Chapter 2, Problem 14PE

(a)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\text{15}\text{.2-2}\text{.75+15}\text{.67}$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Significant figures:  The digits that are measured in a number including one estimated digit are called as significant figures.

The answers for those calculations that involve addition and subtraction will have the same number of decimal places as the original number with the fewest decimal places.

Answer to Problem 14PE

The answer for $\text{15}\text{.2-2}\text{.75+15}\text{.67}$ is $\text{28}\text{.1}\text{.}$

Explanation of Solution

The calculation for $\text{15}\text{.2-2}\text{.75+15}\text{.67}$ is,

  $\begin{array}{l}\text{15}\text{.2}\\ \text{-2}\text{.75}\\ \underset{\_}{\text{15}\text{.67}}\\ \text{28}\text{.12}\end{array}$

The number with the least precision is $\text{15}\text{.2}\text{.}$  Therefore, the answer is rounded off to the nearest tenth: $\text{28}\text{.1}\text{.}$  The answer for $\text{15}\text{.2-2}\text{.75+15}\text{.67}$ is $\text{28}\text{.1}\text{.}$

(b)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\left(\text{4}\text{.68}\right)\left(\text{12}\text{.5}\right)$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Significant figures:  The digits that are measured in a number including one estimated digit are called as significant figures.

The answers for those calculations that involve multiplication and division will have the same number of significant figures as the original number with the fewest number of significant figures.

Answer to Problem 14PE

The answer for the product of $\left(\text{4}\text{.68}\right)\left(\text{12}\text{.5}\right)$ is $\text{58}\text{.5}\text{.}$

Explanation of Solution

The product of $\left(\text{4}\text{.68}\right)\left(\text{12}\text{.5}\right)$ is $\text{58}\text{.5}\text{.}$  The answer should have three significant figures because both the numbers $\left(\text{4}\text{.68}\right)\left(\text{12}\text{.5}\right),$ with the fewest significant figures, has only three significant figures.

The answer for the product of $\left(\text{4}\text{.68}\right)\left(\text{12}\text{.5}\right)$ is $\text{58}\text{.5}\text{.}$

(c)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\frac{\text{182}\text{.6}}{\text{4}\text{.6}}$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

Answer to Problem 14PE

The answer for $\frac{\text{182}\text{.6}}{\text{4}\text{.6}}$ is $\text{40}\mathrm{..}$

Explanation of Solution

The answer for $\frac{\text{182}\text{.6}}{\text{4}\text{.6}}$ is $\text{39}\text{.69}\text{.}$  The answer should have only two significant figures, since the number $\text{4}\text{.6,}$ with the fewest significant figures, has only two significant figures.

In the number $\text{39}\text{.69,}$ the last two digits are dropped and one is added to the next digit $\text{9}\text{.}$  The answer for $\frac{\text{182}\text{.6}}{\text{4}\text{.6}}$ is $\text{40}\mathrm{..}$

(d)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\text{1986+23}\text{.84+0}\text{.012}$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (a).

Answer to Problem 14PE

The answer for $\text{1986+23}\text{.84+0}\text{.012}$ is $\text{2}\text{.010}\times {\text{10}}^3\text{.}$

Explanation of Solution

The sum of $\text{1986+23}\text{.84+0}\text{.012}$ is $\text{2009}\text{.852}\text{.}$  The number with the least precision is $\text{1986}\text{.}$  The answer for $\text{1986+23}\text{.84+0}\text{.012}$ is $\text{2}\text{.010}\times {\text{10}}^3\text{.}$

(e)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\frac{\text{29}\text{.3}}{\left(\text{284}\right)\left(\text{415}\right)}$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

Answer to Problem 14PE

The answer for $\frac{\text{29}\text{.3}}{\left(\text{284}\right)\left(\text{415}\right)}$ is $\text{2}{\text{.49×10}}^{\text{-4}}\text{.}$

Explanation of Solution

The answer for $\frac{\text{29}\text{.3}}{\left(\text{284}\right)\left(\text{415}\right)}$ is $\text{2}{\text{.486×10}}^{\text{-4}}\text{.}$  The answer should have only three significant figures because all the three numbers with fewest significant figures, have only three significant figures.

In the number $\text{2}{\text{.486×10}}^{\text{-4}}\text{,}$ the last digit is dropped and one is added to the next digit $\text{8}\text{.}$  The rounded off three significant number is $\text{2}{\text{.49×10}}^{\text{-4}}\text{.}$  The answer for $\frac{\text{29}\text{.3}}{\left(\text{284}\right)\left(\text{415}\right)}$ is $\text{2}{\text{.49×10}}^{\text{-4}}\text{.}$

(f)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for $\left(\text{2}{\text{.92×10}}^{\text{-3}}\right)\left(\text{6}{\text{.14×10}}^{\text{5}}\right)$ and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

Answer to Problem 14PE

The answer for $\left(\text{2}{\text{.92×10}}^{\text{-3}}\right)\left(\text{6}{\text{.14×10}}^{\text{5}}\right)$ is $\text{1}{\text{.79×10}}^{\text{3}}\text{.}$

Explanation of Solution

The answer for $\left(\text{2}{\text{.92×10}}^{\text{-3}}\right)\left(\text{6}{\text{.14×10}}^{\text{5}}\right)$ is $\text{1792}\text{.88}\text{.}$  The answer should have only three significant figures, since both the numbers $\text{2}{\text{.92×10}}^{\text{-3}}\text{,6}{\text{.14×10}}^{\text{5}}$ with the fewest significant figures, has only three significant figures.

In the number $\text{1792}\text{.88,}$ the last three digits are dropped and the rounded off three significant number is $\text{1}{\text{.79×10}}^{\text{3}}\text{.}$  The answer for $\left(\text{2}{\text{.92×10}}^{\text{-3}}\right)\left(\text{6}{\text{.14×10}}^{\text{5}}\right)$ is $\text{1}{\text{.79×10}}^{\text{3}}\text{.}$