Some part of the 50% solution of acid was replaced with the : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 14:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 43 [1] , given: 0

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

### Show Tags

18 Nov 2006, 19:56
1
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

84% (01:48) correct 16% (01:07) wrong based on 259 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
[Reveal] Spoiler: OA

_________________

Impossible is nothing

Senior Manager
Joined: 08 Jun 2006
Posts: 340
Location: Washington DC
Followers: 1

Kudos [?]: 48 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 242
Followers: 1

Kudos [?]: 11 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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

Is the answer 1/2?
Manager
Joined: 21 Mar 2005
Posts: 115
Location: Basel
Followers: 1

Kudos [?]: 9 [2] , given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

19 Nov 2006, 06:50
2
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
Followers: 1

Kudos [?]: 43 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

20 Nov 2006, 07:18
cool, short way to solve it, tobiastt! Thanks everyone
_________________

Impossible is nothing

Manager
Joined: 25 May 2009
Posts: 144
Concentration: Finance
GMAT Date: 12-16-2011
Followers: 3

Kudos [?]: 113 [0], given: 2

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

20 Jun 2009, 08:38
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: 17
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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%
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14455
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3722

Kudos [?]: 23008 [1] , given: 4514

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

22 Jun 2009, 08:43
1
KUDOS
Expert's post
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

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 05 Sep 2006
Posts: 17
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 608
Location: Kolkata,India
Followers: 13

Kudos [?]: 798 [0], given: 100

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

15 Aug 2009, 18:50
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: 119
Followers: 3

Kudos [?]: 66 [0], given: 10

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 156
Followers: 1

Kudos [?]: 11 [0], given: 9

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

12 Nov 2009, 14:04
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

50X-50Y+30Y=40X, Y=X/2
Manager
Joined: 22 Sep 2009
Posts: 222
Location: Tokyo, Japan
Followers: 2

Kudos [?]: 22 [1] , given: 8

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

12 Nov 2009, 16:19
1
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 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

So the answer is C
Intern
Joined: 10 Nov 2009
Posts: 5
Followers: 0

Kudos [?]: 1 [0], given: 1

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 30
Followers: 0

Kudos [?]: 7 [0], given: 0

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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?
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [0], given: 235

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

16 Apr 2010, 12:37
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.
_________________

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/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Last edited by gurpreetsingh on 16 Apr 2010, 22:57, edited 1 time in total.
Intern
Joined: 27 Jul 2008
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 3

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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

Is the answer 1/2?

Great explanation! Thank you.
Manager
Joined: 13 Dec 2009
Posts: 129
Followers: 6

Kudos [?]: 281 [0], given: 10

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 178
Location: India
Followers: 2

Kudos [?]: 70 [0], given: 11

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

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: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13673 [1] , given: 222

Re: Some part of the 50% solution of acid was replaced with the [#permalink]

### Show Tags

07 Nov 2010, 05:59
1
KUDOS
Expert's post
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, 05:59

Go to page    1   2    Next  [ 34 posts ]

Similar topics Replies Last post
Similar
Topics:
17 10% of a 50% alcohol solution is replaced with water 11 25 Oct 2014, 21:53
A solution of 140 ltrs. of acid contains 10% of acid 3 20 Oct 2013, 23:06
2 Some part of a 50% solution of acid was replaced with an equal amount 3 30 Dec 2009, 14:09
18 Some part of a 50% solution of acid was replaced with an 16 12 Aug 2008, 13:27
38 Some of 50%-intensity red paint is replaced with 25% 23 17 Nov 2007, 07:05
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.