Estimation and sense checks

(a) $3000$

1. Round each number to 1 significant figure.
   $$612 \approx 600, \qquad 4.87 \approx 5$$
2. $$600 \times 5 = 3000$$

• M1 Rounding to $600$ and $5$.
• A1 The correct answer, $3000$.

(b) £400

1. Round each number to 1 significant figure: $38 \approx 40$ and £11.95 £$\approx$10 (to 1 significant figure, 11.95 is 10, not 12).
2. $$40 \times 10 = 400$$

• M1 Rounding both numbers to 1 significant figure and multiplying: $40 \times 10$.
• A1 The correct answer, £400.

(a) Not sensible: $4 \times 20 = 80$, so the answer should be about 80.

1. Round to 1 significant figure.
   $$4.1 \times 19.8 \approx 4 \times 20 = 80$$
2. $811.8$ is about ten times too big, so the decimal point is in the wrong place. (The exact answer is $81.18$.)

• M1 Estimating $4 \times 20 = 80$.
• C1 Saying her answer is not sensible (about ten times too big), with the estimate of 80.

(a) 19.9

1. $$0.5\times 80/2$$
2. $$0.48\times 79.6/1.92$$
3. Since 1.92 = 4 $\times$ 0.48, the exact calculation simplifies to 79.6 $\div$ 4 = 19.9.
4. Check the magnitude: 0.5 $\times$ 80 $\div$ 2 = 20, which is close to 19.9.

• P1 Establishing $0.5\times 80/2$ or an equivalent valid method.
• P1 Establishing $0.48\times 79.6/1.92$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: 19.9

(a) $1000$

1. Round the operands before calculating.
   $$40\times 5/0.2$$
2. Dividing by one fifth is multiplication by five.
   $$40\times 25$$
3. Therefore $1000$.

• M1 Round the operands before calculating.
• M1 Dividing by one fifth is multiplication by five.
• A1 Correct answer: $1000$

(b) The estimate is 1000. The stated answer is far too small compared with the estimate, so a decimal-place error is likely.

1. The estimate is 1000. The stated answer is far too small compared with the estimate, so a decimal-place error is likely.

• C1 Correct conclusion with supporting reasoning: The estimate is 1000. The stated answer is far too small compared with the estimate, so a decimal-place error is likely.

(a) £$80$

1. Round both values and multiply.
   $$2\times 40$$
2. Therefore £$80$.

• M1 Round both values and multiply.
• A1 Correct answer: £$80$

(b) Greater: both numbers were rounded up, so their product is larger than the exact cost.

1. Greater: both numbers were rounded up, so their product is larger than the exact cost.

• C1 Correct conclusion with supporting reasoning: Greater: both numbers were rounded up, so their product is larger than the exact cost.

(a) $500$

1. Round each number to one significant figure.
   $$200\times 50/20$$
2. Therefore $500$.

• P1 Round each number to one significant figure.
• A1 Correct answer: $500$

(b) No. The estimate is about 500 boxes, well above 450.

1. Compare the number bought with the estimate.
   $$450<500$$
2. No. The estimate is about 500 boxes, well above 450.

• M1 Compare the number bought with the estimate.
• C1 Correct conclusion with supporting reasoning: No. The estimate is about 500 boxes, well above 450.

(a) $640$

1. Round every number to 1 significant figure.
   $$\frac{4 \times 80}{0.5}$$
2. $$4 \times 80 = 320, \qquad 320 \div 0.5 = 640$$

• M1 Rounding all three numbers: $4$, $80$ and $0.5$.
• M1 Dividing correctly by $0.5$ (doubling): $320 \div 0.5 = 640$.
• A1 The correct answer, $640$.

(b) Smaller: the numerator was rounded up only slightly, but the denominator was rounded up more, which makes the fraction smaller.

1. $3.94 \to 4$ and $81.2 \to 80$ nearly cancel: $4 \times 80 = 320$ against $3.94 \times 81.2 \approx 320$.
2. $0.487 \to 0.5$ makes the divisor bigger, and dividing by a bigger number gives a smaller answer. So the estimate is smaller than the exact value (about 657).

• C1 "Smaller", because the denominator was rounded up, which makes the quotient smaller (the numerator's rounding almost cancels out).

(a) $12$ tonnes

1. Round each number to 1 significant figure.
   $$385 \approx 400, \qquad 29.7 \approx 30$$
2. $$400 \times 30 = 12\,000 \text{ kg} = 12 \text{ tonnes}$$

• P1 Rounding to $400 \times 30$ (or $400 \times 29.7$ style rounding shown).
• P1 Converting $12\,000$ kg to $12$ tonnes.
• A1 The correct answer, $12$ tonnes.

(b) No: both numbers were rounded up, so the real mass is less than 12 tonnes and one trip is enough.

1. $385$ was rounded up to $400$ and $29.7$ up to $30$. Both roundings increase the product.
2. So the true mass is less than $12$ tonnes, and one trip is enough. (It is $11.4$ tonnes.)

• C1 Stating one trip is enough because both numbers were rounded up, so the estimate is an overestimate.