SOLUTION: How much of a saline solution that is 20% saline should be mixed with 200 ounces of a saline solution that is 50% saline to get a solution that is 30% saline? State your solution i
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: How much of a saline solution that is 20% saline should be mixed with 200 ounces of a saline solution that is 50% saline to get a solution that is 30% saline? State your solution i
Log On
Question 1156222: How much of a saline solution that is 20% saline should be mixed with 200 ounces of a saline solution that is 50% saline to get a solution that is 30% saline? State your solution in a complete sentence. Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52802) (Show Source):
I think, it is totally clear to you, as it is clear to me, that this problem came not from a textbook.
Hence, it was created by somebody, who DOES NOT KNOW the subject.
Tomorrow this incorrect problem will be placed to a somebody's web-site, and then this FALSE INFORMATION / DISINFORMATION
will be spread AS A VIRUS over the Internet.
In couple of years the entire Internet will be one huge GARBAGE BOX with no bounds.
I am capable of forgiving the author of the problem for not knowing that saline solutions of those percentages are not possible. Clearly the purpose of the problem is to learn how to solve mixture problems. So let's change the saline solutions to antifreeze or acid and work the problem.
Algebraically....
20% of x ounces, plus 50% of 200 ounces, makes 30% of (x+200) ounces:
ANSWER: 400 ounces
There is a MUCH easier way to solve problems like this, if a formal algebraic solution is not required.
Here is the complete solution, in two sentences:
30% is "twice as close to 20% as it is to 50%"; therefore the amount of the 20% ingredient is twice the amount of the 50% ingredient".
200 ounces of 50% means 2*200 = 400 ounces of 20%.
