.1818\ldots\) — M1
    • \(99x = 18\) — M1
    • \(\dfrac{2}{11}\) — A1
  4. 4 Show that [3 marks]

    Write \(0.1\dot{6}\) as a fraction in its simplest form. [3 marks]

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    Model answer

    Let \(x = 0.1666\ldots\). Then \(100x = 16.666\ldots\) and \(10x = 1.666\ldots\). Subtracting, \(90x = 15\), so \(x = \dfrac{15}{90} = \dfrac{1}{6}\).

    Mark scheme

    • \(100x\) and \(10x\) seen — M1
    • \(90x = 15\) — M1
    • \(\dfrac{1}{6}\) — A1
  5. 5 Explain [2 marks]

    Here are four numbers: \(\dfrac{22}{7}\), \(\sqrt{25}\), \(\sqrt{11}\) and \(0.\dot{4}\). Which number is irrational? Give a reason. [2 marks]

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    Model answer

    \(\sqrt{11}\). The others can be written as fractions: \(\dfrac{22}{7}\), \(5\) and \(\dfrac{4}{9}\). \(\sqrt{11}\) is not a perfect square, so it cannot be written as a fraction.

    Mark scheme

    • \(\sqrt{11}\) — B1
    • A correct reason — Q1
  6. 6 Show that [3 marks]

    Show that \(0.\dot{5}\dot{4} = \dfrac{6}{11}\). [3 marks]

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    Model answer

    Let \(x = 0.5454\ldots\). Then \(100x = 54.5454\ldots\), so \(99x = 54\) and \(x = \dfrac{54}{99} = \dfrac{6}{11}\).

    Mark scheme

    • \(100x = 54.5454\ldots\) — M1
    • \(99x = 54\) — M1
    • \(\dfrac{54}{99} = \dfrac{6}{11}\) — Q1

Quick check

  1. 1

    What is \(\dfrac{3}{8}\) as a decimal?

    1. A0.38
    2. B0.375
    3. C0.83
    4. D0.\dot{3}
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    B: 0.375

    \(3 \div 8 = 0.375\), which stops.

  2. 2

    What is \(\dfrac{1}{3}\) as a recurring decimal?

    1. A\(0.\dot{3}\)
    2. B0.3
    3. C0.33
    4. D\(0.\dot{1}\dot{3}\)
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    A: \(0.\dot{3}\)

    The 3 repeats for ever.

  3. 3

    What is \(\dfrac{3}{11}\) as a recurring decimal?

    1. A\(0.\dot{3}\)
    2. B\(0.\dot{2}\dot{3}\)
    3. C\(0.2\dot{7}\)
    4. D\(0.\dot{2}\dot{7}\)
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    D: \(0.\dot{2}\dot{7}\)

    \(3 \div 11 = 0.272727\ldots\), where 27 repeats.

  4. 4

    Which fraction gives a terminating decimal?

    1. A\(\dfrac{1}{3}\)
    2. B\(\dfrac{2}{7}\)
    3. C\(\dfrac{7}{20}\)
    4. D\(\dfrac{5}{6}\)
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    C: \(\dfrac{7}{20}\)

    \(20 = 2^2 \times 5\), so the decimal 0.35 stops.

  5. 5

    Is \(\pi\) rational or irrational?

    1. ARational
    2. BIrrational
    3. CBoth
    4. DNeither
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    B: Irrational

    \(\pi\) cannot be written as a fraction.

  6. 6

    Is \(\sqrt{16}\) rational or irrational?

    1. ARational, because it equals 4
    2. BIrrational, because it is a root
    3. CIrrational, because it is not a fraction
    4. DNeither
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    A: Rational, because it equals 4

    \(\sqrt{16} = 4 = \dfrac{4}{1}\).

  7. 7

    Write \(0.\dot{4}\) as a fraction.

    1. A\(\dfrac{4}{10}\)
    2. B\(\dfrac{4}{99}\)
    3. C\(\dfrac{1}{4}\)
    4. D\(\dfrac{4}{9}\)
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    D: \(\dfrac{4}{9}\)

    \(x = 0.\dot{4}\), \(10x = 4.\dot{4}\), so \(9x = 4\).

  8. 8

    Write \(0.\dot{4}\dot{5}\) as a fraction in its simplest form.

    1. A\(\dfrac{45}{100}\)
    2. B\(\dfrac{9}{20}\)
    3. C\(\dfrac{5}{11}\)
    4. D\(\dfrac{4}{9}\)
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    C: \(\dfrac{5}{11}\)

    \(100x - x = 45\), so \(x = \dfrac{45}{99} = \dfrac{5}{11}\).

  9. 9

    Write \(0.1\dot{6}\) as a fraction in its simplest form.

    1. A\(\dfrac{16}{99}\)
    2. B\(\dfrac{1}{6}\)
    3. C\(\dfrac{16}{100}\)
    4. D\(\dfrac{1}{16}\)
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    B: \(\dfrac{1}{6}\)

    \(100x - 10x = 15\), so \(x = \dfrac{15}{90} = \dfrac{1}{6}\).