One solution contains 25% as much water as alcohol and another solution contains 5 times as much water as alcohol. How many gallons of each solution must be mixed in order to obtain 7 gallons of the new solution which contains as much water as alcohol?

Question

One solution contains 25% as much water as alcohol and another solution contains 5 times as much water as alcohol. How many gallons of each solution must be mixed in order to obtain 7 gallons of the new solution which contains as much water as alcohol?

Please help. I don't know how to solve this problem.

Answer 1

Anne/Jeysi/Ben -- please use the same name for your posts.

Answer 2

call the solutions A and B.

If there are x gal of A, then there are 7-x gal of B.

So, if we keep track of the alcohol amounts, we have

4/5 x + 1/6 (7-x) = 1/2 (7)

Now just solve for x, the amount of solution A.

Answer 3

To solve this problem, we can set up a system of equations:

Let's assume x represents the amount of alcohol (in gallons) in the first solution.

According to the problem, the first solution contains 25% water and 75% alcohol. So, the amount of water in the first solution is 0.25x gallons.

The second solution contains 5 times as much water as alcohol. So, the amount of water in the second solution is 5x gallons, and the amount of alcohol is x/5 gallons.

To obtain the new solution containing 7 gallons with equal amounts of water and alcohol, we add the amounts of water and alcohol from the two solutions.

Total water = 0.25x + 5x = 7 gallons
Total alcohol = x + (x/5) = 7 gallons

Simplifying the equations, we have:
1.25x = 7 (equation for total water)
6x/5 = 7 (equation for total alcohol)

We can solve this system of equations to find the value of x.

Multiplying the second equation by 5/6, we have:
x = (35/6)

Now, substituting this value of x back into the equations, we can find the amounts of water and alcohol in each solution.

Amount of water in the first solution = 0.25 * (35/6) = 35/24 gallons
Amount of water in the second solution = 5 * (35/6) = 175/6 gallons

Amount of alcohol in the first solution = (35/6) gallons
Amount of alcohol in the second solution = (35/6) / 5 = 7/6 gallons

Therefore, to obtain 7 gallons of the new solution containing equal amounts of water and alcohol, we need to mix:
35/24 gallons of the first solution and 175/6 gallons of the second solution.