To start the tutorial, push the f5 key in the upper row on your keyboard

To start the tutorial, push the f5 key in the upper row on your keyboard

f5

Then you can use the up and down arrow keys to advance backwards and forwards through the tutorial.

Calculations with significant figures

So now you know how to determine the number of significant figures in a number and how to round numbers off, what good does that do?


Well, let’s consider the rectangle below. If we want to determine the area of the rectangle, the easiest way would be to measure the length and the height, and then multiply these two numbers together. (For rectangles, Area = Length x Height.)


First let’s measure the length: a correct reading might be something like 36.3 cm (…or 36.2 cm or 36.4 cm)

0 10cm

20 30 40 50


Now let’s measure the height: for height, a correct reading might be something like 6.7 cm (…or 6.6 cm or 6.8 cm).

010

cm20


So let’s multiply these two numbers:


So let’s multiply these two numbers:

36.3 x 6.7 = 243.21 (and cm x cm = cm2).

So we have 243.21 cm2. But if we state the area to be 243.21cm2, we are stating a pretty high level of precision.


But if we say the area is 243.21cm2, we are saying we know the area to a very high level of precision.


But if we say the area is 243.21cm2, we are saying we know the area to a very high level of precision. We are saying that we are certain of the “243.2..” and that we are guessing the “1.”


But if 36.2 cm long and 6.6 cm high were also correct measurements, then 36.2 cm x 6.6 cm = 238.92 cm2 would have to be a correct area for the same rectangle.


But if 36.2 cm long and 6.6 cm high were also correct measurements, then 36.2 cm x 6.6 cm = 238.92 cm2 would have to be a correct area for the same rectangle. That implies we are certain of the “238.9…” and only guessing the “2.”


And if 36.4 cm long and 6.8 cm high were also correct measurements, then 36.4 cm x 6.8 cm = 247.52 cm2 would also have to be a correct area for the same rectangle.


And if 36.4 cm long and 6.8 cm high were also correct measurements, then 36.4 cm x 6.8 cm = 247.52 cm2 would also have to be a correct area for the same rectangle. That implies we are certain of the “247.5…” and only guessing the “2.”


243.21 cm2, 238.92 cm2 and 247.52 cm2


243.21 cm2, 238.92 cm2 and 247.52 cm2 These three values cannot all be correct.


243.21 cm2, 238.92 cm2 and 247.52 cm2 These three values cannot all be correct. The only digit that seems to be definite is the first “2” (in the hundreds place). After that the values are not at all consistent with one another.


This would mean that our guess should be the second digit (in the tens place), and that the values should all be rounded there – to two significant figures.


243.21 cm2 rounds to 240 cm2 238.92 cm2 rounds to 240 cm2 and 247.52 cm2 rounds to 250 cm2.


240 cm2, 240 cm2 and 250 cm2.


240 cm2, 240 cm2 and 250 cm2. These are all consistent with one another.


240 cm2, 240 cm2 and 250 cm2. These are all consistent with one another. They all have two significant figures, and they show disagreement only in the guessed digit.


Is there a way we could have known from the beginning that our answer needed to be rounded to only two significant figures?


If we look at the original measurements that went into the calculation, we see a length of 36.3 cm, which has three significant figures, and a height of 6.7 cm, which has two significant figures.


Imagine there is a chain that is made of only two links, and one link is able to hold 3 kg before it breaks and the other is able to hold 2 kg, how much weight can the entire chain hold?

Strong enough to hold 3 kg



If you are thinking that the chain could hold 5 kg (3 kg + 2 kg), then think again!




If you are thinking that the chain could hold 5 kg (3 kg + 2 kg), then think again!

The chain would break at its weakest point. And so, as a whole, the chain would only be able to hold 2 kg before it broke.

Together only strong enough to hold 2 kg


There is an old expression that says: “A chain is only as strong as its weakest link.”


There is an old expression that says: “A chain is only as strong as its weakest link.”

If a chain were made of ten links, and nine of those links could hold 100 kg, but one could only hold 1 kg…


How much weight would the entire chain be able to hold?



Just 1 kg!



Just 1 kg! Essentially, the one weak link ruins it for the rest of the links.


The same holds true for calculations involving measurements. Consider the calculation below.

23.40 cm x 0.47 cm x 6.05 cm =

precise to 4 significant

figures


figures


figures


The calculator answer has 6 significant figures.

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures


The calculator answer has 6 significant figures. But the weakest measurement has only 2 significant figures.

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures


The calculator answer has 6 significant figures. But the weakest measurement has only 2 significant figures. This means the answer must be rounded to only two significant figures:

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures

67 cm3


When you report an answer to be something like “66.5379 cm3”(just because that is what showed up on your calculator), you are claiming a level of precision much higher than the measurements deserve.

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures

67 cm3


An answer of 66.5379 cm3 means that the “66.537..” are definite, and only the “9” is a guess.

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures

67 cm3


An answer of 66.5379 cm3 means that the “66.537..” are definite, and only the “9” is a guess. But if the 0.47 cm could have just as easily been read as 0.46 cm, consider how different the answer would be.

23.40 cm x 0.47 cm x 6.05 cm = 66.5379 cm3


figures


figures


figures

67 cm3


So here is the rule:When multiplying or dividing two or more measurements, always round your answer off to the number of significant figures in the weakest measurement. (The weakest measurement is the one with the fewest significant figures)


So here is the rule:When multiplying or dividing two or more measurements, always round your answer off to the number of significant figures in the weakest measurement. (The weakest measurement is the one with the fewest significant figures)

This ensures that your answer will not be any more or less precise than it should be.


If crude measurements were made, then only crude values can be calculated from them. If more precise measurements were made, then more precise values can be calculated.


So let’s say a student is calculating the average speed of a car as it traveled down the road.


So let’s say a student is calculating the average speed of a car as it traveled down the road. Speed is distance divided by time.


So let’s say a student is calculating the average speed of a car as it traveled down the road. Speed is distance divided by time. The student measures the time with a very precise stop watch and records a time of 146.39 s.


So let’s say a student is calculating the average speed of a car as it traveled down the road. Speed is distance divided by time. The student measures the time with a very precise stop watch and records a time of 146.39 s. Distance is measured rather crudely: 680 m.

0 100 200 300 400 500 600 700 800m


Speed = = =

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

This answer is what appears on the calculator, but it is obviously way too precise.

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

This answer is what appears on the calculator, but it is obviously way too precise.

What should it be rounded to?

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

The distance (680 m) has two significant figures

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

The distance (680 m) has two significant figures… and the time (146.39 s) has five significant figures.

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

The distance (680 m) has two significant figures… and the time (146.39 s) has five significant figures. The weaker measurement is the one with just two significant figures...

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

The distance (680 m) has two significant figures… and the time (146.39 s) has five significant figures. The weaker measurement is the one with just two significant figures… so the answer should be rounded to just two sig. figs.

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s


Speed = = = 4.645126033 m/s

The distance (680 m) has two significant figures… and the time (146.39 s) has five significant figures. The weaker measurement is the one with just two significant figures… so the answer should be rounded to just two sig. figs.

0 100 200 300 400 500 600 700 800m

distance time

680 m146.39 s

4.6 m/s


Now try each of the following 20 problems.


Now try each of the following 20 problems. Use a calculator, and then write your answer on paper.


Now try each of the following 20 problems. Use a calculator, and then write your answer on paper. Make sure to round the answer to the correct number of significant figures…


Now try each of the following 20 problems. Use a calculator, and then write your answer on paper. Make sure to round the answer to the correct number of significant figures…

…and also make sure to include correct units with each answer.


1) 34.6 m x 14.8 m =


1) 34.6 m x 14.8 m =

512 m2


1) 34.6 m x 14.8 m =

512 m2

The 34.6 m and the 14.8 m both have three significant figures, so the answer is rounded to three significant figures.


2) 67 cm x 38 cm =


2) 67 cm x 38 cm =

2500 cm2


2) 67 cm x 38 cm =

2500 cm2

The 67 cm and the 38 cm both have two significant figures, so the answer must be rounded to two significant figures.


3) 0.0057 m x 0.0063 m =


3) 0.0057 m x 0.0063 m =

0.000036 m2


3) 0.0057 m x 0.0063 m =

0.000036 m2

The 0.0057 m and the 0.0063 m both have two significant figures, so the answer must be rounded to two significant figures.


4) 34.0 mm x 72.7 mm =


4) 34.0 mm x 72.7 mm =

2470 mm2


4) 34.0 mm x 72.7 mm =

2470 mm2

The 34.0 mm and the 72.7 mm both have three significant figures, so the answer is rounded to three significant figures.


5) 18.38 m x 23.54 m =


5) 18.38 m x 23.54 m =

432.7 m2


5) 18.38 m x 23.54 m =

432.7 m2

The 18.38 m and the 23.54 m both have four significant figures, so the answer is rounded to four significant figures.


6) 207 m x 64 m =


6) 207 m x 64 m =

13,000 m2


6) 207 m x 64 m =

13,000 m2

The 207 m has three significant figures, but the 64 m has only two. Thus, the answer should be rounded to two significant figures.


7) 296.3 cm x 0.8 cm =


7) 296.3 cm x 0.8 cm =

200 cm2


7) 296.3 cm x 0.8 cm =

200 cm2

The 296.3 cm has four significant figures, but the 0.8 cm only has one. Thus, the answer must be rounded to just one significant figure.


8) 3.4 m x 16.3 m x 25.7 m =


8) 3.4 m x 16.3 m x 25.7 m =

1400 m3


8) 3.4 m x 16.3 m x 25.7 m =

1400 m3

The 3.4 m has two significant figures, and the 16.3 m and 25.7 m each have three significant figures. Thus, the answer is rounded to just two significant figure.


9) 0.48 mL x 9.365 g/mL =


9) 0.48 mL x 9.365 g/mL =

4.5 g


9) 0.48 mL x 9.365 g/mL =

4.5 gThe 0.48 mL has two significant figures, and the 9.365 cm has four. Thus, the answer is rounded to just two significant figures.


10) 25.50 m x 12.00 m =


10) 25.50 m x 12.00 m =

306.0 m2


10) 25.50 m x 12.00 m =

306.0 m2

The 25.50 m and the 12.00 m both have four significant figures, so the answer should have four significant figures. The calculator gives only “306” as the answer. In this situation, the answer must be enhanced up to four significant figures: “306.0”


11) 17.5 km - 3.20 hr =


11) 17.5 km - 3.20 hr =

5.47 km/hr


11) 17.5 km - 3.20 hr =

5.47 km/hrThe 17.5 km and the 3.20 hr both have three significant figures, so the answer should have just three significant figures.


12) 234.53 g - 85 mL =


12) 234.53 g - 85 mL =

2.8 g/mL


12) 234.53 g - 85 mL =

2.8 g/mLThe 234.53 g has five significant figures, but the 85 mL has only two. Thus, the answer should have only two significant figures.


13) 2300 m2 - 725 m =


13) 2300 m2 - 725 m =

3.2 m


13) 2300 m2 - 725 m =

3.2 mThe 2300 m2 has two significant figures and the 725 m has three. The answer should therefore be rounded to just two significant figures.


14) 17.5 km - 3.50 hr =


14) 17.5 km - 3.50 hr =

5.00 km/hr


14) 17.5 km - 3.50 hr =

5.00 km/hrThe 17.5 km and the 3.50 hr both have three significant figures, so the answer should have three significant figures. The calculator gives an answer of simply “5.” So this “5” must be enhanced up to three significant figures: “5.00.”


15) 32.0 m3 – 7.0 m =


15) 32.0 m3 – 7.0 m =

4.6 m2


15) 32.0 m3 – 7.0 m =

4.6 m2

The 32.0 m3 has three significant figures, but the 7.0 m has only two, so the answer should be rounded to just two significant figures.


16) 679.0 g – 678.9 mL =


16) 679.0 g – 678.9 mL =

1.000 g/mL


16) 679.0 g – 678.9 mL =

1.000 g/mLThe 679.0 g and the 678.9 mL both have four significant figures, so the answer should be rounded to four significant figures.


17) (65 m x 17 m) – 4.83 s =


17) (65 m x 17 m) – 4.83 s =

230 m2/s


17) (65 m x 17 m) – 4.83 s =

230 m2/s

The 65 m and 17 m both have two significant figures, and the 4.83 s has three significant figures. Thus the answer should have two significant figures.


18) 56.94 g – (3.42 cm x 7.61 cm x 0.35 cm) =


18) 56.94 g – (3.42 cm x 7.61 cm x 0.35 cm) =

6.3 g/cm3


18) 56.94 g – (3.42 cm x 7.61 cm x 0.35 cm) =

6.3 g/cm3

The 56.94 g has four significant figures, the 3.42 cm and the 7.61 cm each have three significant figures, but the 0.35 cm has only two significant figures. The answer therefore should be rounded to just two significant figures.


19) 215 cm x 372 cm =


19) 215 cm x 372 cm =

80,000 cm2


19) 215 cm x 372 cm =

80,000 cm2

The 215 cm and the 372 cm both have three significant figures, so the answer should be rounded to three significant figures. 79,980 rounds up to 80,000, which appears to have only one significant figure. A line over the second 0 fixes this problem.


20) 2.480 m – 0.124 s =


20) 2.480 m – 0.124 s =

20.0 m/s


20) 2.480 m – 0.124 s =

20.0 m/sThe 2.480 m has four significant figures and the 0.124 s has three, so the answer should have three significant figures. The calculator gives an answer of “20”, which has only one significant figure. It must be enhanced to three significant figures: 20.0.


So… How many did you get correct?


So… How many did you get correct?

Hopefully this tutorial program has helped you understand how to round off answers for these sort of calculations


It is important to remember that this rule of rounding answers off to the fewest number of significant figures applies only to multiplication and division.



There is a different rule that is used for adding and subtracting...



There is a different rule that is used for adding and subtracting… and that will be the topic of our next tutorial.

To start the tutorial, push the f5 key in the upper row on your keyboard

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