Rounding, significant figures and error intervals

(a) $3.48$

1. The second decimal digit is 7; the next digit, 6, is 5 or more, so round up: $3.48$.

• B1 The correct answer, $3.48$.

(b) $48\,000$

1. The first two significant figures are 4 and 8; the next digit is 3, so round down. Keep the place value with zeros: $48\,000$.

• B1 The correct answer, $48\,000$.

(c) $0.006$

1. The zeros after the point are not significant. The first significant figure is 6; the next digit is 0, so $0.006$.

• B1 The correct answer, $0.006$.

(a) $0.0078$

1. Leading zeros do not count as significant figures. The first two significant digits are 7 and 8.
2. The next digit is 4, so the 8 stays unchanged: 0.0078.

• B1 Correct answer: $0.0078$

(a) $13.4745$

1. Work out the top and the bottom.
   $$7.83^2 = 61.3089, \qquad 3.1 + 1.45 = 4.55$$
2. Divide, then round to 4 decimal places: the fifth decimal place is 8, so round up.
   $$61.3089 \div 4.55 = 13.474483\ldots = 13.4745$$

• M1 Finding $61.3089$ or $4.55$.
• A1 $13.4745$ (correct to 4 decimal places).

(b) $13.5$

1. The first three significant figures are 1, 3 and 4; the next digit is 7, so round up to $13.5$.

• B1 The correct answer, $13.5$.

(a) $6.5 \le l < 7.5$

1. Half of the rounding unit (1 cm) is $0.5$ cm. The length is at least $6.5$ and less than $7.5$, which would round to 8.

• B1 $6.5 \le l < 7.5$.

(b) $4.3 \le x < 4.4$

1. Truncating chops off the later digits, so every number from $4.3$ up to (but not including) $4.4$ truncates to $4.3$.

• B1 $4.3 \le x < 4.4$.

(c) $345 \le m < 355$

1. To 2 significant figures, $350$ is rounded to the nearest 10. Half of 10 is 5.
2. $$345 \le m < 355$$

• M1 Using a rounding unit of 10 (half-width 5): one correct limit, 345 or 355.
• A1 $345 \le m < 355$ with the correct signs.

(a) 5.8 $\le$ x < 5.9

1. Truncation removes digits without increasing the last retained digit.
2. The numbers begin at 5.8 and stop before 5.9: 5.8 $\le$ x < 5.9.

• B1 Correct answer: 5.8 $\le$ x < 5.9

(a) $4650$

1. $$4700-50$$
2. Two significant figures here means rounding to the nearest 100. Half of 100 is 50.
3. The lower limit is 4700 $-$ 50 = 4650, and this endpoint is included.
4. So the smallest integer number of visitors is 4650.

• P1 Establishing $4700-50$ or an equivalent valid method.
• A1 Correct answer: $4650$

(a) $3.695\le m<3.705$

1. Half of the rounding unit is 0.005 kg.
   $$0.01/2$$
2. Therefore $3.695\le m<3.705$.

• M1 Half of the rounding unit is 0.005 kg.
• A1 Correct answer: $3.695\le m<3.705$

(b) $29.64$ kg

1. Use the upper endpoint for each of eight bags.
   $$8\times 3.705$$
2. Therefore $29.64$ kg.

• M1 Use the upper endpoint for each of eight bags.
• A1 Correct answer: $29.64$ kg

(a) £402\,500

1. To 2 significant figures, $12\,000$ is rounded to the nearest $1000$. The number of people is a whole number from $11\,500$ upwards.
2. $$11\,500 \times 35 = 402\,500$$

• P1 Finding the least number of people, $11\,500$.
• P1 Multiplying their lower limit by 35.
• A1 The correct answer, £$402\,500$.

(b) $12\,500$ rounds to $13\,000$ to 2 significant figures, so the greatest whole number is $12\,499$.

1. $12\,500$ rounds up to $13\,000$ to 2 significant figures, so it would not be reported as $12\,000$.
2. People are whole numbers, so the greatest possible number is $12\,499$.

• C1 Saying that $12\,500$ would round to $13\,000$.
• C1 Giving $12\,499$ as the greatest possible number, because people are counted in whole numbers.