Applications: Significant Figures and Estimation

Significant figures are counted from the first non-zero digit. Zeros in front of that digit only place the decimal point. A zero between significant digits, or a zero produced by rounding, is significant.

1. The zeros at the front of 0.004 06 only show where the decimal point is, so they are not significant. Counting starts at the first non-zero digit, which is the 4.
2. The digits from the 4 onwards are 4, 0 and 6. The zero between the 4 and the 6 is significant, because it lies between two significant digits. The mass has 3 significant figures.
3. To round to 2 significant figures, keep the 4 and the 0 and look at the next digit, which is 6. Since 6 is 5 or more, the 0 rounds up to 1. (a) The mass is 0.0041 kg, correct to 2 significant figures.
4. In 34 962 the first two significant figures are 3 and 4. The next digit is 9, so the 4 rounds up to 5. Correct to 2 significant figures the number of visitors is 35 000. The three zeros only keep the 3 and the 5 in their places.
5. Correct to 3 significant figures, keep 3, 4 and 9 and look at the next digit, which is 6. The 9 rounds up, so 349 hundreds become 350 hundreds and the number is again 35 000.
6. (b) Both answers are written 35 000, but correct to 3 significant figures the first zero is significant: it says that the number is nearer to 35 000 than to 34 900 or 35 100. Check: 34 962 is 38 away from 35 000 and 62 away from 34 900.

answer(a) 3 significant figures; 0.0041 kg; (b) 35 000 both times, but correct to 3 significant figures the first zero is significant

techniqueSignificant Figures

examsO-Level · GCSE Higher

A length rounded to the nearest centimeter can be up to half a centimeter away from the stated value on either side. The lower bound is a possible length. The upper bound is not, because it rounds up to the next centimeter.

1. The length is rounded to the nearest 1 cm, so the true length is at most half of 1 cm away from 84 cm. Half of 1 cm is 0.5 cm.
2. The lower bound is 84 − 0.5 = 83.5 cm. A length of exactly 83.5 cm rounds up to 84 cm, so 83.5 cm is a possible length.
3. The upper bound is 84 + 0.5 = 84.5 cm. (a) The lower bound is 83.5 cm and the upper bound is 84.5 cm.
4. A length of exactly 84.5 cm rounds up to 85 cm, not to 84 cm. Every length below 84.5 cm, such as 84.49 cm or 84.499 cm, rounds to 84 cm. So L can equal the lower bound, but it must be less than the upper bound.
5. (b) The error interval is 83.5 ≤ L < 84.5, and the shelf cannot be exactly 84.5 cm long. Check: the interval is 84.5 − 83.5 = 1 cm wide, which is the unit the length was rounded to.

answer(a) lower bound 83.5 cm, upper bound 84.5 cm; (b) 83.5 ≤ L < 84.5; no, a length of 84.5 cm rounds to 85 cm

techniqueBounds and Error Intervals

examsGCSE Higher

Each side has a lower bound and an upper bound. A sum is least when every part is least and greatest when every part is greatest, so the bounds of the perimeter come from adding bounds of the same kind.

1. Each side is rounded to the nearest meter, so it can be 0.5 m either way. The length l m satisfies 11.5 ≤ l < 12.5 and the width w m satisfies 6.5 ≤ w < 7.5.
2. The perimeter is least when both sides are at their lower bounds: 2 × (11.5 + 6.5) = 2 × 18 = 36 m.
3. The upper bound of the perimeter uses both upper bounds: 2 × (12.5 + 7.5) = 2 × 20 = 40 m. (a) The least possible perimeter is 36 m and the upper bound is 40 m, so the perimeter P m satisfies 36 ≤ P < 40.
4. (b) 38 − 36 = 2 and 40 − 38 = 2, so the true perimeter can differ from 38 m by up to 2 m. Check: the fence has four sides and each can be 0.5 m out, and 4 × 0.5 = 2 m.

answer(a) least 36 m, upper bound 40 m; (b) 2 m

techniqueBounds and Error Intervals

examsGCSE Higher

The area is a product of two positive lengths, so it is least when both lengths are least and greatest when both are greatest. An error in one side is multiplied by the whole of the other side.

1. Each side can be 0.5 m either way. The length l m satisfies 8.5 ≤ l < 9.5 and the width w m satisfies 5.5 ≤ w < 6.5.
2. The area is least when both sides are at their lower bounds: 8.5 × 5.5 = 46.75 m².
3. The upper bound of the area uses both upper bounds: 9.5 × 6.5 = 61.75 m². (a) The least possible area is 46.75 m² and the upper bound is 61.75 m².
4. The difference between the two areas is 61.75 − 46.75 = 15 m².
5. (b) The larger floor is the smaller floor with two strips added, each 1 m wide. One strip runs along the full length and has area 9.5 × 1 = 9.5 m². The other runs along the rest of the width and has area 1 × 5.5 = 5.5 m². Each strip is narrow but several meters long, and 9.5 + 5.5 = 15 m². Check: 9 × 6 = 54 m² lies between 46.75 m² and 61.75 m².

answer(a) least 46.75 m², upper bound 61.75 m²; (b) 15 m²

techniqueBounds and Error Intervals

examsGCSE Higher

Speed is distance divided by time. A quotient becomes larger when the number on top becomes larger and also when the number underneath becomes smaller, so the two bounds of the speed each pair opposite bounds.

1. The distance d m satisfies 93.5 ≤ d < 94.5, and the time T s satisfies 4.5 ≤ T < 5.5.
2. The speed is dT. It is greatest when the car covers the longest distance in the shortest time, so the upper bound of d goes with the lower bound of T.
3. (a) The upper bound of the speed is 94.54.5 = 21 m/s.
4. The speed is least when the car covers the shortest distance in the longest time. (b) The least possible speed is 93.55.5 = 17 m/s.
5. Check: the rounded values give 945 = 18.8 m/s, which lies between 17 m/s and 21 m/s. Dividing upper bound by upper bound gives 94.55.5 ≈ 17.2 m/s, which is nowhere near the greatest speed.

answer(a) 21 m/s; (b) 17 m/s

techniqueBounds and Error Intervals

examsGCSE Higher

A load is certain to be safe only when its greatest possible mass is within the limit. The greatest possible mass comes from the upper bound of the mass of each crate.

1. The mass of one crate, m kg, satisfies 24.5 ≤ m < 25.5. For safety the bound that matters is the upper bound, 25.5 kg.
2. The upper bound of the mass of 16 crates is 16 × 25.5 = 408 kg.
3. (a) 408 kg is more than 400 kg, so 16 crates are not certain to be safe. The rounded masses give 16 × 25 = 400 kg, which only looks safe because the rounding hides up to 0.5 kg on each crate.
4. A load of n crates is certain to be safe when n × 25.5 ≤ 400. Since 400 ÷ 25.5 ≈ 15.7, the largest whole number of crates is n = 15.
5. (b) The greatest number is 15 crates. Check: 15 × 25.5 = 382.5 kg, which is within 400 kg, and 16 × 25.5 = 408 kg, which is not.

answer(a) No: the load could be almost 408 kg; (b) 15 crates

techniqueBounds and Error Intervals

examsGCSE Higher

The error is the difference between the measured value and the true value. The percentage error compares the error with the true value, so that measurements of different sizes can be compared fairly.

1. The error in the desk measurement is 125 − 120 = 5 cm.
2. Percentage error = errortrue value × 100%. (a) For the desk it is 5125 × 100% = 4%.
3. The error in the corridor measurement is 20 − 19.8 = 0.2 m, which is 20 cm.
4. (b) For the corridor the percentage error is 0.220 × 100% = 1%.
5. The corridor measurement is more accurate, because 1% is less than 4%. Its error of 20 cm is four times the error for the desk, but the corridor is 16 times as long as the desk. Check: 1% of 20 m is 0.2 m, and 4% of 125 cm is 5 cm.

answer(a) 4%; (b) 1%, so the corridor measurement is more accurate, although its error of 20 cm is larger

techniquePercentage Error

Rounding to 1 significant figure keeps only the first non-zero digit of each price, so the prices are rounded to different place values. The percentage error measures the gap between the estimate and the exact bill against the exact bill.

1. Round each price to 1 significant figure. $28.50 becomes $30 because the second digit is 8. $16.40 becomes $20 because the second digit is 6. $4.34 becomes $4 because the second digit is 3. In $0.76 the first significant figure is the 7 and the next digit is 6, so it becomes $0.80.
2. (a) Her estimate is 30 + 20 + 4 + 0.80 = $54.80.
3. The exact bill is 28.50 + 16.40 + 4.34 + 0.76. Add in pairs: 28.50 + 16.40 = 44.90 and 4.34 + 0.76 = 5.10, so the bill is 44.90 + 5.10 = $50.
4. The error of the estimate is 54.80 − 50 = $4.80.
5. (b) The percentage error is 4.8050 × 100% = 9.6%. The estimate is too high because the two largest prices were both rounded up. Check: 9.6% of $50 is $4.80.

answer(a) $54.80; (b) the exact bill is $50 and the percentage error is 9.6%

techniquePercentage Error · Significant Figures · Estimating with One Significant Figure

examsO-Level · GCSE Higher

An estimate with every number rounded to 1 significant figure can be worked out without a calculator, and it has the same size as the exact answer. A calculator answer that is ten times larger or smaller than the estimate has been keyed in wrongly.

1. Round each number to 1 significant figure: 7840 becomes 8000, 0.0285 becomes 0.03 and 0.62 becomes 0.6. The calculation is now 8000 × 0.030.6.
2. Work out the top first: 8000 × 0.03 = 80 × 3 = 240.
3. To divide by 0.6, multiply the top and the bottom by 10: 2400.6 = 24006 = 400. The answer is larger than 240, because dividing by a number less than 1 makes a number larger. (a) About 400 boxes are needed.
4. (b) The clerk's 36.04 is about 10 times too small, so it is not of the right size. A correct answer should be a few hundred. The digits suggest a decimal point in the wrong place, and the calculation does give about 360 boxes.

answer(a) about 400 boxes; (b) no: 36.04 is about 10 times too small, and the calculation gives about 360 boxes

techniqueEstimating with One Significant Figure · Significant Figures

examsO-Level · GCSE Higher

A small rounding error in an intermediate value is multiplied by everything that comes after it. Keeping the intermediate value exact, as a fraction or in the calculator, avoids this.

1. The mass of 1 m of cable is 1.3 ÷ 6 = 0.21666… kg. The digit 6 repeats without end.
2. Correct to 2 significant figures this is 0.22 kg, because the third significant figure is 6 and the 1 rounds up to 2.
3. (a) Ravi's answer is 0.22 × 2400 = 528 kg.
4. Without rounding, keep the mass of 1 m as the fraction 1.36. The mass of the cable is 1.36 × 2400 = 1.3 × 400 = 520 kg, because 2400 ÷ 6 = 400.
5. (b) The mass is 520 kg, and Ravi's answer is 528 − 520 = 8 kg too large. Check: the rounding added 0.22 − 0.21666… = 0.00333… kg to each meter, and 0.00333… × 2400 = 8 kg.

answer(a) 528 kg; (b) 520 kg, so rounding early makes the answer 8 kg too large

techniqueSignificant Figures

examsO-Level · GCSE Higher