Hard GCSE Error intervals Questions

Challenging, exam-style GCSE Error intervals questions with worked solutions. Stretch yourself on the hardest error intervals, rounding to 1 dp, bounds of a calculation, perimeter problems.

error intervalsrounding to 1 dpbounds of a calculationperimeterareaaddition
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A rectangle measures 9.59.5 cm by 4.54.5 cm, each correct to 11 decimal place. Work out the error interval for its area AA (in cm2^2).
Show worked solution

Worked solution

  1. Write down the formula you need

    A=l×wA = l \times w

    Area of a rectangle is length times width.

  2. Find the bounds of ll

    9.45≤l<9.559.45 \le l < 9.55

    ll is 9.5 cm to 1 dp, so it lies half a rounding unit either side.

  3. Find the bounds of ww

    4.45≤w<4.554.45 \le w < 4.55

    ww is 4.5 cm to 1 dp, so it lies half a rounding unit either side.

  4. Decide which bounds to combine

    A=l×wA = l \times w

    Smallest length and width give the smallest area; largest give the largest.

  5. Work out the upper bound

    9.55×4.55=43.45259.55 \times 4.55 = 43.4525

    The largest allowed values give the largest result.

  6. Work out the lower bound

    9.45×4.45=42.05259.45 \times 4.45 = 42.0525

    The smallest allowed values give the smallest result.

  7. State the answer

    42.0525≤A<43.452542.0525 \le A < 43.4525

    Report the required bound with the correct inequality.

  8. Double-check the arithmetic

    9.55×4.55=43.45259.55 \times 4.55 = 43.4525

    Recompute carefully so a slip does not change the bound.

  9. Confirm the correct bounds were chosen

    A=l×wA = l \times w

    A bound of a calculation comes from the extreme values of each measurement.

  10. Explain the strict inequality

    42.0525≤A<43.452542.0525 \le A < 43.4525

    Each measurement is strictly below its upper bound, so the result never quite reaches its upper limit.

  11. State the units

    units: cm2\text{units: }cm^2

    Always attach the correct units to a bound.

  12. Beware the common mistake

    A=l×wA = l \times w

    Substituting the rounded values gives only an estimate, not a true bound.

  13. Interpret the result

    42.0525≤A<43.452542.0525 \le A < 43.4525

    This gives the guaranteed range for the true value.

  14. Check reasonableness

    9.55×4.55=43.45259.55 \times 4.55 = 43.4525

    The bound should sit close to the estimate from the rounded values.

  15. Re-read the question

    A=l×wA = l \times w

    Make sure the reported bound (upper, lower or interval) is the one asked for.

Answer
42.0525≤A<43.452542.0525 \le A < 43.4525
Question 2
6 markschallenging
A rectangle measures 1515 cm by 88 cm, each to the nearest centimetre. Work out the error interval for its area AA (in cm2^2).
Show worked solution

Worked solution

  1. Write down the formula you need

    A=l×wA = l \times w

    Area of a rectangle is length times width.

  2. Find the bounds of ll

    14.5≤l<15.514.5 \le l < 15.5

    ll is 15 cm to the nearest cm, so it lies half a rounding unit either side.

  3. Find the bounds of ww

    7.5≤w<8.57.5 \le w < 8.5

    ww is 8 cm to the nearest cm, so it lies half a rounding unit either side.

  4. Decide which bounds to combine

    A=l×wA = l \times w

    Smallest length and width give the smallest area; largest give the largest.

  5. Work out the upper bound

    15.5×8.5=131.7515.5 \times 8.5 = 131.75

    The largest allowed values give the largest result.

  6. Work out the lower bound

    14.5×7.5=108.7514.5 \times 7.5 = 108.75

    The smallest allowed values give the smallest result.

  7. State the answer

    108.75≤A<131.75108.75 \le A < 131.75

    Report the required bound with the correct inequality.

  8. Double-check the arithmetic

    15.5×8.5=131.7515.5 \times 8.5 = 131.75

    Recompute carefully so a slip does not change the bound.

  9. Confirm the correct bounds were chosen

    A=l×wA = l \times w

    A bound of a calculation comes from the extreme values of each measurement.

  10. Explain the strict inequality

    108.75≤A<131.75108.75 \le A < 131.75

    Each measurement is strictly below its upper bound, so the result never quite reaches its upper limit.

  11. State the units

    units: cm2\text{units: }cm^2

    Always attach the correct units to a bound.

  12. Beware the common mistake

    A=l×wA = l \times w

    Substituting the rounded values gives only an estimate, not a true bound.

  13. Interpret the result

    108.75≤A<131.75108.75 \le A < 131.75

    This gives the guaranteed range for the true value.

  14. Check reasonableness

    15.5×8.5=131.7515.5 \times 8.5 = 131.75

    The bound should sit close to the estimate from the rounded values.

  15. Re-read the question

    A=l×wA = l \times w

    Make sure the reported bound (upper, lower or interval) is the one asked for.

Answer
108.75≤A<131.75108.75 \le A < 131.75
Question 3
5 markschallenging
A rod has length 2424 cm to the nearest centimetre, so 23.5≤L<24.523.5 \le L < 24.5. Why is the upper bound written with << and not ≤\le?
Show worked solution

Worked solution

  1. State the interval

    23.5≤L<24.523.5 \le L < 24.5

    A length of 2424 cm to the nearest cm gives this interval.

  2. Test the value 24.524.5

    24.5→2524.5 \to 25

    A length of exactly 24.524.5 cm rounds up to 2525 cm, not 2424 cm.

  3. Conclude why << is used

    L<24.5L < 24.5

    So 24.524.5 is not part of this interval — it belongs to the next one.

  4. Rule out the rounding distractor

    compare methods\text{compare methods}

    One option uses rounding boundaries instead of the correct method.

  5. Rule out the swapped-symbol distractor

    ≤ vs <\le \text{ vs } <

    An option that swaps the inclusive/exclusive symbols is wrong.

  6. Rule out the add-tops-and-bottoms distractor

    invalid method\text{invalid method}

    Combining figures incorrectly does not give a valid bound.

  7. Re-read each option

    scan options\text{scan options}

    Check exactly which values and symbols each option uses.

  8. Confirm the correct option

    L<24.5L < 24.5

    Only the first option matches the correct interval and reasoning.

  9. Check the boundary logic

    L<24.5L < 24.5

    The lower bound is included; the upper bound is excluded.

  10. Sanity-check with a test value

    L<24.5L < 24.5

    A value inside the interval behaves as the correct option describes.

  11. Eliminate the remaining distractor

    eliminate\text{eliminate}

    The last option relies on a misconception about rounding or truncation.

  12. Write the correct interval clearly

    L<24.5L < 24.5

    State it with the lower bound included and the upper bound excluded.

  13. Name the error in the distractors

    identify error\text{identify error}

    Each wrong option changes either the boundary values or the inequality symbols.

  14. Justify the final choice

    L<24.5L < 24.5

    The correct option matches the boundaries found from first principles.

  15. State the chosen answer

    L<24.5L < 24.5

    Select the option that is both correct and fully justified.

Answer
24.5 would round up to 25, so it is excluded from this interval
Question 4
6 markschallenging
aa and bb are each truncated to 11 decimal place, giving a=3.7a = 3.7 and b=2.4b = 2.4. Work out the error interval for a+ba + b.
Show worked solution

Worked solution

  1. Write down the formula you need

    a+ba + b

    Add the two truncated values.

  2. Find the bounds of aa

    3.7≤a<3.83.7 \le a < 3.8

    aa is truncated to 3.7 (1 dp); truncation gives an interval from the value up to the next.

  3. Find the bounds of bb

    2.4≤b<2.52.4 \le b < 2.5

    bb is truncated to 2.4 (1 dp); truncation gives an interval from the value up to the next.

  4. Decide which bounds to combine

    a+ba + b

    Truncated values are the lower bounds; the next values up are the (excluded) upper bounds.

  5. Work out the upper bound

    3.8+2.5=6.33.8 + 2.5 = 6.3

    The largest allowed values give the largest result.

  6. Work out the lower bound

    3.7+2.4=6.13.7 + 2.4 = 6.1

    The smallest allowed values give the smallest result.

  7. State the answer

    6.1≤a+b<6.36.1 \le a+b < 6.3

    Report the required bound with the correct inequality.

  8. Double-check the arithmetic

    3.8+2.5=6.33.8 + 2.5 = 6.3

    Recompute carefully so a slip does not change the bound.

  9. Confirm the correct bounds were chosen

    a+ba + b

    A bound of a calculation comes from the extreme values of each measurement.

  10. Explain the strict inequality

    6.1≤a+b<6.36.1 \le a+b < 6.3

    Each measurement is strictly below its upper bound, so the result never quite reaches its upper limit.

  11. State the units

    state units\text{state units}

    Always attach the correct units to a bound.

  12. Beware the common mistake

    a+ba + b

    Substituting the rounded values gives only an estimate, not a true bound.

  13. Interpret the result

    6.1≤a+b<6.36.1 \le a+b < 6.3

    This gives the guaranteed range for the true value.

  14. Check reasonableness

    3.8+2.5=6.33.8 + 2.5 = 6.3

    The bound should sit close to the estimate from the rounded values.

  15. Re-read the question

    a+ba + b

    Make sure the reported bound (upper, lower or interval) is the one asked for.

Answer
6.1≤a+b<6.36.1 \le a+b < 6.3
Question 5
5 markschallenging
A number xx is 75007500 to 22 significant figures. Work out the error interval for xx.
Show worked solution

Worked solution

  1. State what you are told

    x=7500x = 7500

    xx has been rounded to 2 significant figures.

  2. Identify the significant figures

    75007500

    Locate the last significant figure to fix the rounding unit.

  3. Find the rounding unit

    unit=100\text{unit} = 100

    This is the place value of the last significant figure.

  4. Halve the unit

    12×100=50\tfrac{1}{2} \times 100 = 50

    Each boundary is half a unit from 75007500.

  5. Lower bound

    7500−50=74507500 - 50 = 7450

    Any value from 74507450 up to 75007500 rounds to 75007500, so the lower bound 74507450 is included.

  6. Why the lower bound is included

    x≥7450x \ge 7450

    74507450 rounds up to 75007500.

  7. Upper bound

    7500+50=75507500 + 50 = 7550

    A value of 75507550 would round up to the next value, so 75507550 is not included.

  8. Why the upper bound is excluded

    x<7550x < 7550

    75507550 rounds up to the next value.

  9. Write the error interval

    7450≤x<75507450 \le x < 7550

    The boundaries lie exactly half a unit (5050) either side of 75007500.

  10. Test the lower boundary

    7450→75007450 \to 7500

    74507450 rounds to 75007500, so it is inside.

  11. Test a value just below the top

    7550−tiny→75007550 - \text{tiny} \to 7500

    Values just under 75507550 still round to 75007500.

  12. Test the upper boundary

    7550→next value7550 \to \text{next value}

    75507550 rounds up, so it is excluded.

  13. State the width of the interval

    7550−7450=1007550 - 7450 = 100

    The interval is exactly one rounding unit wide.

  14. Watch the common error

    7450≤x≤75507450 \le x \le 7550

    Do not use ≤\le on the upper bound.

  15. State the final answer

    7450≤x<75507450 \le x < 7550

    Lower bound inclusive, upper bound exclusive.

Answer
7450≤x<75507450 \le x < 7550

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