Chapter 1, Problem 1.74E

Do the following calculations and use the correct number of significant figures in your answers. Assume all numbers are the results of measurements. In calculations involving both addition/subtraction and multiplication/division, it is usually better to do additions/subtractions first.

a. $\frac{\left(0.0267+0.0019\right)\left(4.626\right)}{28.7794}$

b. $\frac{212.6-21.88}{86.37}$

c. $\frac{27.99-18.07}{4.63-0.88}$

d. $\frac{18.87}{2.46}-\frac{18.07}{0.88}$

(HINT: Do divisions first, then subtract.)

e. $\frac{\left(8.46-2.09\right)\left(0.51+0.22\right)}{\left(3.74+0.07\right)\left(0.16+0.2\right)}$

f. $\frac{12.06-11.84}{0.271}$

Interpretation Introduction

(a)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $0.00460$ and three significant figures are present.

Explanation of Solution

The given problem is,

$\frac{\left(0.0267+0.0019\right)\left(4.626\right)}{28.7794}$

The calculation is as follows:

$\begin{array}{rll}\frac{\left(0.0267+0.0019\right)\left(4.626\right)}{28.7794}& =& \frac{0.0286\left(4.626\right)}{28.7794}\\ & =& \frac{0.1323036}{28.7794}\\ & =& 0.0045971632487126\end{array}$

The lowest number of significant figures in the problem is three. Therefore, the answer is expressed by using three significant figures. The non-significant figure which is dropped is $7$ and it is greater than five. Thus, one unit is added to last digit remained. Hence, the result of a given problem is $0.00460$.

Conclusion

The result of the given calculation is $0.00460$ and three significant figures are present.

Interpretation Introduction

(b)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $2.208$ and four significant figures are present.

Explanation of Solution

The given problem is,

$\frac{212.6-21.88}{86.37}$

The calculation is as follows:

$\begin{array}{rll}\frac{212.6-21.88}{86.37}& =& \frac{190.72}{86.37}\\ & =& 2.208174134537455\end{array}$

The lowest number of significant figures in the problem is four. Therefore, the answer is expressed by using four significant figures. The non-significant figure which is dropped is $1$ and it is lesser than five. Thus, one unit is not added to last digit. Hence, the result of a given problem is $2.208$.

Conclusion

The result of the given calculation is $2.208$ and four significant figures are present.

Interpretation Introduction

(c)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $2.65$ and three significant figures are present.

Explanation of Solution

The given problem is,

$\frac{27.99-18.07}{4.63-0.88}$

The calculation is as follows:

$\begin{array}{rll}\frac{27.99-18.07}{4.63-0.88}& =& \frac{9.92}{3.75}\\ & =& 2.6453333\end{array}$

The lowest number of significant figures in the problem is three. Therefore, the answer is expressed by using three significant figures. The non-significant figure which is dropped is $5$. Thus, one unit is added to last digit. Hence, the result of a given problem is $2.65$.

Conclusion

The result of the given calculation is $2.65$ and three significant figures are present.

Interpretation Introduction

(d)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $-12.86$ and two significant figures are present.

Explanation of Solution

The given problem is,

$\frac{18.87}{2.46}-\frac{18.07}{0.88}$

The calculation is as follows:

$\begin{array}{rll}\frac{18.87}{2.46}-\frac{18.07}{0.88}& =& 7.67-20.53\\ & =& -12.86\end{array}$

The answer is rounded off to that number which contains smallest number of places after the decimal. Hence, the result of a given problem is $-12.86$.

Conclusion

The result of the given calculation is $-12.86$ and $-13.0$ with two significant figures are present.

Interpretation Introduction

(e)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $3.3903$ and five significant figures are present.

Explanation of Solution

The given problem is,

$\frac{\left(8.46-2.09\right)\left(0.51+0.22\right)}{\left(3.74+0.07\right)\left(0.16+0.2\right)}$

The calculation is as follows:

$\begin{array}{rll}\frac{\left(8.46-2.09\right)\left(0.51+0.22\right)}{\left(3.74+0.07\right)\left(0.16+0.2\right)}& =& \frac{6.37\times 0.73}{3.81\times 0.36}\\ & =& \frac{4.6501}{1.3716}\\ & =& 3.390274132400117\end{array}$

The lowest number of significant figures in the problem is five. Therefore, the answer is expressed by using five significant figures. The non-significant figure which is dropped is $7$, it is greater than five. Thus, one unit is added to last digit remained. Hence, the result of a given problem is $3.3903$.

Conclusion

The result of the given calculation is $3.3903$ and five significant figures are present.

Interpretation Introduction

(f)

Interpretation:

The calculations are to be expressed by using correct number of significant figures.

Concept introduction:

Scientific notation is used to write large numbers by shifting the decimal point. It is expressed as the product of a decimal term and $10$ raised to some power. Significant figures are used to express a measurement in scientific notation. The significant figures reflect the closeness of measured values to the standard values.

Answer to Problem 1.74E

The result of the given calculation is $0.81$ and two significant figures are present.

Explanation of Solution

The given problem is,

$\frac{12.06-11.84}{0.271}$

The calculation is as follows:

$\begin{array}{rll}\frac{12.06-11.84}{0.271}& =& \frac{0.22}{0.271}\\ & =& 0.8118081180811808\end{array}$

The lowest number of significant figures in the problem is two. Therefore, the answer is expressed by using two significant figures. The non-significant figure which is dropped is $1$, it is lesser than five. Thus, one unit is not added to last digit remained. Hence, the result of a given problem is $0.81$.

Conclusion

The result of the given calculation is $0.81$ and two significant figures are present.