Chapter 11.2, Problem 18STP

To determine

To find: The value of $\frac{c}{d}$ .

Answer to Problem 18STP

Thevalue of $\frac{c}{d}$ is $\frac{3}{2}$ .

Explanation of Solution

Given:

The average score of a class of $c$ students is $80$ .

The average score of a class of $d$ students is $85$ .

The average score of combined $\left(c+d\right)$ students is $82$ .

The value of $\frac{c}{d}$ is calculated as:

  $\begin{array}{c}\text{Average}\text{score}\text{of}\text{c}\text{students}=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}\text{c}\text{students}}{\text{Total}\text{number}\text{of}\text{}\text{students}}\\ \text{Average}\text{score}\text{of}d\text{students}=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}d\text{students}}{\text{Total}\text{number}\text{of}\text{}\text{students}}\\ \text{Average}\text{score}\text{of}\left(\text{c}+d\right)\text{students}=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}\left(\text{c}+d\right)\text{students}}{\text{Total}\text{number}\text{of}\text{}\text{students}}\end{array}$

Now by substituting the given information in above expression, further simplified as:

  $\begin{array}{c}80=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}\text{c}\text{students}}{c}\\ 85=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}d\text{students}}{d}\\ 82=\frac{\text{sum}\text{of}\text{the}\text{scores}\text{of}\left(\text{c}+d\right)\text{students}}{c+d}\end{array}$

Further simplified as:

  $\begin{array}{c}80c=\text{sum}\text{of}\text{the}\text{scores}\text{of}\text{c}\text{students}\left(1\right)\\ 85d=\text{sum}\text{of}\text{the}\text{scores}\text{of}d\text{students}\left(2\right)\\ 82\left(c+d\right)=\text{sum}\text{of}\text{the}\text{scores}\text{of}\left(\text{c}+d\right)\text{students}\left(3\right)\end{array}$

Solving $\left(1\right),\left(2\right)\&\left(3\right)$ as combined, the solution is given as:

  $\begin{array}{c}82c+82d=80c+85d\\ 2c=3d\\ \frac{c}{d}=\frac{3}{2}\end{array}$

Therefore the value of $\frac{c}{d}$ is $\frac{3}{2}$ .