Upper/lower bounds in calculations

(a) $108.9375\ \text{cm}^2$

1. Upper bounds: $12.45$ and $8.75$.
2. $$12.45 \times 8.75 = 108.9375$$

• M1 Using both upper bounds, $12.45$ and $8.75$.
• A1 The correct answer, $108.9375$.

(b) $42$ cm

1. Lower bounds: $12.35$ and $8.65$.
2. $$2(12.35 + 8.65) = 2 \times 21 = 42$$

• M1 Using both lower bounds in the perimeter.
• A1 The correct answer, $42$ cm.

(a) No: the upper bound of 40 boxes is 102 kg, which is more than 100 kg.

1. Upper bound of one box: $2.55$ kg.
2. Upper bound of 40 boxes: $40 \times 2.55 = 102$ kg.
3. 102 kg is more than 100 kg, so the lift cannot definitely carry them safely.

• P1 Using the upper bound $2.55$ kg.
• P1 Finding the upper bound for 40 boxes, 102 kg.
• C1 "No", because 102 kg is more than 100 kg.

(a) $3.57$

1. Largest top over smallest bottom.
2. $$\frac{6.25}{1.75} = 3.5714\ldots \approx 3.57$$

• M1 Using $\frac{6.25}{1.75}$.
• A1 The correct answer, $3.57$.

(b) $4.5$

1. Largest $x$ minus smallest $y$: $6.25 - 1.75 = 4.5$.

• M1 Using $6.25 - 1.75$.
• A1 The correct answer, $4.5$.

(a) 41/9

1. The largest possible numerator approaches 20.5 and the smallest denominator is 4.5.
2. The upper bound of the quotient is 20.5/4.5 = 41/9.

• B1 Correct answer: 41/9

(a) $2.855$ g/cm³

1. $$35.6-0.05$$
2. $$12.4+0.05$$
3. $$35.55/12.45$$
4. The lower mass bound is 35.55 g and the upper volume bound is 12.45 cm³.
5. To minimise a positive quotient, use the lower numerator and upper denominator: 35.55 $\div$ 12.45 = 2.855421686… .
6. The lower bound is 2.855 g/cm³ to 3 decimal places.

• P1 Establishing $35.6-0.05$ or an equivalent valid method.
• P1 Establishing $12.4+0.05$ or an equivalent valid method.
• P1 Establishing $35.55/12.45$ or an equivalent valid method.
• A1 Correct answer: $2.855$ g/cm³

(a) $148$

1. $$(2.8-0.05)\times 1000$$
2. $$18+0.5$$
3. $$2750/18.5$$
4. The smallest possible bag mass is 2.75 kg = 2750 g. A packet mass is less than 18.5 g.
5. 2750 $\div$ 18.5 = 148.6486… . Rounding down gives 148 packets that are always possible.
6. A packet mass sufficiently close to 18.5 g would make 149 packets need more than 2750 g, so 149 cannot be guaranteed.

• P1 Establishing $(2.8-0.05)\times 1000$ or an equivalent valid method.
• P1 Establishing $18+0.5$ or an equivalent valid method.
• P1 Establishing $2750/18.5$ or an equivalent valid method.
• A1 Correct answer: $148$

(a) $7.55$ m/s

1. Lower bound of speed $= \frac{\text{lower bound of distance}}{\text{upper bound of time}}$.
2. $$\frac{245}{32.45} = 7.5500\ldots$$

• P1 Finding the bounds: 245 or 255 m, and $32.35$ or $32.45$ s.
• P1 Dividing the lower bound of distance by the upper bound of time.
• A1 The correct answer, $7.55$ m/s.

(b) 8 m/s: both bounds round to 8 to 1 significant figure, but they do not agree to 2 significant figures.

1. Lower bound $7.55\ldots$ and upper bound $7.88\ldots$. To 2 significant figures they give $7.6$ and $7.9$, which disagree.
2. To 1 significant figure both give 8, so the speed is 8 m/s to a suitable degree of accuracy.

• M1 Rounding both bounds to the same degree of accuracy and comparing.
• C1 8 m/s, because both bounds round to 8 to 1 significant figure (and they differ to 2 significant figures).

(a) 56.875

1. $$30.5/2$$
2. $$6.5\times (15.25-6.5)$$
3. Let the perimeter be P and width be w. The length is P/2 $-$ w, so area A = w(P/2 $-$ w). The bounds are P < 30.5 and 5.5 $\le$ w < 6.5.
4. For fixed positive w, increasing P increases A, so first use its upper bound 30.5. Then A < w(15.25 $-$ w).
5. Complete the square: w(15.25 $-$ w) = $7.625^{2}$ $-$ (7.625 $-$ w)$^{2}.$ Throughout 5.5 $\le$ w < 6.5, increasing w decreases the squared term, so increases the area.
6. Use the upper width bound: 6.5 $\times$ (15.25 $-$ 6.5) = 56.875 cm². Values can approach this area, so it is the upper bound.

• P1 Establishing $30.5/2$ or an equivalent valid method.
• P1 Establishing $6.5\times (15.25-6.5)$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: 56.875