How many ounces of a 14​% alcohol solution must be mixed with 4 ounces of a 20​% alcohol solution to make a 18​% alcohol​ solution?

Answer 1

Hi Destiney,

How many ounces of a 14​% alcohol solution must be mixed with 4 ounces of a 20​% alcohol solution to make a 18​% alcohol​ solution?

To solve this problem, first organize the information given, and what the question is asking for

Given:

a 14% solution

4 oz of a 20% solution

Solution made is an 18% solution

Find:

The number of ounces of 14% solution needed

Let

x₁ = the amount in ounces of the 14%solution

x₂ = the amount of the 20% solution

x_T = The total amount of the 18% solution

The total amount of 18% solution would be the amount of the 14% solution we add plus the amount of the 20% solution.

So x_T = x₁ + x₂

x₂ was given to us. It is 4 oz

x_T = x₁ + 4 oz

The equation for the amount of solution we need is as follows

x₁(0.14) + x₂(0.20) = x_T(0.18)

Now substitute in for x₂ and x_T = x₁ + 4 oz

x₁(0.14) + (4 oz)(0.20) = (x₁+4 oz)(0.18)

0.14 x₁ + 0.8 = 0.18 x₁ + 0.72

0.08 = 0.04 x₁

x₁= 2

So you would need 2 oz of the 14% solution to add to 4 oz of the 20% solution to get an 18% solution.

We can check to see if this works by determining the percent of alchohol of the final solution.

% of final solution = (amount of 1st solution x percent + amount of 2nd solution x percent) / (total amount of solution)

[ x1 (0.14) + 4oz (0.2) ] / [x_t]

[ 2oz (0.14) + 4oz (0.2) ] / [2oz + 4 oz]

1.08 / 6

= 0.18 which is 18%

So our solution checks out.