Chapter 12, Problem 10SAP

To determine

To find: percentage of cars in the lot that are neither blue nor convertibles.

Answer to Problem 10SAP

50%

Explanation of Solution

Given information:

Let the total number of cars in the parking lot be $x$ , number of convertible car be $c$ , blue cars be $b$ , convertible blue cars be $b_c$ and non-convertible non-blue cars be $n$ .

  $\begin{array}{c}b=\frac{x}{5}\\ b_c=\frac{1}{2}\left(\frac{x}{5}\right)\\ \frac{c}{4}=b_c\end{array}$

Calculation:

Evaluate the value of $b_c\text{ and }c$ as follows:

  $\begin{array}{c}{b}_c=\frac{1}{2}\left(\frac{x}{5}\right)\\ =\frac{x}{10}\end{array}$

Evaluate the value of $c$ as follows:

  $\begin{array}{c}\frac{c}{4}=b_c\\ c=4b_c\\ c=4\left(\frac{x}{10}\right)\left[\text{Put the value of }{b}_c\right]\\ c=\frac{2x}{5}\end{array}$

Now substitute the values of $b,c\text{ and }{b}_c$ in the formula given below to evaluate percentage of non-convertible non-blue cars $\left(n\%\right)$ as follows:

  $\begin{array}{c}x=c+b-b_c+n\\ n=x-c-b+b_c\\ n=x-\frac{2}{5}x-\frac{x}{5}+\frac{x}{10}\\ n=x\left(1-\frac{2}{5}-\frac{1}{5}+\frac{1}{10}\right)\\ \frac{n}{x}=1-\frac{3}{5}+\frac{1}{10}\\ \frac{n}{x}=\frac{10-6+1}{10}\\ \frac{n}{x}=\frac{1}{2}\\ \frac{n}{x}\times 100\%=\frac{1}{2}\times 100\%\\ n\%=50\%\end{array}$

Thus, the percentage of the cars in the parking lot that are neither blue nor convertible is 50%.