Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
18 Nov 2006, 20:56
1
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
83% (01:47) correct
17% (01:11) wrong based on 286 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: somepartofa50solutionofacidwasreplacedwithan68805.html
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Impossible is nothing



Senior Manager
Joined: 08 Jun 2006
Posts: 335
Location: Washington DC

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
18 Nov 2006, 21:22
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: 242

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
18 Nov 2006, 21:28
Suppose originally there was x litres of solution. Out of which y litres was removed.
After removing the amount y, xy litres of solution is left.
Amount of acid in xy = xy/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 = (xy)/2 + 30y/100
(xy)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2
Is the answer 1/2?



Manager
Joined: 21 Mar 2005
Posts: 115
Location: Basel

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
19 Nov 2006, 07:50
2
This post received KUDOS
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
>x= 1/2



Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
20 Nov 2006, 08:18
cool, short way to solve it, tobiastt! Thanks everyone
_________________
Impossible is nothing



Manager
Joined: 25 May 2009
Posts: 140
Concentration: Finance
GMAT Date: 12162011

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
20 Jun 2009, 09:38
Does anyone know how to set this problem up using the box method? Orig Add Remove Result Concent. Amt
Thanks.



Intern
Joined: 05 Sep 2006
Posts: 14

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
22 Jun 2009, 08:20
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%



Founder
Joined: 04 Dec 2002
Posts: 15151
Location: United States (WA)

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
22 Jun 2009, 09:43
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
_________________
Founder of GMAT Club
US News Rankings progression  last 10 years in a snapshot  New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books
Coauthor of the GMAT Club tests



Intern
Joined: 05 Sep 2006
Posts: 14

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
23 Jun 2009, 09:12
Thanks.. yes.. I have been hiding in the lurks but hoping to be an active participant now.. cheers!



Director
Joined: 25 Oct 2008
Posts: 596
Location: Kolkata,India

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
15 Aug 2009, 19:50
50..................40 .........30......... 10..................20 10/20=1/2
_________________
http://gmatclub.com/forum/countdownbeginshasended8548340.html#p649902



Manager
Joined: 10 Sep 2009
Posts: 112

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
12 Nov 2009, 14:08
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: 147

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
12 Nov 2009, 15:04
I3igDmsu wrote: Does anyone know how to set this problem up using the box method? Orig Add Remove Result Concent. Amt
Thanks. O R A D VOLUME X Y Y X CONCENTRATION 50 50 30 40 50X50Y+30Y=40X, Y=X/2



Manager
Joined: 22 Sep 2009
Posts: 215
Location: Tokyo, Japan

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
12 Nov 2009, 17:19
1
This post received KUDOS
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 1X will be the parts not replaced. (1X)*0.5 + 0.3*x = 0.4 0.5  0.5X +0.3X = 0.4 0.2X = 0.1 X=1/2 So the answer is C



Intern
Joined: 10 Nov 2009
Posts: 5

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
02 Feb 2010, 05:19
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: 30

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
09 Apr 2010, 16:04
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?



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
16 Apr 2010, 13:37
Initially 50 ml acid and 100 ml of total suppose x part of it is removed. thus we are left with 50(1x) of acid and 100(1x) 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(1x)+ 30x = 5020x total solution = 100(1x)+100x = 100 since if it equal to 40%acid solution(40ml acid and 100ml total) => (5020x)/100 = 40/100 => 5020x = 40 => 20x=10 => x=1/2 This looks lengthy but we can easily skip these steps while solving it.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html
Last edited by gurpreetsingh on 16 Apr 2010, 23:57, 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]
Show Tags
16 Apr 2010, 14:31
mst wrote: Suppose originally there was x litres of solution. Out of which y litres was removed.
After removing the amount y, xy litres of solution is left. Amount of acid in xy = xy/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 = (xy)/2 + 30y/100
(xy)/2 + 3y/10 = (40 % of x) Solving the equation gives y= x/2
Is the answer 1/2? Great explanation! Thank you.



Manager
Joined: 13 Dec 2009
Posts: 128

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
17 Apr 2010, 07:04
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: 182
Location: India

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
07 Nov 2010, 00:47
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: 7447
Location: Pune, India

Re: Some part of the 50% solution of acid was replaced with the [#permalink]
Show Tags
07 Nov 2010, 06:59
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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews




Re: Some part of the 50% solution of acid was replaced with the
[#permalink]
07 Nov 2010, 06:59



Go to page
1 2
Next
[ 34 posts ]




