Chapter 5.4, Problem 34IP

To determine

To find: The order of group of rates from least to greatest.

Answer to Problem 34IP

The order of rates from least to greatestis $\text{6600}\text{yd/qt}<\text{32}\text{mi/gal}<\text{15}\text{m/mL}$ .

Explanation of Solution

Given information:The given different rates are $\text{32}\text{mi/gal}$ , $\text{15}\text{m/mL}$ and $\text{6600}\text{yd/qt}$ .

Calculation:

For comparing different rate, convert the rates to the same basic unit.

To convert miles and meter into yards, use $\text{1}\text{mi}=\text{1760}\text{yd}$ and $\text{1}\text{m}=\text{1}\text{.09}\text{yd}$ .

To convert gallon and quartinto milliliters, use $\text{1}\text{gal}=\text{3785}\text{mL}$ and $\text{1}\text{qt}=\text{946}\text{.35}\text{mL}$ .

The rate $\text{32}\text{mi/gal}$ into yards per milliliters can be converted as:

  $\begin{array}{c}\text{32}\text{mi/gal}=\frac{\text{32}\text{mi}}{\text{1 gal}}\cdot \frac{\text{1760 yd}}{\text{1}\text{mi}}\cdot \frac{\text{1 gal}}{\text{3785 mL}}\\ =\text{14}\text{.88 yd/mL}\end{array}$

The rate $\text{15}\text{m/mL}$ into yards per milliliters can be converted as:

  $\begin{array}{c}\text{15 m/mL}=\frac{\text{15 m}}{\text{1 mL}}\cdot \frac{\text{1}\text{.09 yd}}{\text{1 m}}\\ =\text{16}\text{.35 yd/mL}\end{array}$

The rate $\text{6600}\text{yd/qt}$ into yards per milliliters can be converted as:

  $\begin{array}{c}\text{6600 yd/qt}=\frac{\text{6600}\text{yd}}{\text{1}\text{qt}}\cdot \frac{\text{1 qt}}{\text{946}\text{.35 mL}}\\ =\text{6}\text{.974}\text{yd/mL}\end{array}$

Compare the calculated values.

  $\text{6600}\text{yd/qt}<\text{32}\text{mi/gal}<\text{15}\text{m/mL}$

Therefore, the order of rates from least to greatest as required are $\text{6600}\text{yd/qt}<\text{32}\text{mi/gal}<\text{15}\text{m/mL}$ .