Chapter 3.1, Problem 51PPS

To determine

ToOrder the group of numbers $\frac{2}{25}$ , $\frac{1}{13}$ , $0.089$ from least to greatest.

Answer to Problem 51PPS

Ordering the given numbers from least to greatest, we have $0.\overline{076923}$ , $0.08$ , $0.089$ i.e., $\frac{1}{13}$ , $\frac{2}{25}$ , $0.089$

Explanation of Solution

Given:

The numbers $\frac{2}{25}$ , $\frac{1}{13}$ , $0.089$

Concept Used:

A fraction can be converted into decimal form by dividing numerator of the fraction with its denominator.

Calculation:

Given the numbers $\frac{2}{25}$ , $\frac{1}{13}$ , $0.089$

Converting the fractions into decimals and then comparing them all

Converting $\frac{2}{25}$ into decimals

Dividing numerator with denominator, we have

i.e., $25\begin{array}{c}\hfill 0.08\\ \hfill \overline{)\begin{array}{l}2\\ \underset{\_}{0}\\ 200\\ \underset{\_}{200}\\ 0\end{array}}\end{array}$

Thus, $\frac{2}{25}=0.08$

Converting $\frac{1}{13}$ into decimals

Dividing numerator with denominator, we have

i.e., $13\begin{array}{c}\hfill 0.0769230\\ \hfill \overline{)\begin{array}{l}1\\ \underset{\_}{\text{0 }}\\ 100\\ \underset{\_}{91}\\ \text{ 9}0\\ \underset{\_}{78}\\ \text{ 120}\\ \underset{\_}{117}\\ \text{ 30}\\ \underset{\_}{26}\\ \text{ 40}\\ \underset{\_}{39}\\ \text{ 1}00\\ \underset{\_}{91}\\ \text{ 9}\end{array}}\end{array}$

Thus, $\frac{1}{13}=0.0769230\text{. }\text{. }\text{.}$

The repeating decimal can be written as $\frac{1}{13}=0.\overline{076923}$

After converting the numbers $2\frac{3}{5}$ , $2.67$ , $2\frac{2}{3}$ into decimal form, we have

$\frac{2}{25}=0.08$ , $\frac{1}{13}=0.\overline{076923}$ , $0.089=0.089$

Ordering them from least to greatest, we have $0.\overline{076923}$ , $0.08$ , $0.089$ i.e., $\frac{1}{13}$ , $\frac{2}{25}$ , $0.089$

Conclusion:

Ordering the given numbers from least to greatest, we have $0.\overline{076923}$ , $0.08$ , $0.089$ i.e., $\frac{1}{13}$ , $\frac{2}{25}$ , $0.089$