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# m25,#30

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VP
Joined: 18 May 2008
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01 Dec 2008, 06:39
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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) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can sum1 give a simpler explanation?
SVP
Joined: 17 Jun 2008
Posts: 1547

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01 Dec 2008, 07:11
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Look at it this way.

Total part is 1. Out of this, x part was replaced.

Hence, (1-x)*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
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03 Feb 2009, 11:27
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.

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?
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Joined: 29 Aug 2007
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03 Feb 2009, 12:27
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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.

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.
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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?
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Joined: 18 Aug 2010
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Concentration: Entrepreneurship, Finance
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14 Sep 2010, 17:06
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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) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

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.
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Joined: 18 Aug 2010
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Concentration: Entrepreneurship, Finance
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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.

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
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Joined: 02 Apr 2010
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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.
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Joined: 12 Apr 2010
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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.

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 re-wording 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

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15 Sep 2011, 07:16
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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, 50-50 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
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17 Sep 2011, 02:22
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Yup its C. Since (50+30) /2= 40 => The resulting solution should have half of each of the solutions.
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http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
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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.
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Joined: 16 Feb 2011
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26 Mar 2012, 14:19
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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!
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17 Sep 2012, 05:14
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Alligative Method is such a time-saving 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.
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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

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26 Sep 2012, 12:07
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Expert's post
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.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

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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.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

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"?
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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.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

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".
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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 $$0.5x + 0.3(1-x)=0.4$$ 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
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17 Sep 2013, 03:44
obs23 wrote:
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".

But then, mathematically, we seem to treat it identically? My mathematical equation $$0.5x + 0.3(1-x)=0.4$$ 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 [/quote]

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