SOLUTION: You have 20 gallons of a 45% antifreeze solution. How many gallons of a 57% antifreeze solution needs to be added to make a 51% antifreeze solution?

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Question 1170291: You have 20 gallons of a 45% antifreeze solution. How many gallons of a 57% antifreeze solution needs to be added to make a 51% antifreeze solution?
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Add v gallons of the 57% antifreeze.

%280.45%29%2820%29%2B%280.57%29v=%280.51%29%28v%2B20%29
Solve.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Short answer....

51% is exactly halfway between 45% and 57%, so the two ingredients need to be mixed in equal amounts.

ANSWER: 20 gallons.

With formal algebra....

45% of 20 gallons, plus 57% of x gallons, makes 51% of (20+x) gallons:

.45%2820%29%2B.57%28x%29+=+.51%2820%2Bx%29

Solve using basic algebra....

Answer by ikleyn(52848) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x be the volume of the 57% antifreeze solution to add.


The governing equation is


    0.45*20 + 0.57x = 0.51*(20 + x)


saying that sum of the antifreeze volumes in ingredient is equal to the antifreeze volume in the final mixture.


From the equation


    x = %280.51%2A20+-+0.45%2A20%29%2F%280.57-0.51%29 = 20.


ANSWER.  20 gallons of the 57% antifreeze solution should be added.


CHECK.  I am checking for the concentration  C = %280.45%2A20%2B0.57%2A2.312%29%2F%2820%2B20%29 = 0.51 = 51%.

Solved.

From the other side,  the answer is  OBVIOUS since  51%  is half-way between  45%  and  57%.