Chapter 5.2, Problem 29HP

To determine

To find: The correct situation in which rate $\frac{x\text{miles}}{y\text{hours}}$ decreases.

Answer to Problem 29HP

The correct situation is situation (b):x is unchanged; y increases.

Explanation of Solution

Given information:

The given situations are as follows.

1. x increases; y is unchanged.
2. xis unchanged; yincreases.

Calculation:

Let’s consider the situation (a) by taking an example.

Person A drives a car 120 miles in 6 hours and Person B drives a car 140 miles in 6 hours.Who drives faster?

Calculate the unit rate of Person A.

  $\begin{array}{c}\frac{120\text{miles}}{6\text{hours}}=\frac{\frac{120}{6}}{\frac{6}{6}}\\ =\frac{20\text{miles}}{1\text{hour}}\end{array}$

Calculate the unit rate of Person B.

  $\begin{array}{c}\frac{140\text{miles}}{6\text{hours}}=\frac{\frac{140}{6}}{\frac{6}{6}}\\ =\frac{23.33\text{miles}}{1\text{hour}}\end{array}$

From the above, it is clear that the unit rate increases when the distance increases.

Let’s consider the situation (b) by taking an example.

Person A drives a car 120 miles in 6 hours and Person B drives a car 140 miles in 8 hours. Who drives faster?

Calculate the unit rate of Person A.

  $\begin{array}{c}\frac{120\text{miles}}{6\text{hours}}=\frac{\frac{120}{6}}{\frac{6}{6}}\\ =\frac{20\text{miles}}{1\text{hour}}\end{array}$

Calculate the unit rate of Person B.

  $\begin{array}{c}\frac{140\text{miles}}{4\text{hours}}=\frac{\frac{140}{8}}{\frac{8}{8}}\\ =\frac{17.5\text{miles}}{1\text{hour}}\end{array}$

From the above, it is clear that the unit rate decreases when the time increases.

Therefore, the correct situation is situation (b):x is unchanged; y increases.