Linked representations and multistep reasoning

(a) 12

1. The four numbers add up to $4 \times 10 = 40$.
2. $$40 - (7 + 9 + 12) = 40 - 28 = 12$$

• P1 The total, $4 \times 10 = 40$.
• P1 Subtracting the three known numbers from 40.
• A1 Correct answer: 12.

(a) Courier $A$: both means are 3.2 days, but $A$'s range (2) is smaller than $B$'s (5).

1. Means: $A$: $\frac{16}{5} = 3.2$. $B$: $\frac{16}{5} = 3.2$.
2. Ranges: $A$: $4 - 2 = 2$. $B$: $6 - 1 = 5$.
3. The averages match, so compare spread: $A$ is more consistent.

• P1 Both means (or medians) worked out.
• P1 Both ranges, 2 and 5.
• C1 Courier $A$, because the averages are equal and $A$ has the smaller range, so its times are more consistent.

(a) $\frac{181}{12}$

1. Recover the old total, then add the new score.
   $$11\times 15+16$$
2. Divide by the increased number of scores.
   $$181/12$$
3. Therefore $\frac{181}{12}$.

• P1 Recover the old total, then add the new score.
• P1 Divide by the increased number of scores.
• A1 Correct answer: $\frac{181}{12}$

(b) It increased because the new score exceeds the old mean.

1. It increased because the new score exceeds the old mean.

• C1 Correct conclusion with supporting reasoning: It increased because the new score exceeds the old mean.

(a) $24/49$

1. Count the red-blue route.
   $$6/14\times 8/14$$
2. Include the blue-red route, which has the same probability.
   $$2\times 6\times 8/14^{2}$$
3. Therefore $24/49$.

• P1 Count the red-blue route.
• P1 Include the blue-red route, which has the same probability.
• A1 Correct answer: $24/49$

(b) $\frac{2400}{49}$

1. Multiply the probability of one mixed pair by 100 trials.
   $$100\times 2\times 6\times 8/14^{2}$$
2. Therefore $\frac{2400}{49}$.

• P1 Multiply the probability of one mixed pair by 100 trials.
• A1 Correct answer: $\frac{2400}{49}$

(a) $60$

1. Divide the total by the 4 parts of the ratio.
   $$240/4$$
2. Therefore $60$.

• M1 Divide the total by the 4 parts of the ratio.
• A1 Correct answer: $60$

(b) $40$%

1. Find the takings from shrubs and from flowers.
   $$60\times 10$$
2. Divide the shrub takings by the total takings and multiply by 100.
   $$600/1500\times 100$$
3. Therefore $40$%.

• P1 Find the takings from shrubs and from flowers.
• P1 Divide the shrub takings by the total takings and multiply by 100.
• A1 Correct answer: $40$%

(a) $1.55$

1. Multiply each number of goals by its frequency and add.
   $$0\times 3+1\times 7+2\times 6+3\times 4$$
2. Divide by the number of matches.
   $$31/20$$
3. Therefore $1.55$.

• M1 Multiply each number of goals by its frequency and add.
• M1 Divide by the number of matches.
• A1 Correct answer: $1.55$

(b) $1.5$

1. The 10th and 11th values are 1 and 2.
   $$\frac{1+2}{2}$$
2. Therefore $1.5$.

• M1 The 10th and 11th values are 1 and 2.
• A1 Correct answer: $1.5$

(c) It increases, because 4 is greater than the current mean of 1.55.

1. It increases, because 4 is greater than the current mean of 1.55.

• C1 Correct conclusion with supporting reasoning: It increases, because 4 is greater than the current mean of 1.55.

(a) 13.6 years

1. Boys: $\frac{3}{5} \times 40 = 24$. Girls: $16$.
2. Total ages.
   $$24 \times 13 + 16 \times 14.5 = 312 + 232 = 544$$
3. $$544 \div 40 = 13.6$$

• P1 Splitting 40 in the ratio: 24 and 16.
• P1 Total ages for each group.
• P1 Adding the totals and dividing by 40.
• A1 Correct answer: 13.6.

(a) 16

1. Let there be $3k$ red and $5k$ blue.
   $$\frac{3k + 4}{8k + 4} = \frac{1}{2}$$
2. $$6k + 8 = 8k + 4 \;\Rightarrow\; k = 2$$
3. At the start: $6 + 10 = 16$ counters.

• P1 Writing red and blue as $3k$ and $5k$.
• P1 An equation for the new probability.
• A1 Correct answer: 16.

(b) $\frac{9}{38}$

1. The new bag has 10 red and 10 blue.
2. $$\tfrac{10}{20} \times \tfrac{9}{19} = \tfrac{90}{380} = \tfrac{9}{38}$$

• M1 $\frac{10}{20} \times \frac{9}{19}$.
• A1 $\frac{9}{38}$ or equivalent.