Plugging in is a wonderful way to wrap your head around a problem that is initially difficult to conceptualize. My initial approach was plugging in and common sense.
Let's say we have 100 of 50% acid, so there are 50 liters of pure acid in that mixture. After replacing a set amount, we're down to 40% - 40 liters of acid. So in the process, we should lose 10 liters of acid.
How do we do that? for every 10 liters of 50% (which hold five liters of acid) we replace with 10 liters of 30% (3 liters), we lose 2 liters of acid. So if we need to lose a total of 10 liters of acid, we need to replace 5*10 liters, at 2 liters lost for every batch. Bottom line, replace 50 liters of acid out of 100, or 1/2.
alternative approach, slightly more difficult to see, but it works: replacing 50% with 30% creates a mixture that is a weighted average of the new 30% solution and the remaining 50% solution. If the result is a straight-down-the-middle 40% average, it means we need equal amount of 30% and 50% - or half and half.
GOOD MIXTURE SUM
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Geva@EconomistGMAT
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Geva@EconomistGMAT
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Plug in 100 for the solution, and count the real number of liters in every step.
We begin with 100 liters solution, out of which 50% is acid (and the rest water, for instance) - 50 liters of acid.
The end result after the replacement is 40%, or 40 liters of acid out of 100 liters. This means we need replace enough 50% with 30% to lose 50-40 = 10 liters of acid.
Let's see what happens when we do the replacement:
Take a liter of 50% solution out - half of that liter is acid, so you have just lost 0.5 liter of acid.
Replace it with one liter of 30% solution - 0.3 of that liter is real acid, so you have just added 0.3 liter of real acid back. You still have 100 liters of solution, but you have less acid in there - you've taken 0.5 liters, but replaced only 0.3 liters.
thus, for every liter of solution you replace, you are effectively taking out 0.5-0.3=0.2 liters of acid, counted towards your goal of taking out 10 liters of acid.
Work in batches of replacing 10 liters to avoid the decimals:
if for every liter you replace you lose 0.2 liters of acid,
then for every 10 liters your replace you lose 2 liters of acid.
Since you want to lose 10 liters of acid overall, you need to replace five batches of 10 liters: every batch of 10 loses 2 liters of acid, so you need 5 batches of 10 to lose the required 10 liters of acid. 50 liters overall will get you to 40 liters of real acid from the same 100 liters of solution, or 40%.
We begin with 100 liters solution, out of which 50% is acid (and the rest water, for instance) - 50 liters of acid.
The end result after the replacement is 40%, or 40 liters of acid out of 100 liters. This means we need replace enough 50% with 30% to lose 50-40 = 10 liters of acid.
Let's see what happens when we do the replacement:
Take a liter of 50% solution out - half of that liter is acid, so you have just lost 0.5 liter of acid.
Replace it with one liter of 30% solution - 0.3 of that liter is real acid, so you have just added 0.3 liter of real acid back. You still have 100 liters of solution, but you have less acid in there - you've taken 0.5 liters, but replaced only 0.3 liters.
thus, for every liter of solution you replace, you are effectively taking out 0.5-0.3=0.2 liters of acid, counted towards your goal of taking out 10 liters of acid.
Work in batches of replacing 10 liters to avoid the decimals:
if for every liter you replace you lose 0.2 liters of acid,
then for every 10 liters your replace you lose 2 liters of acid.
Since you want to lose 10 liters of acid overall, you need to replace five batches of 10 liters: every batch of 10 loses 2 liters of acid, so you need 5 batches of 10 to lose the required 10 liters of acid. 50 liters overall will get you to 40 liters of real acid from the same 100 liters of solution, or 40%.
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Asif
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Hi rb90: I have a different approach, we can derive the following equation from the question stem
50% * (1-x) + 30%x = 40%*1 ( where x is the fraction amount replaced and 1 is the entire solution)
Solving for x:
.5 - .5x + .3x = .4 or,
-.2x = -.1 or,
x = .1/.2 or,
x=1/2
I think it works.
Thanks.
50% * (1-x) + 30%x = 40%*1 ( where x is the fraction amount replaced and 1 is the entire solution)
Solving for x:
.5 - .5x + .3x = .4 or,
-.2x = -.1 or,
x = .1/.2 or,
x=1/2
I think it works.
Thanks.
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GMATGuruNY
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An easy approach would be to plug in the answer choices, which represent the fraction of 30% solution that will be needed. We should start with C, the answer choice in the middle.
Answer choice C: 1/2 30% solution, 1/2 50% solution
Let's plug in that we have 50 gallons of the 30% solution and 50 gallons of the 50% solution.
.3*50 = 15 alcohol.
.5*50 = 25 alcohol.
Total alcohol = 25+15 = 40.
Total solution = 50+50 = 100.
40/100 = 40%. Success!
The correct answer is C.
Answer choice C: 1/2 30% solution, 1/2 50% solution
Let's plug in that we have 50 gallons of the 30% solution and 50 gallons of the 50% solution.
.3*50 = 15 alcohol.
.5*50 = 25 alcohol.
Total alcohol = 25+15 = 40.
Total solution = 50+50 = 100.
40/100 = 40%. Success!
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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