How many ounces of 10% alcohol would need to be mixed with pure rubbing alcohol to get 16

ounces of a 14% alcohol mix?

Suppose you add x oz of 10% alcohol to y oz of pure alcohol. Then the mixture has a total volume of x + y oz, and we want to end up with 16 oz so that

x + y = 16

For each oz of the solution 10% used, 0.1 oz of alcohol is contributed, and each oz of pure alcohol contributes 1 oz of alcohol. The mixture is supposed to have a concentration of 14%, which comes out to 0.14*16 = 2.24 oz of alcohol. Then

0.1 x + 1 y = 2.24

Solve for y in both equations:

y = 16 - x

y = 2.24 - 0.1 x

Set them equal to one another and solve for x, then for y.

16 - x = 2.24 - 0.1 x

13.76 = 0.9 x

x = 13.76/0.9 ≈ 15.29

y = 16 - 15.29 ≈ 0.71

So you need about 15.29 oz of 10% alcohol and 0.71 oz of pure alcohol to get the desired mixture.

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.

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