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

It is currently 21 Nov 2018, 06:10

LIVE NOW!

LBS is Calling R1 Admits - Join Chat Room to Catch the Latest Action


Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

     November 22, 2018

     November 22, 2018

     10:00 PM PST

     11:00 PM PST

    Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
  • Key Strategies to Master GMAT SC

     November 24, 2018

     November 24, 2018

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

M06-12

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

Hide Tags

Intern
Intern
avatar
B
Joined: 27 Mar 2017
Posts: 5
GMAT 1: 720 Q50 V38
Re: M06-12  [#permalink]

Show Tags

New post 18 May 2017, 02:01
It is always so time taking to arrive at the integer values of the equations like this by picking numbers! Is there any material on how we can quicken this step?
Thank you!
Intern
Intern
avatar
Joined: 30 Oct 2017
Posts: 1
Re: M06-12  [#permalink]

Show Tags

New post 30 Oct 2017, 06:14
(5,25) clearly violates the condition that the total number of heads and feet between the cows and chickens equals 100, so C cannot be a possible answer. (5,25) = 5(3) + 25 (5)
= 15 + 125 = 140.

Sachin07 wrote:
A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17


Let \(C\) be a number of chickens, \(W\) - cows, and \(S\) - sheep.

Chickens and cows together have 100 feet and heads: \(100=1C+1W+4W+2C\), assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per \(C\) and \(W\) accordingly. Therefore: \(100 = 3C + 5W\).

It is given that \(W \gt C\), and we also see that \(C\) should be a multiple of 5 for \(3C + 5W\) to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

\(S = \frac{(C+W)}{3}\); \(5 + 17 = 22\) which is not divisible by 3, but \(10 + 14 = 24\) which is divisible by 3, therefore only the second pair will work. So, \(S = 8\).


Answer: B



another pair of (5,25) is possible and as per this the answer can be C.10
[/quote]
Intern
Intern
avatar
B
Joined: 16 Jul 2017
Posts: 1
Re: M06-12  [#permalink]

Show Tags

New post 17 Nov 2017, 17:58
First post here, so sorry if it is put in the wrong place.

This is a solution to - M06-12

Chicken + Cows = 3 times Sheep
=> Chicken + Cows + Sheep = 3 Sheep + Sheep = 4 Sheep

Therefore, the answer need to be a multiple of 4. Only one choice i.e. 8.

Thanks!
Manager
Manager
avatar
G
Joined: 27 Dec 2016
Posts: 243
Re: M06-12  [#permalink]

Show Tags

New post 03 Mar 2018, 18:09
Hi Bunuel,

I was wondering could you please explain this concept one more time? You mentioned in your OE that, "3C + 5W = 3C + (a multiple of 5) = 100 = (a multiple of 5), thus 3C must also be a multiple of 5, which means that C must be a multiple of 5". I understood that 100 = (a multiple of 5) so we have, 3C + (a multiple of 5) = (a multiple of 5). What I couldn't understand was how did we deduce that 3C is also a multiple of 5 and then concluded that C must be a multiple of 5 as well? Could you please explain this part? Would greatly appreciate it?
Intern
Intern
avatar
B
Joined: 10 May 2018
Posts: 23
CAT Tests
Re: M06-12  [#permalink]

Show Tags

New post 21 Oct 2018, 05:54
seems impossible to solve under 2 minutes... It's unlikely to encounter something like this on the GMAT
GMAT Club Bot
Re: M06-12 &nbs [#permalink] 21 Oct 2018, 05:54

Go to page   Previous    1   2   [ 25 posts ] 

Display posts from previous: Sort by

M06-12

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.