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

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

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New post 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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New post 30 Jan 2017, 17:45
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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.

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

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New post 15 Mar 2017, 07:14
Super Good Question!!!
can be easily done through weighted average concept as explained by "Veritasprepkarishma" as explained under:
25 % of original solution is replaced, therefore only 75 % 0f original solution remains
:i.e. (volume of 10% solution): (volume of 2nd solution) = 75:25
Putting in the values in the weghted average formula (let x be the percentage to be found)
75/25 = (x-16)/16-10.........pl obseve that 16 is the average percentage of the final mixture
3=(x-16)/6
or 18=x-16
or x=34%

Hope it helps..the concept pf the weighted average formula has been properly explained by karishma-veritas prep...
Kudos pl if u liked the above explanation
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 15 Mar 2017, 07:19
VeritasPrepKarishma wrote:
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!



Ditto...read thoroughly the concept explained by you under the weighted av and mixture problems....I am also getting all the questions right.....
5000 Kudos to u..thanx a ton

Can u send ur link of modulus problems as well....went through the the inequalities though but it does not cover much about modulus..
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 16 Mar 2017, 10:23
saurabhsavant wrote:
VeritasPrepKarishma wrote:
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!



Ditto...read thoroughly the concept explained by you under the weighted av and mixture problems....I am also getting all the questions right.....
5000 Kudos to u..thanx a ton

Can u send ur link of modulus problems as well....went through the the inequalities though but it does not cover much about modulus..



Great to hear this Saurabh! Wonderful to know that you have understood the concept very well. Here are some links to guide you through absolute values:

https://www.veritasprep.com/blog/2011/0 ... edore-did/
http://www.veritasprep.com/blog/categor ... ?s=modulus
http://www.veritasprep.com/blog/categor ... ute+values
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 17 Mar 2017, 03:38
let the solution be of 40 L and the concentration for second solution be x

after removal of 1/4th of soln, left soln = 30 L with sugar volume be 3L

equating the content of sugar we get

3 +10x/100 = 0/16*40
3 + 0.1x = 6.4
x = 34%
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One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 10 Jun 2017, 03:52
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%

1. Let x be the initial quantity and let percent sugar by weight of second solution be y.
2.(Quantity of the 10% solution - Quantity replaced by second solution)*(percent sugar of the solution) + (Quantity of the second solution added)*(percent sugar of second solution) / (initial quantity) = 16/100
3.( (x-x/4)*1/10 +(x/4)*y ) / x = 16/100
i.e., 3/4 *1/10 + (1/4)*y = 16/100
y=34/100=34%
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 19 Sep 2018, 02:52
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Re: One-fourth of a solution that was 10% by weight was replaced &nbs [#permalink] 19 Sep 2018, 02:52

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