Sharing quantities and equivalent ratios

(a) £15

1. There are $3 + 5 = 8$ parts. One part is $40 \div 8 = £5$.
2. Kim gets $3 \times 5 = £15$ (and Leo gets £25).

• M1 Finding one part: $40 \div 8 = 5$.
• A1 The correct answer, £15.

(b) £3.00

1. One pen costs $1.80 \div 3 = £0.60$.
2. $5$ pens cost $5 \times 0.60 = £3.00$.

• M1 Finding the cost of one pen, 60p.
• A1 £3.00 (or 300p).

(a) £$27$

1. $$72/(5+3)$$
2. $$9\times 3$$
3. There are 5 + 3 = 8 equal parts.
4. Each part is £72 $\div$ 8 = £9; music gets 3 £$\times$9 = £27.

• P1 Establishing $72/(5+3)$ or an equivalent valid method.
• P1 Establishing $9\times 3$ or an equivalent valid method.
• A1 Correct answer: £$27$

(a) £120

1. $2 + 3 + 4 = 9$ parts; one part is $270 \div 9 = £30$.
2. The largest share is $4 \times 30 = £120$.

• M1 Finding one part: $270 \div 9 = 30$.
• A1 The correct answer, £120.

(b) $22.2\%$

1. The smallest share is $2 \times 30 = £60$.
2. $$\frac{60}{270} \times 100 = 22.22\ldots\% \approx 22.2\%$$

• M1 Writing $\frac{60}{270} \times 100$ or $\frac{2}{9} \times 100$.
• A1 The correct answer, $22.2\%$.

(a) $99$

1. $$27/(7-4)$$
2. $$9\times 11$$
3. The difference represents 7 $-$ 4 = 3 parts.
4. One part is 27 $\div$ 3 = 9 tiles.
5. The total is (7 + 4) $\times$ 9 = 99 tiles.

• P1 Establishing $27/(7-4)$ or an equivalent valid method.
• P1 Establishing $9\times 11$ or an equivalent valid method.
• A1 Correct answer: $99$

(a) £$63$

1. $$210\times 5/10$$
2. $$105\times 0.6$$
3. There are 10 parts, so the largest share is 5/10 £$\times$210 = £105.
4. 60% remains: £105 $\times$ 0.6 = £63.

• P1 Establishing $210\times 5/10$ or an equivalent valid method.
• P1 Establishing $105\times 0.6$ or an equivalent valid method.
• A1 Correct answer: £$63$

(a) $1250$ ml

1. If Tom uses all 500 ml of yellow, he needs $500 \div 2 \times 3 = 750$ ml of blue. He has 900 ml, so that works.
2. If he used all 900 ml of blue, he would need $900 \div 3 \times 2 = 600$ ml of yellow, but he only has 500 ml.
3. So yellow runs out first: $500 + 750 = 1250$ ml of green.

• P1 Finding the blue needed for all the yellow (750 ml), or the yellow needed for all the blue (600 ml).
• P1 Deciding that the yellow paint runs out first.
• A1 The correct answer, $1250$ ml.

(b) $150$ ml

1. $900 - 750 = 150$ ml.

• B1 The correct answer, $150$ ml.

(a) $34.5$

1. Use a common B value: A : B : C = 12 : 21 : 35, totalling 68 parts.
2. $$136/68$$
3. $$24+42/4$$
4. Use a common B value: A : B : C = 12 : 21 : 35, totalling 68 parts.
5. Each part is 136/68 = 2, so initially A = 24 and B = 42.
6. The transfer is 42/4 = 10.5, giving new A = 24 + 10.5 = 34.5.

• P1 Use a common B value: A : B : C = 12 : 21 : 35, totalling 68 parts.
• P1 Establishing $136/68$ or an equivalent valid method.
• P1 Establishing $24+42/4$ or an equivalent valid method.
• A1 Correct answer: $34.5$

(a) $180$

1. $$3(3k-12)=7k$$
2. $$2k=36$$
3. Let the original numbers be 3k and 7k.
4. The new ratio gives 3(3k $-$ 12) = 7k, so 2k = 36 and k = 18.
5. The original total was 10k = 180 counters.

• P1 Establishing $3(3k-12)=7k$ or an equivalent valid method.
• P1 Establishing $2k=36$ or an equivalent valid method.
• A1 Correct answer: $180$