Author 
Message 
VP
Joined: 18 May 2008
Posts: 1260

2
This post received KUDOS
4
This post was BOOKMARKED
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced? (A) \(\frac{1}{5}\) (B) \(\frac{1}{4}\) (C) \(\frac{1}{2}\) (D) \(\frac{3}{4}\) (E) \(\frac{4}{5}\) Source: GMAT Club Tests  hardest GMAT questions Can sum1 give a simpler explanation?



SVP
Joined: 17 Jun 2008
Posts: 1547

Re: m25,#30 [#permalink]
Show Tags
01 Dec 2008, 07:11
4
This post received KUDOS
2
This post was BOOKMARKED
Look at it this way.
Total part is 1. Out of this, x part was replaced.
Hence, (1x)*50 (percent) + x*30 = 1*40 or, 50  20x = 40 or, 10 = 20x or, x = 1/2....or, 50% of original solution was replaced.



Intern
Joined: 19 Jan 2009
Posts: 22

A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?
(A) 20% of all chips in the basket are green (B) The ratio of the number of red chips to the number of green chips is 4:1
I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.
The correct answer is E.
Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: m25, 30 [#permalink]
Show Tags
03 Feb 2009, 12:27
2
This post received KUDOS
1
This post was BOOKMARKED
jairus wrote: A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?
(A) 20% of all chips in the basket are green (B) The ratio of the number of red chips to the number of green chips is 4:1
I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.
The correct answer is E.
Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we? A: you need the total number of chips. ok, 20% of all chips in the basket are green. total = 100 green = 20 red = 80 the prob = (20/100) x (19/99) = 19/(5x99) suppose the total = 20 green = 4 red = 16 the prob = (4/20) x (3/19) = 3/(5x19) see how the prob differes with the number of chips in the bowl. (B) again, its same with B too even if the ration of red to green is 4:1. try with some number. togather also not clear. Therefore, it is E.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Current Student
Joined: 04 Nov 2009
Posts: 78
Schools: NYU Stern '12
WE 1: Business Development
WE 2: Superhero shhhhhh

Re: m25,#30 [#permalink]
Show Tags
17 Dec 2009, 23:55
ritula wrote: Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(C) 2008 GMAT Club  m25#30
Can sum1 give a simpler explanation? I understand the alegbraic approach here but can this question be solved using more basic math principles? If a solution goes from 50% to 40% in a mixture with 30% of some proportion, doesnt logic dictate that the mixture be 50/50? i understand this approach doesnt work with many other mixture problems, but in this case where equal amounts are used and it is asking about proportions and not the overall amounts used, wouldnt it work? thoughts?
_________________
"Any school that meets you and still lets you in is not a good enough school to go to"  my mom upon hearing i got in Thanks mom.



Intern
Joined: 18 Aug 2010
Posts: 29
Concentration: Entrepreneurship, Finance
GPA: 3.6
WE: Engineering (Telecommunications)

Re: m25,#30 [#permalink]
Show Tags
14 Sep 2010, 17:06
2
This post received KUDOS
ritula wrote: Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced? (A) \(\frac{1}{5}\) (B) \(\frac{1}{4}\) (C) \(\frac{1}{2}\) (D) \(\frac{3}{4}\) (E) \(\frac{4}{5}\) Can sum1 give a simpler explanation? Number and Answer back substitution works quite well for these kind of problems. First convert % to a number(100 usually) and also choose a quantity, like ml. Say you had 100ml of the solution with 50ml acid and 50ml water, and you want to end up with 40ml of acid(40%) Pick Choice A  1/5 1/5 solution = 1/5(100ml) = 20ml. Because you remove a a 50/50 solution, you remove 10ml acid and 10ml water. Because you add a 30/70 solution, it's 6ml acid(30% of 20ml) and 14ml(70% of 20ml) water. So you have 50ml(original 50/50)  10ml(50/50) + 6ml(30/70) = 46ml acid  this is different from the 40% You also know that you didn't remove enough of the original solution, so skip B and jump to choice C. (I usually start with choice C anyway for answer substitution) Pick Choice A  1/2 1/2 solution = 1/2(100ml) = 50ml. Because you remove a a 50/50 solution, you remove 25ml acid and 25ml water. Because you add a 30/70 solution, it's 15ml acid(30% of 50ml) and 35ml water(70% of 20ml). So you have 50ml(original 50/50)  25ml(50/50) + 15ml(30/70) = 40ml acid  this is 40%, and is hence the correct answer.



Intern
Joined: 18 Aug 2010
Posts: 29
Concentration: Entrepreneurship, Finance
GPA: 3.6
WE: Engineering (Telecommunications)

Re: m25, 30 [#permalink]
Show Tags
14 Sep 2010, 17:10
jairus wrote: A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?
(A) 20% of all chips in the basket are green (B) The ratio of the number of red chips to the number of green chips is 4:1
I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.
The correct answer is E.
Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we? You came up with the answer 0 for 4:1. Now if you choose red:green=8:2 (which is equivalent to the ratio 4:1), you get an answer(2/10*1/9) which is different from 0. So its E



Intern
Joined: 02 Apr 2010
Posts: 47
Location: Mumbai

Re: m25,#30 [#permalink]
Show Tags
15 Sep 2010, 04:18
mustafaj wrote: ritula wrote: Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(C) 2008 GMAT Club  m25#30
Can sum1 give a simpler explanation? I understand the alegbraic approach here but can this question be solved using more basic math principles? If a solution goes from 50% to 40% in a mixture with 30% of some proportion, doesnt logic dictate that the mixture be 50/50? i understand this approach doesnt work with many other mixture problems, but in this case where equal amounts are used and it is asking about proportions and not the overall amounts used, wouldnt it work? thoughts? In this particular question, before replacement, solution is 50% and after replacement it is 40%. Since we are given that the added solution is 30%, we can think of basic average formula which is (50+30)/2 = 40. Hence both 50% and 30% solutions should be same and is equal to half.
_________________
Consider kudos for good explanations.



Intern
Joined: 12 Apr 2010
Posts: 4

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2010, 13:54
A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?
(A) 20% of all chips in the basket are green (B) The ratio of the number of red chips to the number of green chips is 4:1
I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.
The correct answer is E.
Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?
Addressing your problem  your intent is right, however what you have to realise is that you are only provided with the ratios in the quesion and not specefic values. It is easy assume that 20 out of 100 or 2 out of 10 chips are green, which is not the case here.
Option 1 says  20% of chips are green. Now this info is clearly not enough as we don't know total number of chips. So we can't say whether number of green chips are 20 or 200 or 200000. So choice 1 is clearly insufficient.
Now if you get my point above, you'd realise that Option 2 is just a rewording of the option 1, i.e. if the ratio is 4:1, then green chips are certainly 20% of the total.
I hope you understand why D is wrong. D is incorrect since we do not have values and only have ratios. If the question had asked us to pick just one chip, D wld have been correct.
Cheers!



Senior Manager
Joined: 11 May 2011
Posts: 361
Location: US

Re: m25,#30 [#permalink]
Show Tags
15 Sep 2011, 07:16
1
This post received KUDOS
krishnasty wrote: Although i got the ans through conventional method, can somebody please solve this question through amalgation method used in mixture problems? This one is cake. Answer C. Can be solved on 30 sec. It is just becasue data is very simple. 50% solution of acid and 30% solution of acid = 40% solution of acid  Straight average of both acid is required. Hence, 5050 of both. Answer C. Hope it helps. Cheers! P.S.  This is applicable for all the questions in which final concentration is becoming average concentration of 2 solutions which are getting mixed. 10% solution of acid and 30% solution of acid = 20% solution of acid 20% solution of acid and 40% solution of acid = 30% solution of acid
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Manager
Joined: 20 Nov 2010
Posts: 218

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2011, 02:22
1
This post was BOOKMARKED
Yup its C. Since (50+30) /2= 40 => The resulting solution should have half of each of the solutions.
_________________
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ MGMAT 6 650 (51,31) on 31/8/11 MGMAT 1 670 (48,33) on 04/9/11 MGMAT 2 670 (47,34) on 07/9/11 MGMAT 3 680 (47,35) on 18/9/11 GMAT Prep1 680 ( 50, 31) on 10/11/11
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CR notes http://gmatclub.com/forum/massivecollectionofverbalquestionsscrcandcr106195.html#p832142 http://gmatclub.com/forum/1001dsquestionsfile106193.html#p832133 http://gmatclub.com/forum/gmatprepcriticalreasoningcollection106783.html http://gmatclub.com/forum/howtoget60awamyguide64327.html http://gmatclub.com/forum/howtoget60awamyguide64327.html?hilit=chineseburned



Manager
Joined: 27 Oct 2011
Posts: 186
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
WE: Account Management (Consumer Products)

Re: m25,#30 [#permalink]
Show Tags
22 Mar 2012, 20:23
you can easily solve this question by just seeing that 40% is also in the mid of the two numbers and to be in the middle you must have an equal amount of both. Caution though that this will only work in this situation where it is perfectly in the middle of the two numbers... it would be harder to tell if the numbers were larger and you couldnt see the middle number of the two.
_________________
DETERMINED TO BREAK 700!!!



Intern
Joined: 16 Feb 2011
Posts: 4

Re: m25,#30 [#permalink]
Show Tags
26 Mar 2012, 14:19
1
This post received KUDOS
I think you can look at this conceptually to safe a lot of time. It's clearly a weighted average question.
30%40%50%
40% is perfectly in the middle, therefore you need 1 part 30% solution to each part 50% solution. The mixture will contain 1/2 30% mix and 1/2 50% mix. You have removed 1/2 of the original 50% solution and replaced it with 30% solution.
The tricky thing here is that if you didn't see this and you were looking to eliminate answers using the method described in Mgmat where you look for fractional values that add to 1, you would eliminate answer (C) for not having a match.
Done!



Senior Manager
Joined: 15 Sep 2009
Posts: 265

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2012, 05:14
2
This post received KUDOS
Alligative Method is such a timesaving method. Here we go: 30%40%50% Distances 1:1 This means that both contribute an equal amount of the final solution. Straight C: 1/2 Cheers, Der alte Fritz.
_________________
+1 Kudos me  I'm half Irish, half Prussian.



Intern
Joined: 07 Aug 2012
Posts: 14

Re: m25,#30 [#permalink]
Show Tags
26 Sep 2012, 11:49
I solved this question by following method without being sure if it was true, but it gave a correct answer and took some times ...
0.5x  a0.5x + a0.3x = 0.4x (1)
a is the same amount of solutions replaced
from (1) we have: a0.2x = 0.1x thus, a = 1/2
and the answer is C



Math Expert
Joined: 02 Sep 2009
Posts: 39709

Re: m25,#30 [#permalink]
Show Tags
26 Sep 2012, 12:07



Manager
Joined: 06 Feb 2013
Posts: 59

Re: m25,#30 [#permalink]
Show Tags
16 Sep 2013, 23:18
Bunuel wrote: Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{2}\) D. \(\frac{3}{4}\) E. \(\frac{4}{5}\)
Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.
Answer: C.
Alternately you can do the following, say \(x\) part of the original solution was replaced then: \((1x)*0.5+x*0.3=1*0.4\) > \(x=\frac{1}{2}\).
Answer: C. I wonder if anybody was confused with the word "replaced" at all? I was trying to come up with some different math only because of that word. Had the question said the solution was "added" to the original it would make it as simple as Bunuel put it. So it seems to me that "replace" = "add"in this context...disappointed how you can know the solution but wording messes up with you to the point you begin doubting what you can... Bunuel  are there instances in mixtures when "replace" is not equal to "add"?
_________________
There are times when I do not mind kudos...I do enjoy giving some for help



Math Expert
Joined: 02 Sep 2009
Posts: 39709

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2013, 00:58
obs23 wrote: Bunuel wrote: Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{2}\) D. \(\frac{3}{4}\) E. \(\frac{4}{5}\)
Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.
Answer: C.
Alternately you can do the following, say \(x\) part of the original solution was replaced then: \((1x)*0.5+x*0.3=1*0.4\) > \(x=\frac{1}{2}\).
Answer: C. I wonder if anybody was confused with the word "replaced" at all? I was trying to come up with some different math only because of that word. Had the question said the solution was "added" to the original it would make it as simple as Bunuel put it. So it seems to me that "replace" = "add"in this context...disappointed how you can know the solution but wording messes up with you to the point you begin doubting what you can... Bunuel  are there instances in mixtures when "replace" is not equal to "add"? Replacing and adding are not the same. Here we are told that "some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid".
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 06 Feb 2013
Posts: 59

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2013, 03:27
Quote: Replacing and adding are not the same. Here we are told that "some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid". [/quote] But then, mathematically, we seem to treat it identically? My mathematical equation \(0.5x + 0.3(1x)=0.4\) was identical to yours right away, but then I started questioning, thinking since we are "replacing" we could not express this relationship the same way, in the weighted average manner...I am not sure what I am missing, but it does not feel right
_________________
There are times when I do not mind kudos...I do enjoy giving some for help



Math Expert
Joined: 02 Sep 2009
Posts: 39709

Re: m25,#30 [#permalink]
Show Tags
17 Sep 2013, 03:44







Go to page
1 2
Next
[ 25 posts ]



