Chapter 1, Problem 1.57QE

(a)

Interpretation Introduction

Interpretation:

The following expression has to be evaluated and the answer has to be expressed to the correct number of significant figures.

  $\frac{\left(\text{25}\text{.12}\text{-}\text{1}\text{.75}\right)\text{×}\text{0}\text{.01920}}{\left(\text{24}\text{.339}\text{-}\text{23}\text{.15}\right)}$

Concept Introduction:

Rules for significant figure:

• All nonzero digits are significant.  Example:  123.4 in which four significant figures are there.
• Zeros between nonzero digits are significant.  Example: 105 in which digit zero is a significant figure.  Therefore, three significant figures.
• In a number with no decimal point, zeros at the end of the number (“trailing zeros”) are not necessarily significant.  Example:  602,000 in which there are only three significant figures.
• If a number contains a decimal point, zeros at the beginning (“leading zeros”) are not significant, but zeros at the end of the number are significant.  Example:  0.0044 in which number of significant figure is two.

In multiplication and division, the result has the same number of significant figures as the multiplier or divisor with fewest number of significant figures.

In addition, and subtraction, the result is expressed to the smallest number of decimal places of the numbers involved.

Explanation of Solution

Given,

    $\frac{\left(\text{25}\text{.12}\text{-}\text{1}\text{.75}\right)\text{×}\text{0}\text{.01920}}{\left(\text{24}\text{.339}\text{-}\text{23}\text{.15}\right)}$

The given calculation involves subtraction, multiplication and division.

First, the subtraction part of the expression is evaluated.

  $\begin{array}{l}\text{=}\frac{\left(\text{25}\text{.12}\text{-}\text{1}\text{.75}\right)\text{×}\text{0}\text{.01920}}{\left(\text{24}\text{.339}\text{-}\text{23}\text{.15}\right)}\\ \text{=}\frac{\left(\text{23}\text{.37}\right)\text{×}\text{0}\text{.01920}}{\left(\text{1}\text{.19}\right)}\end{array}$

Now, the multiplication part is evaluated and then the result is divided to get the final answer.

  $\begin{array}{l}\text{=}\frac{\left(\text{25}\text{.12}\text{-}\text{1}\text{.75}\right)\text{×}\text{0}\text{.01920}}{\left(\text{24}\text{.339}\text{-}\text{23}\text{.15}\right)}\\ \text{=}\frac{\left(\text{23}\text{.37}\right)\text{×}\text{0}\text{.01920}}{\left(\text{1}\text{.19}\right)}\\ \text{=}\frac{\text{0}\text{.4487}}{\text{1}\text{.19}}\\ \text{=}\text{0}\text{.377}\text{or}\text{3}\text{.77}\text{×}{\text{10}}^{\text{-1}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The following expression has to be evaluated and the answer has to be expressed to the correct number of significant figures.

  $\frac{\text{55}\text{.4}}{\left(\text{26}\text{.3}\text{-}\text{18}\text{.904}\right)}$

Concept Introduction:

Refer sub part-a.

Explanation of Solution

Given,

    $\frac{\text{55}\text{.4}}{\left(\text{26}\text{.3}\text{-}\text{18}\text{.904}\right)}$

The given calculation involves subtraction and division.

The first calculation is of subtraction.  The significant figures have to be noted.

  $\begin{array}{l}\text{=}\frac{\text{55}\text{.4}}{\left(\text{26}\text{.3}\text{-}\text{18}\text{.904}\right)}\\ \text{=}\frac{\text{55}\text{.4}}{\text{7}\text{.4}}\end{array}$

The second calculation is of division.

The first digit consists of three significant figures and second digit consists of three significant figures.  The answer consists of three significant figures.

  $\begin{array}{l}\text{=}\frac{\text{55}\text{.4}}{\left(\text{26}\text{.3}\text{-}\text{18}\text{.904}\right)}\\ \text{=}\frac{\text{55}\text{.4}}{\text{7}\text{.4}}\\ \text{=}\text{7}\text{.5}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The following expression has to be evaluated and the answer has to be expressed to the correct number of significant figures.

  $\left(\text{0}\text{.921}\text{×}\text{27}\text{.977}\right)\text{+}\left(\text{0}\text{.470}\text{×}\text{28}\text{.976}\right)\text{+}\left(\text{3}\text{.09}\text{×}\text{29}\text{.974}\right)$

Concept Introduction:

Refer sub part-a.

Explanation of Solution

Given,

    $\left(\text{0}\text{.921}\text{×}\text{27}\text{.977}\right)\text{+}\left(\text{0}\text{.470}\text{×}\text{28}\text{.976}\right)\text{+}\left(\text{3}\text{.09}\text{×}\text{29}\text{.974}\right)$

The given calculation involves multiplication and addition.

The operation present inside the parentheses (multiplication) is performed first.  Then, addition of the products gives the final answer.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\left(\text{0}\text{.921}\text{×}\text{27}\text{.977}\right)\text{=}\text{25}\text{.8}\\ \text{+}\left(\text{0}\text{.470}\text{×}\text{28}\text{.976}\right)\text{=}\text{13}\text{.6}\\ \text{+}\left(\text{3}\text{.09}\text{×}\text{29}\text{.974}\right)\text{=}\text{92}\text{.6}\end{array}}\\ \text{=}\text{132}\text{.0}\end{array}$

The least decimal place that is common in three numbers is first or tenths place.  The final answer is limited to that place.