Hard GCSE Negative numbers Questions

Challenging, exam-style GCSE Negative numbers questions with worked solutions. Stretch yourself on the hardest order of operations, subtracting a negative, multiplying, multiplying a negative problems.

order of operationssubtracting a negativemultiplyingmultiplying a negativesquaring negativessubtraction
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
Work out ∣−7∣−∣3∣−(−2)|-7| - |3| - (-2), where ∣x∣|x| means the distance of xx from zero.
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Worked solution

  1. Meaning of the modulus

    ∣x∣=distance from 0|x|=\text{distance from }0

    The bars mean the distance of the number from zero, which is never negative.

  2. Work out the first modulus

    ∣−7∣=7|-7|=7

    −7 is 7 away from zero, so ∣−7∣=7|-7| = 7.

  3. Work out the second modulus

    ∣3∣=3|3|=3

    3 is 3 away from zero, so ∣3∣=3|3| = 3.

  4. Rewrite the calculation

    7−3−(−2)7-3-(-2)

    Replace the moduli with their values.

  5. Work left to right: first part

    7−37-3

    Start with 7 − 3.

  6. Evaluate

    7−3=47-3=4

    7−3=47 - 3 = 4.

  7. Now the last part

    4−(−2)4-(-2)

    We still have to subtract −2.

  8. Deal with the double sign

    −(−2)=+2-(-2)=+2

    Subtracting −2 becomes adding 2.

  9. Rewrite

    4+24+2

    So we compute 4 + 2.

  10. Evaluate

    4+2=64+2=6

    4+2=64 + 2 = 6.

  11. Watch the key trap

    ∣−7∣=7, not −7|-7|=7,\ \text{not}\ -7

    The modulus removes the minus sign, so |−7| is +7.

  12. And the double sign

    −(−2)=+2-(-2)=+2

    The final −(−2) turns into +2.

  13. Combine as a check

    7−3+27-3+2

    Written without the moduli or double sign, it is 7 − 3 + 2.

  14. Evaluate the check

    7−3+2=67-3+2=6

    This gives 6, matching.

  15. State the answer

    66

    So the value is 6.

Answer
66
Question 2
6 markschallenging
Design a real-life word problem about temperature or money that uses all four operations with negative numbers, then give the full solution and final answer.
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Worked solution

  1. Understand the task

    +,−,×,÷ with negatives+,-,\times,\div\ \text{with negatives}

    Design a real-life problem using all four operations, involving a negative number, then solve it.

  2. Choose a context

    an overdrawn bank account\text{an overdrawn bank account}

    A balance that starts negative is a natural directed-number setting.

  3. Set the starting balance

    −£20-\pounds20

    The account starts £20 overdrawn, i.e. −£20.

  4. Use multiplication

    4×£84\times\pounds8

    Four payments of £8 are made into the account.

  5. Work it out

    4×8=324\times8=32

    So £32 is paid in.

  6. Use addition

    −20+32-20+32

    Add the £32 to the starting balance.

  7. Work it out

    −20+32=12-20+32=12

    The balance becomes £12.

  8. Use subtraction

    12−412-4

    A £4 bank charge is taken off.

  9. Work it out

    12−4=812-4=8

    The balance is now £8.

  10. Use division

    8÷28\div2

    The £8 is shared equally between 2 people.

  11. Work it out

    8÷2=48\div2=4

    Each person gets £4.

  12. Check all four operations used

    ×,+,−,÷\times,+,-,\div

    Multiplication, addition, subtraction and division each appear once.

  13. Check a negative is involved

    −£20 start-\pounds20\ \text{start}

    The negative starting balance makes it a directed-number problem.

  14. Write the calculation

    (−20+4×8−4)÷2(-20+4\times8-4)\div2

    The whole thing in one line, respecting order of operations.

  15. State the problem and answer

    =£4 each=\pounds4\ \text{each}

    So the designed problem gives £4 for each person.

Answer
e.g. start −£20, +4×£8, −£4, ÷2⇒£4 each\text{e.g. start }-\pounds20,\ +4\times\pounds8,\ -\pounds4,\ \div2 \Rightarrow \pounds4\text{ each}
Question 3
5 markschallenging
A sequence begins −20,−14,−8,…-20, -14, -8, \ldots. Find the rule, the next two terms, and the first positive term.
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Worked solution

  1. Look at the differences

    −20, −14, −8, …-20,\ -14,\ -8,\ \ldots

    Find how the sequence changes from one term to the next.

  2. First gap

    −14−(−20)-14-(-20)

    Subtract the first term from the second.

  3. Evaluate it

    −14+20=6-14+20=6

    Subtracting −20 becomes adding 20, giving +6.

  4. Second gap

    −8−(−14)=6-8-(-14)=6

    The next gap is also +6.

  5. State the rule

    add 6 each time\text{add }6\ \text{each time}

    The sequence goes up by 6 each step.

  6. Find the next term

    −8+6-8+6

    Add 6 to the last known term −8.

  7. Evaluate

    −8+6=−2-8+6=-2

    So the next term is −2.

  8. Find the term after

    −2+6-2+6

    Add 6 again.

  9. Evaluate

    −2+6=4-2+6=4

    So the following term is 4.

  10. Next two terms

    −2, 4-2,\ 4

    The next two terms are −2 and 4.

  11. Look for the first positive term

    −20,−14,−8,−2,4-20,-14,-8,-2,4

    List the terms until one is above zero.

  12. Is −2 positive?

    −2<0-2<0

    −2 is still negative.

  13. Is 4 positive?

    4>04>0

    4 is the first term above zero.

  14. So the first positive term

    44

    The first positive term in the sequence is 4.

  15. State the answers

    +6; −2,4; 4+6;\ -2,4;\ 4

    Rule: add 6; next two terms −2 and 4; first positive term 4.

Answer
rule +6; next: −2, 4; first positive term =4\text{rule }+6;\ \text{next: }-2,\ 4;\ \text{first positive term }=4
Question 4
5 markschallenging
Two numbers satisfy a+b=−2a + b = -2 and a−b=8a - b = 8. Find aa and bb.
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Worked solution

  1. Write the two facts

    a+b=−2, a−b=8a+b=-2,\ a-b=8

    We are given the sum and the difference of the two numbers.

  2. Idea: add the equations

    (a+b)+(a−b)(a+b)+(a-b)

    Adding them will cancel the b terms.

  3. Left-hand side

    a+b+a−b=2aa+b+a-b=2a

    The +b and −b cancel, leaving 2a.

  4. Right-hand side

    −2+8=6-2+8=6

    Add the two right-hand sides: −2+8=6-2 + 8 = 6.

  5. So

    2a=62a=6

    This gives an equation for a alone.

  6. Solve for a

    a=6÷2=3a=6\div2=3

    Divide both sides by 2.

  7. First value

    a=3a=3

    So a=3a = 3.

  8. Use a fact to find b

    a+b=−2a+b=-2

    Substitute a=3a = 3 into the sum equation.

  9. Substitute

    3+b=−23+b=-2

    Replace a with 3.

  10. Rearrange for b

    b=−2−3b=-2-3

    Subtract 3 from both sides.

  11. Work it out

    b=−5b=-5

    −2−3=−5-2 - 3 = -5.

  12. Second value

    b=−5b=-5

    So b=−5b = -5.

  13. Check the difference

    a−b=3−(−5)=8 ✓a-b=3-(-5)=8\ \checkmark

    3−(−5)=3+5=83 - (-5) = 3 + 5 = 8, which matches.

  14. Check the sum

    a+b=3+(−5)=−2 ✓a+b=3+(-5)=-2\ \checkmark

    3+(−5)=−23 + (-5) = -2, which also matches.

  15. State the answer

    a=3, b=−5a=3,\ b=-5

    So a=3a = 3 and b=−5b = -5.

Answer
a=3, b=−5a=3,\ b=-5
Question 5
5 markschallenging
A climber descends from +150+150 m to −40-40 m relative to a base camp, resting every 5050 m of descent. How far does she descend in total, and how many rests does she take?
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Worked solution

  1. Total descent == start −- end

    150−(−40)150-(-40)

    How far she descends is the starting height minus the finishing height.

  2. Deal with the double sign

    −(−40)=+40-(-40)=+40

    Subtracting −40 becomes adding 40.

  3. Rewrite

    150+40150+40

    So the descent is 150 + 40.

  4. Work it out

    150+40=190150+40=190

    She descends 190 m in total.

  5. First answer

    190 m190\text{ m}

    The total descent is 190 m.

  6. Rests every 50 m

    rest each 50 m\text{rest each }50\text{ m}

    Now find how many 50 m stretches fit into 190 m.

  7. Set up the division

    190÷50190\div50

    Divide the total descent by 50.

  8. Divide

    190÷50=3 r 40190\div50=3\ \text{r }40

    50 goes into 190 three times (150) with 40 left over.

  9. Interpret the whole-number part

    3 full 50 m stretches3\ \text{full }50\text{ m stretches}

    There are 3 complete 50 m descents.

  10. Where the rests happen

    50, 100, 150 m50,\ 100,\ 150\text{ m}

    She rests after 50 m, 100 m and 150 m of descent.

  11. The remaining 40 m

    40<5040<50

    The last 40 m is less than 50 m, so no extra rest there.

  12. Count the rests

    3 rests3\ \text{rests}

    So she takes 3 rests in total.

  13. Sense check the descent

    3×50=150≤1903\times50=150\le190

    Three full stretches is 150 m, within the 190 m descent.

  14. Combine the answers

    190 m, 3 rests190\text{ m},\ 3\ \text{rests}

    Descent 190 m and 3 rests.

  15. State the answer

    190 m; 3190\text{ m};\ 3

    So she descends 190 m in total and takes 3 rests.

Answer
descends 190 m; 3 rests\text{descends }190\text{ m; }3\text{ rests}

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