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N2

Apply the four operations, including formal written methods, to integers, decimals and simple fractions (proper and improper), and mixed numbers, positive and negative; understand and use place value

Four operations

Worked answers, methods and verified real exam appearances for N2 on Edexcel GCSE Maths 1MA1.

Definition

Specification N2: apply the four operations, including formal written methods, to integers, decimals and simple fractions (proper and improper) and mixed numbers, all both positive and negative, and understand and use place value.

Explanation

  • Written methods score for the method, then for the answer. In a decimal division such as 818.4÷1.2818.4\div1.2, multiply both numbers by the same power of 1010 so the divisor is an integer, divide, then place the decimal point by comparing the size of the numbers: 818.4÷1.2818.4\div1.2 is a little less than 818818, so 682682 is sensible and 68.268.2 is not.
  • For fraction addition or subtraction, write both fractions over a common denominator and keep the answer in the form requested, usually a mixed number. For multiplication or division of mixed numbers, convert every mixed number to an improper fraction first; to divide, multiply by the reciprocal of the second fraction only.
  • A show-that question must display each stage, including the unsimplified product.
  • Track signs separately: like signs give a positive result, unlike signs a negative one.
  • Cancelling common factors before multiplying keeps the arithmetic small.

Worked example

1

Work out −214÷35-2\dfrac{1}{4}\div\dfrac{3}{5}. Give your answer as a mixed number.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Convert the mixed number to an improper fraction: −214=−94-2\dfrac{1}{4}=-\dfrac{9}{4}.
  2. 2.Divide by multiplying by the reciprocal of 35\dfrac{3}{5}: −94×53-\dfrac{9}{4}\times\dfrac{5}{3}.
  3. 3.Cancel the common factor 33 and multiply: −3×54×1=−154-\dfrac{3\times5}{4\times1}=-\dfrac{15}{4}.
  4. 4.Write as a mixed number: −154=−334-\dfrac{15}{4}=-3\dfrac{3}{4}.

Answer: −334-3\dfrac{3}{4}

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for conversion to an improper fraction, eg −94-\dfrac{9}{4}M1Need not be shown with operators
(dep) for a method to divide by multiplying by the reciprocal, eg −94×53-\dfrac{9}{4}\times\dfrac{5}{3} or −4512-\dfrac{45}{12}M1oe
for −334-3\dfrac{3}{4}A1Accept −154-\dfrac{15}{4} only if the mixed number is not required

Common mistakes

  • Don't add fraction denominators as well as numerators, writing ab+cd=a+cb+d\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{a+c}{b+d}.
  • Don't divide by a fraction without inverting it.
  • Don't align the last digits of decimals instead of aligning their decimal points.

Mark-scheme wording to use

  • “M1 for a start to a method, eg 8184 ÷ 12 (or 818.4 ÷ 1.2) that leads to 6 as the first digit or for a complete method with no more than one arithmetic error … A1 for digits 682 A1 ft (dep M1) for correct placement of the decimal point into their final answer” — Pearson 1MA1, November 2024 Paper 1H, Q1
  • “M1 for conversion to improper fractions, eg 21/4 or 7/3 or 9/4 M1 (dep) for method to divide by a fraction, eg 21/4 × 3/7 or 63/12 ÷ 28/12 C1 for complete work showing each stage as far as 9/4 or 2 7/28 Must see an intermediate step, eg 63/28” — Pearson 1MA1, November 2024 Paper 1H, Q3(b)
  • “M1 for finding two fractions with a correct common denominator (multiple of 20), with at least one correct corresponding numerator, eg 12/20, 5/20 or 32/20, 45/20 … A1 for 3 17/20 or an equivalent mixed number” — Pearson 1MA1, November 2022 Paper 1H, Q2(a)

What the examiner wants

Sensible place value and a visible reciprocal step. The November 2024 Paper 1H report notes that after correctly finding 682682 many wrote 68.268.2 or 6.826.82, and that in the division of mixed numbers some flipped the wrong fraction, flipped both, or multiplied whole numbers and fractions separately, scoring nothing.

Exam tip

Keep an exact fraction until the final line and show the reciprocal explicitly for the division method mark.

Worked practice

Q1
Tier 1 · Easy

1

Work out −17+29-17+29.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • 1212
1The signs are different, so find 29−17=1229-17=12 and use the sign of the number with the larger magnitude. The result is 1212.
Q2
Tier 2 · Standard

2

Work out 7.2÷0.067.2\div0.06.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • 120120
2Multiply both numbers by 100100 so the divisor is an integer: 7.2÷0.06=720÷6=1207.2\div0.06=720\div6=120.
Q3
Tier 3 · Hard

3

Work out 3.6−(−1.75)+56÷(−59)3.6-(-1.75)+\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right). Give your answer as a decimal.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • 3.853.85
3First 56÷(−59)=56×(−95)=−32=−1.5\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right)=\dfrac{5}{6}\times\left(-\dfrac{9}{5}\right)=-\dfrac{3}{2}=-1.5. Then 3.6+1.75−1.5=5.35−1.5=3.853.6+1.75-1.5=5.35-1.5=3.85.
Q4
Tier 1 · Easy

4

Work out 42.7−18.9542.7-18.95.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • 23.7523.75
1Align the decimal points and write 42.70−18.9542.70-18.95. Column subtraction gives 23.7523.75.
Q5
Tier 2 · Standard

5

Work out 213+1562\dfrac{1}{3}+1\dfrac{5}{6}.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • 4164\dfrac{1}{6} (or 256\dfrac{25}{6})
2213=1462\dfrac{1}{3}=\dfrac{14}{6} and 156=1161\dfrac{5}{6}=\dfrac{11}{6}. Their sum is 256=416\dfrac{25}{6}=4\dfrac{1}{6}.
Q6
Tier 3 · Hard

6

A container holds 12.512.5 litres. Then 2342\dfrac{3}{4} litres are removed and 1.61.6 litres are added. Work out the final volume.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • 11.3511.35 litres
3Convert 2342\dfrac{3}{4} to 2.752.75. After the removal there are 12.5−2.75=9.7512.5-2.75=9.75 litres, and after the addition there are 9.75+1.6=11.359.75+1.6=11.35 litres.
Q7
Tier 2 · Standard

7

A roll contains 8358\dfrac{3}{5} metres of ribbon. Six pieces, each 0.750.75 metres long, are cut from the roll. Work out the length of ribbon left.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • 4.14.1 metres
3835=8.68\dfrac35=8.6. The six pieces use 6×0.75=4.56\times0.75=4.5 metres, so the length left is 8.6−4.5=4.18.6-4.5=4.1 metres.
Q8
Tier 3 · Hard

8

An account has a balance of −£18.40-\pounds18.40. A payment of £75.75\pounds75.75 is added. A fee of 2252\dfrac{2}{5} pounds and three payments of £16.85\pounds16.85 are then taken from the account. Work out the final balance.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • £4.40\pounds4.40
4After the payment, the balance is −18.40+75.75=57.35-18.40+75.75=57.35. The fee is 225=2.402\dfrac25=2.40, leaving 54.9554.95. The three later payments total 3×16.85=50.553\times16.85=50.55, so the final balance is 54.95−50.55=£4.4054.95-50.55=\pounds4.40.
Q9
Tier 3 · Hard

9

Work out 8.03−2.476+0.095×168.03-2.476+0.095\times16.

(3)

(Total for Question 9 is 3 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • 7.0747.074
3Do the multiplication first: 0.095×16=1.520.095\times16=1.52. Written subtraction gives 8.030−2.476=5.5548.030-2.476=5.554, so the result is 5.554+1.52=7.0745.554+1.52=7.074.
Q10
Tier 3 · Hard

10

Work out (415−2.85)÷38\left(4\dfrac{1}{5}-2.85\right)\div\dfrac{3}{8}. Give your answer as a decimal.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • 3.63.6
4415=4.24\dfrac15=4.2, so the value in brackets is 4.2−2.85=1.35=27204.2-2.85=1.35=\dfrac{27}{20}. Dividing by 38\dfrac38 means multiplying by 83\dfrac83: 2720×83=185=3.6\dfrac{27}{20}\times\dfrac83=\dfrac{18}{5}=3.6.

Verified exam appearances

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