word problem

Question

You need 1200 mL of a 30% alcohol solution. On hand, you have a 20% alcohol mixture. How much of the 20% alcohol mixture and pure alcohol will you need to obtain the desired solution?

You will need ____mL of the 20% solution
and _____mL of pure alcoh

Answer 1

Let X represent the amount of the 20% alcohol solution.

Let Y represent the amount of pure alcohol.

 

X + Y = 1200

 

0.2*X + 1*Y = 1200 * 0.3; 20% = 0.2,  100% = 1, and 30% = 0.3 <--- amount alcohol

 

0.2*X + Y = 360

 

 

Solving the first equation for Y,  Y = 1200 - X

Substitutes this into the second equation:

0.2X + (1200 - X) = 360

 

1200 - 0.8X = 360

 

12000 - 8x = 3600

 

12000 - 3600 = 8x

 

8400 = 8x

 

x = 8400/8 = 4200/4 = 2100/2 = 1050

 

y = 1200 - 1050  = 150

 

You will need 1050 mL of the 20% alcohol solution

and 150 mL of pure alcohol.

 

CHECK:

   20% of 1050 = 0.2 x 1050 = 210 mL of alcohol in the first solution.

 

  So there is a total of 210 + 150 = 360 mL of alcohol.

 The total volume of the mixture is 1050+150 = 1200 mL

 

 360/1200 = 36/120 = 18/60 = 9/30 = 3/10 = 0.3  = 30%

 

The answer is n bold above, checked and PROVEN to be correct

 

 

Answer 2

a represents the amount of pure alcohol

1200 - a represents the amount of 20%

 

0.20(1200 - a ) + 1.00a = 0.30(1200)

 

Can you solve for a and answer?