Hard GCSE Four operations and place value Questions

Challenging, exam-style GCSE Four operations and place value questions with worked solutions. Stretch yourself on the hardest long multiplication, three-digit by two-digit, long division, two-digit divisor problems.

long multiplicationthree-digit by two-digitlong divisiontwo-digit divisordividing by a decimalequivalent calculation
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Design a single word problem that uses all four operations (+,−,×,÷+, -, \times, \div) exactly once, then provide the full solution and final answer.
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Worked solution

  1. Understand the task

    +,−,×,÷ each once+,-,\times,\div\ \text{each once}

    We must invent a real-life problem that uses each of the four operations exactly once, then solve it.

  2. Choose a context

    buying tickets\text{buying tickets}

    A group of friends buying event tickets is a natural setting.

  3. Set a ticket price

    £12 each\pounds12\ \text{each}

    Say each ticket costs £12.

  4. Use multiplication

    12×512\times5

    The group buys 5 tickets, so multiply.

  5. Work it out

    12×5=6012\times5=60

    Five tickets cost £60.

  6. Use subtraction

    60−860-8

    They receive an £8 discount, so subtract.

  7. Work it out

    60−8=5260-8=52

    The bill comes down to £52.

  8. Use division

    52÷452\div4

    The cost is split equally between 4 friends.

  9. Work it out

    52÷4=1352\div4=13

    Each friend pays £13 for their share of tickets.

  10. Use addition

    13+313+3

    Each friend also buys a £3 programme, so add.

  11. Work it out

    13+3=1613+3=16

    So each friend pays £16 in total.

  12. Check all four operations appear

    ×,−,÷,+\times,-,\div,+

    Multiplication, subtraction, division and addition are each used exactly once.

  13. Write the calculation clearly

    ((12×5−8)÷4)+3((12\times5-8)\div4)+3

    Brackets show the order the operations are applied.

  14. Evaluate to confirm

    ((60−8)÷4)+3=13+3((60-8)\div4)+3=13+3

    Following the brackets gives 13+313+3.

  15. State the problem and answer

    =£16=\pounds16

    The designed problem is solved: each friend pays £16.

Answer
e.g. ((12×5−8)÷4)+3=£16\text{e.g. }((12\times5-8)\div4)+3=\pounds16
Question 2
5 markschallenging
Place the operations ++, −-, ×\times, ÷\div (one each) into   6  □  2  □  4  □  3  □  1  \;6 \;\square\; 2 \;\square\; 4 \;\square\; 3 \;\square\; 1\; so that the result equals 1010, using left-to-right evaluation.
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Worked solution

  1. Understand the task

    6 □ 2 □ 4 □ 3 □ 1=106\,\square\,2\,\square\,4\,\square\,3\,\square\,1=10

    Place +, −, ×, ÷ (one each) in the gaps so the value is 10, working left to right.

  2. Rule for evaluating

    left to right\text{left to right}

    We evaluate strictly left to right, not using BIDMAS, as the question says.

  3. Strategy

    try a promising first op\text{try a promising first op}

    There are 24 arrangements, so use reasoning to cut them down rather than trying all.

  4. Try dividing first

    6÷2=36\div2=3

    6÷26\div 2 gives a neat 3 to build on.

  5. Then multiply

    3×4=123\times4=12

    Next 3×4=123\times 4=12, overshooting 1010 a little.

  6. Now use − and +

    on 3 and 1\text{on }3\ \text{and}\ 1

    We still have − and + to place before the 3 and the 1.

  7. Subtract next

    12−3=912-3=9

    12−312-3 brings us down to 9.

  8. Add last

    9+1=109+1=10

    9+19+1 lands exactly on 10.

  9. The operations used

    ÷,×,−,+\div,\times,-,+

    We used division, multiplication, subtraction and addition — each exactly once.

  10. Write the full expression

    6÷2×4−3+16\div2\times4-3+1

    Putting the operations into the gaps.

  11. Verify step 1

    6÷2=36\div2=3

    Recheck left to right: 6÷2=36\div 2=3.

  12. Verify step 2

    3×4=123\times4=12

    3×4=123\times 4=12.

  13. Verify step 3

    12−3=912-3=9

    12−3=912-3=9.

  14. Verify step 4

    9+1=109+1=10

    9+1=109+1=10, as required.

  15. State the answer

    6÷2×4−3+1=106\div2\times4-3+1=10

    So this arrangement of the four operations makes 10.

Answer
6÷2×4−3+1=106\div2\times4-3+1=10
Question 3
5 markschallenging
A ribbon 6236\tfrac{2}{3} m long is cut into pieces each 56\tfrac{5}{6} m long. How many complete pieces are cut, and how much ribbon is left over?
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Worked solution

  1. Set up the calculation

    623÷566\tfrac{2}{3}\div\tfrac{5}{6}

    The number of pieces is the total length divided by the length of each piece.

  2. Convert the mixed number

    623=2036\tfrac{2}{3}=\tfrac{20}{3}

    6×3+2=206\times 3+2=20 over 33.

  3. Rewrite the division

    203÷56\tfrac{20}{3}\div\tfrac{5}{6}

    Now both parts are fractions.

  4. Recall the rule

    ÷ → × reciprocal\div\ \to\ \times\ \text{reciprocal}

    Dividing by a fraction means multiplying by its reciprocal.

  5. Flip the second fraction

    56→65\tfrac{5}{6}\to\tfrac{6}{5}

    Turn 56\frac{5}{6} upside down.

  6. Rewrite as multiplication

    203×65\tfrac{20}{3}\times\tfrac{6}{5}

    Replace division with multiplication by 65\frac{6}{5}.

  7. Cancel where possible

    20 & 5→4 & 120\ \&\ 5\to4\ \&\ 1

    20 and 5 share a factor of 5.

  8. Cancel again

    6 & 3→2 & 16\ \&\ 3\to2\ \&\ 1

    6 and 3 share a factor of 3.

  9. Simplified multiplication

    41×21\tfrac{4}{1}\times\tfrac{2}{1}

    After cancelling we have 41×21\frac{4}{1} \times \frac{2}{1}.

  10. Multiply

    4×2=84\times2=8

    So the division equals 8.

  11. Interpret the whole number

    8 exactly8\ \text{exactly}

    8 is a whole number, so the pieces fit exactly.

  12. Check by multiplying back

    8×56=4068\times\tfrac{5}{6}=\tfrac{40}{6}

    Eight pieces of 56\frac{5}{6} m give 406\frac{40}{6} m.

  13. Simplify the check

    406=203=623\tfrac{40}{6}=\tfrac{20}{3}=6\tfrac{2}{3}

    That equals 6236\frac{2}{3} m, the whole ribbon.

  14. Work out the leftover

    623−623=06\tfrac{2}{3}-6\tfrac{2}{3}=0

    Nothing is left because the pieces use up the ribbon exactly.

  15. State the answer

    8 pieces, 0 left8\ \text{pieces},\ 0\ \text{left}

    So 8 complete pieces are cut with no ribbon left over.

Answer
8 pieces, 0 m left over8 \text{ pieces, } 0\text{ m left over}
Question 4
4 markschallenging
Work out (12−20)÷4+3×(−2)\left(12 - 20\right) \div 4 + 3 \times (-2), using the correct order of operations.
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Worked solution

  1. Order of operations

    () then ×÷ then +−()\ \text{then}\ \times\div\ \text{then}\ +-

    Do brackets first, then multiply and divide, then add and subtract.

  2. Work out the bracket

    12−2012-20

    Start with the bracket (12−2012-20).

  3. Subtract past zero

    12−20=−812-20=-8

    20 is bigger than 12, so the result is negative: −8.

  4. Rewrite the expression

    (−8)÷4+3×(−2)(-8)\div4+3\times(-2)

    Put the bracket value back in.

  5. Do the division

    −8÷4-8\div4

    Division comes before addition.

  6. Divide the magnitudes

    8÷4=28\div4=2

    Ignoring signs, 8÷4=28\div 4=2.

  7. Apply the sign

    −8÷4=−2-8\div4=-2

    A negative divided by a positive is negative.

  8. Do the multiplication

    3×(−2)3\times(-2)

    Multiplication also comes before addition.

  9. Multiply the magnitudes

    3×2=63\times2=6

    Ignoring signs, 3×2=63\times 2=6.

  10. Apply the sign

    3×(−2)=−63\times(-2)=-6

    A positive times a negative is negative.

  11. Rewrite what is left

    −2+(−6)-2+(-6)

    The expression is now −2 plus −6.

  12. Adding a negative

    +(−6)=−6+(-6)=-6

    Adding a negative number is the same as subtracting.

  13. Combine

    −2−6=−8-2-6=-8

    Two negatives combine to −8.

  14. Sense check the signs

    both terms negative\text{both terms negative}

    Both parts were negative, so a negative total makes sense.

  15. State the answer

    −8-8

    So the whole expression equals −8.

Answer
−8-8
Question 5
5 markschallenging
Shop A sells 66 eggs for £1.44\pounds1.44. Shop B sells 1010 eggs for £2.30\pounds2.30. Using division, decide which shop offers better value per egg.
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Worked solution

  1. Decide how to compare

    cost per egg\text{cost per egg}

    The fair way to compare is the cost of a single egg at each shop.

  2. Shop A information

    6 eggs=£1.446\ \text{eggs}=\pounds1.44

    Shop A sells 6 eggs for £1.44.

  3. Set up Shop A's unit cost

    1.44÷61.44\div6

    Divide the price by the number of eggs.

  4. Divide

    144÷6=24144\div6=24

    144 pence divided by 6 is 24.

  5. Shop A cost per egg

    £0.24\pounds0.24

    So each egg at Shop A costs £0.24.

  6. Shop B information

    10 eggs=£2.3010\ \text{eggs}=\pounds2.30

    Shop B sells 10 eggs for £2.30.

  7. Set up Shop B's unit cost

    2.30÷102.30\div10

    Divide the price by the number of eggs.

  8. Divide by 10

    2.30→0.232.30\to0.23

    Dividing by 10 moves the point one place left.

  9. Shop B cost per egg

    £0.23\pounds0.23

    So each egg at Shop B costs £0.23.

  10. Compare the two costs

    £0.24 vs £0.23\pounds0.24\ \text{vs}\ \pounds0.23

    Now compare the cost of one egg at each shop.

  11. Which is smaller

    0.23<0.240.23<0.24

    Shop B's egg is cheaper by 1p.

  12. Lower unit cost is better value

    cheaper per egg=better\text{cheaper per egg}=\text{better}

    Better value means less money for the same single item.

  13. Identify the winner

    Shop B\text{Shop B}

    Shop B gives the lower price per egg.

  14. Sense check

    1p cheaper each1\text{p cheaper each}

    Every egg is 1p cheaper at Shop B, so it is better value however many you buy.

  15. State the answer

    Shop B is better value\text{Shop B is better value}

    So Shop B offers better value at £0.23 per egg versus £0.24.

Answer
Shop B (£0.23/egg<£0.24/egg)\text{Shop B}\ (\pounds0.23\text{/egg} < \pounds0.24\text{/egg})

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