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rvc27
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May I propose another approach here? That, because coming from a psychology background and pretty much learning math through my gmat practice, the more mathematical approaches are still sometimes difficult for me.

So, as we said above, we have 5 gallons of solution, with 0.4 gallons of vinegar (8% of solution) and 4.6 gallons of water (92% of solution). Now, we can say that if the vinegar is the new solution is 4% of the total solution, then water must be 98%.

Then, we can say: 4.6 gallons of water is 92% of the solution - x gallons are 98%? Or..

4.6 ---> 92
x ---> 98

92x=4.6*98
92x=450.8
x=450.8/92
x= 4.9, so rounding up to 5.
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Initial .................. Vinegar .................... Water ............... Total

8% Vinegar ............ \(\frac{40}{100}\) ........................ \(\frac{460}{100}\) .............. 5

"x" water added .... \(\frac{40}{100}\) ....................... \(\frac{460}{100} + x\) ......... 5+x

Given that post addition of water, vinegar concentration should decrease to 4%

\(\frac{4}{100} (5+x) = \frac{40}{100}\)

x = 5

Answer = B
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rvc27
Five gallons of a solution of vinegar and water with 8% vinegar is to be diluted with water to make a 4% vinegar mixture. How many gallons of water should be added?

A. 10
B. 5
C. 4
D. 2
E. 1

This is probably a very easy question but I have an extreme aversion to raitos, so any help is appreciated. :|

Source: Bell Curves

Yes, the question can be very easily solved by considering it a mixture problem where 8% vinegar solution is mixed with 0% vinegar solution (pure water) to make a 4% vinegar solution.


w1/w2 = (A2 - Aavg)/(Aavg - A1) = (0 - 4)/(4 - 8) = 1/1
So weights of both solutions should be same. We should put 5 gallons of water in 5 gallons of 8% vinegar solution.
Answer (B)
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rvc27
Five gallons of a solution of vinegar and water with 8% vinegar is to be diluted with water to make a 4% vinegar mixture. How many gallons of water should be added?

A. 10
B. 5
C. 4
D. 2
E. 1

let w=gallons of water to be added
.08*5=.04(5+w)
w=5 gallons
B
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rvc27
Five gallons of a solution of vinegar and water with 8% vinegar is to be diluted with water to make a 4% vinegar mixture. How many gallons of water should be added?

A. 10
B. 5
C. 4
D. 2
E. 1

This is probably a very easy question but I have an extreme aversion to raitos, so any help is appreciated. :|

Source: Bell Curves
Let x = amount of water to be added
A = solution with 8% vinegar
B = water, which 0.0 percent vinegar

(% A)(Vol A) + (% B)(Vol B) = (% of A+B)(Vol A+B)

You do not need the 0x on LHS. It's a reminder that water IS being added, but that its concentration /percent vinegar has no weight.

\(.08(5) + 0.0(x) = .04(5 + x)\)
\(.40 + 0x = .20 + .04x\)
\(.20 = .04x\)
\(x = \frac{.20}{.04}=\frac{20}{5}=5\)

Answer B
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rvc27
Five gallons of a solution of vinegar and water with 8% vinegar is to be diluted with water to make a 4% vinegar mixture. How many gallons of water should be added?

A. 10
B. 5
C. 4
D. 2
E. 1

This is probably a very easy question but I have an extreme aversion to raitos, so any help is appreciated. :|

Source: Bell Curves


Initial vinegar content = 8%

Final vinegar content = 4%

Vinegar content in the additional water that is added = 0% (the water is 100% water and 0% vinegar)

Thus the ratio of the initial mixture and water that is added = (4-0) : (8 - 4) = 1 : 1

Thus in the 5 gallons of initial mixture, you will have to add another 5 gallons of water to get a 4% vinegar in the final mixture.

Correct answer is Option B.

Note: The method that is used here is Alligation, which is similar to weighted averages, but faster to calculate. :)


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Saquib
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