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

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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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Re: 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 05 Dec 2017, 09:05
3
*NO ALGEBRA SOLUTION:*

There is one solution with 10% sugar and another solution with x% sugar, which when combined give you 16% sugar

This is a very simple question and while generally people use algebra for mixtures questions, I would advise to STAY AWAY FROM ANY ALGEBRA and use a simpler method:

If 1/4th of the 10% solution is replaced, that means only 3/4th of the final mixture is the 10% solution (S1) and 1/4th is the x% solution (S2).
Therefore we can say that the two solutions S1 and S2 are mixed IN THE RATIO 3 : 1

S1____________S2
10%__________x%

_______16%______
__3a____:____1a__

Now, we need to find the actual value in which the solutions were mixed. To get the value of this, you need to subtract Solution 1 from the middle value and Solution 2 from the middle value. Then, you take the absolute value of the resulting numbers to give you 1a and 3a.

So
1a = 10 - 16 = 6
3a = x - 16

If 1a = 6, 3a is therefore = 18.

Now, x% - 16% = 18, therefore x% = 34. Answer is B.

As a note: This method can be difficult to grasp in the beginning but once you are able to understand it, you will never use any other method to solve mixtures questions.
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Re: One-fourth of a solution that was 10% by weight was replaced  [#permalink]

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New post 26 Oct 2019, 18:38
Since one fourth of the solution was replaced, it means that the ratio of the weight of the first solution to that of the second solution was 3/1.

We know that the ratio of weights w1/w2 = (c2-a)/(a-c1).
c1 = 10%
c2=x
a = 16%
w1/w2=3/1
Therefore, 3/1 = (x-16)/(16-10)
x-16 = 3(6)
x=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 27 Oct 2019, 03:12
mbassmbass04 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%

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 "

0.1*0.75+0.25*Y=0.16
Y=0.34
A:)
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Re: One-fourth of a solution that was 10% by weight was replaced   [#permalink] 27 Oct 2019, 03:12

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