Hard GCSE Upper and lower bounds Questions

Challenging, exam-style GCSE Upper and lower bounds questions with worked solutions. Stretch yourself on the hardest compound measures, speed, upper bound, density problems.

compound measuresspeedupper bounddensitylower boundbounds of a quotient
GCSE Higher34 questionsStep-by-step solutions
Question 1
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 9 and 11 and bb between 4 and 6 to test the rule.

  12. Compute the test minimum

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

    Smallest top over largest bottom gives 1.5.

  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 1.5.

  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}
Question 2
6 markschallenging
A rectangle measures 19.619.6 cm by 10.210.2 cm, each to 11 decimal place. Work out the upper and lower bounds of its area, and hence state the area to a suitable degree of accuracy.
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Worked solution

  1. Recall the area formula

    A=l×wA = l \times w

    A rectangle's area is length times width.

  2. Note the rounding

    1 d.p.⇒±0.05\text{1 d.p.} \Rightarrow \pm 0.05

    Each side can vary by 0.05 cm.

  3. Upper bound of the length

    19.6+0.05=19.6519.6 + 0.05 = 19.65

    The longest the 19.6 cm side can be is 19.65 cm.

  4. Upper bound of the width

    10.2+0.05=10.2510.2 + 0.05 = 10.25

    The longest the 10.2 cm side can be is 10.25 cm.

  5. Compute the maximum area

    19.65×10.25=201.412519.65 \times 10.25 = 201.4125

    Multiply the two upper bounds for the largest area.

  6. Lower bound of the length

    19.6−0.05=19.5519.6 - 0.05 = 19.55

    The shortest the 19.6 cm side can be is 19.55 cm.

  7. Lower bound of the width

    10.2−0.05=10.1510.2 - 0.05 = 10.15

    The shortest the 10.2 cm side can be is 10.15 cm.

  8. Compute the minimum area

    19.55×10.15=198.432519.55 \times 10.15 = 198.4325

    Multiply the two lower bounds for the smallest area.

  9. State the two bounds

    198.4325≤A<201.4125198.4325 \le A < 201.4125

    The true area lies between these values.

  10. Round the upper bound to 1 s.f.

    201.4125→200201.4125 \to 200

    To 1 significant figure this is 200.

  11. Round the lower bound to 1 s.f.

    198.4325→200198.4325 \to 200

    To 1 significant figure this is also 200.

  12. Check 2 s.f.

    201.4→200, 198.4→200201.4 \to 200,\ 198.4 \to 200

    At 2 s.f. both give 200 as well, but at 3 s.f. they differ (201 vs 198).

  13. Apply the accuracy rule

    safe to 2 s.f.\text{safe to 2 s.f.}

    The bounds agree to 2 significant figures, giving 200 cm².

  14. Attach the units

    cm2\text{cm}^2

    Area is measured in square centimetres.

  15. State the answer

    UB=201.4125, LB=198.4325, A≈200 cm2\text{UB}=201.4125,\ \text{LB}=198.4325,\ A \approx 200 \text{ cm}^2

    Bounds 198.4325 cm² and 201.4125 cm² give a safe value of 200 cm² (2 s.f.).

Answer
UB=201.4125 cm2, LB=198.4325 cm2, A≈200 cm2 (2 s.f.)\text{UB} = 201.4125 \text{ cm}^2,\ \text{LB} = 198.4325 \text{ cm}^2,\ A \approx 200 \text{ cm}^2 \text{ (2 s.f.)}
Question 3
6 markschallenging
Block P has mass 6060 g to the nearest gram and volume 88 cm3^3 to the nearest cm3^3. Block Q has a density of exactly 7.67.6 g/cm3^3. Determine whether block P could be denser than block Q.
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Worked solution

  1. Write the density formula

    density=massvolume\text{density} = \frac{\text{mass}}{\text{volume}}

    Density is mass divided by volume.

  2. Bounds of P's mass

    59.5≤mass<60.559.5 \le \text{mass} < 60.5

    60 g to the nearest gram varies by 0.5 g.

  3. Bounds of P's volume

    7.5≤volume<8.57.5 \le \text{volume} < 8.5

    8 cm³ to the nearest cm³ varies by 0.5 cm³.

  4. Rule for P's maximum density

    max=UBmassLBvol\text{max} = \frac{\text{UB}_{mass}}{\text{LB}_{vol}}

    Largest mass over smallest volume gives P's largest possible density.

  5. Substitute for the maximum

    60.57.5\frac{60.5}{7.5}

    Use 60.5 g and 7.5 cm³.

  6. Compute P's maximum density

    60.57.5=8.066…\frac{60.5}{7.5} = 8.066\ldots

    60.5÷7.5=8.0760.5 \div 7.5 = 8.07 g/cm³ (2 d.p.).

  7. Compare with Q

    8.07>7.68.07 > 7.6

    P's greatest possible density exceeds Q's density.

  8. Rule for P's minimum density

    min=LBmassUBvol\text{min} = \frac{\text{LB}_{mass}}{\text{UB}_{vol}}

    Smallest mass over largest volume gives P's smallest possible density.

  9. Substitute for the minimum

    59.58.5\frac{59.5}{8.5}

    Use 59.5 g and 8.5 cm³.

  10. Compute P's minimum density

    59.58.5=7\frac{59.5}{8.5} = 7

    59.5÷8.5=759.5 \div 8.5 = 7 g/cm³ exactly.

  11. State P's density range

    7≤densityP<8.077 \le \text{density}_P < 8.07

    P's density lies between 7 and 8.07 g/cm³.

  12. Locate Q within P's range

    7<7.6<8.077 < 7.6 < 8.07

    Q's density lies inside P's possible range.

  13. Interpret the comparison

    P could be denser or less dense\text{P could be denser or less dense}

    Because 7.6 is inside P's range, either outcome is possible.

  14. Answer the question asked

    Yes, P could be denser\text{Yes, P could be denser}

    Since P's density can reach 8.07 g/cm³, P could indeed be denser than Q.

  15. State the conclusion

    Yes — up to 8.07 g/cm3>7.6\text{Yes — up to } 8.07 \text{ g/cm}^3 > 7.6

    Block P could be denser than Q, because its density could be as high as 8.07 g/cm³.

Answer
Yes: P’s density could reach 8.07 g/cm3>7.6 g/cm3.\text{Yes: P's density could reach } 8.07 \text{ g/cm}^3 > 7.6 \text{ g/cm}^3.
Question 4
6 markschallenging
A car travels at 2525 m/s, to the nearest m/s, for 1818 s, to the nearest second. Work out the upper and lower bounds of the distance travelled, and state whether the car definitely travels more than 450450 m.
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Worked solution

  1. Write the formula

    distance=speed×time\text{distance} = \text{speed} \times \text{time}

    Distance is speed multiplied by time.

  2. Bounds of the speed

    24.5≤speed<25.524.5 \le \text{speed} < 25.5

    25 m/s to the nearest m/s varies by 0.5 m/s.

  3. Bounds of the time

    17.5≤time<18.517.5 \le \text{time} < 18.5

    18 s to the nearest second varies by 0.5 s.

  4. Rule for the maximum distance

    max=UBspeed×UBtime\text{max} = \text{UB}_{speed} \times \text{UB}_{time}

    Largest speed times largest time gives the largest distance.

  5. Compute the maximum distance

    25.5×18.5=471.7525.5 \times 18.5 = 471.75

    25.5×18.5=471.7525.5 \times 18.5 = 471.75 m.

  6. Rule for the minimum distance

    min=LBspeed×LBtime\text{min} = \text{LB}_{speed} \times \text{LB}_{time}

    Smallest speed times smallest time gives the smallest distance.

  7. Compute the minimum distance

    24.5×17.5=428.7524.5 \times 17.5 = 428.75

    24.5×17.5=428.7524.5 \times 17.5 = 428.75 m.

  8. State the two bounds

    428.75≤distance<471.75428.75 \le \text{distance} < 471.75

    The distance lies between these values.

  9. Compare the lower bound with 450

    428.75<450428.75 < 450

    The distance could be as little as 428.75 m.

  10. Compare the upper bound with 450

    471.75>450471.75 > 450

    The distance could also be as much as 471.75 m.

  11. Locate 450 within the range

    428.75<450<471.75428.75 < 450 < 471.75

    450 m lies inside the interval of possible distances.

  12. Interpret

    not guaranteed\text{not guaranteed}

    Because 450 is inside the range, we cannot be sure the distance exceeds it.

  13. State the condition for certainty

    need LB>450\text{need LB} > 450

    Only if the lower bound exceeded 450 could we be certain.

  14. Attach the units

    metres\text{metres}

    Both bounds are distances in metres.

  15. State the answer

    UB=471.75, LB=428.75; not guaranteed\text{UB}=471.75,\ \text{LB}=428.75; \text{ not guaranteed}

    Distance is between 428.75 m and 471.75 m; it is not certain to exceed 450 m.

Answer
UB=471.75 m, LB=428.75 m; not guaranteed to exceed 450 m\text{UB} = 471.75 \text{ m},\ \text{LB} = 428.75 \text{ m}; \text{ not guaranteed to exceed } 450 \text{ m}
Question 5
5 markschallenging
Each coin has a mass of 5.55.5 g to 11 decimal place. A pile of these coins weighs 500500 g to the nearest 1010 g. Work out the greatest possible number of whole coins in the pile.
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Worked solution

  1. Set up the calculation

    coins=total massmass of one coin\text{coins} = \frac{\text{total mass}}{\text{mass of one coin}}

    The number of coins is the total divided by one coin's mass.

  2. Decide what maximises the count

    max=UBtotalLBcoin\text{max} = \frac{\text{UB}_{total}}{\text{LB}_{coin}}

    The most coins come from the largest total and the lightest coin.

  3. Upper bound of the total mass

    500+5=505500 + 5 = 505

    500 g to the nearest 10 g has upper bound 505 g.

  4. Lower bound of one coin's mass

    5.5−0.05=5.455.5 - 0.05 = 5.45

    5.5 g to 1 d.p. has lower bound 5.45 g.

  5. Substitute

    5055.45\frac{505}{5.45}

    Largest total over lightest coin.

  6. Carry out the division

    5055.45=92.66…\frac{505}{5.45} = 92.66\ldots

    505÷5.45=92.66505 \div 5.45 = 92.66…

  7. Interpret the decimal

    92.66…⇒92 whole coins92.66\ldots \Rightarrow 92 \text{ whole coins}

    Only whole coins count, so round down.

  8. Explain why we round down

    a fraction of a coin is impossible\text{a fraction of a coin is impossible}

    The leftover 0.66 is not a whole coin.

  9. Guard against the common error

    not 4955.55\text{not } \frac{495}{5.55}

    That pairing gives the minimum number of coins.

  10. Compute the minimum for contrast

    4955.55=89.1…→89\frac{495}{5.55} = 89.1\ldots \to 89

    At worst there would be 89 coins.

  11. Confirm the maximum

    92.66…→9292.66\ldots \to 92

    The greatest whole number below 92.66 is 92.

  12. Sanity check

    92×5.45=501.4<50592 \times 5.45 = 501.4 < 505

    92 lightest coins weigh 501.4 g, which fits inside 505 g.

  13. Check 93 is too many

    93×5.45=506.85>50593 \times 5.45 = 506.85 > 505

    93 coins would exceed the largest possible total, so 92 is the most.

  14. Attach the interpretation

    92 coins92 \text{ coins}

    The greatest possible number of whole coins is 92.

  15. State the answer

    92 coins92 \text{ coins}

    The pile can contain at most 92 whole coins.

Answer
92 coins92 \text{ coins}

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