Question

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?

Answer 1

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 

Answer 2

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) = $\frac{13(x-100)}{100}$ = 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 = $\frac{360}{0.30}$ = 1200

Hence, the quantity of solution b is 1200 ml.