GCSE Four operations and place value Practice Questions

Free GCSE Four operations and place value practice questions with full step-by-step worked solutions. Covers place value, digit value, column addition, carrying. Practise exam-style problems and check your method.

place valuedigit valuecolumn additioncarryingcolumn subtractionborrowing
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Write down the value of the digit 77 in the number 4,7384{,}738.
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Worked solution

  1. Find the column of the 7

    4 738: Th ∣ H ∣ T ∣ U4\,738:\ \text{Th}\,|\,\text{H}\,|\,\text{T}\,|\,\text{U}

    Reading columns from the right they are units, tens, hundreds, thousands. The 7 sits in the hundreds column.

  2. Multiply the digit by its place value

    7×100=7007\times100=700

    The value of a digit is the digit multiplied by its column value. Here that is 7 hundreds.

  3. State the value

    700700

    So the 7 stands for 700.

Answer
700700
Question 2
1 markeasy
Work out 112+121\tfrac{1}{2} + \tfrac{1}{2}.
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Worked solution

  1. Add the fraction parts

    12+12=22=1\tfrac{1}{2}+\tfrac{1}{2}=\tfrac{2}{2}=1

    The two halves combine to make one whole.

  2. Add to the whole number

    1+1=21+1=2

    Add that new whole to the 1 already there.

  3. State the answer

    22

    So the answer is 2.

Answer
22
Question 3
2 marksintermediate
Work out 3×12+453 \times 12 + 45.
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Worked solution

  1. Decide the order of operations

    × before +\times\ \text{before}\ +

    Using BIDMAS, multiplication is done before addition.

  2. Identify the multiplication

    3×123\times12

    Do the 3×123\times 12 part first.

  3. Work it out

    3×12=363\times12=36

    3 lots of 12 is 36.

  4. Now do the addition

    36+4536+45

    Add 45 to the result.

  5. Work it out

    36+45=8136+45=81

    36+45=8136+45=81 using column addition.

  6. State the answer

    8181

    So 3×12+45=813\times 12+45=81.

Answer
8181
Question 4
4 markshard
In the addition   2 7 □+1 □ 6=4 3 1  \;2\,7\,\square + 1\,\square\,6 = 4\,3\,1\; each □\square is a missing digit. Find the two missing digits.
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Worked solution

  1. Label the missing digits

    27a+1b6=43127a + 1b6 = 431

    Let the missing units digit be a and the missing tens digit be b.

  2. Write each number using place value

    270+a, 100+10b+6270+a,\ 100+10b+6

    Break each number into its place-value parts.

  3. Add the known parts

    270+100+6=376270+100+6=376

    Add up everything except the unknown digits.

  4. Form an equation

    376+a+10b=431376+a+10b=431

    The total must equal 431.

  5. Rearrange

    a+10b=55a+10b=55

    Subtract 376 from both sides.

  6. Use the fact that a is a single digit

    0≤a≤90\le a\le9

    Since a is one digit, 10b must be within 9 of 55.

  7. Find 10b

    10b=50⇒b=510b=50\Rightarrow b=5

    The only multiple of 10 that leaves a single-digit a is 50.

  8. Find a

    a=55−50=5a=55-50=5

    Then a is 5.

  9. Check the addition

    275+156=431 ✓275+156=431\ \checkmark

    The numbers are 275 and 156, and they do add to 431.

  10. State the answer

    both missing digits=5\text{both missing digits}=5

    So both missing digits are 5.

Answer
both digits are 5 (275+156=431)\text{both digits are } 5\ (275+156=431)
Question 5
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

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