One fourth of a solution that was 10 percent sugar by weight : GMAT Problem Solving (PS)
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# One fourth of a solution that was 10 percent sugar by weight

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One fourth of a solution that was 10 percent sugar by weight [#permalink]

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18 Oct 2004, 07:31
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One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

OPEN DISCUSSION OF THIS QUESTION IS HERE: one-fourth-of-a-solution-that-was-10-by-weight-was-replaced-149134.html
[Reveal] Spoiler: OA
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18 Oct 2004, 09:53
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I am awful at these but here is what I got

75(0.1) + 25(x) = 100(0.16)
7.5 + 25x = 16
x = 8.5/25 = 0.34

34% it is
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18 Oct 2004, 10:03
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3(16-10) = 1(X-16)

48-30=X-16
18+16=X
X = 34%
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24 Jan 2010, 20:49
I was never quite good with chemistry, the mixture problems are tricky for me. This helps! Thanks.
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14 Jul 2010, 13:40
mbassmbass04 wrote:
One fourth of a solution that was 10 percent sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight ?

(A) 34%

(B) 24%

(C) 22%

(D) 18%

(E) 8.5%

3/4*.1x+1/4y = .16x --> .075x+1/4y=.16x --> 1/4y=.085x -> y=.34x

34%.
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19 Jul 2011, 07:53
Question: 10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand?

Please explain how to solve this problem using tabular approach.
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19 Jul 2011, 16:55
Ok here goes my first gmat club post! After spending a good 6 mins on this problem I can provide some good insight

So the total is 10 kg. We are told to remove from the mixture and replace with pure sand. This means that the total is still 10 kg.

Now, let X be the amount of pure sand that is added. This is what we are solving for.

Now the sand in the mixture is going to be (10-X)*0.3 + X. The first part is the sand that is in the mixture now WITHOUT the addition of the additional sand. To make this more clear, (10-X) is the amount of sand and clay mixture with the subtraction of the pure sand that will be added. Multiplying it by 0.3 gives us the amount of sand in this part of the whole mixture. the second part is the additional amount of pure sand that is added. Note that we do not multiply this by any percent number because it is pure sand!

So now the whole equation will look like this:

(10-X)*0.3+X/((10-X)+X) = 0.5.

The numerator is the amount of sand in the mixture. (10-X)*0.3 = amount of sand that is not pure in the mixture. X is the amount of pure sand.

(10-X)*0.3 + X = 5.
3-0.3X+X = 5.
0.7X = 2.
X = 20/7.
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19 Jul 2011, 16:56
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hopewillprevail wrote:
Ok here goes my first gmat club post! After spending a good 6 mins on this problem I can provide some good insight

So the total is 10 kg. We are told to remove from the mixture and replace with pure sand. This means that the total is still 10 kg.

Now, let X be the amount of pure sand that is added. This is what we are solving for.

Now the sand in the mixture is going to be (10-X)*0.3 + X. The first part is the sand that is in the mixture now WITHOUT the addition of the additional sand. To make this more clear, (10-X) is the amount of sand and clay mixture with the subtraction of the pure sand that will be added. Multiplying it by 0.3 gives us the amount of sand in this part of the whole mixture. the second part is the additional amount of pure sand that is added. Note that we do not multiply this by any percent number because it is pure sand!

So now the whole equation will look like this:

(10-X)*0.3+X/((10-X)+X) = 0.5.

The numerator is the amount of sand in the mixture. (10-X)*0.3 = amount of sand that is not pure in the mixture. X is the amount of pure sand.

(10-X)*0.3 + X = 5.
3-0.3X+X = 5.
0.7X = 2.
X = 20/7.

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13 Sep 2011, 05:16
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let 100 be the total volume.
total salt = 10
salt taken out = 10/4 = 2.5

to make solution 16%, total salt = 16
salt added = 16-7.5 = 8.5

solution had = 8.5/25*100 = 34% sugar
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14 Sep 2011, 17:48
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Here is another method:

Consider that solution is made of sugar and water:
Solution A: Total 1000gm (10% sugar)
1000= 900 (water) + 100 (Sugar)

Expected Solutin B: Total 1000gm (16% sugar)
1000= 840 (water) + 160 (Sugar)

Now, as said in the question 25% of the solution A is replaced with another solution, which means:
75% of the solution A in untouched and only 25% of the solution (250gm) is touched

Total Sugar in Solution B= 160 = (75% of sugar in solution A) + (sugar replaced in 25% of Solution A with a new solution)

Sugar replaced in 25% Solution A with a new solution = 160- 75 =85

Sugar in new solution: 85/250 = 340/1000 = 34% (Answer)
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Last edited by chawlavinu on 14 Sep 2011, 22:11, edited 1 time in total.
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23 Sep 2013, 02:42
nzgmat wrote:
3(16-10) = 1(X-16)

48-30=X-16
18+16=X
X = 34%

Why did you multiply by 3?
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23 Sep 2013, 04:54
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igotthis wrote:
nzgmat wrote:
3(16-10) = 1(X-16)

48-30=X-16
18+16=X
X = 34%

Why did you multiply by 3?

nzgmat used allegation method to solve. Check Karishma's post: one-fourth-of-a-solution-that-was-10-by-weight-was-replaced-149134.html#p1195701

One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

This is a weighted average question.

Say the second solution (which was 1/4 th of total) was x% sugar, then 3/4*0.1+1/4*x=1*0.16 --> x=0.34. Alternately you can consider total solution to be 100 liters and in this case you'll have: 75*0.1+25*x=100*0.16 --> x=0.34.

OPEN DISCUSSION OF THIS QUESTION IS HERE: one-fourth-of-a-solution-that-was-10-by-weight-was-replaced-149134.html
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Re: Re:   [#permalink] 23 Sep 2013, 04:54
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