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# Some part of a 50% solution of acid was replaced with an

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Some part of a 50% solution of acid was replaced with an  [#permalink]

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Updated on: 19 Aug 2015, 12:12
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81% (01:15) correct 19% (02:00) wrong based on 301 sessions

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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. $$\frac{4}{5}$$

M25Q30

Originally posted by chrissy28 on 12 Aug 2008, 14:27.
Last edited by ENGRTOMBA2018 on 19 Aug 2015, 12:12, edited 2 times in total.
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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12 Aug 2008, 15:10
chrissy28 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?

(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5

Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.

in the new solution.. you added x of 30% solution. i.e 30x acid

(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%

[(1-x)50 + 30x] / 100 = 40/100

solve to get x=1/2
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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12 Aug 2008, 15:16
The reason that 1/2 is correct is this:

What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).

The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.

Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.

chrissy28 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?

(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5

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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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16 Jun 2013, 16:55
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If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. $$\frac{1}{2}$$ of 50% solution and $$\frac{1}{2}$$ of 30% solution. If the solution must be equally weighted, then 30% solution must replace $$\frac{1}{2}$$ of 50% solution (assuming original solution was 50% solution of acid).

Just another way of saying the same thing. Cheers
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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16 Jun 2013, 23:07
1
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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. $$\frac{4}{5}$$

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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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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17 Jun 2013, 21:57
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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17 Jun 2013, 23:14
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....

(100-x)*0.5+0.3x=100*0.4 --> x=50 --> 50/100=1/2.

Hope it's clear.
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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18 Jun 2013, 00:03
1
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.

Let total solution =100
Replacement = x (say)

equation:
(100-x)*50% + X*30% = 40% Solving this equation i am getting x= 248

I looked every way possible but could not figure out the problem in the equation....

Try using below equation.
50% of (100-x) + 30% of x = 40% of 100
x=50litres.
So 50 of 100 litres got replaced would amount to 1/2 in ratio.
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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18 Jun 2013, 00:33
Ahhhh!! I got it....Thank you both of you guys....I hope silly mistakes won't trouble me on the Real Test...
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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10 Jul 2014, 02:09
2
Did using allegation method

Refer diagram below

Bunuel, kindly update the OA.

Attachments

all.png [ 3.61 KiB | Viewed 5115 times ]

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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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20 Jul 2015, 06:11
3
chrissy28 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. $$\frac{4}{5}$$

M25Q30

Responding to a pm:

It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.

You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.

The first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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19 Aug 2015, 11:41
Can you please explain to me how you got the answer to this problem using the allegation methos?

Thanks
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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19 Aug 2015, 11:44
tinnyshenoy wrote:
Can you please explain to me how you got the answer to this problem using the allegation methos?

Thanks

Look at: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html#p1381700
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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19 Aug 2015, 12:02
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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19 Aug 2015, 12:22
tinnyshenoy wrote:
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?

It is from the alligation method

As shown in the post above, per the alligation method, the distrbution will be:

50 10 (=40-30) parts

40

30 10 (=50-40) parts

Thus, you see that you need 10 parts out of 20 (=(10+10)) parts of 50% solution, giving you 10/20 = 1/2 or 50% as your answer.

Hope this helps.
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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09 Apr 2016, 00:38
chrissy28 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. $$\frac{4}{5}$$

M25Q30

w1 is the weight 50% acid solution which has concentration C1 = 50%
w2 is the weight of 20% acid solution which has concentration C1 = 30%
Concentration of mixture (average) = 40%

using weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)
w1/w2 = (30-40)/(40-50)
w1/w2 = 1/1

w1 is 1 part and w2 is 1 part in a total 2 part of solution.

question is What part of the original solution was replaced?
w2 is what was replaced, hence 1 out of 2 i.e. 1/2 part was replaced.
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Re: Some part of a 50% solution of acid was replaced with an  [#permalink]

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19 Sep 2018, 02:52
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