Foundation numerical fluency clinic

(a) $4\frac{7}{12}$

1. Add the whole numbers, then the fractions over a common denominator of 12.
   $$2 + 1 = 3, \qquad \frac{3}{4} + \frac{5}{6} = \frac{9}{12} + \frac{10}{12} = \frac{19}{12} = 1\tfrac{7}{12}$$
2. $3 + 1\frac{7}{12} = 4\frac{7}{12}$.

• M1 Writing the fractions over a common denominator, $\frac{9}{12} + \frac{10}{12}$ (or both as improper fractions $\frac{33}{12} + \frac{22}{12}$).
• M1 Reaching $\frac{19}{12}$ or $\frac{55}{12}$.
• A1 $4\frac{7}{12}$.

(a) $391$

1. Divide the distances.
   $$\frac{1.5 \times 10^8}{3.84 \times 10^5} = 390.625$$
2. To 3 significant figures: $391$.

• M1 Dividing the Sun's distance by the Moon's.
• A1 The correct answer, $391$.

(b) $384\,000$

1. Move the digits five places: $384\,000$.

• B1 The correct answer, $384\,000$.

(a) £$16.45$

1. Multiply the unit price by the number bought.
   $$13\times 67/20$$
2. Subtract the total cost from the payment.
   $$60-871/20$$
3. Therefore £$16.45$.

• P1 Multiply the unit price by the number bought.
• P1 Subtract the total cost from the payment.
• A1 Correct answer: £$16.45$

(b) The cost is close to £44, so the change should be close to £16.

1. The cost is close to £44, so the change should be close to £16.

• C1 Correct conclusion with supporting reasoning: The cost is close to £44, so the change should be close to £16.

(a) £$-58$

1. Add the incoming payment to the balance.
   $$-135+81$$
2. Deduct the fee.
   $$-54-4$$
3. Therefore £$-58$.

• P1 Add the incoming payment to the balance.
• P1 Deduct the fee.
• A1 Correct answer: £$-58$

(b) £$68$

1. Calculate the difference between the target and the negative balance.
   $$10-(-58)$$
2. Therefore £$68$.

• P1 Calculate the difference between the target and the negative balance.
• A1 Correct answer: £$68$

(a) $\frac{66}{13}$

1. Reverse the final division first.
   $$13x-6=60$$
2. Add 6 before dividing by the coefficient.
   $$x=66/13$$
3. Therefore $\frac{66}{13}$.

• M1 Reverse the final division first.
• M1 Add 6 before dividing by the coefficient.
• A1 Correct answer: $\frac{66}{13}$

(b) The outermost operation is division by 3, so it must be undone before the earlier subtraction.

1. The outermost operation is division by 3, so it must be undone before the earlier subtraction.

• C1 Correct conclusion with supporting reasoning: The outermost operation is division by 3, so it must be undone before the earlier subtraction.

(a) $9.29\times10^{11}$ bytes/s

1. Divide distance by time, dividing coefficients and subtracting indices.
   $$\frac{13\times 10^{14}}{14\times 10^{2}}$$
2. Therefore $9.29\times10^{11}$ bytes/s.

• P1 Divide distance by time, dividing coefficients and subtracting indices.
• A1 Correct answer: $9.29\times10^{11}$ bytes/s

(b) A coefficient from 1 up to but not including 10 gives one consistent power of ten for the size of a positive number.

1. A coefficient from 1 up to but not including 10 gives one consistent power of ten for the size of a positive number.

• C1 Correct conclusion with supporting reasoning: A coefficient from 1 up to but not including 10 gives one consistent power of ten for the size of a positive number.

(a) $84$

1. Find the quantity left after the first sale.
   $$336\times 3/4$$
2. One third of that remainder survives the second sale.
   $$252/3$$
3. Therefore $84$.

• P1 Find the quantity left after the first sale.
• P1 One third of that remainder survives the second sale.
• A1 Correct answer: $84$

(b) $\frac{3}{4}$

1. Compare the total sold with the original quantity.
   $$\frac{336-84}{336}$$
2. Therefore $\frac{3}{4}$.

• P1 Compare the total sold with the original quantity.
• A1 Correct answer: $\frac{3}{4}$

(a) 7 loaves, with $\frac{1}{4}$ kg left over

1. Divide by $\frac{3}{4}$.
   $$5\tfrac{1}{2} \div \frac{3}{4} = \frac{11}{2} \times \frac{4}{3} = \frac{44}{6} = 7\tfrac{1}{3}$$
2. So 7 whole loaves.
3. Flour used and left.
   $$7 \times \frac{3}{4} = 5\tfrac{1}{4}, \qquad 5\tfrac{1}{2} - 5\tfrac{1}{4} = \frac{1}{4} \text{ kg}$$

• P1 Dividing $5\frac{1}{2}$ by $\frac{3}{4}$ (or repeatedly subtracting $\frac{3}{4}$).
• P1 Deciding 7 whole loaves can be made.
• A1 $\frac{1}{4}$ kg left over (with 7 loaves).