Signed arithmetic and number lines

(a) $-1.45$

1. Convert the fraction to compare like forms: $-7$/5 = $-1.4.$
2. For negative numbers, the number furthest below zero is the smallest. The answer is $-1.45.$

• B1 Correct answer: $-1.45$

(a) $-1$

1. $$-8+13$$
2. Add first: $-8$ + 13 = 5.
3. Then subtract 6: 5 $-$ 6 = $-1.$

• M1 Establishing $-8+13$ or an equivalent valid method.
• A1 Correct answer: $-1$

(a) $-\frac{2}{3}$

1. $\frac{2}{3} \approx 0.667$, so compare $-0.667$ with $-0.7$. The number closer to zero is larger, so $-\frac{2}{3}$ is larger.

• B1 Choosing $-\frac{2}{3}$.

(b) $-1.8$

1. Subtracting a negative is adding.
   $$-4.2 - (-1.8) = -4.2 + 1.8 = -2.4$$
2. $$-2.4 + 0.6 = -1.8$$

• M1 Rewriting $-(-1.8)$ as $+1.8$, reaching $-2.4$.
• A1 The correct answer, $-1.8$.

(c) $14$

1. $3.5 \times 4 = 14$, and a negative times a negative is positive.

• B1 The correct answer, $14$.

(a) $-30$

1. A negative answer needs one negative card and one positive card. To make it as far below zero as possible, use the largest sizes: $-6 \times 5$.
2. $$-6 \times 5 = -30$$

• M1 Choosing one negative and one positive card with the largest sizes, $-6 \times 5$.
• A1 The correct answer, $-30$.

(b) $5.5$

1. Test each card as the divisor. Dividing by a positive card needs the largest positive numerator; dividing by a negative card needs the most negative numerator.
2. Dividing by $3$: the best numerator is $5 - (-6) = 11$.
   $$\frac{11}{3} \approx 3.67$$
3. Dividing by $-2$: the most negative numerator is $-6 - 5 = -11$.
   $$\frac{-11}{-2} = 5.5$$
4. Dividing by $5$ gives at most $\frac{9}{5}$ and by $-6$ at most $\frac{7}{6}$, so the largest answer is $5.5$.

• P1 Finding a large candidate with a positive divisor, such as $\frac{5 - (-6)}{3} = \frac{11}{3}$.
• P1 Using a negative divisor with a negative numerator, such as $\frac{-6 - 5}{-2}$.
• A1 $5.5$ (or $\frac{11}{2}$) with the comparison that shows it is the largest.

(a) $-8$ °C

1. Add the rise to the negative starting temperature.
   $$-9+5$$
2. Subtract the fall from the new temperature.
   $$-4-4$$
3. Therefore $-8$ °C.

• P1 Add the rise to the negative starting temperature.
• P1 Subtract the fall from the new temperature.
• A1 Correct answer: $-8$ °C

(b) $13$ °C

1. Find the distance on the number line from the final temperature to 5.
   $$5-(-8)$$
2. Therefore $13$ °C.

• M1 Find the distance on the number line from the final temperature to 5.
• A1 Correct answer: $13$ °C

(a) £$-22$

1. Add the incoming payment to the balance.
   $$-45+27$$
2. Deduct the fee.
   $$-18-4$$
3. Therefore £$-22$.

• P1 Add the incoming payment to the balance.
• P1 Deduct the fee.
• A1 Correct answer: £$-22$

(b) £$32$

1. Calculate the difference between the target and the negative balance.
   $$10-(-22)$$
2. Therefore £$32$.

• P1 Calculate the difference between the target and the negative balance.
• A1 Correct answer: £$32$

(a) $-2$

1. Combine the two moves with their signs.
   $$3-7+2$$
2. Therefore $-2$.

• M1 Combine the two moves with their signs.
• A1 Correct answer: $-2$

(b) $13$

1. Subtract the lower floor number from the higher one.
   $$9-(-4)$$
2. Therefore $13$.

• P1 Subtract the lower floor number from the higher one.
• A1 Correct answer: $13$

(a) $-3\,^{\circ}\text{C}$

1. The six temperatures add up to six times the mean.
   $$6 \times (-1.5) = -9$$
2. Add the five known temperatures.
   $$-6.5 + 2 - 4 + 3.5 - 1 = -6$$
3. The sixth is the difference.
   $$-9 - (-6) = -3$$

• M1 Finding the total of all six, $-9$.
• M1 Finding the sum of the five known values, $-6$.
• A1 The correct answer, $-3$.

(b) $-5$ and $2$

1. The range of the six is from $-6.5$ to $3.5$, which is $10$. For the range to stay the same, the new value $t$ must satisfy
   $$-6.5 \le t \le 3.5$$
2. The total of all seven is $-9 + t$. For a whole-number mean, $-9 + t$ must be a multiple of $7$.
   $$-9 + t \in \{\dots, -21, -14, -7, 0, \dots\} \Rightarrow t \in \{\dots, -12, -5, 2, 9, \dots\}$$
3. Only $-5$ and $2$ lie between $-6.5$ and $3.5$.

• P1 Finding the range of the six, $10$, and so the limits $-6.5 \le t \le 3.5$.
• P1 Writing the total of seven as $-9 + t$ and requiring it to be a multiple of 7.
• A1 Both $-5$ and $2$, and no other values.