SOLUTION: Manuel has 400 mL of a 100% antifreeze solution. how many mL of a 30% antifreeze solution should be added to get a 50% solution of the antifreeze?
a. 250 mL
b. 300 mL
c. 450
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a. 250 mL
b. 300 mL
c. 450
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Question 1205616: Manuel has 400 mL of a 100% antifreeze solution. how many mL of a 30% antifreeze solution should be added to get a 50% solution of the antifreeze?
a. 250 mL
b. 300 mL
c. 450 mL
d. 1000 mL
e. 1500 mL Found 2 solutions by ikleyn, math_tutor2020:Answer by ikleyn(52921) (Show Source):
You can put this solution on YOUR website! .
Manuel has 400 mL of a 100% antifreeze solution.
how many mL of a 30% antifreeze solution should be added
to get a 50% solution of the antifreeze?
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Let x be the volume of the 30% antifreeze solution to add.
Then you have this equation
= 0.5.
The denominator id the volume of the mixture; the numerator is the volume of the pure antifreeze.
Simplify and solve to find x
400 + 0.3x = 0.5*(400+x)
400 + 0.3x = 200 + 0.5x
400 - 200 = 0.5x - 0.3x
200 = 0.2x
x = 200/0.2 = 2000/2 = 1000.
ANSWER. 1000 mL of the 30% antifreeze solution should be added.
Solved.
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It is a standard and typical mixture word problem.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
Batch A = 100% antifreeze
Batch B = 30% antifreeze
Goal is to mix A and B to form a new batch C that is 50% antifreeze.
400 = number of mL of batch A
x = number of mL of batch B
0.30x = number of mL of pure antifreeze from batch B.
m = 400+0.30x = number of mL of pure antifreeze from both batches mixed together.
n = 400+x = total amount of solution (antifreeze plus other chemicals).
m/n = 0.50 to represent getting a 50% solution
(400+0.30x)/(400+x) = 0.50
400+0.30x = 0.50(400+x)
400+0.30x = 200+0.50x
400-200 = 0.50x-0.30x
200 = 0.20x
x = 200/0.20
x = 1000 mL is the final answer
The check section is at the bottom.
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Method 2
I'll use the concept of alligation (not to be confused with the word "allegation") which is often used in pharmacy contexts.
Here's an article talking about it along with some examples
https://en.wikipedia.org/wiki/Alligation
The gap from A to the target is 50% (since 100-50 = 50)
The gap from B to the target is 20% (since 50-30 = 20)
Ignoring the percent signs the ratio of 50:20 reduces to 5:2
It means we have 5 parts of A and 2 parts of B.
Or we might have 5 parts of B and 2 parts of A.
The order at this point isn't clear.
Let's assume it's the first scenario.
If so then
5/2 = 400/x
solves to
x = 160
I'll leave the steps for the student to do.
Luckily this isn't one of the answer choices so we can eliminate this scenario.
But if you didn't have multiple choice answers to rely on, then here's how we can eliminate this scenario.
If we had 400 mL of A and 160 mL of B then we'll have 1*400+0.30*160 = 448 mL of pure antifreeze out of 400+160 = 560 mL total solution.
The ratio is 448/560 = 0.8 which converts to 80% antifreeze instead of the 50% we want.
This is why we can eliminate the first scenario.
The other scenario must be the case.
5/2 = x/400
solves to
x = 1000
I'll leave the steps for the student to do.
If we had 400 mL of A and 1000 mL of B then we'll have 1*400+0.30*1000 = 700 mL of pure antifreeze out of 400+1000 = 1400 mL total solution.
The ratio of pure antifreeze to total solution is 700/1400 = 0.50 converting to the 50% solution we want.
The answer is confirmed.
More practice with a similar question can be found here:
https://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.1204463.html
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