GCSE Error intervals Practice Questions

Free GCSE Error intervals practice questions with full step-by-step worked solutions. Covers error intervals, rounding to nearest 10, rounding to nearest 100, rounding to nearest integer. Practise exam-style problems and check your method.

error intervalsrounding to nearest 10rounding to nearest 100rounding to nearest integerrounding to 1 dptruncation
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A number xx is 7070 when rounded to the nearest 1010. Write down the error interval for xx.
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Worked solution

  1. Interpret the rounded value

    x70x \approx 70

    xx has been rounded to the nearest 10, so it lies in an interval around 7070.

  2. Find the lower and upper bounds

    705=65and70+5=7570 - 5 = 65 \quad\text{and}\quad 70 + 5 = 75

    The boundaries lie exactly half a unit (55) either side of 7070.

  3. Write the error interval

    65x<7565 \le x < 75

    Any value from 6565 up to 7070 rounds to 7070, so the lower bound 6565 is included. A value of 7575 would round up to the next value, so 7575 is not included.

Answer
65x<7565 \le x < 75
Question 2
1 markeasy
A number xx is 6060 when rounded to the nearest 1010. Write down the error interval for xx.
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Worked solution

  1. Interpret the rounded value

    x60x \approx 60

    xx has been rounded to the nearest 10, so it lies in an interval around 6060.

  2. Find the lower and upper bounds

    605=55and60+5=6560 - 5 = 55 \quad\text{and}\quad 60 + 5 = 65

    The boundaries lie exactly half a unit (55) either side of 6060.

  3. Write the error interval

    55x<6555 \le x < 65

    Any value from 5555 up to 6060 rounds to 6060, so the lower bound 5555 is included. A value of 6565 would round up to the next value, so 6565 is not included.

Answer
55x<6555 \le x < 65
Question 3
2 marksintermediate
The number xx rounds to 3.93.9 to 11 decimal place. Write down the error interval for xx.
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Worked solution

  1. State what you are told

    x=3.9x = 3.9

    xx has been rounded to 1 decimal place.

  2. Find the size of the rounding unit

    unit=0.1\text{unit} = 0.1

    This is the place value xx was rounded to.

  3. Halve the unit

    12×0.1=0.05\tfrac{1}{2} \times 0.1 = 0.05

    Each boundary is half a unit from 3.93.9.

  4. Lower bound

    3.90.05=3.853.9 - 0.05 = 3.85

    Any value from 3.853.85 up to 3.93.9 rounds to 3.93.9, so the lower bound 3.853.85 is included.

  5. Upper bound

    3.9+0.05=3.953.9 + 0.05 = 3.95

    A value of 3.953.95 would round up to the next value, so 3.953.95 is not included.

  6. Write the error interval

    3.85x<3.953.85 \le x < 3.95

    Lower bound with \le, upper bound with <<.

Answer
3.85x<3.953.85 \le x < 3.95
Question 4
4 markshard
A population xx is 2400024000, correct to 22 significant figures. Work out the error interval for xx.
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Worked solution

  1. State what you are told

    x=24000x = 24000

    xx has been rounded to 2 significant figures.

  2. Find the rounding unit

    unit=1000\text{unit} = 1000

    The value was rounded to this place value.

  3. Halve the unit

    12×1000=500\tfrac{1}{2} \times 1000 = 500

    Each boundary is half a unit from 2400024000.

  4. Lower bound

    24000500=2350024000 - 500 = 23500

    Any value from 2350023500 up to 2400024000 rounds to 2400024000, so the lower bound 2350023500 is included.

  5. Why the lower bound is included

    x23500x \ge 23500

    2350023500 rounds up to 2400024000, so it belongs to the interval.

  6. Upper bound

    24000+500=2450024000 + 500 = 24500

    A value of 2450024500 would round up to the next value, so 2450024500 is not included.

  7. Why the upper bound is excluded

    x<24500x < 24500

    2450024500 rounds up to the next value, so it is left out.

  8. Write the error interval

    23500x<2450023500 \le x < 24500

    The boundaries lie exactly half a unit (500500) either side of 2400024000.

  9. Check the boundaries

    23500240002450023500 \to 24000 \to 24500

    2350023500 rounds to 2400024000; 2450024500 rounds to the next value.

  10. State the final answer

    23500x<2450023500 \le x < 24500

    Lower bound inclusive, upper bound exclusive.

Answer
23500x<2450023500 \le x < 24500
Question 5
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).
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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

    9.45l<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

    4.45w<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

    42.0525A<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

    42.0525A<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

    42.0525A<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
42.0525A<43.452542.0525 \le A < 43.4525

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