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

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

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New post 15 Feb 2015, 14:25
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:
[Reveal] Spoiler:
B


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

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New post 15 Feb 2015, 15:28
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.

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

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

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let x=fraction of solution replaced
.15-.15x+.25x=.21
x=3/5

Last edited by gracie on 11 Nov 2015, 19:22, 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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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.

Answer (B)

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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New post 24 Feb 2015, 01:49
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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Re: A certain fraction of a 15% salt solution by weight was replaced [#permalink]

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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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pacifist85 wrote:

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


Check out this post. It will help you understand weighted average.

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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Answer = B. 3/5

Using allegation method, refer diagram below:

Attachment:
alli.png
alli.png [ 2.83 KiB | Viewed 2299 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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New post 28 Feb 2015, 02:20
hi though I have got till the point where we have got the ratio ..4:6 i.e 2:3
but can`t figure out then why we have taken the fraction as 3/5..
as in ..in 3/5 ..3/5th is the part of the new mixture..instead ques is asking for -part of the "original" mixture.
3/5th is the part of the new mixture but not of the original mixture..i guess..
please help me..
thanku

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

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New post 28 Feb 2015, 02:20
hi though I have got till the point where we have got the ratio ..4:6 i.e 2:3
but can`t figure out then why we have taken the fraction as 3/5..
as in ..in 3/5 ..3/5th is the part of the new mixture..instead ques is asking for -part of the "original" mixture.
3/5th is the part of the new mixture but not of the original mixture..i guess..
please help me..
thanku

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

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New post 28 Feb 2015, 08:19
hi thanks for your valuable reply but couldnt understand your last line.
still can`t 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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shreygupta3192 wrote:
hi thanks for your valuable reply but couldnt understand your last line.
still can`t 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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Re: A certain fraction of a 15% salt solution by weight was replaced [#permalink]

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New post 28 Feb 2015, 17:15
thanks ..but please check my reasoning if I have understood your answer..
we can say that the amount replaced by the 25%sol is equal to the "certain fraction of 15%sol" replaced(replaced for in ques)
thus ..3/5th (25%sol) of the new mixture would be equal to the 3/5th of the original mixture(as the total amount of the mixture still remains the same, only the proportion has changed as in the total of the mixture still remains the same.

As in the your example too,12 of the blue balls are equal to the 12 of the initial red balls replaced for but the total still remains the same i.e 20.
so red balls 12/20 (replaced)of the original mixture(out of total 20 balls;red balls 12,blue balls=0)=blue balls 12/20(in place of replaced red balls) of the new mixture(out of total 20 balls;red balls =8, blue balls=12)..

please check..

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

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New post 19 Nov 2017, 07:07
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 = \(\frac{6}{10}=\frac{3}{5}\)

Answer B

--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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New post 22 Nov 2017, 11:28
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

Answer: B
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Re: A certain fraction of a 15% salt solution by weight was replaced   [#permalink] 22 Nov 2017, 11:28
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