Error intervals and bounds

(a) $1 \div 0.8 = 1.25$

(b) 2 s.f. here means to the nearest 100, so half of 100 is 50 either side.

$4700 - 50 = 4650$ and $4700 + 50 = 4750$

$4650 \le x < 4750$

Rounded to the nearest metre, so go half a metre either side.

Lower bound: $90 - 0.5 = 89.5$

Upper bound: $90 + 0.5 = 90.5$

$89.5 \le \text{length} < 90.5$

Half of 0.1 is 0.05.

Lower bound $9.4 - 0.05 = 9.35$, upper bound $9.4 + 0.05 = 9.45$

$9.35 \le m < 9.45$

Nearest millimetre means half a millimetre either side.

Lower bound: $128 - 0.5 = 127.5$

Upper bound: $128 + 0.5 = 128.5$

$127.5 \le \text{length} < 128.5$

To the nearest cm, the true length is within half a centimetre either side of 13.

$13 - 0.5 = 12.5$ and $13 + 0.5 = 13.5$

So $12.5 \le L < 13.5$.

(a) Numbers that round to 7.3 (2 s.f.) are from 7.25 up to (not including) 7.35, so the least value is $7.25$.

(b) No. Values such as 7.3495 are bigger than 7.349 but still round to 7.3; $N$ can be any value below 7.35.

The calculator display began 10.2, so the number was cut off (truncated), not rounded.

It could be anything from 10.2 up to but not including 10.3:

$10.2 \le t < 10.3$

Rounding to 1 d.p. means the value is within $0.05$ of $12.7$.

Lower bound: $12.7 - 0.05 = 12.65$

Upper bound: $12.7 + 0.05 = 12.75$

$12.65 \le d < 12.75$

Rounding to 2 significant figures here means to the nearest 0.1, so half of that is 0.05.

Lower bound: $1.9 - 0.05 = 1.85$

Upper bound: $1.9 + 0.05 = 1.95$

$1.85 \le x < 1.95$

To 1 decimal place, the length could be up to half of 0.1 either side.

$7.2 - 0.05 = 7.15$ and $7.2 + 0.05 = 7.25$

$7.15 \le p < 7.25$

Rounding to 1 decimal place means $x$ is within $0.05$ of $6.4$.

Lower bound $= 6.4 - 0.05 = 6.35$, upper bound $= 6.4 + 0.05 = 6.45$

$6.35 \le x < 6.45$

The display began 8.3, so the digits after 8.3 have been cut off (truncated), not rounded.

The smallest value is 8.3 and $y$ can get as close as you like to 8.4 without reaching it.

$8.3 \le y < 8.4$

Martin wrote down the first two digits, so he truncated rather than rounded.

Any number from 1.2 up to (but not including) 1.3 would show as 1.2

$1.2 \le P < 1.3$

Truncating chops off the digits after the first one, so N can be 7 itself or anything up to (but not including) 8.

$7 \le N < 8$

$P$ lower bound $= 5.875 \times 10^8$

$Q$ upper bound $= 3.65 \times 10^5$

Lower bound of $R = \dfrac{5.875 \times 10^8}{3.65 \times 10^5} = 1609.6 \approx 1610$

Bounds: $6.75 \le e < 6.85$ and $0.045 \le f < 0.055$

For the largest $p$, make the top big and the bottom small:

$p = \sqrt{\frac{2 \times 6.85}{0.045}} = \sqrt{304.44\ldots} = 17.448\ldots \approx 17.4$

$x$: upper bound 99.75

$y = 67$ to 2 s.f.: lower bound 66.5

Upper bound of $D = \frac{99.75}{66.5} = 1.5$

Bounds: $w$: 432.5 to 437.5; $a$: 9.75 to 9.85; $c$: 2.45 to 2.55

Lower bound of $T = \frac{432.5}{9.85 - 2.45} = 58.44…$

Upper bound of $T = \frac{437.5}{9.75 - 2.55} = 60.76…$

Both round to 60 (1 sf), so $T = 60$.

Bounds: $AB$ from 9.25 to 9.35; $AC$ from 12.55 to 12.65.

$\sin x = \frac{AB}{AC}$ is smallest with the smallest $AB$ and largest $AC$.

$\sin x = \frac{9.25}{12.65} = 0.7312...$

$x = 46.989...°$

To find the least possible rate, use the smallest takings, the largest costs and the longest time.

Takings LB $= 872.5$, costs UB $= 192.5$, time UB $= 72.5$

LB of rate $= \dfrac{872.5 - 192.5}{72.5} = \dfrac{680}{72.5} = 9.379...$

Even the lowest possible rate, £9.379..., is more than £8.20, so her hourly rate was definitely more than £8.20.

Rearranging: $g = \frac{4\pi^2 l}{T^2}$.

Bounds: $51.95 \le l < 52.05$ and $1.445 \le T < 1.455$.

Upper bound: $g = \frac{4 \times 3.142^2 \times 52.05}{1.445^2} = 984.4$ (to 1 d.p.).

Lower bound: $g = \frac{4 \times 3.142^2 \times 51.95}{1.455^2} = 969.0$ (to 1 d.p.).

Lower bound of $AH = 11.25$ cm

For a cube of edge $a$, the space diagonal satisfies $a^2 + a^2 + a^2 = AH^2$

$3a^2 = 11.25^2$, so $a = \frac{11.25}{\sqrt{3}} \approx 6.495$ cm

Bounds of $c$: $10.85 \le c < 10.95$

Lower bound of $d = \frac{1}{8} \times 10.85^3 = 159.66...$

Upper bound of $d = \frac{1}{8} \times 10.95^3 = 164.11...$

Both round to 160 to 2 significant figures, so $d = 160$.

(a) $u$: 26.15 to 26.25. $a$: 4.25 to 4.35.

Largest $D$: biggest top, smallest bottom.

$D_{max} = \frac{26.25^2}{2 \times 4.25} = \frac{689.0625}{8.5} = 81.0662$

(b) Lower bound 78.6003, upper bound 81.0662. They differ at 1 s.f. in the units but both round to 80 to the nearest 10, so $D = 80$.

Lengths are to the nearest mm, mass to the nearest 5 g.

Lower bound: $\frac{1967.5}{13.25 \times 16.05 \times 21.75} = 0.4254$

Upper bound: $\frac{1972.5}{13.15 \times 15.95 \times 21.65} = 0.4344$

Both round to $0.43$ (2 d.p.), so the density is $0.43\text{ g/cm}^3$.

Time: 3 hours $= 180$ minutes, bounds $179.5$ and $180.5$.

Distance bounds: $486.5$ and $487.5$ km.

Upper bound of speed $= \frac{487.5}{179.5} = 2.7158\ldots$

Lower bound of speed $= \frac{486.5}{180.5} = 2.6953\ldots$

Both round to $2.7$ (1 d.p.) so the speed is $2.7$ km/minute.

Distance: $10.55 \le d < 10.65$ km

Time: $1907.5 \le t < 1908.5$ s

Upper bound $= 10.65 \div \frac{1907.5}{3600} = 20.099...$ km/h

Lower bound $= 10.55 \div \frac{1908.5}{3600} = 19.900...$ km/h

Both round to 20 (2 s.f.), so $V = 20$ km/h.