In the lab, Keith has two solutions that contain alcohol and is mixing them with each other. He uses twice as much Solution A as Solution B. Solution A is 19% alcohol and Solution B is 14% alcohol. How many milliliters of Solution B does he use, if the resulting mixture has 156 milliliters of pure alcohol

Quantity of solution B required is 300 ml

Step-by-step explanation:

Let "a" be quantity of solution A required

Let "b" be quantity of solution B required

We are given Solution A is 19% alcohol and Solution B is 14% alcohol and that the resulting mixture has 156 milliliters of pure alcohol

Thus;

0.19a + 0.14b = 156 - - - (eq 1)

Also, He uses twice as much Solution A as Solution B.

Thus;

a = 2b

Putting 2b for a in eq 1 gives;

0.19(2b) + 0.14b = 156

0.38b + 0.14b = 156

0.52b = 156

b = 156/0.52

b = 300 ml

Quantity of solution B required is 300 ml

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

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step-by-step explanation:

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