Chapter 1.6, Problem 59E

Mixture Problem A bottle contains 750 mL of fruit punch with a concentration of 50% pure fruit juice. Jill drinks 100 mL of the punch and then refills the bottle with an equal amount of a cheaper brand of punch. If the concentration of juice in the bottle is now reduced to 48%, what was the concentration in the punch that Jill added?

To determine

To find: The concentration of cheaper brand in the punch.

Answer to Problem 59E

The concentration of cheaper brand is 35% in the punch.

Explanation of Solution

Given:

The concentration of fruit juice is 50% in a bottle of 750 ml fruit punch.

Jill drinks 100 ml of fruit punch and refills the bottle with equal amount of cheaper brand of punch.

The concentration of juice now in bottle is 48%.

Calculation:

Let the concentration of cheaper brand be x.

Tabulate the given information into the language of algebra.

In words	In Algebra
Concentration of cheaper brand	x
Amount of fruit juice before refill	$0.50\left(750-100\right)$
48% of concentration of juice	$0.48\left(750\right)$

Model the equation for the above information.

$\begin{array}{c}\left(\begin{array}{l}\text{Amount}\text{juice}\text{from}\\ \text{cheaper}\text{brand}\end{array}\right)+\left(\begin{array}{l}\text{Amount}\text{of}\text{juice}\\ \text{before}\text{refill}\end{array}\right)=\text{48\%}\text{of}\text{concentrartion}\\ 100x+0.50\left(750-100\right)=0.48\left(750\right)\end{array}$

Simplify the above equation for x,

$\begin{array}{c}100x+0.50\left(100\right)=360\\ 100x+325=360\\ 100x=35\\ x=35\%\end{array}$

Thus, the concentration of cheaper brand is 35% in the punch.