Ordering numbers and inequality symbols - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 20 exam-style questions on Ordering numbers and inequality symbols for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

20 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Write the correct symbol, <, > or =, between each pair of numbers.

−7 ... −9
0.3 ... \(\dfrac{3}{10}\)
\(\dfrac{2}{7}\) ... 0.3
\((-3)^2\) ... \(-3^2\)

Model answer

−7 > −9 (−7 is nearer to zero)
0.3 = \(\dfrac{3}{10}\)
\(\dfrac{2}{7} = 0.285...\), so \(\dfrac{2}{7}\) < 0.3
\((-3)^2 = 9\) and \(-3^2 = -9\), so \((-3)^2 > -3^2\)

Explanation

For negative integers, the one closer to zero is larger. To compare a fraction with a decimal, convert the fraction to a decimal (or note that 0.3 × 7 = 2.1, which is more than 2). In \(-3^2\) the power is worked out before the minus sign is applied, so \(-3^2 = -9\), but \((-3)^2 = 9\).

Mark scheme

2 marks
B2 for all four correct: −7 > −9, 0.3 = \(\dfrac{3}{10}\), \(\dfrac{2}{7}\) < 0.3, \((-3)^2 > -3^2\)
B1 for two or three correct. Accept statements written the other way round with the correct symbol, e.g. −9 < −7

Question 2

2 marks

Find a fraction that lies between \(\dfrac{3}{8}\) and \(\dfrac{2}{5}\).

Model answer

\(\dfrac{3}{8} = \dfrac{15}{40} = \dfrac{30}{80}\) and \(\dfrac{2}{5} = \dfrac{16}{40} = \dfrac{32}{80}\)
\(\dfrac{31}{80}\) lies between \(\dfrac{30}{80}\) and \(\dfrac{32}{80}\).
Answer: \(\dfrac{31}{80}\)

Explanation

Write both fractions with the same denominator. If the numerators are next to each other (15 and 16), double the numerator and denominator so there is a whole number between them. You could also use decimals: any decimal between 0.375 and 0.4, such as 0.39, can be written as a fraction.

Mark scheme

1 mark
M1 for writing both fractions with a common denominator, e.g. \(\dfrac{15}{40}\) and \(\dfrac{16}{40}\) (or \(\dfrac{30}{80}\) and \(\dfrac{32}{80}\)), or as decimals 0.375 and 0.4
1 mark
A1 for any correct fraction strictly between them, e.g. \(\dfrac{31}{80}\), \(\dfrac{39}{100}\), \(\dfrac{19}{50}\), \(\dfrac{77}{200}\)
the answer must be written as a fraction; do not accept \(\dfrac{3}{8}\) or \(\dfrac{2}{5}\) themselves; a correct fraction with no working scores 2 marks

Question 3

3 marks

\(x\) is a number such that \(-1 < x < 0\).

Write these in order of size. Start with the smallest.

\(x \qquad x^2 \qquad x^3 \qquad \dfrac{1}{x}\)

You must show your working.

Model answer

Try \(x = -\frac{1}{2}\):
\(x = -\frac{1}{2}\)
\(x^2 = \frac{1}{4}\)
\(x^3 = -\frac{1}{8}\)
\(\dfrac{1}{x} = -2\)
\(-2 < -\frac{1}{2} < -\frac{1}{8} < \frac{1}{4}\)
Order: \(\dfrac{1}{x}, x, x^3, x^2\)

Explanation

\(x^2\) is positive, so it is the largest. \(x^3\) is negative but closer to zero than \(x\) (its size is smaller), so \(x^3 > x\). The reciprocal of a number between −1 and 0 is less than −1, so \(\dfrac{1}{x}\) is the smallest. A common error is to think \(x^3 < x\) because the power is bigger.

Mark scheme

1 mark
M1 for substituting a value of \(x\) with \(-1 < x < 0\), e.g. \(x = -\frac{1}{2}\), and evaluating at least two of \(x^2\), \(x^3\), \(\dfrac{1}{x}\) correctly
or a correct general argument about the sign and size of each expression
1 mark
M1 for all three correct for their value, e.g. \(x^2 = \frac{1}{4}\), \(x^3 = -\frac{1}{8}\), \(\frac{1}{x} = -2\)
1 mark
A1 for \(\dfrac{1}{x}, x, x^3, x^2\)
cao; correct order with no working scores 0 (working required)

Question 4

4 marks

\(a\) and \(b\) are positive integers with \(a < b\).

Prove that \(\dfrac{a}{b} < \dfrac{a+1}{b+1}\)

Model answer

\(\dfrac{a+1}{b+1} - \dfrac{a}{b} = \dfrac{b(a+1) - a(b+1)}{b(b+1)}\)
\(= \dfrac{ab + b - ab - a}{b(b+1)}\)
\(= \dfrac{b - a}{b(b+1)}\)
Since \(a < b\), \(b - a > 0\).
Since \(b\) is positive, \(b(b+1) > 0\).
So \(\dfrac{a+1}{b+1} - \dfrac{a}{b} > 0\), which means \(\dfrac{a}{b} < \dfrac{a+1}{b+1}\).

Explanation

To prove one expression is bigger than another, show that their difference is positive. Writing the difference over the common denominator \(b(b+1)\) leaves the numerator \(b - a\). You must explain why both the numerator and the denominator are positive; checking with numbers (e.g. \(\frac{2}{3} < \frac{3}{4}\)) is not a proof.

Mark scheme

1 mark
M1 for considering the difference \(\dfrac{a+1}{b+1} - \dfrac{a}{b}\) with a common denominator, e.g. \(\dfrac{b(a+1) - a(b+1)}{b(b+1)}\)
or cross-multiplying: comparing \(a(b+1)\) with \(b(a+1)\), with a statement that \(b\) and \(b + 1\) are positive (may be seen later)
1 mark
M1 for expanding correctly, e.g. \(\dfrac{ab + b - ab - a}{b(b+1)}\), or \(ab + a\) and \(ab + b\)
1 mark
A1 for \(\dfrac{b - a}{b(b+1)}\), or for reducing the comparison to \(a < b\)
1 mark
C1 for a complete argument: \(b - a > 0\) because \(a < b\), and \(b(b+1) > 0\) because \(b\) is positive, so the difference is positive and \(\dfrac{a}{b} < \dfrac{a+1}{b+1}\)
for the cross-multiplying method, the reasons must include that multiplying by the positive numbers \(b\) and \(b + 1\) keeps the inequality the same way round

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