Question

section 3.4 motion, money, and mixture problems

1. how many milliliters of a 30% salt solution must be added to 7500 milliliters of a 5% salt solution to yield 10% salt solution?
2. how many liters each of a 6% acid solution and a 30% acid solution should be mixed in order to obtain 10 liters of a 15% acid solution?

Explanation:

Step1: Set up equations for problem 6

Let $x$ be the volume (in milliliters) of the 30% salt - solution to be added. The amount of salt in the 30% solution is $0.3x$, the amount of salt in 7500 milliliters of 5% salt - solution is $0.05\times7500 = 375$ milliliters, and the final volume is $(7500 + x)$ milliliters with a 10% salt - concentration. So the equation is $0.3x+375 = 0.1(7500 + x)$.

Step2: Solve the equation for problem 6

Expand the right - hand side: $0.3x+375=750 + 0.1x$.
Subtract $0.1x$ from both sides: $0.3x-0.1x+375=750 + 0.1x-0.1x$, which gives $0.2x+375 = 750$.
Subtract 375 from both sides: $0.2x+375 - 375=750 - 375$, so $0.2x = 375$.
Divide both sides by 0.2: $x=\frac{375}{0.2}=1875$ milliliters.

Step3: Set up equations for problem 7

Let $x$ be the volume (in liters) of the 6% acid - solution and $y$ be the volume (in liters) of the 30% acid - solution. We have two equations: $x + y=10$ (total volume) and $0.06x+0.3y = 0.15\times10$ (total amount of acid).

Step4: Solve the system of equations for problem 7

From $x + y=10$, we get $x = 10 - y$.
Substitute $x = 10 - y$ into $0.06x+0.3y = 1.5$: $0.06(10 - y)+0.3y = 1.5$.
Expand: $0.6-0.06y+0.3y = 1.5$.
Combine like terms: $0.24y=1.5 - 0.6$.
$0.24y = 0.9$.
$y=\frac{0.9}{0.24}=3.75$ liters.
Then $x = 10 - 3.75 = 6.25$ liters.

Answer:

1. 1875 milliliters
2. 6.25 liters of 6% acid - solution and 3.75 liters of 30% acid - solution