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# One-fourth of a solution that was 10% by weight was replaced

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Manager
Joined: 25 Aug 2004
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One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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18 Oct 2004, 08:31
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63% (02:23) correct 37% (02:26) wrong based on 474 sessions

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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%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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13 Mar 2013, 20:52
14
13
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

The allegation formula is this:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 is the ratio of weights of the two solutions. Here, since 1/4 of the first solution is replaced, the weight of the first solution is 3/4 and that of the second solution is 1/4 so w1/w2 = 3/1

A2 - Concentration of second solution which we have to find here
Aavg - Concentration of mixture which is 16%
A1 - Concentration of first solution which is 10%

3/1 = (A2 - 16)/(16 - 10)
A2 = 34%

To find out how you get this formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
For more on mixtures and replacement, check:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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18 Oct 2004, 10:53
11
1
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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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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21 Dec 2009, 13:41
2
1
1
Let the original solution be 100ml
1/4 of this solution is removed
sugar remaining = 7.5 ML
25 ml of x% solution is added.
New volume of sugar = 7.5+25*x/100
Since resulting solution is 16% sugar by weight, volume of sugar = 16ml

equating the 2

x=(16-7.5)*4
=34

A
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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21 Dec 2009, 17:52
2

Simply,
.75*.1+.25*y = .16
Solving for y would give the value 34%

Explanation:
.75% is still 10% solution
.25% is the y% solution

16% solution is 100%
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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22 Dec 2009, 06:59
1
3/4 * 10 + 1/4 * x = 16
x = 16*4 - (30/4)*4
x = 64 - 30
x = 34

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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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19 Jul 2011, 17:55
2
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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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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13 Sep 2011, 06:16
8
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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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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Updated on: 14 Sep 2011, 23:11
6
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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Originally posted by chawlavinu on 14 Sep 2011, 18:48.
Last edited by chawlavinu on 14 Sep 2011, 23:11, edited 1 time in total.
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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13 Mar 2013, 18:30
5
2
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

This is the first question I trying to explain here, so please bear with my mistakes.

The best I could think of solving this is using nos.

Lets say we have a 100 ml of solution and it has 10 gm of sugar.

100 ml ---- 10 gm
75 ml ------- 7.5 gm
25 ml ------- 2.5 gm

Once we remove 25 ml, we have only 7.5 gm sugar left to make 16% of 100 ml solution we need 8.5 gm more sugar.
so we need 8.5 gm in 25 ml, which is 8.5 * 4 = 34 gm/ (25 * 4).

Ans A) 34 %
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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13 Mar 2013, 22:22
8
2
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.
34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

Instead of using complex calculations and remembering formulae, why dont u directly get to weighted average.

3 parts of 10% + 1 part of x (unknown) % = 4 parts of 16%
=> x% = 64%-30% = 34%

ans A it is.
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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14 Mar 2013, 02:16
5
3
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.

Hope it helps.
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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19 Apr 2014, 06:41
2
1
Alligation is the KEY

3/4 of 10%------------------------------1/4 of X%

-------------------16%----------------------------

(X-16)%------------------------------------------6%

(X-16)/6=3/1

X=34%
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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04 Aug 2014, 22:46
TGC wrote:
Alligation is the KEY

3/4 of 10%------------------------------1/4 of X%

-------------------16%----------------------------

(X-16)%------------------------------------------6%

(X-16)/6=3/1

X=34%

Is it called cross method? I think it is the fastest solution
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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05 Aug 2014, 00:05
TGC wrote:
Alligation is the KEY

3/4 of 10%------------------------------1/4 of X%

-------------------16%----------------------------

(X-16)%------------------------------------------6%

(X-16)/6=3/1

X=34%

Is it called cross method? I think it is the fastest solution

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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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05 Aug 2014, 00:50
guerrero25 wrote:
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%

Original - sugar=0.1*s1, other = 0.9*s1
Removed - sugar= -0.1 * s1/4, other= - 0.9*s1/4
Added - sugar= x * s1/4, other= (1-x)*s1/4
Resultant sugar= 0.16*s1, other=0.84*s1

(0.1*s1 - 0.1*s1/4 + x*s1/4) / (0.9*s1 - 0.9*s1/4 + (1-x)*s1/4) = 16/84
x=0.34 or percent sugar by weight is 34%

The advantage with this approach is that you can use it for any mixture problem.
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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05 Aug 2014, 03:07
2
Total Solution ........................ Sugar Concentration

100 ............................................ 10
(Assumed)

25% solution removed

100-25 ......................................... $$10-\frac{10}{4}$$

75 ................................................... $$\frac{30}{4}$$

Same quantity has been replaced. Say the new solution sugar weight = x%

75 + 25 ......................................... $$\frac{30}{4} + \frac{25x}{100}$$

Given that the resultant solution has sugar concentration of 16%

So equating the above

$$\frac{30}{4} + \frac{25x}{100} = 16$$

x = 34%

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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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18 Jan 2016, 16:02
1
Thank you SO MUCH VeritasPrepKarishma !!! You made this topic look so easy in your blogs below. I am getting all the mixture questions correct now with the help of your blog.

VeritasPrepKarishma wrote:
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

The allegation formula is this:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 is the ratio of weights of the two solutions. Here, since 1/4 of the first solution is replaced, the weight of the first solution is 3/4 and that of the second solution is 1/4 so w1/w2 = 3/1

A2 - Concentration of second solution which we have to find here
Aavg - Concentration of mixture which is 16%
A1 - Concentration of first solution which is 10%

3/1 = (A2 - 16)/(16 - 10)
A2 = 34%

To find out how you get this formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
For more on mixtures and replacement, check:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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18 Jan 2016, 22:33
neeraj609 wrote:
Thank you SO MUCH VeritasPrepKarishma !!! You made this topic look so easy in your blogs below. I am getting all the mixture questions correct now with the help of your blog.

Great! Keep practicing!
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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30 Jan 2017, 17:45
1
guerrero25 wrote:
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%

Let’s let x = the amount of solution, which initially contains 10% sugar. We remove one-fourth of that solution (.25x), which has 10% sugar. Then we replace that .25x with a solution of unknown percentage (p) of sugar. These actions result in our still having x amount of solution, but now with 16% sugar. We can summarize this in the following equation:

x(0.10) - (0.25x)(0.10) + (0.25x)(p) = x(0.16)

0.10x - 0.025x + 0.25xp = 0.16x

100x - 25x + 250xp = 160x

250xp = 85x

p = 85/250 = .34

Thus, the second solution was 34% sugar by weight.

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Re: One-fourth of a solution that was 10% by weight was replaced   [#permalink] 30 Jan 2017, 17:45

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