Table of Contents
C1.10 — Limits of Accuracy — Answers #
IGCSE Mathematics | Worked Solutions | 25 Questions
Section A — Recall
Questions 1–10
1.
What does the term lower bound mean for a rounded measurement?
Answer
The lower bound is the smallest value that would round to the given number.
Example: If a length is given as $7\text{ m}$ to the nearest metre, the lower bound is $6.5\text{ m}$ — this is the smallest value that rounds to $7$.
2.
What does the term upper bound mean for a rounded measurement?
Answer
The upper bound is the smallest value that would round to the next number up. The true value must be less than the upper bound — it is not included.
Example: For $7\text{ m}$ to the nearest metre, the upper bound is $7.5\text{ m}$. A true value of $7.5\text{ m}$ would round to $8\text{ m}$, not $7\text{ m}$, so it is excluded.
3.
State the rule for finding the lower bound and upper bound of a rounded value.
Answer
Find the degree of accuracy, then divide it by 2.
Lower bound = given value − (degree of accuracy ÷ 2)
Upper bound = given value + (degree of accuracy ÷ 2)
Upper bound = given value + (degree of accuracy ÷ 2)
4.
A measurement is given correct to the nearest metre. What is the degree of accuracy?
Answer
Degree of accuracy $= 1\text{ m}$
Remember: The degree of accuracy is the unit being rounded to — in this case, the nearest $1\text{ m}$.
5.
A number is given correct to 1 decimal place. What is the degree of accuracy?
Answer
Degree of accuracy $= 0.1$
Remember: 1 decimal place means rounded to the nearest $0.1$. So the degree of accuracy is $0.1$, and you add or subtract $0.1 \div 2 = 0.05$.
6.
A length is $12\text{ m}$, correct to the nearest metre. Write down the upper bound.
Answer
- Degree of accuracy $= 1\text{ m}$ → $1 \div 2 = 0.5\text{ m}$
- Upper bound $= 12 + 0.5 = 12.5\text{ m}$
Upper bound $= 12.5\text{ m}$
7.
A length is $12\text{ m}$, correct to the nearest metre. Write down the lower bound.
Answer
- Degree of accuracy $= 1\text{ m}$ → $1 \div 2 = 0.5\text{ m}$
- Lower bound $= 12 – 0.5 = 11.5\text{ m}$
Lower bound $= 11.5\text{ m}$
8.
A mass is $5.3\text{ kg}$, correct to 1 decimal place. Write down the lower bound.
Answer
- 1 decimal place means the nearest $0.1$, so degree of accuracy $= 0.1\text{ kg}$ → $0.1 \div 2 = 0.05\text{ kg}$
- Lower bound $= 5.3 – 0.05 = 5.25\text{ kg}$
Lower bound $= 5.25\text{ kg}$
9.
A number is $600$, correct to the nearest $100$. Write down the upper bound.
Answer
- Degree of accuracy $= 100$ → $100 \div 2 = 50$
- Upper bound $= 600 + 50 = 650$
Upper bound $= 650$
10.
A measurement has a lower bound $L$ and an upper bound $U$. Write the inequality that the true value $x$ must satisfy.
Answer
$$L \leq x < U$$
Key point: The lower bound uses $\leq$ because the true value can equal the lower bound — it would still round to the given number. The upper bound uses $<$ because the true value cannot equal the upper bound — a value at exactly $U$ would round to the next number up.
Section B — Application
Questions 11–20
11.
A length is $27\text{ cm}$, correct to the nearest centimetre. Find the upper and lower bounds.
Answer
- Degree of accuracy $= 1\text{ cm}$ → $1 \div 2 = 0.5\text{ cm}$
- Lower bound $= 27 – 0.5 = 26.5\text{ cm}$
- Upper bound $= 27 + 0.5 = 27.5\text{ cm}$
Lower bound $= 26.5\text{ cm}$, Upper bound $= 27.5\text{ cm}$
12.
A mass is $480\text{ g}$, correct to the nearest $10\text{ g}$. Find the upper and lower bounds.
Answer
- Degree of accuracy $= 10\text{ g}$ → $10 \div 2 = 5\text{ g}$
- Lower bound $= 480 – 5 = 475\text{ g}$
- Upper bound $= 480 + 5 = 485\text{ g}$
Lower bound $= 475\text{ g}$, Upper bound $= 485\text{ g}$
13.
A temperature is $36.8^\circ\text{C}$, correct to 1 decimal place. Find the upper and lower bounds.
Answer
- 1 decimal place means the nearest $0.1$, so degree of accuracy $= 0.1^\circ\text{C}$ → $0.1 \div 2 = 0.05^\circ\text{C}$
- Lower bound $= 36.8 – 0.05 = 36.75^\circ\text{C}$
- Upper bound $= 36.8 + 0.05 = 36.85^\circ\text{C}$
Lower bound $= 36.75^\circ\text{C}$, Upper bound $= 36.85^\circ\text{C}$
14.
A number is $7.25$, correct to 2 decimal places. Find the upper and lower bounds.
Answer
- 2 decimal places means the nearest $0.01$, so degree of accuracy $= 0.01$ → $0.01 \div 2 = 0.005$
- Lower bound $= 7.25 – 0.005 = 7.245$
- Upper bound $= 7.25 + 0.005 = 7.255$
Lower bound $= 7.245$, Upper bound $= 7.255$
15.
A time is $90\text{ s}$, correct to the nearest $10\text{ s}$. Find the upper and lower bounds. Then write the inequality for the true time $t$.
Answer
- Degree of accuracy $= 10\text{ s}$ → $10 \div 2 = 5\text{ s}$
- Lower bound $= 90 – 5 = 85\text{ s}$
- Upper bound $= 90 + 5 = 95\text{ s}$
- Write the inequality: $85 \leq t < 95$
Lower bound $= 85\text{ s}$, Upper bound $= 95\text{ s}$ → $85 \leq t < 95$
16.
A height is $2.0\text{ m}$, correct to 1 decimal place. Find the upper and lower bounds.
Answer
- 1 decimal place means the nearest $0.1$, so degree of accuracy $= 0.1\text{ m}$ → $0.1 \div 2 = 0.05\text{ m}$
- Lower bound $= 2.0 – 0.05 = 1.95\text{ m}$
- Upper bound $= 2.0 + 0.05 = 2.05\text{ m}$
Lower bound $= 1.95\text{ m}$, Upper bound $= 2.05\text{ m}$
Note: Notice that the lower bound ($1.95\text{ m}$) is below $2\text{ m}$. This is correct — a height of $1.95\text{ m}$ would still round to $2.0\text{ m}$ when given to 1 decimal place.
17.
A price is $\$8.40$, correct to the nearest $\$0.10$. Find the upper and lower bounds.
Answer
- Degree of accuracy $= \$0.10$ → $0.10 \div 2 = \$0.05$
- Lower bound $= \$8.40 – \$0.05 = \$8.35$
- Upper bound $= \$8.40 + \$0.05 = \$8.45$
Lower bound $= \$8.35$, Upper bound $= \$8.45$
18.
A distance is $5000\text{ m}$, correct to the nearest $1000\text{ m}$. Find the upper and lower bounds.
Answer
- Degree of accuracy $= 1000\text{ m}$ → $1000 \div 2 = 500\text{ m}$
- Lower bound $= 5000 – 500 = 4500\text{ m}$
- Upper bound $= 5000 + 500 = 5500\text{ m}$
Lower bound $= 4500\text{ m}$, Upper bound $= 5500\text{ m}$
19.
A mass $m$ is recorded as $63\text{ kg}$, correct to the nearest kilogram. Write the inequality for $m$.
Answer
- Degree of accuracy $= 1\text{ kg}$ → $1 \div 2 = 0.5\text{ kg}$
- Lower bound $= 63 – 0.5 = 62.5\text{ kg}$
- Upper bound $= 63 + 0.5 = 63.5\text{ kg}$
$62.5 \leq m < 63.5$
20.
A number is $50$, correct to the nearest $10$. A student says: “The upper bound is $60$.” Is the student correct? Show your working and explain any error.
Answer
- Degree of accuracy $= 10$ → $10 \div 2 = 5$
- Upper bound $= 50 + 5 = 55$
The student is incorrect. The correct upper bound is $55$, not $60$.
Error explained: The student added the full degree of accuracy ($10$) instead of half of it ($5$). The rule is always to add or subtract half the degree of accuracy. Adding $10$ to $50$ gives $60$, which is the next rounded value — not the upper bound.
Section C — Challenge
Questions 21–25
21.
(a) A length is given as $9.4\text{ cm}$, correct to 1 decimal place. Find the upper and lower bounds.
(b) The same length is now given as $9.40\text{ cm}$, correct to 2 decimal places. Find the upper and lower bounds.
(c) Even though $9.4 = 9.40$ as numbers, the bounds in (a) and (b) are different. Explain why.
Answer
(a) 1 decimal place
- Degree of accuracy $= 0.1\text{ cm}$ → $0.1 \div 2 = 0.05\text{ cm}$
- Lower bound $= 9.4 – 0.05 = 9.35\text{ cm}$
- Upper bound $= 9.4 + 0.05 = 9.45\text{ cm}$
Lower bound $= 9.35\text{ cm}$, Upper bound $= 9.45\text{ cm}$
(b) 2 decimal places
- Degree of accuracy $= 0.01\text{ cm}$ → $0.01 \div 2 = 0.005\text{ cm}$
- Lower bound $= 9.40 – 0.005 = 9.395\text{ cm}$
- Upper bound $= 9.40 + 0.005 = 9.405\text{ cm}$
Lower bound $= 9.395\text{ cm}$, Upper bound $= 9.405\text{ cm}$
(c) Why are the bounds different?
The value $9.4$ and $9.40$ are equal as numbers, but they tell us different things about how precisely the measurement was made.
- “$9.4\text{ cm}$ correct to 1 d.p.” means rounded to the nearest $0.1\text{ cm}$ — the true value could be up to $0.05\text{ cm}$ away.
- “$9.40\text{ cm}$ correct to 2 d.p.” means rounded to the nearest $0.01\text{ cm}$ — the true value could only be up to $0.005\text{ cm}$ away.
The degree of accuracy is different. The more precise the rounding, the smaller the amount added or subtracted, and the narrower (tighter) the bounds.
22.
The true value $x$ of a measurement satisfies $24.5 \leq x < 25.5$.
(a) Write down the lower bound. (b) Write down the upper bound. (c) Write down the rounded value. (d) What was the degree of accuracy? Explain how you know.
Answer
(a) Lower bound
Lower bound $= 24.5$
(b) Upper bound
Upper bound $= 25.5$
(c) Rounded value
Rounded value $= 25$
How: The rounded value sits exactly halfway between the lower and upper bound. $\frac{24.5 + 25.5}{2} = 25$.
(d) Degree of accuracy
- The amount added or subtracted $= 25.5 – 25 = 0.5$
- This is half the degree of accuracy, so degree of accuracy $= 0.5 \times 2 = 1$
Degree of accuracy $= 1$ (rounded to the nearest whole number)
23.
A distance is recorded as $7\text{ m}$, correct to the nearest metre.
(a) Find the upper and lower bounds. (b) Write the inequality for the true distance $d$. (c) Is $7.4\text{ m}$ possible? (d) Is $7.5\text{ m}$ possible?
Answer
(a) Upper and lower bounds
- Degree of accuracy $= 1\text{ m}$ → $1 \div 2 = 0.5\text{ m}$
- Lower bound $= 7 – 0.5 = 6.5\text{ m}$
- Upper bound $= 7 + 0.5 = 7.5\text{ m}$
Lower bound $= 6.5\text{ m}$, Upper bound $= 7.5\text{ m}$
(b) Inequality
$6.5 \leq d < 7.5$
(c) Could the true distance be $7.4\text{ m}$?
Check: is $7.4$ inside the range $6.5 \leq d < 7.5$?
$6.5 \leq 7.4 < 7.5$ → Yes, $7.4$ satisfies the inequality.
Yes, $7.4\text{ m}$ is possible. It lies within the bounds and rounds to $7\text{ m}$.
(d) Could the true distance be $7.5\text{ m}$?
Check: is $7.5$ inside the range $6.5 \leq d < 7.5$?
The inequality requires $d < 7.5$. The value $7.5$ is not less than $7.5$ — it is equal to it, so it fails the condition.
No, $7.5\text{ m}$ is not possible. A distance of exactly $7.5\text{ m}$ would round up to $8\text{ m}$, not $7\text{ m}$.
Key rule: The upper bound is never included. Always use a strict $<$ sign for the upper bound.
24.
The number $800$ has been rounded in two different ways: Version A (nearest $100$) and Version B (nearest $10$).
(a) Find bounds for Version A. (b) Find bounds for Version B. (c) Which version gives a narrower range? Explain.
Answer
(a) Version A — nearest 100
- Degree of accuracy $= 100$ → $100 \div 2 = 50$
- Lower bound $= 800 – 50 = 750$
- Upper bound $= 800 + 50 = 850$
Lower bound $= 750$, Upper bound $= 850$ (range of $100$)
(b) Version B — nearest 10
- Degree of accuracy $= 10$ → $10 \div 2 = 5$
- Lower bound $= 800 – 5 = 795$
- Upper bound $= 800 + 5 = 805$
Lower bound $= 795$, Upper bound $= 805$ (range of $10$)
(c) Which is narrower?
- Version A range: $850 – 750 = 100$
- Version B range: $805 – 795 = 10$
Version B gives a narrower range. Rounding to the nearest $10$ is more precise than rounding to the nearest $100$, so less information is lost. Less information lost means the true value is known more accurately, and the bounds are tighter.
25.
A bag of flour has a mass of $2.00\text{ kg}$, correct to 2 decimal places.
(a) Write down the degree of accuracy. (b) Find the upper and lower bounds. (c) Write the inequality for $m$. (d) Explain why a true mass of exactly $2.005\text{ kg}$ would not be recorded as $2.00\text{ kg}$.
Answer
(a) Degree of accuracy
Degree of accuracy $= 0.01\text{ kg}$
Remember: 2 decimal places means rounded to the nearest $0.01$.
(b) Upper and lower bounds
- Degree of accuracy $= 0.01\text{ kg}$ → $0.01 \div 2 = 0.005\text{ kg}$
- Lower bound $= 2.00 – 0.005 = 1.995\text{ kg}$
- Upper bound $= 2.00 + 0.005 = 2.005\text{ kg}$
Lower bound $= 1.995\text{ kg}$, Upper bound $= 2.005\text{ kg}$
(c) Inequality
$1.995 \leq m < 2.005$
(d) Why is $2.005\text{ kg}$ excluded?
By the rounding rule, if the digit after the last kept decimal place is 5 or more, we round up.
$2.005$ to 2 decimal places: the third decimal digit is $5$, so we round up → $2.01\text{ kg}$.
A true mass of $2.005\text{ kg}$ would be rounded to $2.01\text{ kg}$, not $2.00\text{ kg}$. So it falls outside the range. This is why the upper bound uses a strict $<$ sign — $2.005$ itself is not included.
General rule: The upper bound is always the value that would just tip over into the next rounded value. It marks the boundary but is never actually included in the range of possible true values.
