GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Aug 2018, 21:15

### 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

# If 200 lb of a mixture contain 80% husk and 20% sand. Then

Author Message
TAGS:

### Hide Tags

Director
Joined: 25 Oct 2008
Posts: 546
Location: Kolkata,India
If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

Updated on: 02 Dec 2013, 08:21
10
00:00

Difficulty:

35% (medium)

Question Stats:

78% (01:23) correct 22% (01:51) wrong based on 462 sessions

### HideShow timer Statistics

If 200 lb of a mixture contain 80% husk and 20% sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk?

A. 1/4
B. 20/3
C. 1/2
D. 40
E. 60

_________________

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

Originally posted by tejal777 on 13 Aug 2009, 03:27.
Last edited by Bunuel on 02 Dec 2013, 08:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

22 Apr 2014, 02:46
7
1
Mixture has 160 lbs husk & 40 lbs sand

After removal of husk, its percentage should be 75%, it means

percentage of sand should be 25%

25% of What?? = 40

What comes out to be 160

200 - 160 = 40

40lbs husk should be removed to have its share 75%

Algebraic Method:

Husk............ Sand........... Total
160............... 40.................. 200

Lets say quantity x husk is removed

160-x ............. 40 .................. 200-x

Setting up the equation

$$160-x = \frac{75}{100} (200-x)$$
x = 40

_________________

Kindly press "+1 Kudos" to appreciate

##### General Discussion
Senior Manager
Joined: 25 Jun 2009
Posts: 285

### Show Tags

13 Aug 2009, 05:10
3
1
tejal777 wrote:
if 200 lb of a mixture contain 80% husk and 20%sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk?

1/4
20/3
1/2
40
60

We need to extract 40 lb of husk.

80 % of 200 = 160
20 % of 200 = 40

Now we have t remove husk to have 75 % of Husk ( key point sand is not supposed to be removed)

Let X be the new quantity after removing husk

So, 40 = 25 % of X
X = 160

Husk will 120 lbs , hence we need to remove 1/4 of the total husk or 40 lbs
Manager
Joined: 15 Jun 2009
Posts: 133

### Show Tags

26 Aug 2009, 03:00
1
40 it is...
ONE MORE WAY TO APPROACH ..ALTHOUGH CONCEPT IS SIMILAR....

(160-x) = (200-x)75/100

Solve for x...
x = 40
Intern
Joined: 12 Oct 2009
Posts: 14

### Show Tags

26 Jan 2010, 22:50
Good question. When I first tried it, I kind of overlooked the word extracted. Once I realized the answer i came up with was not an answer choice...I decided to look at the question again. I need to figure out how to overcome this problem I sometimes have with reading a question and not catching certain key words.
Intern
Joined: 19 Dec 2009
Posts: 10

### Show Tags

27 Jan 2010, 02:11
4
I would suggest a generalized strategy:
1) reading the question in parts and not the whole in one go.
2) writing down the necessary numbers to make it simpler (i.e. converting the English into Numbers).
In the above example, the key is to understand that 'sand' should not be extracted, which means that the 'sand' quantity in the mixture will not change. Only the proportion(% in terms of the mixture) will change.
First 'sand' was 20% of the mixture i.e. 40lbs. Later-on this 40lbs 'sand' becomes 25% of the mixture. Now the rest 75% of the mixture will be 120lbs of husk. Which is a reduction of 40lbs of 'husk', which is your answer.
Intern
Joined: 10 Jan 2011
Posts: 14
Location: India
Schools: ISB, IIM-A
WE 1: 4 yrs in finance

### Show Tags

26 Jan 2011, 10:40
one more way to approach

200*80/100-x=75/100*(200-x)

solve for x=40
_________________

Rahul

Retired Moderator
Joined: 16 Nov 2010
Posts: 1458
Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

12 Mar 2011, 02:49
160-x/200-x = 3/4

So x = 40 and the answer is D.
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 11 Apr 2011
Posts: 261

### Show Tags

17 May 2011, 07:50
2
1
80% of 200 = 160 lb of husk ---> Total Husk in 200 lb.
75% concentration of total husk = 75% of 160 lb = (3/4) * 160 = 120

Total amount to be extracted = Initial Amount of Husk - 75% concentration of Husk
Total amount to be extracted = 160 lb - 120 lb = 40 lb
_________________

Powerscore CR Bible Full Chapter Notes | Easily Extend Vocabulary List with Google Dictionary

Please kudo me if you found my post useful. Thanks!!!

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1124

### Show Tags

17 May 2011, 11:55
(160-x)/(200-x) = 75/100=3/4

x=640-600 = 40
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Director
Joined: 01 Feb 2011
Posts: 671

### Show Tags

17 May 2011, 16:53
let x be the removed husk quantity

=> $$(80/100)(200)-x = (75/100)(200-x)$$

=> x = 40
Director
Joined: 03 Aug 2012
Posts: 822
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

18 Apr 2014, 21:25
Total=200lb

Husk=160
Other=40

Consider X is taken out.

(160-X) = .75 (200-X) {Since the new solution has 75% conc.}

X=40
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Intern
Joined: 26 Oct 2013
Posts: 22
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

28 May 2014, 19:13
Easy approach:

From the question you know that
Husk = 160lbs (80% of 200)
Sand = 40lbs (20% of 200)

After removing some Husk the concentration will be 75%. We need to find how much Husk to remove

Remember that the quantity of sand doesn't change, we still have 40lbs of sand, but now they represent 25% of the new mix.
If 25% = 40lbs, 75% equals 120lbs (simple math)

So we need to remove: 160-120 lbs = 40!!
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

01 Jun 2014, 21:51
PareshGmat wrote:
Mixture has 160 lbs husk & 40 lbs sand

After removal of husk, its percentage should be 75%, it means

percentage of sand should be 25%

25% of What?? = 40

What comes out to be 160

200 - 160 = 40

40lbs husk should be removed to have its share 75%

Algebraic Method:

Husk............ Sand........... Total
160............... 40.................. 200

Lets say quantity x husk is removed

160-x ............. 40 .................. 200-x

Setting up the equation

$$160-x = \frac{75}{100} (200-x)$$
x = 40

One more method to setup the equation:

$$40 = \frac{25}{100} * (200-x)$$

160 = 200 - x

x = 40
_________________

Kindly press "+1 Kudos" to appreciate

Director
Joined: 23 Jan 2013
Posts: 598
Schools: Cambridge'16
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

03 Aug 2015, 03:40
(160-x/200-x)*100=75

1000=25x => x=40

OR

0.8*(200)-x=0.75*(200-x)

10=0.25x => x=40

D
Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 397
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

25 Sep 2016, 09:28
200(80/100) - x = (200-x)(75/100)

160 - x = 150 - 3/4 (x)

10 = -3/4 (x) + 4/4

10 = 1/4 (x) --> x = 40
Non-Human User
Joined: 09 Sep 2013
Posts: 7754
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then  [#permalink]

### Show Tags

06 Dec 2017, 20:15
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then &nbs [#permalink] 06 Dec 2017, 20:15
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.