Chapter A.9, Problem 119AYU

(a)

To determine

To find:the range of score needed in the last date to get B.

Answer to Problem 119AYU

The range of marks needed is $74\le x<124$ . And the minimum marks needed is $74$ .

Explanation of Solution

Given information:

The first four scores are $68,82,87$ and $89$ . The average of first five test scores should be greater than or equal to $80$ and less than $90$ .

Calculation:

Assume that the score in fifth test is $x$ .

Calculate the average of the scores in all the five tests.

  $\begin{array}{c}\text{avg}=\frac{68+82+87+89+x}{5}\\ =\frac{326+x}{5}\end{array}$

For grade B average should be greater than or equal to $80$ and less than $90$ .

  $80\le \frac{326+x}{5}<90$

Multiply each side of inequality by $5$ .

  $\begin{array}{l}80\left(5\right)\le \frac{326+x}{5}\left(5\right)<90\left(5\right)\\ 400\le 326+x<45\end{array}$

Add $-326$ to each part of inequality.

  $\begin{array}{l}400-326\le 326+x-326<450-326\\ 74\le x<124\end{array}$

So the minimum score in the fifth test should be $74$ .

Therefore, the minimum score in the fifth test should be $74$ .

(b)

To determine

To find:

The score needed in the fifth test if it counts double.

Answer to Problem 119AYU

The score needed in the fifth test if it counts double is $77$ .

Explanation of Solution

Given information:

The first four scores are $68,82,87$ and $89$ . The average of first five test scores should be greater than or equal to $80$ and less than $90$ .

Calculation:

If the fifth test doubles the average calculated in part (a) will be expressed as,

  $avg=\frac{326+2x}{6}$

For grade B average should be greater than or equal to $80$ and less than $90$ .

  $80\le \frac{326+2x}{6}<90$

Multiply each side of inequality by $65$ .

  $\begin{array}{l}80\left(6\right)\le \frac{326+2x}{6}\left(6\right)<90\left(6\right)\\ 480\le 326+2x<540\end{array}$

Add $-326$ to each part of inequality.

  $\begin{array}{l}480-326\le 326+2x-326<540-326\\ 154\le 2x<214\end{array}$

Multiply each part of the inequality by $\frac{1}{2}$ .

  $\begin{array}{l}154\left(\frac{1}{2}\right)\le 2x\left(\frac{1}{2}\right)<214\left(\frac{1}{2}\right)\\ 77\le x<107\end{array}$

So the minimum score in the fifth test should be $77$ .

Therefore, the minimum score in the fifth test should be $77$ .