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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.

Explanation

  • The four operations must work with positive and negative integers, decimals, proper and improper fractions, and mixed numbers. For written decimal addition or subtraction, align decimal points so digits of equal place value share a column.
  • When adding fractions, create a common denominator; when multiplying, cancel common factors where useful.
  • Convert a mixed number to an improper fraction before multiplication or division, and divide by a fraction by multiplying by its reciprocal.
  • Track signs separately: equal signs give a positive product or quotient and different signs give a negative one.
  • Examiners award method marks for these conversions even if the final arithmetic slips.

Worked example

Work out −214÷35-2\dfrac{1}{4}\div\dfrac{3}{5}.

  1. 1.Convert the mixed number: −214=−94-2\dfrac{1}{4}=-\dfrac{9}{4}.
  2. 2.Multiply by the reciprocal: −94×53-\dfrac{9}{4}\times\dfrac{5}{3}.
  3. 3.Cancel the factor 33 and evaluate to get −154-\dfrac{15}{4}.

Answer: −154=−334-\dfrac{15}{4}=-3\dfrac{3}{4}.

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.

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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