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Approximation and Errors: Paper 1 Challenge

This is the Challenge tier: ten harder Paper 1 questions that state a condition and leave the method to you. Most of them turn on choosing the right limit of each measurement, and on whether that limit can actually be reached. Work the standard Paper 1 set first. Detailed solutions can be requested by email.

10 questions · 49 marks · about 55 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · New technique — two rounding conditions on the same number, to be satisfied together

When a number $x$ is rounded off to 3 significant figures, the result is $2.50$ . When $x$ is rounded down to 2 decimal places, the result is $2.49$ . Find the range of values of $x$ .

(4 marks)

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$2.495 \leq x < 2.5$

Each condition gives its own range. The answer is the part the two ranges share.

Question 2 · New technique — rounding in two stages is not the same as rounding once

A number $x$ is rounded off to 3 decimal places and the result is $4.245$ . Mary then rounds off $4.245$ to 2 decimal places and claims that $x = 4.25$ , correct to 2 decimal places.

(a) Find the range of values of $x$ .

(b) Must Mary's claim be true? Explain your answer.

(c) Find the range of values of $x$ for which Mary's claim is true.

(5 marks)

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(a) $4.2445 \leq x < 4.2455$ (b) No. For example, $x = 4.2447$ gives $4.24$ . (c) $4.245 \leq x < 4.2455$

The first rounding may already have pushed x up to 4.245. Rounding again pushes it up a second time.

Question 3 · Extends Paper 1 Question 8 — the percentage error gives the reading, and then the same object is measured again on a coarser scale

The weight of a letter is measured as $w g$ correct to the nearest $0.1 g$ . The percentage error of the measurement is $0.4%$ .

(a) Find $w$ .

(b) The letter is weighed again, this time correct to the nearest g. Someone claims that the new reading must be $13 g$ . Do you agree? Explain your answer.

(5 marks)

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(a) $w = 12.5$ (b) Disagreed. The actual weight lies from $12.45 g$ up to $12.55 g$ . A letter of $12.47 g$ reads $12 g$ .

The trap is rounding the reading 12.5 again. The actual weight can be just under 12.5 g.

Question 4 · Modelled on 2012 P2 Q14 — the division comes out exact, so the obvious answer is one too many, and the least value is asked as well

The length of a roll of ribbon is measured as $34 m$ correct to the nearest m. The whole roll is cut into $n$ pieces such that the length of each piece is measured as $8 cm$ correct to the nearest cm.

(a) Find the greatest possible value of $n$ .

(b) Find the least possible value of $n$ .

(5 marks)

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(a) $459$ (b) $395$

3450 ÷ 7.5 is exactly 460, but the roll is shorter than 3450 cm, so 460 pieces cannot be cut.

Question 5 · Modelled on 2023 P1 Q3 — run backwards: the reading of the total is known and the number of items is not

A coin is termed standard if its weight is measured as $7.1 g$ correct to the nearest $0.1 g$ . A bag contains $n$ standard coins. The total weight of the coins is measured as $1.00 kg$ correct to the nearest $0.01 kg$ . Find all possible values of $n$ .

(5 marks)

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$n = 140, 141$ or $142$

1000 ÷ 7.1 = 140.8… suggests only 140 or 141. The limits allow 142 as well.

Question 6 · New technique — subtracting two nearly equal measurements leaves a small answer carrying the whole error

The lengths of two rods are measured as $50.2 cm$ and $49.8 cm$ , each correct to the nearest $0.1 cm$ . Let $d cm$ be the actual difference of the lengths of the two rods.

(a) Find the range of values of $d$ .

(b) The percentage error of each measurement is about $0.1%$ . Someone claims that the percentage error of the calculated difference, $0.4 cm$ , is also about $0.1%$ . Do you agree? Explain your answer.

(5 marks)

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(a) $0.3 < d < 0.5$ (b) Disagreed. The calculated difference can be out by almost $0.1 cm$ , which is $25%$ of $0.4 cm$ .

The errors of the two rods add, while the quantity they are compared with shrinks from about 50 cm to 0.4 cm.

Question 7 · Modelled on 2014 P2 Q11 — the trap is adding the two separate errors, which gives a clean 17.5 and misses a small product term

The length and the width of a rectangular garden are measured as $20 m$ and $15 m$ respectively, correct to the nearest m. Let $A m^{2}$ be the actual area of the garden.

(a) Find the range of values of $A$ .

(b) Someone claims that the actual area and the calculated area, $300 m^{2}$ , differ by at most $17.5 m^{2}$ . Do you agree? Explain your answer.

(5 marks)

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(a) $282.75 \leq A < 317.75$ (b) Disagreed. A garden of $20.498 m$ by $15.498 m$ has area $317.678 \dots m^{2}$ , which is more than $17.5 m^{2}$ above $300 m^{2}$ .

17.5 comes from 0.5 × 20 + 0.5 × 15. The true upper gap is 17.75, because of the extra 0.5 × 0.5.

Question 8 · Modelled on 2018 P2 Q14 — the area is found by subtraction, so a lower limit has to be paired with an upper limit

In the figure, a square hole is cut from a square card. The sides of the card and of the hole are measured as $10 cm$ and $6 cm$ respectively, correct to the nearest cm. Let $A cm^{2}$ be the actual area of the shaded region.

(a) Find the range of values of $A$ .

(b) A student claims that $60 \leq A < 68$ . Do you agree? Explain your answer.

A square card with a smaller square hole cut from its centre. The side of the card is marked 10 cm and the side of the hole is marked 6 cm. The remaining card is shaded.

(5 marks)

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(a) $48 < A < 80$ (b) Disagreed. A card of side $9.6 cm$ with a hole of side $6.4 cm$ leaves $51.2 cm^{2}$ .

60 and 68 come from using both lower limits, then both upper limits. For a difference the limits must be crossed.

Question 9 · New technique — a quotient of two measurements, with a claim that is settled by whether a limit can be reached

A cyclist rides a distance which is measured as $140 m$ correct to the nearest $10 m$ . The time taken is measured as $13 s$ correct to the nearest s. Let $v m/s$ be the actual average speed of the cyclist.

(a) Find the range of values of $v$ .

(b) Someone claims that the average speed of the cyclist must be higher than $36 km/h$ . Do you agree? Explain your answer.

(6 marks)

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(a) $10 < v < 11.6$ (b) Agreed. $36 km/h = 10 m/s$ , and $v$ is always greater than $10$ .

The slowest case needs a time of exactly 13.5 s, which the reading 13 s rules out. That is why v = 10 is never reached.

Question 10 · New technique — the direction of rounding decides what an estimate is able to prove

Amy has $$500$ . She wants to buy 23 notebooks at $$18.7$ each and 4 pens at $$16.4$ each.

(a) By rounding up each price to the nearest dollar, estimate the total cost.

(b) Using the result of (a), Ben claims that Amy does not have enough money. Is the claim correct? Explain your answer.

(4 marks)

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(a) $$505$ (b) Not correct. $$505$ is an overestimate, so it cannot show that the cost is above $$500$ . The exact cost is $$495.7$ .

An overestimate below the budget proves there is enough. An overestimate above the budget proves nothing.

Continue practising this topic: Go back to the Paper 1 long questions for the standard set, or move to Paper 2 multiple-choice — also available as Paper 2 Challenge MC.

Where this sits in the syllabus

Approximation and Errors is junior secondary content that the HKDSE keeps examining. From 2012 to 2026, Paper 2 carried a rounding question in most years and a range-of-values question in many, and Paper 1 Section A(1) has several times set a three-part rounding question or a claim about a total weight or capacity. The marks are reachable. They are usually lost by using the measured value where a limit is needed, or by pairing the wrong two limits.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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