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Re: GMAT Diagnostic Test Question 31 [#permalink]
10 Jun 2009, 18:21

6

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Explanation Official Answer: B

Alcohol liter = 10% of 40 liters = 4 liters Water liter = 90% of 40 liters = 36 liters New ratio of alcohol = 20% of total solution. Water will be 80% of total. Therefore total = 36/0.8 = 45 liters The added alcohol is 5 liters.
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Re: GMAT Diagnostic Test Question 31 [#permalink]
21 Aug 2009, 11:12

5

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Another way to solve this: 4 + x = 0.2 (40+x), where 4= 40 litres*10% - current number of litres of alcohol x - number of litres of 100% alcohol to add 40+x - is our original solution plus x litres of 100% alcohol that we added since we need 20% solution, we mutiply 40+x by 0.2

Re: GMAT Diagnostic Test Question 31 [#permalink]
21 Aug 2009, 12:00

Expert's post

CasperMonday wrote:

Another way to solve this: 4 + x = 0.2 (40+x), where 4= 40 litres*10% - current number of litres of alcohol x - number of litres of 100% alcohol to add 40+x - is our original solution plus x litres of 100% alcohol that we added since we need 20% solution, we mutiply 40+x by 0.2

Thank you and welcome to GMAT Club!
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Re: GMAT Diagnostic Test Question 31 [#permalink]
26 Aug 2009, 21:31

2

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Expert's post

Another 10sec approach:

The solution contains 40*0.1=4 L of alcohol. In order to double the concentration of the solution, we can add 4L but doing so we increase slightly total volume. So, our answer will be something a bit larger than 4L. Let's look at our options... 5L is exactly what we need.
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Re: GMAT Diagnostic Test Question 31 [#permalink]
19 Dec 2009, 15:08

walker wrote:

Another 10sec approach:

The solution contains 40*0.1=4 L of alcohol. In order to double the concentration of the solution, we can add 4L but doing so we increase slightly total volume. So, our answer will be something a bit larger than 4L. Let's look at our options... 5L is exactly what we need.

Re: GMAT Diagnostic Test Question 31 [#permalink]
22 Dec 2009, 09:24

How many liters of pure alcohol must be added to a 40-liter solution that is 10% alcohol by volume in order to double the alcohol proportion?

A. 4 B. 5 C. 10 D. 20 E. 40

For me it helped to think of the question in terms of fractions and then plug in the answer choices find the new one.

So we know we have the fraction of alcohol to the rest of the solution is 10% or 4/40 and we need the new one to be 20% or 1/5, so the only one that works is B because 4+5/ 40+5 = 9/45=1/5. Not much math needed for this one.

Good ideas to answer this quicker. I noticed A was a trap and skipped it, but then planned to calculate the percentage using the remaining answer choices. I was just lucky that the first one I did was the right answer.

Re: GMAT Diagnostic Test Question 31 [#permalink]
09 Oct 2010, 09:14

3

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Amount of alcohol in 40 L of solution = 4L Say we add xL of alcohol to it, then amount of alcohol = (4 + x)L and amount of solution now becomes = (40 + x)L

The addition of xL of alcohol should increase the %age of alcohol to 20% (4+x)/(40+x) = 0.2

Re: GMAT Diagnostic Test Question 31 [#permalink]
02 Nov 2010, 23:32

My way of doing this was:

Since only the content of alcohol is changing(10%--->20%),volume of water remains same i.e 36.And now this volume of water as percentage would be 80% of solution.

So to get to 20% its 8 original mixture : 1 new mixture

40 ltrs : 5 ltrs

Ans: 5

Please explain this further. I don't follow.

I will try to explain this to my best.

Step 1, Basically you put the two proportions on each side, 10% vs. 100% (pure alcohol). Step 2, you put the desired ratio (20%) in the middle. Step 3, you subtract the 10% from 20% which gives you 10% (don't worry about signs), and also 100% - 20%. Step 4, after getting 80% and 10%, simplify the ratio. In this case, 8 to 1. Step 5, put an X in front of both 8 and 1. Step 6, solve for X, in this case we solve the 8x since we know that represents the amount in 40 Gallon. So, 8x=40, x=5. Step 7, jump to the other side (1x) and plug in X. 1(5)= 5, answer is B.

Its a nice solution. I solved it in traditional way of 4+x/40+x = 0.2 => x=5

soundofmusic wrote:

My way of doing this was:

Since only the content of alcohol is changing(10%--->20%),volume of water remains same i.e 36.And now this volume of water as percentage would be 80% of solution.

==> 80/100*(New Volume) = 36 ==> New Vol = 45.

Added alcohol amount is 45-40 = 5

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