Hard GCSE Estimation and checking Questions

Challenging, exam-style GCSE Estimation and checking questions with worked solutions. Stretch yourself on the hardest estimating, fractions, decimals, estimating square roots problems.

estimatingfractionsdecimalsestimating square rootsreal-life estimationarea
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Estimate the value of 104×390.021\dfrac{\sqrt{104} \times 39}{0.021}.
Show worked solution

Worked solution

  1. Plan the estimate

    root first, then round\text{root first, then round}

    Estimate the square root, then round the other numbers.

  2. Round the number under the root

    104≈100104 \approx 100

    104 to 1 s.f. is 100.

  3. Take the square root

    100=10\sqrt{100} = 10

    So √104 ≈ 10.

  4. Round 39

    39≈4039 \approx 40

    To 1 s.f. this is 40.

  5. Round 0.021

    0.021≈0.020.021 \approx 0.02

    To 1 s.f. this is 0.02.

  6. Rewrite the calculation

    10×400.02\frac{10 \times 40}{0.02}

    Substitute the rounded values.

  7. Work out the numerator

    10×40=40010 \times 40 = 400

    Multiply the top first.

  8. Rewrite the fraction

    4000.02\frac{400}{0.02}

    Now divide by the decimal.

  9. Divide by a decimal

    ÷0.02=×50\div 0.02 = \times 50

    Dividing by 0.02 is the same as multiplying by 50.

  10. Work it out

    400×50=20000400 \times 50 = 20000

    So the estimate is 20000.

  11. Check the size

    0.02×20000=4000.02 \times 20000 = 400

    Reversing the division returns 400.

  12. Note the effect

    0.02<10.02 < 1

    Dividing by a small decimal greatly increases the value.

  13. Sense-check the root

    102=100≈10410^{2} = 100 \approx 104

    The square root estimate is reliable.

  14. Reflect

    answer in the tens of thousands\text{answer in the tens of thousands}

    A sensible size given the small divisor.

  15. State the estimate

    ≈20000\approx 20000

    So the estimate is 20000.

Answer
2000020000
Question 2
6 markschallenging
A builder needs 48704870 bricks costing £0.62\pounds 0.62 each, plus a delivery charge of £198\pounds 198. Estimate the total cost.
Show worked solution

Worked solution

  1. Plan the estimate

    cost=bricks×price+delivery\text{cost} = \text{bricks} \times \text{price} + \text{delivery}

    Estimate the brick cost, then add the delivery.

  2. Round the number of bricks

    4870≈50004870 \approx 5000

    4870 to 1 s.f. is 5000.

  3. Round the price per brick

    £0.62≈£0.6\pounds 0.62 \approx \pounds 0.6

    0.62 to 1 s.f. is 0.6.

  4. Estimate the brick cost

    5000×0.65000 \times 0.6

    Multiply the number of bricks by the price.

  5. Multiply

    5000×0.6=30005000 \times 0.6 = 3000

    5×0.6=35 \times 0.6 = 3, then attach the zeros.

  6. Round the delivery

    £198≈£200\pounds 198 \approx \pounds 200

    198 to 1 s.f. is 200.

  7. Add the delivery

    3000+2003000 + 200

    Add the delivery to the brick cost.

  8. Work it out

    3000+200=32003000 + 200 = 3200

    The total is 3200.

  9. Attach the units

    £3200\pounds 3200

    The cost is in pounds.

  10. State the estimate

    ≈£3200\approx \pounds 3200

    The total is about £3200.

  11. Compare the bricks rounding

    5000>4870, 0.6<0.625000 > 4870,\ 0.6 < 0.62

    One number up, one down for the bricks.

  12. Compare the delivery rounding

    200>198200 > 198

    The delivery was rounded up slightly.

  13. Reason about direction

    errors roughly balance\text{errors roughly balance}

    So £3200 is close to the true cost.

  14. Reflect

    answer in the thousands\text{answer in the thousands}

    A sensible size for a bulk order.

  15. State the result

    ≈£3200\approx \pounds 3200

    So the total cost is about £3200.

Answer
£3200\pounds 3200
Question 3
6 markschallenging
Estimate the value of 6.1×1051.9×102\dfrac{6.1 \times 10^{5}}{1.9 \times 10^{2}}.
Show worked solution

Worked solution

  1. Plan the estimate

    round the number parts\text{round the number parts}

    Round the front numbers to 1 s.f. and keep the powers of 10.

  2. Round 6.1

    6.1≈66.1 \approx 6

    To 1 s.f. this is 6.

  3. Round 1.9

    1.9≈21.9 \approx 2

    To 1 s.f. this is 2.

  4. Keep the powers

    105 and 10210^{5} \text{ and } 10^{2}

    Deal with the powers of 10 separately.

  5. Rewrite the calculation

    6×1052×102\frac{6 \times 10^{5}}{2 \times 10^{2}}

    Substitute the rounded values.

  6. Divide the number parts

    62=3\frac{6}{2} = 3

    Six divided by two is three.

  7. Divide the powers

    105102=103\frac{10^{5}}{10^{2}} = 10^{3}

    Subtract the indices: 5−2=35 - 2 = 3.

  8. Combine

    3×1033 \times 10^{3}

    Put the two results together.

  9. Work out the value

    3×103=30003 \times 10^{3} = 3000

    10³ is 1000, so 3×1000=30003 \times 1000 = 3000.

  10. Sense-check

    2×3000=6000≈6.1×105?2 \times 3000 = 6000 \approx 6.1 \times 10^{5}?

    Check the front and power together.

  11. Confirm the power

    3000=3×1033000 = 3 \times 10^{3}

    The power of 10 is correct.

  12. Note index laws

    10a10b=10a−b\frac{10^{a}}{10^{b}} = 10^{a-b}

    The powers of 10 follow the index law.

  13. Reflect

    answer in the thousands\text{answer in the thousands}

    A sensible size.

  14. Check the front number

    6÷2=36 \div 2 = 3

    The front number is 3, as expected.

  15. State the estimate

    ≈3000\approx 3000

    So the estimate is 3000.

Answer
30003000
Question 4
6 markschallenging
A train travels 388388 km in 44 hours 5555 minutes. Estimate its average speed in km/h.
Show worked solution

Worked solution

  1. Plan the estimate

    speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}

    Round the distance and time, then divide.

  2. Round the distance

    388≈400388 \approx 400

    388 to 1 s.f. is 400.

  3. Convert the time

    4h 55m≈5 hours4\text{h }55\text{m} \approx 5 \text{ hours}

    4 hours 55 minutes is almost 5 hours.

  4. Rewrite the calculation

    4005\frac{400}{5}

    Substitute the rounded values.

  5. Divide

    4005=80\frac{400}{5} = 80

    Work out the division.

  6. Attach the units

    80 km/h80 \text{ km/h}

    Speed is in kilometres per hour.

  7. Sense-check

    5×80=4005 \times 80 = 400

    Reversing the division returns the distance.

  8. Compare the distance rounding

    400>388400 > 388

    The distance was rounded up.

  9. Compare the time rounding

    5>4h 55m5 > 4\text{h }55\text{m}

    The time was rounded up too.

  10. Reason about direction

    both up, ratio steady\text{both up, ratio steady}

    Rounding both up keeps the ratio close.

  11. Check exactly

    3884.917≈78.9\tfrac{388}{4.917} \approx 78.9

    The true speed is about 78.9 km/h.

  12. Compare

    80≈78.980 \approx 78.9

    The estimate is close to the true value.

  13. Reflect

    good estimate\text{good estimate}

    80 km/h is a sensible average speed.

  14. Confirm

    ≈80 km/h\approx 80 \text{ km/h}

    The estimate is 80 km/h.

  15. State the result

    ≈80 km/h\approx 80 \text{ km/h}

    So the average speed is about 80 km/h.

Answer
80 km/h80 \text{ km/h}
Question 5
6 markschallenging
Estimate the value of 0.0384×51201.97\dfrac{0.0384 \times 5120}{1.97}.
Show worked solution

Worked solution

  1. Plan the estimate

    round each number to 1 s.f.\text{round each number to 1 s.f.}

    Replace every number by its 1 s.f. value.

  2. Round 0.0384

    0.0384≈0.040.0384 \approx 0.04

    To 1 s.f. this is 0.04.

  3. Round 5120

    5120≈50005120 \approx 5000

    To 1 s.f. this is 5000.

  4. Round 1.97

    1.97≈21.97 \approx 2

    To 1 s.f. this is 2.

  5. Rewrite the calculation

    0.04×50002\frac{0.04 \times 5000}{2}

    Substitute the rounded values.

  6. Work out the numerator

    0.04×5000=2000.04 \times 5000 = 200

    Multiply the top first.

  7. Show a method

    ×0.04=÷25\times 0.04 = \div 25

    Multiplying by 0.04 is dividing by 25.

  8. Confirm the numerator

    5000÷25=2005000 \div 25 = 200

    Both methods give 200.

  9. Rewrite the fraction

    2002\frac{200}{2}

    Now divide by the denominator.

  10. Divide

    2002=100\frac{200}{2} = 100

    Work out the division.

  11. Attach the units

    100100

    No units are needed here.

  12. Sense-check

    2×100=2002 \times 100 = 200

    Reversing the last step returns the numerator.

  13. Compare the rounding

    mixed up and down\text{mixed up and down}

    Some numbers went up and some down.

  14. Reflect

    answer near 100\text{answer near 100}

    A sensible size.

  15. State the estimate

    ≈100\approx 100

    So the estimate is 100.

Answer
100100

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