TMUA: Number, units and bounds

Number: primes, factors, recurring decimals, standard form and bounds.

Section 1 Part 2 · M1-M7 GCSE-level knowledge (Part 2) · spec M1-M2
Papers 1 and 2
8 practice questions
No calculator

Practise number, units and bounds →Timed TMUA paper

What the specification covers

Primes, HCF, LCM, prime factorisation
Recurring decimals to fractions; standard form
Upper and lower bounds; systematic listing (m × n rule)

Key ideas

Every integer \(>1\) has a unique prime factorisation; HCF and LCM come from it.
Recurring decimal to fraction: multiply by powers of 10 so the repeating parts line up, then subtract.
Upper and lower bounds: \(3.6\) to 1 d.p. lies in \([3.55,3.65)\). For a quotient, the largest value is (largest numerator) \(\div\) (smallest denominator).
Systematic listing: \(m\) ways then \(n\) ways gives \(mn\) ways.

Common mistakes

Bounds for a difference: largest \(a-b\) uses the largest \(a\) and the smallest \(b\).
1 is not prime.

Exam tip

For bounds questions, write the interval for each quantity first.

Worked example

Which fraction is equal to the recurring decimal \(0.1\dot{3}\dot{6}=0.1363636\ldots\)?

\(\frac{7}{50}\)
\(\frac{13}{99}\)
\(\frac{3}{22}\)
\(\frac{136}{999}\)
\(\frac{5}{33}\)

Answer: C: \(\frac{3}{22}\)

Let \(d=0.13636\ldots\). Then \(1000d=136.3636\ldots\) and \(10d=1.3636\ldots\); subtracting, \(990d=135\), \(d=\tfrac{135}{990}=\tfrac{3}{22}\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

\(x=3.6\) and \(y=1.2\), each correct to 1 decimal place. What is the upper bound of \(\dfrac{x}{y}\)?

\(\frac{73}{23}\)
\(\frac{71}{25}\)
\(\frac{73}{25}\)
\(3\)
\(\frac{71}{23}\)

Show the answer and solution

Answer: A: \(\frac{73}{23}\)

The quotient is largest for the largest \(x\) and smallest \(y\): \(\tfrac{3.65}{1.15}=\tfrac{365}{115}=\tfrac{73}{23}\).

Question 2

Which fraction is equal to the recurring decimal \(0.9818181\ldots\), in which the digits 81 repeat for ever after the first decimal place?

\(\frac{9}{11}\)
\(\frac{54}{55}\)
\(\frac{36}{37}\)
\(\frac{981}{1000}\)
\(\frac{109}{111}\)

Show the answer and solution

Answer: B: \(\frac{54}{55}\)

Let \(z=0.9818181\ldots\). Then \(1000z-10z=990z=981-9=972\), so \(z=\tfrac{972}{990}=\frac{54}{55}\).

Question 3

Which fraction is equal to the recurring decimal \(0.3151515\ldots\), in which the digits 15 repeat for ever after the first decimal place?

\(\frac{104}{333}\)
\(\frac{35}{111}\)
\(\frac{63}{200}\)
\(\frac{5}{33}\)
\(\frac{52}{165}\)

Show the answer and solution

Answer: E: \(\frac{52}{165}\)

Let \(z=0.3151515\ldots\). Then \(1000z-10z=990z=315-3=312\), so \(z=\tfrac{312}{990}=\frac{52}{165}\).

Question 4

Which fraction is equal to the recurring decimal \(0.4454545\ldots\), in which the digits 45 repeat for ever after the first decimal place?

\(\frac{49}{111}\)
\(\frac{5}{11}\)
\(\frac{445}{999}\)
\(\frac{89}{200}\)
\(\frac{49}{110}\)

Show the answer and solution

Answer: E: \(\frac{49}{110}\)

Let \(z=0.4454545\ldots\). Then \(1000z-10z=990z=445-4=441\), so \(z=\tfrac{441}{990}=\frac{49}{110}\).

Question 5

Which fraction is equal to the recurring decimal \(0.9636363\ldots\), in which the digits 63 repeat for ever after the first decimal place?

\(\frac{53}{55}\)
\(\frac{7}{11}\)
\(\frac{106}{111}\)
\(\frac{963}{1000}\)
\(\frac{107}{111}\)

Show the answer and solution

Answer: A: \(\frac{53}{55}\)

Let \(z=0.9636363\ldots\). Then \(1000z-10z=990z=963-9=954\), so \(z=\tfrac{954}{990}=\frac{53}{55}\).

Question 6

Which fraction is equal to the recurring decimal \(0.7242424\ldots\), in which the digits 24 repeat for ever after the first decimal place?

\(\frac{724}{999}\)
\(\frac{8}{33}\)
\(\frac{239}{330}\)
\(\frac{181}{250}\)
\(\frac{239}{333}\)

Show the answer and solution

Answer: C: \(\frac{239}{330}\)

Let \(z=0.7242424\ldots\). Then \(1000z-10z=990z=724-7=717\), so \(z=\tfrac{717}{990}=\frac{239}{330}\).

Question 7

Which fraction is equal to the recurring decimal \(0.7545454\ldots\), in which the digits 54 repeat for ever after the first decimal place?

\(\frac{83}{110}\)
\(\frac{83}{111}\)
\(\frac{754}{999}\)
\(\frac{6}{11}\)
\(\frac{377}{500}\)

Show the answer and solution

Answer: A: \(\frac{83}{110}\)

Let \(z=0.7545454\ldots\). Then \(1000z-10z=990z=754-7=747\), so \(z=\tfrac{747}{990}=\frac{83}{110}\).

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