In the lab, frank has two solutions that contain alcohol and is mixing them with each other. he uses 100 milliliters less of solution a than solution
b. solution a is 13% alcohol and solution b is 17% alcohol. how many milliliters of solution b does he use, if the resulting mixture has 347 milliliters of pure alcohol?

Solution B (17% alcohol) = x ml
Solution A (13% alcohol) = (x-100) ml

0.17x ml alcohol in Solution B
0.13(x-100) ml alcohol in Solution A

0.17x + 0.13(x-100) = 347
0.17x+0.13x-13=347
0.3x = 360
x=360/0.3=3600/3=1200 ml solution B

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slicergiza slicergiza · Answer 2 · 2019-09-20T10:43:39+02:00

Answer:

1200 ml

Explanation:

Let x be the quantity ( in ml ) of b solution,

∵ solution a is 100 milliliters less of than solution b.

So, the quantity of solution a = 100 - x,

Now, solution a has 13% alcohol,

∴ Alcohol in solution a = 13% of (100-x) = [tex]\frac{13(x-100)}{100}[/tex] = 0.13(x-100),

While solution b has 17% alcohol,

∴ Alcohol in solution b = 17% of x = 0.17x,

So, the quantity of alcohol in the mixture of a and b = 0.13(x-100) + 0.17x

= 0.13x - 13 + 0.17x

= 0.30x - 13

According to the question,

Total quantity of alcohol in the mixture = 347 ml,

⇒ 0.30x - 13 = 347

⇒ 0.30x = 347 + 13

⇒ 0.30x = 360

⇒ x = [tex]\frac{360}{0.30}[/tex] = 1200

Hence, the quantity of solution b is 1200 ml.