GCSE Upper and lower bounds Practice Questions

Free GCSE Upper and lower bounds practice questions with full step-by-step worked solutions. Covers upper and lower bounds, rounding to the nearest 10, rounding to the nearest cm, rounding to the nearest 100. Practise exam-style problems and check your method.

upper and lower boundsrounding to the nearest 10rounding to the nearest cmrounding to the nearest 100rounding to 1 decimal placerounding to the nearest 1000
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
A number nn is 4040 rounded to the nearest 1010. Write down the lower bound of nn.
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Worked solution

  1. Identify the rounding

    40 to the nearest 1040 \text{ to the nearest } 10

    The true value has been rounded to the nearest 1010, so it can lie up to half of 1010 away.

  2. Halve the rounding unit

    102=5\frac{10}{2} = 5

    The value can be at most 55 below or 55 above the rounded figure.

  3. Subtract to find the lower bound

    40−5=3540 - 5 = 35

    The smallest value that still rounds to 4040 is 3535.

Answer
3535
Question 2
2 markseasy
A measurement is 0.500.50 m to 22 decimal places. Write down the lower and upper bounds.
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Worked solution

  1. Identify the rounding

    0.50 m to 2 d.p.0.50 \text{ m to } 2 \text{ d.p.}

    Rounded to 2 decimal places, so it is within 0.005 of 0.50 m.

  2. Halve the rounding unit

    0.012=0.005\frac{0.01}{2} = 0.005

    Half of 0.010.01 is 0.0050.005.

  3. State both bounds

    0.50−0.005=0.495,0.50+0.005=0.5050.50 - 0.005 = 0.495,\quad 0.50 + 0.005 = 0.505

    The measurement lies from 0.495 m up to (but not including) 0.505 m.

Answer
0.495 m and 0.505 m0.495 \text{ m and } 0.505 \text{ m}
Question 3
2 marksintermediate
A row is made from 55 tiles, each 3030 cm wide to the nearest centimetre. Work out the upper bound of the total width.
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Worked solution

  1. Recall the rule

    max total=5×UBtile\text{max total} = 5 \times \text{UB}_{tile}

    The row is widest when every tile is as wide as possible.

  2. Upper bound of one tile

    30+0.5=30.530 + 0.5 = 30.5

    3030 cm to the nearest cm has upper bound 30.530.5 cm.

  3. Decide which bound to use

    use UBtile\text{use } \text{UB}_{tile}

    Wider tiles give a wider row.

  4. Multiply by 55

    5×30.5=152.55 \times 30.5 = 152.5

    Five tiles of 30.530.5 cm give 152.5152.5 cm.

  5. Check the units

    152.5 cm152.5 \text{ cm}

    Total width is a length in centimetres.

  6. State the upper bound

    152.5 cm152.5 \text{ cm}

    The greatest possible total width is 152.5152.5 cm.

Answer
152.5 cm152.5 \text{ cm}
Question 4
4 markshard
a=72a = 72 and b=9b = 9, each to the nearest whole number. Work out the upper and lower bounds of a÷ba \div b. Give each answer to 33 significant figures.
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Worked solution

  1. Recall the rules for a quotient

    max=UBaLBb,min=LBaUBb\text{max} = \frac{\text{UB}_a}{\text{LB}_b},\quad \text{min} = \frac{\text{LB}_a}{\text{UB}_b}

    Largest over smallest gives the maximum; smallest over largest gives the minimum.

  2. Bounds of aa

    71.5≤a<72.571.5 \le a < 72.5

    7272 to the nearest whole varies by 0.50.5.

  3. Bounds of bb

    8.5≤b<9.58.5 \le b < 9.5

    99 to the nearest whole varies by 0.50.5.

  4. Substitute for the maximum

    72.58.5\frac{72.5}{8.5}

    Largest top over smallest bottom.

  5. Compute the maximum

    72.58.5=8.529…\frac{72.5}{8.5} = 8.529\ldots

    72.5÷8.5=8.52972.5 \div 8.5 = 8.529…

  6. Round the maximum to 3 s.f.

    8.529…≈8.538.529\ldots \approx 8.53

    The first three significant figures are 88, 55 and 33.

  7. Substitute for the minimum

    71.59.5\frac{71.5}{9.5}

    Smallest top over largest bottom.

  8. Compute the minimum

    71.59.5=7.526…\frac{71.5}{9.5} = 7.526\ldots

    71.5÷9.5=7.52671.5 \div 9.5 = 7.526…

  9. Round the minimum to 3 s.f.

    7.526…≈7.537.526\ldots \approx 7.53

    The first three significant figures are 77, 55 and 33.

  10. State both bounds

    UB=8.53, LB=7.53\text{UB} = 8.53,\ \text{LB} = 7.53

    The value of a÷ba \div b lies between 7.53 and 8.53 (3 s.f.).

Answer
UB=8.53, LB=7.53\text{UB} = 8.53,\ \text{LB} = 7.53
Question 5
5 markschallenging
To find the lower bound of the quotient ab\dfrac{a}{b}, where aa and bb are positive measured quantities, which pair of bounds should be used?
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Worked solution

  1. Recall the meaning of a quotient

    ab\frac{a}{b}

    This is aa shared into bb, so its size depends on both the top and the bottom.

  2. Think about the effect of the numerator

    larger top⇒larger quotient\text{larger top} \Rightarrow \text{larger quotient}

    Increasing aa increases the value of the fraction.

  3. Think about the effect of the denominator

    larger bottom⇒smaller quotient\text{larger bottom} \Rightarrow \text{smaller quotient}

    Increasing bb decreases the value of the fraction.

  4. State the goal

    we want the smallest value\text{we want the smallest value}

    We are asked for the lower bound of the quotient.

  5. Make the numerator as small as possible

    use LBa\text{use } \text{LB}_a

    The smallest possible top helps make the value small.

  6. Make the denominator as large as possible

    use UBb\text{use } \text{UB}_b

    Dividing by the largest possible bottom makes the value small.

  7. Combine the choices

    min=LBaUBb\text{min} = \frac{\text{LB}_a}{\text{UB}_b}

    Smallest top over largest bottom gives the minimum.

  8. Reject the tempting wrong pairing

    LBaLBb is wrong\frac{\text{LB}_a}{\text{LB}_b} \text{ is wrong}

    A small denominator would make the quotient larger, not smaller.

  9. Reject another wrong pairing

    UBaUBb is wrong\frac{\text{UB}_a}{\text{UB}_b} \text{ is wrong}

    A large numerator pushes the value up, so this is not the minimum.

  10. Contrast with the maximum

    max=UBaLBb\text{max} = \frac{\text{UB}_a}{\text{LB}_b}

    The maximum uses the opposite bounds, confirming the pattern.

  11. Set up a numerical check

    a=10±1, b=5±1a = 10 \pm 1,\ b = 5 \pm 1

    Take aa between 99 and 1111 and bb between 44 and 66 to test the rule.

  12. Compute the test minimum

    96=1.5\frac{9}{6} = 1.5

    Smallest top over largest bottom gives 1.51.5.

  13. Compare with other pairings

    94=2.25, 116=1.83\frac{9}{4} = 2.25,\ \frac{11}{6} = 1.83

    Every other pairing gives a larger result than 1.51.5.

  14. Confirm the rule works

    1.5 is indeed the smallest1.5 \text{ is indeed the smallest}

    The numerical check agrees with the reasoning.

  15. State the answer

    LBaUBb\frac{\text{LB}_a}{\text{UB}_b}

    Lower bound of aa divided by upper bound of bb.

Answer
LBaUBb\frac{\text{LB}_a}{\text{UB}_b}

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