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

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Manager
Joined: 13 Aug 2009
Posts: 149

Kudos [?]: 321 [1], given: 7

WE 1: 4 years in IT
One fourth of a solution that was 10 percent sugar by weight [#permalink]

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21 Dec 2009, 06:17
1
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3
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Difficulty:

55% (hard)

Question Stats:

69% (01:45) correct 31% (02:02) wrong based on 176 sessions

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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, 07:13, edited 1 time in total.
Edited the question and added the OA

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Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4981

Kudos [?]: 5483 [0], given: 112

Re: Tricky-mixture problem [#permalink]

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21 Dec 2009, 07:00
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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5483 [0], given: 112

Manager
Joined: 25 Aug 2009
Posts: 165

Kudos [?]: 201 [2], given: 3

Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Re: Tricky-mixture problem [#permalink]

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21 Dec 2009, 13:41
2
KUDOS
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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Manager
Joined: 23 Nov 2009
Posts: 52

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Re: Tricky-mixture problem [#permalink]

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21 Dec 2009, 17:52
1
KUDOS

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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A kudos would greatly help

Tuhin

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Manager
Joined: 17 Dec 2009
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Re: Tricky-mixture problem [#permalink]

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

Therefore the answer is A

Kudos [?]: 18 [0], given: 4

Manager
Joined: 02 Oct 2009
Posts: 193

Kudos [?]: 22 [0], given: 4

Re: Tricky-mixture problem [#permalink]

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22 Dec 2009, 13:57
A

Like ludwigfraboulet simpler approach.

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Manager
Joined: 28 Jul 2011
Posts: 233

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Re: Tricky-mixture problem [#permalink]

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01 Jun 2012, 21:26
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

Kudos [?]: 160 [0], given: 16

Director
Joined: 07 Dec 2014
Posts: 814

Kudos [?]: 248 [0], given: 12

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

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

Kudos [?]: 248 [0], given: 12

One fourth of a solution that was 10 percent sugar by weight   [#permalink] 02 Oct 2015, 16:51
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# One fourth of a solution that was 10 percent sugar by weight

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