Sig Fig Math
Solve the following problems answering with the appropriate number of sig figs:
1. 3.519 x 213.04 x 0.0134 =
2. 0.3020 + 23.01 – 12.4 =
3. 2.145 – 0.0300 + 1.98 =
4. 1.34 x 1012 x 6.231 x 10-22 =
5. 0.0123 + 2.14 =
6.
(9.12 ´10 8 )(1.93 ´10 -19 )
=
5.38 ´1012
4.61 ´10 -8
7.
=
(1.23 ´1014 )(3.92 ´10 -11 )
8. 45.99 – 0.03745 + 0.28449 =
9. 12.00 + 485.2 – 1100 =
10.
(1.040 ´10 8 )(0.00318)
=
2.43 ´10 -12
Coin diameter
A gold coin has an ‘accepted’ diameter of 28.054 mm.
Two students are asked to measure the diameter of four gold coins. Student A uses a
simple plastic ruler. Student B uses a precision measuring tool called a micrometer.
Student A – plastic ruler
27.9 mm
28.0 mm
27.8 mm
28.1 mm
Student B – micrometer
28.246 mm
28.244 mm
28.246 mm
28.248 mm
1. Calculate the average value for each set of measurements
2. Calculate the % error for each set of measurements.
3. Compare the average value for each set with the accepted value:


Which student’s data is more accurate?
Which student’s data is more precise?
4. Compare the percentage error for each set:


Which student’s data is more accurate?
Which student’s data is more precise?
5. Explain any odd findings:
Sig Fig Math Answers
Solve the following problems answering with the appropriate number of sig figs:
1. 3.519 x 213.04 x 0.0134 = 10.0458159

10.0
2. 0.3020 + 23.01 – 12.4 = 10.912

10.9
3. 2.145 – 0.0300 + 1.98 = 4.095

4.10
4. 1.34 x 1012 x 6.231 x 10-22 = 8.34954 x 10-10

8.35 x 10-10
5. 0.0123 + 2.14 = 2.1523

2.15
(9.12 ´10 8 )(1.93 ´10 -19 )
6.
= 3.2716728 x 10-23
5.38 ´1012

3.27 x 10-23
4.61 ´10 -8
= 9.5611415 x 10-12
(1.23 ´1014 )(3.92 ´10 -11 )

9.56 x 10-12
8. 45.99 – 0.03745 + 0.28449 = 46.23704

46.24
9. 12.00 + 485.2 – 1100 = -602.8

-603

1.36 x 1017
7.
10.
(1.040 ´10 8 )(0.00318)
= 1.36098765432 x 1017
2.43 ´10 -12
Coin diameter
A gold coin has an ‘accepted’ diameter of 28.054 mm.
1. Calculate the average value for each set of measurements
27.9+28.0+27.8+28.1 111.8
=
= 27.95 mm
Student A
4
4
28.246+28.244 +28.246+28.248 112.984
=
Student B
= 28.2460 mm
4
4
2. Calculate the % error for each set of measurements.
(28.054 - 27.95)
0.104
´100 =
´100= 0.37% error
Student A
28.054
28.054
(28.054 - 28.2460)
-0.192
´100=
´100 = 0.684% error
Student B
28.054
28.054
3. Compare the average value for each set with the accepted value:
 Which student’s data is more accurate?
Student A, because the average value of 27.95 mm is closer to the accepted
value of 28.054 mm.

Which student’s data is more precise?
Student B, because each measured value sits closer to the average value.
4. Compare the percentage error for each set:

Which student’s data is more accurate?
Student A, because the percentage error of 0.37% is smaller than Student B’s
value of 0.684%.

Which student’s data is more precise?
It is not possible to decide this from percentage error alone. The range of values
is important in establishing precision.
5. Explain any odd findings:
Although Student B had access to a precision measuring tool, called a micrometer,
it did not have high accuracy in this experiment. Perhaps it was not calibrated
correctly. One would have expected Student B’s accuracy and precision to both be
better than Student A.
Aluminium bar density
Two students are given a small cylinder of aluminium of known mass and asked to
determine its density. (The ‘accepted’ density of aluminium is 2.702 g/cm3.) Since
density is mass/volume the students need to calculate the volume of the cylinder. To
do this, the height and diameter of the cylinder need to be measured.
Student A is told to use a simple plastic ruler and to make four independent
measurements for each dimension. Student B is told to use a precision measuring
tool called a micrometer.
Student A – plastic ruler
2.21 g/cm3
2.29 g/cm3
2.72 g/cm3
2.38 g/cm3
Student B – micrometer
2.703 g/cm3
2.701 g/cm3
2.705 g/cm3
5.811 g/cm3
1. Calculate the average value for each set of density values, making sure that any
‘outliers’ are not included.
Student A
Student B
2.21+2.29+2.72+ 2.38 9.60
=
= 2.40 g/cm3
4
4
2.703+ 2.701+2.705 8.109
=
= 2.703 g/cm3
3
3
2. Calculate the % error for each set of values.
Student A
Student B
2.702- 2.40
0.302
(100) =
(100) = 11.1769 
2.702
2.702
2.702- 2.703
0.001
(100) =
(100) = 0.0370 
2.702
2.702
11% error
0.04% error
3. Compare the average value for each set with the accepted value:

Which student’s data is more accurate?
Student B as the average is close to the accepted value.

Which student’s data is more precise?
If the outlier result of 5.811 g/cm3 is excluded (it is very far removed from the
other results), Student B. However, if all results are taken into account, Student
A’s measurements would be more precise.
4. Compare the percentage error for each set:

Which student’s data is more accurate?
The percentage error for Student A is large by comparison with the very low
value for Student B, only if the outlier value of 5.811 g/cm3 is excluded.

Which student’s data is more precise?
It is not possible to decide this from percentage error alone. The range of values
is important in establishing precision.
5. Explain any odd findings:
When analysing data sets, it is important to recognise outlier results. Clearly, the
value 5.811 g/cm3 is not in keeping with the other measured results and can be
confidently eliminated.