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# Some part of the 50% solution of acid was replaced with the

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Senior Manager
Joined: 23 May 2005
Posts: 260
Location: Sing/ HK
Some part of the 50% solution of acid was replaced with the  [#permalink]

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18 Nov 2006, 19:56
1
1
8
00:00

Difficulty:

5% (low)

Question Stats:

85% (00:47) correct 15% (00:58) wrong based on 359 sessions

### HideShow timer Statistics

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

A. 1/5
B. 1/4
C. 1/2
D. 3/4
E. 4/5

OPEN DISCUSSION OF THIS QUESTION IS HERE: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html

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Senior Manager
Joined: 08 Jun 2006
Posts: 321
Location: Washington DC
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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18 Nov 2006, 20:22
X be the amount of 50% solution
n be the amount taken out from 50% solution and the same amount of 30% solution is added later
In the resultant solution, acid amount is X/2 â€“ n/2 + 3n/10 = 2X/5
N = Â½
Manager
Joined: 01 Oct 2006
Posts: 234
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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18 Nov 2006, 20:28
Suppose originally there was x litres of solution. Out of which y litres was removed.

After removing the amount y, x-y litres of solution is left.
Amount of acid in x-y = x-y/2

y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.

In the new mixture, total acid = (x-y)/2 + 30y/100

(x-y)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2

Manager
Joined: 21 Mar 2005
Posts: 112
Location: Basel
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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19 Nov 2006, 06:50
4
1
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2
Senior Manager
Joined: 23 May 2005
Posts: 260
Location: Sing/ HK
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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20 Nov 2006, 07:18
1
cool, short way to solve it, tobiastt! Thanks everyone
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Manager
Joined: 25 May 2009
Posts: 132
Concentration: Finance
GMAT Date: 12-16-2011
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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20 Jun 2009, 08:38
Does anyone know how to set this problem up using the box method?
Concent.
Amt

Thanks.
Intern
Joined: 05 Sep 2006
Posts: 13
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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22 Jun 2009, 07:20
Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50%
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Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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22 Jun 2009, 08:43
2
parimal wrote:
Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50%

Wow - you registered in 2006 and this is your first post! Better late than never!
Welcome back
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Intern
Joined: 05 Sep 2006
Posts: 13
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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23 Jun 2009, 08:12
Thanks.. yes.. I have been hiding in the lurks but hoping to be an active participant now.. cheers!
Director
Joined: 25 Oct 2008
Posts: 504
Location: Kolkata,India
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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15 Aug 2009, 18:50
1
50..................40
.........30.........
10..................20

10/20=1/2
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Manager
Joined: 10 Sep 2009
Posts: 70
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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12 Nov 2009, 13:08
Let's assume original amount = 1L

Equation is :

0,5 - 0,5x + 0,3x = 0,4
=> x = 0,5 = 1/2
Manager
Joined: 23 Jun 2009
Posts: 96
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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12 Nov 2009, 14:04
I3igDmsu wrote:
Does anyone know how to set this problem up using the box method?
Concent.
Amt

Thanks.

O R A D
VOLUME X Y Y X
CONCENTRATION 50 50 30 40

50X-50Y+30Y=40X, Y=X/2
Manager
Joined: 22 Sep 2009
Posts: 188
Location: Tokyo, Japan
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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12 Nov 2009, 16:19
1
Hermione wrote:
Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5

Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.

(1-X)*0.5 + 0.3*x = 0.4
0.5 - 0.5X +0.3X = 0.4
-0.2X = -0.1
X=1/2

Intern
Joined: 10 Nov 2009
Posts: 4
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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02 Feb 2010, 04:19
pierrealexandre77 wrote:
Let's assume original amount = 1L

Equation is :

0,5 - 0,5x + 0,3x = 0,4
=> x = 0,5 = 1/2

Thanks! This explanation really worked for me
Intern
Joined: 27 Jan 2010
Posts: 26
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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09 Apr 2010, 15:04
Wow the box method makes these problems so easy. can you explain a bit more on the box method. ORAD?can we use this for most of the misture problems .any limitations?
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Status: Nothing comes easy: neither do I want.
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Location: Malaysia
Concentration: Technology, Entrepreneurship
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GMAT 1: 670 Q49 V31
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Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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Updated on: 16 Apr 2010, 22:57
Initially
50 ml acid and 100 ml of total
suppose x part of it is removed.
thus we are left with 50(1-x) of acid and 100(1-x) of total

if we add x part of 30% acid solution (30ml acid and 100ml total)
acid will be 30x and total = 100x

so total acid = 50(1-x)+ 30x = 50-20x
total solution = 100(1-x)+100x = 100

since if it equal to 40%acid solution(40ml acid and 100ml total)

=> (50-20x)/100 = 40/100 => 50-20x = 40 => 20x=10 => x=1/2

This looks lengthy but we can easily skip these steps while solving it.
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Originally posted by gurpreetsingh on 16 Apr 2010, 12:37.
Last edited by gurpreetsingh on 16 Apr 2010, 22:57, edited 1 time in total.
Intern
Joined: 27 Jul 2008
Posts: 9
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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16 Apr 2010, 13:31
mst wrote:
Suppose originally there was x litres of solution. Out of which y litres was removed.

After removing the amount y, x-y litres of solution is left.
Amount of acid in x-y = x-y/2

y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.

In the new mixture, total acid = (x-y)/2 + 30y/100

(x-y)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2

Great explanation! Thank you.
Manager
Joined: 13 Dec 2009
Posts: 112
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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17 Apr 2010, 06:04
tobiastt wrote:
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2

cool, it's nice way to answer quickly .
Manager
Joined: 14 May 2007
Posts: 180
Location: India
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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06 Nov 2010, 23:47
I am using a chart to solve this problem. Can anyone point out the error I am making?

See image attached below.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8882
Location: Pune, India
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

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07 Nov 2010, 05:59
1
greatchap wrote:
I am using a chart to solve this problem. Can anyone point out the error I am making?

See image attached below.

x/2 + 3x/10 is the total stuff which is equal to 4
Solving x/2 + 3x/10 = 4 gives x = 5

Since x is the amount you assumed was replaced out of 10, you get 50%.
Nothing wrong with what you did.

Note: Simply put, this question says that some 50% solution was mixed with some 30% solution to get 40% solution. Using weighted averages, you can find out that they must have been mixed in equal quantities or in other words, 50% of the original solution must have been replaced.
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Veritas Prep GMAT Instructor

Re: Some part of the 50% solution of acid was replaced with the   [#permalink] 07 Nov 2010, 05:59

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