Chapter 6.7, Problem 25E

To determine

To calculate: the number of items purchased by 50 cents.

Answer to Problem 25E

The number of items purchased is $\frac{5n}{(5-2n)}\text{items}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\text{Cost of }n\text{ items}=50\text{ cents}\\ \text{Price per itme is decreased by }10\text{cents}.\end{array}$

Formula used:

  $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$ .

  $\text{Speed}=\frac{\text{Time}}{\text{Distance}}$

Proof:

We have ,

The cost of n items is 50 cents.

This implies the cost of 1 item is $\frac{50}{n}$ .

The price per item is decreased by 10 cents

  $\begin{array}{l}=\frac{50}{n}-10\\ =\frac{50}{n}-\frac{10}{1}\\ =\left(\frac{50}{n}\times \frac{1}{1}\right)-\left(\frac{10}{1}\times \frac{n}{n}\right)\\ =\frac{50-10n}{n}\end{array}$

Number of items brought by 50 cents = n

  $\begin{array}{l}\text{For}\frac{50-10n}{n}\text{cents},1\text{ item is purcahsed}\\ \text{For 50 cents},\left(\frac{1}{\frac{50-10n}{n}}\times \frac{50}{1}\right)\text{item is purcahsed}\end{array}$

  $\begin{array}{l}=\left(\frac{1}{\frac{50-10n}{n}}\times \frac{50}{1}\right)\\ =\frac{n}{50-10n}\times \frac{50}{1}\\ =\frac{50n}{10(5-2n)}\\ =\frac{5n}{(5-2n)}\text{items}\end{array}$

Hence, the number of items purchased are $\frac{5n}{(5-2n)}\text{items}$ .