Chapter 5.4, Problem 32IP

To determine

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

Answer to Problem 32IP

The order of rates from least to greatestis $\text{2500 g/min}<\text{100 oz/min}<\text{10 lb/min}$ .

Explanation of Solution

Given information: Different weight measurement units per minute are $\text{100 oz/min}$ , $\text{2500 g/min}$ and $\text{10 lb/min}$ .

Calculation:

For comparing different weight measurement units, convert the rates to the same basic unit.

Use $\text{1}\text{oz}\approx \text{28}\text{gm}$ and $\text{1lb}=\text{454}\text{g}$ .

The rate $\text{100 oz/min}$ into grams per min can be converted as:

  $\begin{array}{c}\text{100 oz/min}=\frac{\text{100 oz}}{\text{1}\text{min}}\cdot \frac{\text{28 g}}{\text{1 oz}}\\ =\text{2800 g/min}\end{array}$

The rate $\text{10 lb/min}$ into grams per min can be converted as:

  $\begin{array}{c}\text{10}\text{lb/min}=\frac{\text{10}\text{lb}}{\text{1}\text{min}}\cdot \frac{\text{454}\text{g}}{\text{1}\text{lb}}\\ =\text{4540}\text{g/min}\end{array}$

Compare the calculated value with 2500 gram per minute.

  $\text{2500 g/min}<\text{100 oz/min}<\text{10 lb/min}$

Therefore, the order of rates from least to greatestas required are $\text{2500 g/min}<\text{100 oz/min}<\text{10 lb/min}$ .