Place value, ordering, negative numbers & BIDMAS – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator2 marks

(a) Write these numbers in order of size. Start with the smallest number.

$0.53\quad 0.503\quad 0.5\quad 0.35\quad 0.535$ [1]

(b) Write down the value of the digit 4 in the number $12.0465$ [1]

(a) $0.35,\ 0.5,\ 0.503,\ 0.53,\ 0.535$ (all in correct order)B1
(b) $\frac{4}{100}$ oe, e.g. $0.04$ or 4 hundredths (do not accept 'hundredths' alone)B1
Answer: (a) 0.35, 0.5, 0.503, 0.53, 0.535 (b) 4 hundredths (0.04)
Examiner tip: Students often put 0.5 last because it 'looks short' – write every number to 3 decimal places (0.500, 0.503, …) before comparing.

2Grade 1–3FoundationNon-calculator2 marks

At 6 am the temperature in Oslo was $-7^\circ\text{C}$. By midday the temperature had risen by $11^\circ\text{C}$.

(a) Work out the temperature in Oslo at midday. [1]

At 6 am the temperature in Moscow was $-13^\circ\text{C}$.

(b) How many degrees colder was Moscow than Oslo at 6 am? [1]

(a) $4\,^\circ\text{C}$B1
(b) $6$ (degrees)B1
Answer: (a) 4 °C (b) 6 degrees
Examiner tip: For 'how many degrees colder' find the difference, −7 − (−13) = 6; writing −20 or −6 scores nothing.

3Grade 1–3FoundationNon-calculator3 marks

Work out

(a) $3 + 4 \times 5$ [1]

(b) $(-6)\times(-4) - 30 \div (-5)$ [2]

(a) $23$B1
(b) for $24$ or $-6$ seen (one product/quotient correctly evaluated with correct sign)M1
(b) $30$ caoA1
Answer: (a) 23 (b) 30
Examiner tip: Subtracting a negative is adding: 24 − (−6) = 30. Many students write 18 by losing the sign.

4Grade 4–5Non-calculator3 marks

(a) Copy the statement and insert two pairs of brackets to make it correct. [1]

$6 + 2 \times 5 - 1 = 32$

(b) Work out $\dfrac{3^2 + 7 \times 3}{12 - 2 \times 3}$ [2]

(a) $(6 + 2) \times (5 - 1) = 32$B1
(b) for $30$ (numerator) or $6$ (denominator) correctly evaluatedM1
(b) $5$ caoA1
Answer: (a) (6 + 2) × (5 − 1) = 32 (b) 5
Examiner tip: A fraction line acts as a bracket: work out the whole top and the whole bottom before dividing.

5Grade 4–5Calculator3 marks

(a) Use your calculator to work out

$$\frac{\sqrt{18.4 + 3.7^2}}{2.6 \times 0.85}$$

Write down all the figures on your calculator display. [2]

(b) Write your answer to part (a) correct to 3 significant figures. [1]

(a) for $\sqrt{32.09}$ or $5.6648…$ or $2.21$ seenM1
(a) $2.563259554…$A1
(b) $2.56$ (ft from (a) provided (a) has more than 3 sf)B1
Answer: (a) 2.563259554… (b) 2.56
Examiner tip: Put the whole numerator and the whole denominator in brackets on the calculator, or use the fraction button.

6Grade 4–5Non-calculator5 marks

In a quiz there are 20 questions. A correct answer scores $+3$ points, a wrong answer scores $-2$ points and a question not answered scores $0$ points.

(a) Sam gets 12 correct, 5 wrong and does not answer 3 questions. Work out Sam's total score. [2]

(b) Priya does not answer 4 questions. Her total score is 13 points. How many questions did Priya get correct? [3]

(a) for $12 \times 3$ or $5 \times -2$ (or $36$ or $-10$ seen)M1
(a) $26$A1
(b) recognises 16 questions answered and makes a trial, e.g. $3c - 2(16 - c)$ evaluated for one value of $c$P1
(b) a systematic approach or a complete method, e.g. $5c - 32 = 13$ or $5c = 45$P1
(b) $9$A1
Answer: (a) 26 (b) 9 correct
Examiner tip: In (b) remember Priya answered only 16 questions, so wrong = 16 − correct; check your answer: 9×3 − 7×2 = 13.

7Grade 1–3FoundationNon-calculator2 marks

You are given that $23 \times 47 = 1081$

Use this to write down the value of

(a) $2.3 \times 4.7$ [1]

(b) $1081 \div 4.7$ [1]

(a) $10.81$B1
(b) $230$B1
Answer: (a) 10.81 (b) 230
Examiner tip: Count the decimal places: 2.3 × 4.7 has two in total, so the answer has two decimal places.

8Grade 1–3FoundationNon-calculator3 marks

Write one of the symbols $<$, $>$ or $=$ between each pair of numbers to make a true statement.

(a) $-8 \;\square\; -3$ [1]

(b) $0.7 \;\square\; \dfrac{7}{10}$ [1]

(a) $-8 < -3$ caoB1
(b) $0.7 = \dfrac{7}{10}$ caoB1
(c) $-2.5 < -2.05$ cao (accept equivalent correct statements written the other way round, e.g. $-2.05 > -2.5$)B1
Answer: (a) $<$ (b) $=$ (c) $<$
Examiner tip: With negatives, the number further from zero is smaller: −2.5 is further left on the number line than −2.05, so it is less.

9Grade 1–3FoundationNon-calculator3 marks

The diagram shows part of a number line. The arrows A, B and C point to three numbers.

(a) Write down the number that arrow A points to. [1]

(b) Write down the number that arrow B points to. [1]

(a) $-2.4$ (accept $-2\frac{2}{5}$)B1
(b) $-0.6$ (accept $-\frac{3}{5}$)B1
(c) $3.8$ (ft their value of A: $1.4 -$ their A, provided A is negative)B1
Answer: (a) −2.4 (b) −0.6 (c) 3.8
Examiner tip: Count the small divisions first: 5 gaps between whole numbers means each gap is 0.2, not 0.1.

10Grade 4–5Non-calculator4 marks

Work out

(a) $4.8 \times 0.35$ [2]

(b) $7.56 \div 0.18$ [2]

(a) complete method for $48 \times 35$ (e.g. grid, column or lattice with no more than one arithmetic error) or $1680$ seenM1
(a) $1.68$ caoA1
(b) correctly rewrites as an equivalent division by a whole number, e.g. $756 \div 18$ (or $7560 \div 180$), with a correct method started (e.g. short or long division, or $18 \times 40 = 720$)M1
(b) $42$ caoA1
Answer: (a) 1.68 (b) 42
Examiner tip: Count decimal places in a multiplication (1 + 2 = 3), but in a division multiply BOTH numbers by 100 first so you divide by a whole number.

11Grade 4–5Non-calculator3 marks

Kai works out $18 - 6 \div 3 \times 2$

He writes: $18 - 6 = 12,\quad 12 \div 3 = 4,\quad 4 \times 2 = 8$

(a) Explain what Kai has done wrong. [1]

(b) Work out the correct value of $18 - 6 \div 3 \times 2$ [1]

(a) explains that the division (and multiplication) must be done before the subtraction, e.g. 'he subtracted first' or 'he did not use BIDMAS' (must refer to the order; 'he got it wrong' alone is not enough)C1
(b) $14$ caoB1
(c) No, with a reason, e.g. the power is worked out before the negative sign is applied so $-3^2 = -(3^2) = -9$; or 'only $(-3)^2 = 9$'C1
Answer: (a) He subtracted before dividing and multiplying (b) 14 (c) No, $-3^2 = -9$
Examiner tip: Division and multiplication come before subtraction, and a power applies only to the number it is attached to, so −3² means −(3²) = −9.

12Grade 4–5Non-calculator4 marks

Work out the following. Give each answer in its simplest form.

(a) $-2\dfrac{1}{3} + 4\dfrac{3}{4}$ [2]

(b) $\left(-\dfrac{3}{5}\right) \div 1\dfrac{1}{5}$ [2]

(a) converts to improper fractions or fractions with a common denominator, e.g. $-\dfrac{28}{12} + \dfrac{57}{12}$, or works with whole numbers and parts, e.g. $2 + \dfrac{9}{12} - \dfrac{4}{12}$M1
(a) $2\dfrac{5}{12}$ (accept $\dfrac{29}{12}$)A1
(b) $-\dfrac{3}{5} \times \dfrac{5}{6}$ or $-\dfrac{3}{5} \div \dfrac{6}{5}$ followed by multiplying by the reciprocalM1
(b) $-\dfrac{1}{2}$ (accept $-0.5$; must be negative)A1
Answer: (a) $2\frac{5}{12}$ (b) $-\frac{1}{2}$
Examiner tip: Turn the mixed numbers into improper fractions before you start, and keep track of the sign: negative ÷ positive is negative.

13Grade 4–5Non-calculator3 marks

(a) Work out the value of

$$\frac{(-3)^3 + 6^2}{\sqrt{49} - 2^2}$$ [2]

(b) Write down the reciprocal of $-\dfrac{2}{7}$ [1]

(a) numerator $9$ or denominator $3$ correctly evaluated (e.g. $-27 + 36$ or $7 - 4$ seen)M1
(a) $3$ caoA1
(b) $-\dfrac{7}{2}$ (accept $-3\frac{1}{2}$ or $-3.5$; the sign must be negative)B1
Answer: (a) 3 (b) $-\frac{7}{2}$
Examiner tip: A fraction line acts like brackets: finish the top and the bottom separately, and remember $(-3)^3$ is negative because there are three negative factors.

14Grade 4–5Non-calculator4 marks

Here is a number machine.

(a) The input is $-6$. Work out the output. [1]

(b) The output is $-4$. Work out the input. [2]

(c) Amir says, 'When the input is a positive number, the output is always positive.' Give an example to show that Amir is wrong. [1]

(a) $6.5$ (accept $\dfrac{13}{2}$ or $6\frac{1}{2}$)B1
(b) uses inverse operations in reverse order, e.g. $-4 \times 4 = -16$ then $-16 - 8 = -24$M1
(b) $8$ caoA1
(c) any positive input greater than $\dfrac{8}{3}$ with its correct negative output shown, e.g. input $4$ gives $-1$ (input $8$ gives $-4$ is fine); also allow input $\dfrac{8}{3}$ giving output $0$ (0 is not positive)C1
Answer: (a) 6.5 (b) 8 (c) e.g. input 4 gives output −1
Examiner tip: To work backwards, undo each step with its inverse AND in the reverse order: ×4 first, then −8, then ÷(−3).

15Grade 6–7HigherNon-calculator4 marks

Here are four number cards.

$\boxed{-2}\quad\boxed{3}\quad\boxed{-4}\quad\boxed{6}$

Each card is placed in one of the boxes below, so that each card is used once.

$\square \times (\square - \square) \div \square$

(a) Work out the greatest possible value of the calculation. You must show how you place the cards. [3]

(b) Write down the least possible value of the calculation. [1]

(a) places the cards and correctly evaluates at least one arrangement, e.g. $6 \times (3 - (-4)) \div (-2) = -21$P1
(a) makes the bracket as large in size as possible by subtracting a negative from a positive or a positive from a negative (e.g. $-4 - 3 = -7$ or $3 - (-4) = 7$) and uses $6$ as the multiplierP1
(a) $21$ from $6 \times (-4 - 3) \div (-2)$A1
(b) $-21$ (ft their arrangement in (a) with the two cards in the bracket swapped)B1
Answer: (a) 21, from $6 \times (-4 - 3) \div (-2)$ (b) −21
Examiner tip: Two negatives in a product or quotient give a positive, so aim for a negative bracket divided by the −2 card; the biggest bracket is −4 − 3 = −7.

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