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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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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%
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Jun 2012, 06:13, edited 1 time in total.
Edited the question and added the OA

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Re: Tricky-mixture problem [#permalink]

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New post 21 Dec 2009, 06:00
EQ can be written as ...
let sol be s... so (3/4)s*(.1)+(1/4)s*t=s*(.16).... we get t(% of sugar in 2nd sol)=34%
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Re: Tricky-mixture problem [#permalink]

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New post 21 Dec 2009, 12:41
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Let the original solution be 100ml
It had 10ml of sugar
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: Tricky-mixture problem [#permalink]

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New post 21 Dec 2009, 16:52
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Answer A: 34%

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: Tricky-mixture problem [#permalink]

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New post 22 Dec 2009, 05:59
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3/4 * 10 + 1/4 * x = 16
x = 16*4 - (30/4)*4
x = 64 - 30
x = 34

Therefore the answer is A

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Re: Tricky-mixture problem [#permalink]

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New post 01 Jun 2012, 20:26
A

Total solution : 16 Liters

Initial Solution (V1) : 12 Liters
Concentration of Sugar (C1) = 10%

1/4 replaced with new solution (V2)= 4 liters
Concentration of Sugar (C2) = x% = ?

Final Solution (Vf): 16 Liters
Concentration of Sugar (Cf) = 16%



v1/v2 = (C2 - cf
) / (cf-c1)

12/4 = (x-16) / (16-10)

18 = (x-16)
34 = x

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

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New post 02 Oct 2015, 15:51
let x=sugar % of solution 2
.1-.1/4+x/4=.16
x=.34=34%

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

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New post 05 Dec 2017, 08:05
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*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 percent sugar by weight   [#permalink] 05 Dec 2017, 08:05
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