Upper and Lower Bounds Speed Problem

Why Bounds and Speed Questions Are Tricky

Upper and lower bounds questions are a standard part of the IGCSE Extended syllabus, and when they are combined with speed, distance, and time, they become particularly challenging. The difficulty lies not in the arithmetic — which is usually straightforward — but in deciding which bounds to use for each quantity.

The fundamental principle is this: every measured quantity has an upper and a lower bound determined by the precision of the measurement. When you combine measurements in a calculation, you need to think about which combination of bounds gives you the largest possible result and which gives the smallest.

The Problem

A car travels a distance of 240 km, measured to the nearest 10 km. The journey takes 3.0 hours, measured to the nearest 0.1 hours.

(a) Write down the upper and lower bounds for the distance.

(b) Write down the upper and lower bounds for the time.

(c) Calculate the upper bound of the speed.

(d) Calculate the lower bound of the speed.

Part (a): Bounds for Distance

The distance is 240 km, measured to the nearest 10 km. This means the actual distance could be anywhere from half a unit below to half a unit above:

Lower bound = 240 − 5 = 235 km
Upper bound = 240 + 5 = 245 km

The actual distance d satisfies: 235 ≤ d < 245

Note the strict inequality on the upper end. The value 245 would round up to 250, so it is not included.

Part (b): Bounds for Time

The time is 3.0 hours, measured to the nearest 0.1 hours. Half of 0.1 is 0.05:

Lower bound = 3.0 − 0.05 = 2.95 hours
Upper bound = 3.0 + 0.05 = 3.05 hours

The actual time t satisfies: 2.95 ≤ t < 3.05

Part (c): Upper Bound of Speed

Speed = distance ÷ time.

To get the maximum speed, we want to divide the largest possible distance by the smallest possible time:

Upper bound of speed = 245 ÷ 2.95 = 83.050… ≈ 83.1 km/h

Think about it intuitively: if you travel a longer distance in less time, you must be going faster. So maximum distance and minimum time give maximum speed.

Part (d): Lower Bound of Speed

To get the minimum speed, we want to divide the smallest possible distance by the largest possible time:

Lower bound of speed = 235 ÷ 3.05 = 77.049… ≈ 77.0 km/h

If you travel a shorter distance in more time, you are going slower. So minimum distance and maximum time give minimum speed.

The General Rules for Combining Bounds

Here are the rules for determining which bounds to use in different operations:

For the maximum value of:

A + B: use upper bound of A + upper bound of B
A − B: use upper bound of A − lower bound of B
A × B: use upper bound of A × upper bound of B
A ÷ B: use upper bound of A ÷ lower bound of B

For the minimum value of:

A + B: use lower bound of A + lower bound of B
A − B: use lower bound of A − upper bound of B
A × B: use lower bound of A × lower bound of B
A ÷ B: use lower bound of A ÷ upper bound of B

The key insight is that for division, you maximise the result by making the numerator as large as possible and the denominator as small as possible. For subtraction, you maximise the result by making the first term large and the second term small.

Why Does This Matter in Real Life?

Bounds are not just an exam topic — they have real-world significance. Engineers use bounds calculations when designing bridges, buildings, and vehicles. If a bridge must support a certain weight, the calculations use the worst-case scenario (upper bound of weight, lower bound of strength) to ensure safety.

In the context of speed, knowing the bounds tells you the range within which the actual speed must lie. If the speed limit is 80 km/h and the upper bound of the calculated speed is 83.1 km/h, you cannot be certain the driver was within the limit — even though the measured values suggest an average of 80 km/h.

A Second Example

A sprinter runs 100 m (measured to the nearest metre) in 11.6 seconds (measured to the nearest 0.1 seconds). Find the upper and lower bounds of the sprinter’s speed.

Distance bounds: 99.5 ≤ d < 100.5 m Time bounds: 11.55 ≤ t < 11.65 s

Upper bound of speed = 100.5 ÷ 11.55 = 8.70 m/s (3 s.f.)

Lower bound of speed = 99.5 ÷ 11.65 = 8.54 m/s (3 s.f.)

The actual speed lies between 8.54 m/s and 8.70 m/s.

Common Mistakes

Confusing which bounds to use for division: This is the number one mistake. Students often use upper ÷ upper for the maximum, which is wrong. Remember: to maximise a fraction, maximise the top and minimise the bottom.
Getting the degree of accuracy wrong: “To the nearest 10” means the bounds are ±5. “To the nearest 0.1” means the bounds are ±0.05. Half the unit of accuracy gives the error bound.
Including the upper bound: Strictly speaking, the upper bound is not included (we use <, not ≤). In practice, this rarely affects the calculation, but it is good to state the inequality correctly.
Not giving enough decimal places: The question may ask for the answer to a certain number of significant figures or decimal places. Follow these instructions precisely.
Applying bounds to an exact value: If a value is exact (for example, π or a defined constant), it has no bounds. Only measured or rounded values have bounds.

Practice Questions

A rectangle measures 15 cm by 8 cm, each measured to the nearest cm. Find the upper and lower bounds of the area.
A car uses 45 litres of fuel (nearest litre) to travel 520 km (nearest 10 km). Find the upper and lower bounds of the fuel consumption in km per litre.
A student’s test score is 72 out of 80 marks. If both values are exact, do bounds apply? Explain why or why not.
The circumference of a circle is 31.4 cm, measured to the nearest 0.1 cm. Find the upper and lower bounds of the diameter.

Exam Strategy

On the exam, draw a small table like this before calculating:

Quantity	Lower Bound	Upper Bound
Distance	235	245
Time	2.95	3.05

Then clearly state which combination you are using for each part. This earns method marks even if your arithmetic is slightly off.

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