- Meaning of Significant Figures
- How to Write Significant Figures
- Multiplication and Division of Significant Figures
- Operation of Addition and Subtraction of Significant Figures
- Operations of Roots and Powers of Significant Figures
- Rounding Significant Figures
EXAMPLES OF QUESTIONS AND SOLUTIONS OF IMPORTANT FIGURES
Problem . 1 . In the measurement of length, the data obtained is 761.5 dm. The number of significant figures from the measurement results is ...
A. 1
B. 2
C. 3
D. 4
E. 5
Solution of Significant Figures: D
The number 761.5 dm has 4 significant figures, namely the numbers 7,6,1 and 5.
Problem . 2. A child measures the length of the rope and gets 0.123 km. Then the number of significant figures from the measurement is...
A. 6
B. 5
C. 4
D. 3
E. 2
Solution of Significant Figures: D
The number 0.123 km has 3 significant figures, namely numbers 1,2 and 3.
Problem . 3 . The sum of 2.30 cm + 1.1 cm according to the rules of significant figures is ...
A. 2,40 cm
B. 2,41 cm
C. 2,4 cm
D. 3,40 cm
E. 3,4 cm
Solution of Significant Figures: D
$ \begin{align*} &2,30 \text{(3 Significant Figures)}\\&1,1 \text{ (2 Significant Figures)}+\\\hline &3,40\textbf{ (3 Significant Figures)}\end{align*}$
has 1 approximation digit, which is 0.
Problem . 4 . The result of subtraction from 4.551 grams - 1.21 grams according to the rules of significant figures is ...
A. 3,3 gram
B. 3,34 gram
C. 3,341 gram
D. 4,67 gram
E. 4,671 gram
Solution of Significant Figures: B
$\small \begin{align*} &4,551 \text{(4 Significant Figures)}\\&1,21 \text{ (3 Significant Figures)}-\\\hline &3,341\textbf{ (Calculator result) }\end{align*}$
the maximum result has 1 estimated digit, so the number 3,341 must be written as 3.34 where 4 is the estimated number.
Problem . 5. The result of measuring the length of an object is 0.025 cm. The measurement results have significant figures as many as...
has 1 approximation digit, which is 0.
Problem . 4 . The result of subtraction from 4.551 grams - 1.21 grams according to the rules of significant figures is ...
A. 3,3 gram
B. 3,34 gram
C. 3,341 gram
D. 4,67 gram
E. 4,671 gram
Solution of Significant Figures: B
$\small \begin{align*} &4,551 \text{(4 Significant Figures)}\\&1,21 \text{ (3 Significant Figures)}-\\\hline &3,341\textbf{ (Calculator result) }\end{align*}$
the maximum result has 1 estimated digit, so the number 3,341 must be written as 3.34 where 4 is the estimated number.
Problem . 5. The result of measuring the length of an object is 0.025 cm. The measurement results have significant figures as many as...
A. Two
B. Three
C. Four
D. Five
E. Six
Solution of Significant Figures:A
The number 0.025 cm has 2 significant figures, namely the numbers 2 and 5.
Problem . 6. The results of measuring the height of grade 10 students are as follows: 154 cm, 158 cm, 163 cm, 165 cm, and 160 cm. According to the rule of significant figures, the average height of the five students is...
A. 160 cm
B. 160.2 cm
C. 160.20 cm
D. 1.6 x 10-2 cm
E. 16.02 x 10-1 cm
Solution of Significant Figures: A
We use the operation of Addition and Division of Significant Figures, but because all the measurement figures in the problem have 3 significant figures, while the number 5 is an exact number, the result must have 3 significant figures, which is 160 cm.
$\small \begin{align*} \overline{x}&=\frac{154+158+163+165+160}{5}\\\overline{x}&=\frac{800\, \text{ (3 Significant Figures)}}{5\, \text{(exact number)}}\\\overline{x}&=160\, \textbf{(3 Significant Numbers)} \end{align*}$
Problem . 7. The results of measuring the length and width of a floor are 12.61 m and 5.2 m. According to the rule of significant figures, the area of the floor is...
A. 65 m2
B. 65.5 m2
C. 65,572 m2
D. 65.6 m2
E. 66 m2
Solution of Significant Figures: E
$\small \begin{align*} &A=\textrm{pxl}\\&p=12.61 \text{(4 Significant Figures)}\\&l=5.2\text{ (2 Significant Figures)} \times \\\hline &A=65,572 \textbf{(Calculator result)}\end{align*}$
B. Three
C. Four
D. Five
E. Six
Solution of Significant Figures:A
The number 0.025 cm has 2 significant figures, namely the numbers 2 and 5.
Problem . 6. The results of measuring the height of grade 10 students are as follows: 154 cm, 158 cm, 163 cm, 165 cm, and 160 cm. According to the rule of significant figures, the average height of the five students is...
A. 160 cm
B. 160.2 cm
C. 160.20 cm
D. 1.6 x 10-2 cm
E. 16.02 x 10-1 cm
Solution of Significant Figures: A
We use the operation of Addition and Division of Significant Figures, but because all the measurement figures in the problem have 3 significant figures, while the number 5 is an exact number, the result must have 3 significant figures, which is 160 cm.
$\small \begin{align*} \overline{x}&=\frac{154+158+163+165+160}{5}\\\overline{x}&=\frac{800\, \text{ (3 Significant Figures)}}{5\, \text{(exact number)}}\\\overline{x}&=160\, \textbf{(3 Significant Numbers)} \end{align*}$
Problem . 7. The results of measuring the length and width of a floor are 12.61 m and 5.2 m. According to the rule of significant figures, the area of the floor is...
A. 65 m2
B. 65.5 m2
C. 65,572 m2
D. 65.6 m2
E. 66 m2
Solution of Significant Figures: E
$\small \begin{align*} &A=\textrm{pxl}\\&p=12.61 \text{(4 Significant Figures)}\\&l=5.2\text{ (2 Significant Figures)} \times \\\hline &A=65,572 \textbf{(Calculator result)}\end{align*}$
Remember: The result of the Multiplication and Division of Significant Figures operations must have the least significant figures in the numbers involved in the Multiplication and Division of Significant Figures operations. So, the number 65.572 m2 must have 2 significant figures, becoming = 66 m2
Problem . 8. From the measurement results, the iron plate is 1.5 m long and 1.20 m wide. According to the rules for writing significant figures, the area of the iron plate is...
A. 1.8012 m2
B. 1,801 m2
C. 1,800 m2
D. 1.80 m2
E. 1.8 m2
Solution of Significant Figures: E
$\small \begin{align*} &A=\textrm{pxl}\\&p=1.5 \text{(2 Significant Figures)}\\&l=1.20\text{ (3 Significant Figures)} \times \\\hline &A=1.8 \textbf{ (Calculator result)}\end{align*}$
Problem . 8. From the measurement results, the iron plate is 1.5 m long and 1.20 m wide. According to the rules for writing significant figures, the area of the iron plate is...
A. 1.8012 m2
B. 1,801 m2
C. 1,800 m2
D. 1.80 m2
E. 1.8 m2
Solution of Significant Figures: E
$\small \begin{align*} &A=\textrm{pxl}\\&p=1.5 \text{(2 Significant Figures)}\\&l=1.20\text{ (3 Significant Figures)} \times \\\hline &A=1.8 \textbf{ (Calculator result)}\end{align*}$
Remember: The result of the Multiplication and Division of Significant Figures operations must have the fewest Significant Figures in the numbers involved in the Multiplication and Division of Significant Figures operations. = 1.8 m2
Problem . 9. A student measures the diameter of a circle and the result is 8.50 cm. The circumference of a circle according to the rule of significant figures is... ($\pi = 3.14$)
A. 267 cm
B. 26.7 cm
C. 2.67 cm
D. 0.267 cm
E. 0.0267 cm
Solution of Significant Figures: B
$\small \begin{align*} &K=\pi \times d\\&\pi=3.14 \text{(3 Significant Figures)}\\&d=8.50 \text{ (3 Significant Figures) )} \times \\\hline &K=1,826.69 \textbf{ (Calculator result)}\end{align*}$
Problem . 9. A student measures the diameter of a circle and the result is 8.50 cm. The circumference of a circle according to the rule of significant figures is... ($\pi = 3.14$)
A. 267 cm
B. 26.7 cm
C. 2.67 cm
D. 0.267 cm
E. 0.0267 cm
Solution of Significant Figures: B
$\small \begin{align*} &K=\pi \times d\\&\pi=3.14 \text{(3 Significant Figures)}\\&d=8.50 \text{ (3 Significant Figures) )} \times \\\hline &K=1,826.69 \textbf{ (Calculator result)}\end{align*}$
Remember: The result of the Multiplication and Division of Significant Figures operations must have the fewest Significant Figures in the numbers involved in the Multiplication and Division of Significant Figures operations, so the number 26.69 cm becomes 26.7 cm.
Problem . 10. From the measurement results, the thin plate is 15.35 cm long and 8.24 cm wide. Then the area of the plate is...
A. 126 cm2
B. 126.5 cm2
C. 126.48 cm2
D. 126.484 cm2
E. 126.4840 cm2
Solution of Significant Figures: A
$\small \begin{align*} &A=p \times l\\&p=15.35 \text{(4 Significant Figures)}\\&l=8.24 \text{ (3 Significant Figures)} \times \\\hline &A=126.484\textbf{ (Calculator result)}\end{align*}$
Problem . 10. From the measurement results, the thin plate is 15.35 cm long and 8.24 cm wide. Then the area of the plate is...
A. 126 cm2
B. 126.5 cm2
C. 126.48 cm2
D. 126.484 cm2
E. 126.4840 cm2
Solution of Significant Figures: A
$\small \begin{align*} &A=p \times l\\&p=15.35 \text{(4 Significant Figures)}\\&l=8.24 \text{ (3 Significant Figures)} \times \\\hline &A=126.484\textbf{ (Calculator result)}\end{align*}$
Remember: The result of the Multiplication and Division of Significant Figures operations must have the fewest Significant Figures in the numbers involved in the Multiplication and Division of Significant Figures operations. So, the number 126.484 cm2 must have 3 Significant Figures, to be 126 cm2.
