Upper and Lower Bounds

The rounding interval extends half of the rounding unit on either side of the stated length.

1. A length correct to the nearest 0.1 cm can be up to half of 0.1 cm, which is 0.05 cm, either side of the recorded value.
2. So the lower bound is 6.4-0.05=6.35 and the upper bound is 6.4+0.05=6.45.
3. A length of exactly 6.35 cm rounds up to 6.4 cm, so it is allowed: use \le.
4. A length of exactly 6.45 cm would round up to 6.5 cm, so it is not allowed: use <. The error interval is 6.35 \le l < 6.45.
5. The option 6.35 < l \le 6.45 has the signs the wrong way round. 6.3 \le l < 6.5 goes a whole 0.1 cm either side of 6.4 instead of 0.05 cm, and 6.4 \le l < 6.5 is not centred on 6.4.

Use half of the rounding unit to find the lower endpoint.

1. Half of 1 kg is 0.5 kg.
2. The lower bound is 49-0.5=48.5 kg.

Identify the place value of the second significant figure.

1. The second significant figure is in the thousandths column, so the length has been rounded to the nearest 0.001 m.
2. Half of 0.001 is 0.0005, so the upper bound is 0.067+0.0005=0.0675 m.

The last significant figure shows the rounding unit.

1. 7300 to 2 significant figures: the second significant figure (the 3) is in the hundreds column, so the mass was rounded to the nearest 100 kg.
2. The mass can be up to half of 100 kg, which is 50 kg, below 7300 kg.
3. The lower bound is 7300-50=7250 kg.

The true length can be at most half of the rounding unit away from the recorded value.

1. The length was rounded to the nearest 0.01 cm, so the true length lies between 3.42-0.005 and 3.42+0.005 cm.
2. The true length can be as far as half of 0.01 cm from 3.42 cm.
3. The greatest possible difference is 0.01\div2=0.005 cm.

Use both lower bounds when finding the smallest possible total.

1. The lower bounds are 8.55 cm and 4.15 cm.
2. Add them to get the lower bound for the total: 8.55+4.15=12.7 cm.

The largest area comes from using the largest length and the largest width.

1. The upper bounds for length and width are 12.45 cm and 5.65 cm.
2. Multiply both upper bounds: 12.45\times5.65=70.3425.
3. The upper bound for area is 70.3425 cm^2.

Make the original cord as long as possible and the piece removed as short as possible.

1. Remaining length = original length - length cut off. To make this as large as possible, use the largest original length and the smallest piece cut off.
2. The upper bound for the original length is 7.8+0.05=7.85 m (half of 0.1 m is 0.05 m).
3. The lower bound for the piece cut off is 2.35-0.005=2.345 m (half of 0.01 m is 0.005 m).
4. Upper bound for the remaining length =7.85-2.345=5.505 m.

To make a difference as small as possible, use the lower first value and the upper second value.

1. To make a-b as small as possible, take the smallest possible a and subtract the largest possible b.
2. Each value is correct to the nearest 0.1, so it can be up to 0.05 either side: the lower bound for a is 4.45 and the upper bound for b is 6.25.
3. Lower bound for a-b =4.45-6.25=-1.8.
4. Using 4.45-6.15 (both lower bounds) is a common mistake: subtracting a smaller b makes the answer larger, not smaller.

Find the midpoint of the interval and the width of the rounding unit.

1. The recorded value is in the middle of the interval: halfway between 4.75 and 4.85 is 4.8.
2. Each end is 0.05 from 4.8. That is half of the rounding unit, so the rounding unit is 0.05\times2=0.1.
3. So x is 4.8 correct to the nearest 0.1.
4. The others give different intervals: 4.8 correct to the nearest 0.01 gives 4.795\le x<4.805, and 5 correct to 1 significant figure gives 4.5\le x<5.5. 4.75 cannot be a value correct to the nearest 0.1, as it has two decimal places.

Use the largest force and the smallest area.

1. The upper bound for force is 245 N, and the lower bound for area is 1.75 m^2.
2. The upper bound for pressure is 245\div1.75=140 N/m^2.

A side length is the positive square root of the area.

1. The lower bound for area is 48.5 cm^2.
2. Taking the positive square root gives the lower side-length bound: \sqrt{48.5}=6.96419\ldots cm.
3. The lower bound is 6.96 cm correct to 3 significant figures.

Compare the whole possible mass interval with the acceptance limit.

1. The possible masses satisfy 4.95\le m<5.05.
2. For example, 4.98 kg would be accepted, but 5.02 kg would be rejected.
3. The rounded measurement does not determine whether this parcel will be accepted.

Use the shortest possible strip before deciding how many complete pieces fit.

1. The lower bound for the strip length is 95.5 cm.
2. Seven pieces need 7\times12=84 cm, which fits even the shortest possible strip.
3. Eight pieces need 96 cm, but the strip could be shorter than this.
4. Therefore exactly 7 whole pieces can be guaranteed.

For the smallest density, use the smallest mass and the largest volume.

1. Both measurements have been rounded to the nearest 10, so the half-interval is 5.
2. The lower bound for mass is 835 g and the upper bound for volume is 125 cm^3.
3. The lower bound for density is 835\div125=6.68 g/cm^3.

Write \cos x in terms of AB and AC, then decide which bounds make \cos x as small as possible.

1. AB is adjacent to angle x and AC is the hypotenuse, so \cos x=\dfrac{AB}{AC}.
2. The bounds are 8.35\le AB<8.45 and 11.65\le AC<11.75 (half of 1 mm is 0.05 cm).
3. Between 0^\circ and 90^\circ, the bigger the angle, the smaller its cosine. So x is largest when \dfrac{AB}{AC} is smallest: use the smallest AB and the largest AC.
4. \cos x=\dfrac{8.35}{11.75}=0.71063\ldots
5. x=\cos^{-1}(0.71063\ldots)=44.713\ldots^\circ
6. The upper bound for x is 44.7^\circ correct to 1 decimal place. (Using both upper bounds, \frac{8.45}{11.75}, gives 44.0^\circ, which is not the largest possible angle.)

Square the bound that makes the positive number as large as possible.

1. The bounds are 335\le p<345.
2. Because p is positive, squaring preserves the order, so p^2<345^2=119025.
3. Since 119025<120000, the claim is guaranteed.
4. The recorded value alone would not prove a statement about every possible true value.

Use the largest distance and the smallest time, and convert the time to hours.

1. The upper bound for distance is 18.65 km and the lower bound for time is 21.5 minutes.
2. The lower bound for time in hours is \dfrac{21.5}{60}.
3. Divide the upper distance bound by the lower time bound: 18.65\div\dfrac{21.5}{60}=52.0465\ldots.
4. The upper bound for average speed is 52.0 km/h correct to 3 significant figures.

For the smallest pressure, use the smallest force and the largest radius.

1. The lower bound for force is 95.5 N, and the upper bound for radius is 2.45 cm.
2. The lower bound for pressure is \dfrac{95.5}{\pi\times2.45^2}=5.06432\ldots N/cm^2.
3. The lower bound is 5.06 N/cm^2 correct to 3 significant figures.

The volume increases when either positive measurement increases, and a justified rounded answer must agree at both bounds.

Part (a)

1. The upper bounds for radius and height are 3.805 cm and 12.405 cm.
2. Use both upper bounds because increasing either measurement increases the volume: V_{\mathrm{upper}}=\pi\times3.805^2\times12.405.
3. This gives 564.229726\ldots cm^3, so the upper bound is 564.2 cm^3 correct to 4 significant figures.

Part (b)

1. The lower bound 560.815443\ldots and upper bound 564.229726\ldots both round to 560 cm^3 to 2 significant figures.
2. The whole interval lies between 555 and 565, so every possible volume rounds to 560 cm^3 to 2 significant figures.
3. To 3 significant figures, values near the lower bound round to 561 while values near the upper bound round to 564.
4. Therefore 560 cm^3 to 2 significant figures is the most precise guaranteed answer.