Place value, ordering and decimal operations

(a) $0.12$

1. $3 \times 4 = 12$. There are two decimal places in the question, so there are two in the answer: $0.12$.

• B1 The correct answer, $0.12$.

(b) $8$

1. Multiply both numbers by 10: $48 \div 6 = 8$.

• M1 Scaling to $48 \div 6$.
• A1 The correct answer, $8$.

(c) $0.47$

1. Compare with the same number of decimal places: $0.407$, $0.470$, $0.400$. The largest is $0.470$.

• B1 The correct answer, $0.47$.

(a) £$12.05$

1. $$3\times 2.65$$
2. The notebooks cost 3 £$\times$2.65 = £7.95.
3. Change = amount paid $-$ cost = £20 £$-$7.95 = £12.05.

• P1 Establishing $3\times 2.65$ or an equivalent valid method.
• A1 Correct answer: £$12.05$

(a) $32.4$

1. $1.35$ is $135 \div 100$, so the answer is $3240 \div 100 = 32.4$.

• B1 The correct answer, $32.4$.

(b) $3.24$

1. $13.5 = 135 \div 10$ and $0.24 = 24 \div 100$, so divide $3240$ by $1000$: $3.24$.

• B1 The correct answer, $3.24$.

(c) $1350$

1. $3240 \div 24 = 135$, and dividing by $2.4$ (ten times smaller than 24) gives an answer ten times bigger: $1350$.

• M1 A valid method: $3240 \div 24 = 135$ then $\times 10$, or scaling both numbers to $32\,400 \div 24$, or dividing $3240$ by $2.4$ directly (a long division with one arithmetic slip still shows the method).
• A1 The correct answer, $1350$.

(a) $36$

1. $$72\times 5$$
2. Write the factors in standard form: 7.2 $\times$ $10^{-5}$ and 5 $\times$ $10^{5}.$
3. Multiply: 7.2 $\times$ 5 $\times$ $10^{0}$ = 36.

• M1 Establishing $72\times 5$ or an equivalent valid method.
• A1 Correct answer: $36$

(a) $15$

1. $$360/24$$
2. Multiply both numbers by 100 without changing the quotient: 3.6 $\div$ 0.24 = 360 $\div$ 24.
3. Since 24 $\times$ 15 = 360, the answer is 15.

• M1 Establishing $360/24$ or an equivalent valid method.
• A1 Correct answer: $15$

(a) £$12.85$

1. Multiply the unit price by the number bought.
   $$7\times 49/20$$
2. Subtract the total cost from the payment.
   $$30-343/20$$
3. Therefore £$12.85$.

• P1 Multiply the unit price by the number bought.
• P1 Subtract the total cost from the payment.
• A1 Correct answer: £$12.85$

(b) The cost is close to £17, so the change should be close to £13.

1. The cost is close to £17, so the change should be close to £13.

• C1 Correct conclusion with supporting reasoning: The cost is close to £17, so the change should be close to £13.

(a) $15$

1. $$12.6-18\times 0.45$$
2. $$4.2/0.28$$
3. The first cuts use 18 $\times$ 0.45 = 8.10 m, leaving 12.6 $-$ 8.1 = 4.5 m.
4. Reserve the required 0.30 m: at most 4.5 $-$ 0.3 = 4.2 m is available.
5. 4.2 $\div$ 0.28 = 15, so 15 more pieces can be cut.

• P1 Establishing $12.6-18\times 0.45$ or an equivalent valid method.
• P1 Establishing $4.2/0.28$ or an equivalent valid method.
• A1 Correct answer: $15$

(a) £247.60

1. Number of panels:
   $$14.4 \div 1.8 = 144 \div 18 = 8$$
2. Posts: one more than the number of panels.
   $$8 + 1 = 9$$
3. Costs:
   $$8 \times 23.75 = 190, \qquad 9 \times 6.40 = 57.60$$
4. Total:
   $$190 + 57.60 = 247.60$$

• P1 Finding 8 panels ($14.4 \div 1.8$).
• P1 Using 9 posts (one more than the panels).
• P1 Working out the cost of the panels (£190) or the posts (£57.60).
• A1 The correct answer, £247.60.