Let the replaced solution containing 80% alcohol be X gallons
So, 10-X gallons contain 15% alcohol
X gallon contain 80% alcohol
(10-X)*(15/100) + X*(80/100) = 10*70/100
150-15x+80x = 700
65x = 550
X = 110/13

Answer is B

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Think of it like this:

You start with 1.5 gallons of alcohol, and you want to end with 7 gallons of alcohol.
To achieve this, you have 2 levers:
1. take out 15% volume alcohol
2. add 80% volume alcohol

Since the question says how much liquid should be 'replaced', you have the same amount getting added.

Thus 1.5 - .15X + .8X = 7
X = 110/13

Solution:

We see that currently there are 1.5 gallons of alcohol in the 10-gallon solution. Let x be the number of gallons of an 80% alcohol solution needed to replace the 15% alcohol solution to yield a 70% alcohol solution. So 0.15x gallons of alcohol from the original solution will be removed and 0.8x gallons of alcohol from the new solution will be added. We can create the equation:

(1.5 - 0.15x + 0.8x)/10 = 70/100

(1.5 + 0.65x)/10 = 7/10

1.5 + 0.65x = 7

0.65x = 5.5

x = 5.5/0.65 = 550/65 = 110/13

Alternate Solution:

We have 10 gallons of a 15% solution. From the 10 gallons, we will remove x gallons of this 15% solution and replace the removed solution with x gallons of 80% solution, and the result will be 10 gallons of a 70% solution. We can summarize these actions in the following equation:

10 (0.15) - x (0.15) + x (0.80) = 10 (0.70)

Now, let’s solve for x.

1.5 - 0.15x + 0.80x = 7

0.65x = 5.5

65x = 550

x = 550/65 = 110/13 gallons


Answer: B
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