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# A certain fraction of a 15% salt solution by weight was replaced

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A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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15 Feb 2015, 13:02
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A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4
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Posts: 7764
Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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15 Feb 2015, 19:55
5
1
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

a. 1/2

b. 3/5

c.2/3

d. 7/10

e. 3/4

hi turkish and celestial,

a simpler approach to these type of problems....

whenever you see mixture problem, you can generally solve by weighted mixture..
see how apart are the two %solutions from final sol...
15% sol is 6% away and 25% sol is 4%away from 21%sol...
so the two sol(15%:25%) are in a mix of 4:6...
so 25% sol is 6/10 or 3/5 of the new solution...
since initially 25% sol was 0 and now it is 3/5of the solution, 3/5 of 15% solution was changed
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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15 Feb 2015, 14:28
2
2
Hi Turkish,

This question is essentially a "weighted average" question, which means that it can be solved in a variety of ways. Here's an approach that uses the Weighted Average Formula:

We're mixing a certain amount of a 15% salt solution with a certain amount of a 25% salt solution and ending up with a total solution that is 21% salt.

X = # of ounces of 15% solution
Y = # of ounces of 25% solution

(.15X + .25Y)/(X+Y) = .21

.15X + .25Y = .21X + .21Y
.04Y = .06X
4Y = 6X
2Y = 3X

This ratio tells us that for every 2 ounces of 15% solution, we have 3 ounces of 25% solution. So, we can have ANY totals that maintain this ratio. For the sake of simplicity, I'm going to choose a TOTAL of 5 ounces (2 ounces of 15% and 3 ounces of 25%)

The original question is phrased in a way that implies that ALL the 5 ounces were 15% salt, but some of those ounces were removed/replaced with 25% salt. If we start with 5 total ounces and replace 3 of them with a 25% salt solution, then we get the 21% solution above. Thus 3/5 of the original solution was replaced.

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Retired Moderator Status: I Declare War!!! Joined: 02 Apr 2014 Posts: 234 Location: United States Concentration: Finance, Economics GMAT Date: 03-18-2015 WE: Asset Management (Investment Banking) Re: A certain fraction of a 15% salt solution by weight was replaced [#permalink] ### Show Tags 15 Feb 2015, 15:25 Dear Rich, I don't need to say every-time that your explanations are really simple to understand. Now this time I am stuck with last para written by you... "The original question is phrased in a way that implies that ALL the 5 ounces were 15% salt, but some of those ounces were removed/replaced with 25% salt. If we start with 5 total ounces and replace 3 of them with a 25% salt solution, then we get the 21% solution above. Thus 3/5 of the original solution was replaced." could you explain me in more easy way... as I arrived till 2/3 thing.. then lost the track Thanks Celestial EMPOWERgmatRichC wrote: Hi Turkish, This question is essentially a "weighted average" question, which means that it can be solved in a variety of ways. Here's an approach that uses the Weighted Average Formula: We're mixing a certain amount of a 15% salt solution with a certain amount of a 25% salt solution and ending up with a total solution that is 21% salt. X = # of ounces of 15% solution Y = # of ounces of 25% solution (.15X + .25Y)/(X+Y) = .21 .15X + .25Y = .21X + .21Y .04Y = .06X 4Y = 6X 2Y = 3X This ratio tells us that for every 2 ounces of 15% solution, we have 3 ounces of 25% solution. So, we can have ANY totals that maintain this ratio. For the sake of simplicity, I'm going to choose a TOTAL of 5 ounces (2 ounces of 15% and 3 ounces of 25%) The original question is phrased in a way that implies that ALL the 5 ounces were 15% salt, but some of those ounces were removed/replaced with 25% salt. If we start with 5 total ounces and replace 3 of them with a 25% salt solution, then we get the 21% solution above. Thus 3/5 of the original solution was replaced. Final Answer: GMAT assassins aren't born, they're made, Rich EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14553 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A certain fraction of a 15% salt solution by weight was replaced [#permalink] ### Show Tags 15 Feb 2015, 16:28 Hi Celestial09, In the first sentence of the question, we're told that a fraction of a solution was REPLACED, so the original number of ounces stays the same. We're NOT adding or subtracting ounces - we're replacing them. In my example, the original mixture is 5 ounces total, so some fraction "out of 5" was replaced.... To get the 21% solution, 3 of those original ounces are replaced by the 25% solution, thus 3 of the 5 (or 3/5) of the original solution is replaced. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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Updated on: 11 Nov 2015, 20:22
1
3
let x=fraction of solution replaced
.15-.15x+.25x=.21
x=3/5

Originally posted by gracie on 17 Feb 2015, 18:46.
Last edited by gracie on 11 Nov 2015, 20:22, edited 3 times in total.
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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17 Feb 2015, 23:30
2
1
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4

This "replacement" problem is just like a regular mixture problem and should be treated as such.

15% salt solution was mixed with 25% salt solution to give 21% salt solution. So we can find the ratio of the weights in which the two were mixed.

w1/w2 = (25 - 21)/(21 - 15) = 4/6 = 2/3

So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.

Check this post for an explanation of the formula: http://www.veritasprep.com/blog/2011/03 ... -averages/
and this one for another similar question: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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24 Feb 2015, 02:49
chetan2u wrote:
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

a. 1/2

b. 3/5

c.2/3

d. 7/10

e. 3/4

hi turkish and celestial,

a simpler approach to these type of problems....

whenever you see mixture problem, you can generally solve by weighted mixture..
see how apart are the two %solutions from final sol...
15% sol is 6% away and 25% sol is 4%away from 21%sol...
so the two sol(15%:25%) are in a mix of 4:6...
so 25% sol is 6/10 or 3/5 of the new solution...
since initially 25% sol was 0 and now it is 3/5of the solution, 3/5 of 15% solution was changed

I am sorry if it sounds stupid, but why is it 4:6 and not 6:4?
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Posts: 7764
Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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24 Feb 2015, 03:57
1
pacifist85 wrote:
chetan2u wrote:
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

a. 1/2

b. 3/5

c.2/3

d. 7/10

e. 3/4

hi turkish and celestial,

a simpler approach to these type of problems....

whenever you see mixture problem, you can generally solve by weighted mixture..
see how apart are the two %solutions from final sol...
15% sol is 6% away and 25% sol is 4%away from 21%sol...
so the two sol(15%:25%) are in a mix of 4:6...
so 25% sol is 6/10 or 3/5 of the new solution...
since initially 25% sol was 0 and now it is 3/5of the solution, 3/5 of 15% solution was changed

I am sorry if it sounds stupid, but why is it 4:6 and not 6:4?

hi pacifist...
the reason is that the average will be closer to the value which has more number....
that is .. an example..
x class has 20 students with 3 books each and y class has 50 students with 10 books each...
so avg books per student=(20*3+10*50)/(20+50)=560 / 70 = 8 books...
so look at the spread from average of 8.... (8-3):(10-8)=5:2...
so students x:Y=20:50=2:5 and books =5:2... both are inverse..
simmilarly for this question
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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25 Feb 2015, 19:57
2
pacifist85 wrote:

I am sorry if it sounds stupid, but why is it 4:6 and not 6:4?

http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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27 Feb 2015, 02:28
3

Using allegation method, refer diagram below:

Attachment:

alli.png [ 2.83 KiB | Viewed 3989 times ]

Answer $$= \frac{6}{6+4} =\frac{6}{10} = \frac{3}{5}$$
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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28 Feb 2015, 09:19
still cant figure out how come 3/5th of "original" mixture.
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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28 Feb 2015, 09:28
1
shreygupta3192 wrote:
still cant figure out how come 3/5th of "original" mixture.

ok shrey,
say you have 20 red balls .. you take out few of them and replace with blue balls..
now blue balls are 12 ie 3/5 of 20 ... what portion of initial mixture was changed..
blue balls have gone from 0 to 12... that is 12 of initial red balls have been changed... or in other words 12/20 (3/5) of initial mixture has been changed..
it is the same with this question.. red balls are 15% and blue balls are 25%
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A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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19 Nov 2017, 08:07
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4

Weighted average with one variable approach.

A is the original solution
B is the resultant solution
x = the amount (volume) replaced:

1) x is a volume of salt, both removed and replaced. The ingredient is the same. You don't need a second variable;

2) Volume removed = volume replaced;

3) However, the concentration (percentage) of salt is different in the volume removed and the volume replaced. Same volume, different concentrations of salt;

4) x, the volume removed and replaced is a part of the whole, A. Part x of whole A = fraction of A. Answers are all 0 < x < 1. Assume A's volume = 1

5) Volume of A and volume of B are equal.

Original, (salt concentration)(volume): (.15)(1)
Removed: (.15)(x)
Replaced: (.25)(x)
Resultant solution, (salt concen.)(volume): (.21)(1)

Original - removed + replaced = Resultant solution

(.15)(1) - .15(x) + .25(x) = .21(1)

Multiply all by 100 to make arithmetic easier.

(15)(1) - 15x + 25x = 21(1)
10x = 6
x = $$\frac{6}{10}=\frac{3}{5}$$

--10x = 6 might seem odd. Fractional replacements are hard. Think in volumes first, then in fractions.
--Amount x = a volume. A has a volume of 1. x is the unknown volume removed and replaced. And it is not all of A.
--Fractions: all of A did not go. Only part of A's volume got removed and replaced. (Part) x of (whole) A got removed and replaced. $$\frac{part}{whole}$$ = fraction.

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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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22 Nov 2017, 12:28
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4

We can let the original amount of salt in the solution = x, and the amount taken out and replaced = y. Thus we have:

0.15x - 0.15y + 0.25y = 0.21(x - y + y)

0.15x + 0.1y = 0.21x

15x + 10y = 21x

10y = 6x

y/x = 6/10 = 3/5

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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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27 Dec 2017, 15:37
@karsihma , can you kindly explain the last statement " So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.
"
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A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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Updated on: 27 Dec 2017, 19:57
Alligation:

15_______21________25

4__________________6

The ratio of the 15 percent solution to the 25 percent solution is 2:3

When you replace the same total amount remains. At first there were 5x grams of 15 percent solution. Then there were 2x grams of the 15 percent solution and 3x of the 25 percent solution. So 3/5 of the solution was replaced.

Alternative solution:
Assume there are 100 grams of the 15 percent solution. After a certain amount is removed there will be 100x grams left and 15x grams of the 15 percent solution. There needs to be 100-100x grams replaced. That is the 25 percent solution so 25-25x grams of alcohol get added.
Since the final mixture is 100 grams and has 21 percent alcohol there aren 21 grams of alcohol.

15x+25-25x=21

-10x=-4

X=.4

So 10x= 4

That implies 6 grams were removed before it was replaced. So 3/5 was replaced.

Alternate solution:

.15x+.25y=.21(x+y)

Posted from my mobile device

Originally posted by Eulerelements on 27 Dec 2017, 19:51.
Last edited by Eulerelements on 27 Dec 2017, 19:57, edited 1 time in total.
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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27 Dec 2017, 19:55
qazi11 wrote:
@karsihma , can you kindly explain the last statement " So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.
"

Originally there were 5 parts of only the 15 percent solution. At at the end there were 2 parts of the 15 percent solution and 3 parts of the 25 percent solution. So 3 parts out of
5 were replaced. 3/5 ths was replaced.

Posted from my mobile device
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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29 Dec 2017, 06:40
Eulerelements wrote:
qazi11 wrote:
@karsihma , can you kindly explain the last statement " So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.
"

Originally there were 5 parts of only the 15 percent solution. At at the end there were 2 parts of the 15 percent solution and 3 parts of the 25 percent solution. So 3 parts out of
5 were replaced. 3/5 ths was replaced.

Posted from my mobile device

Thank you. This kept bugging me
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Re: A certain fraction of a 15% salt solution by weight was replaced  [#permalink]

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10 Jan 2018, 19:48
Turkish wrote:
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4

Let the total solution be 100 ml. Suppose x ml was removed from Solution A.. So the remaining salt solution in A is (100-x) * 15%... Now in this solution 0.25x is added... So (100-x)*0.15 + 0.25x = 0.21
X = 60.. So the fraction of original solution replaced was 60/100 or 3/5.
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Re: A certain fraction of a 15% salt solution by weight was replaced   [#permalink] 10 Jan 2018, 19:48

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