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

It is currently 22 Feb 2019, 05:30

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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
  • Free GMAT RC Webinar

     February 23, 2019

     February 23, 2019

     07:00 AM PST

     09:00 AM PST

    Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
  • FREE Quant Workshop by e-GMAT!

     February 24, 2019

     February 24, 2019

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score.

On Monday, a certain animal shelter housed 55 cats and dogs.

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

Hide Tags

 
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2584
Location: New York City
On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 16 Feb 2007, 04:03
14
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

65% (02:22) correct 35% (02:29) wrong based on 267 sessions

HideShow timer Statistics

On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday.

A. 11
B. 12
C. 13
D. 14
E. 20
Most Helpful Community Reply
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1768
  [#permalink]

Show Tags

New post 16 Feb 2007, 04:22
4
4
Let set some vaiables
o C : the number of cats
o D : the number of dogs

C + D = 55

In this problem, we do not care which one from dogs or cats is in a greater number. So, arbitrarily, we say that the cats are in a bigger number.

To maximise the number of sold pents, we have to maximise C and to respect:
o C/4 = integer
o D/5 = integer

Here, we can turn it to those equations
o C = 4*k
o D = 5*i

4*k + 5*i = 55
<=> k = 5*(11-i)/4

Therefore, (11-i)/4 must be a positive integer and the biggest possible. It's i = 3.

Then, k + i = 10 + 3 = 13 :)
General Discussion
Intern
Intern
User avatar
Joined: 04 Apr 2006
Posts: 35
  [#permalink]

Show Tags

New post 16 Feb 2007, 04:28
3
2
You can not solve this mathematically sinds you have 2 equations with 3 unknowns

c+d=55
1/5*c+1/4*d=x

Maximize x

You have to choose c and d so that c is divisible with 5 and d divisible with 4, because 3.5 dogs or cats can not be adopted

So, c can be
55,50,45,40,35,30,25,20,15,10,5,0

d can be
0,4,8,12,16,20,24,28,32,36,40,44,48,52------(1)

Look then the combination of theese numers which gives 55

If c is
55,50,45,40,35,30,25,20,15,10,5,0
then d must be
0,5,10,15,20,25,30,35,40,45,50,55-----(2)

Numbers that are repeating in (1) and (2) are 0,20,40

d is 40
c is 15

15/5+40/4=3+10=13
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2584
Location: New York City
Re:  [#permalink]

Show Tags

New post 14 Feb 2008, 09:32
Fig wrote:
Let set some vaiables
o C : the number of cats
o D : the number of dogs

C + D = 55

In this problem, we do not care which one from dogs or cats is in a greater number. So, arbitrarily, we say that the cats are in a bigger number.

To maximise the number of sold pents, we have to maximise C and to respect:
o C/4 = integer
o D/5 = integer

Here, we can turn it to those equations
o C = 4*k
o D = 5*i

4*k + 5*i = 55
<=> k = 5*(11-i)/4

Therefore, (11-i)/4 must be a positive integer and the biggest possible. It's i = 3.

Then, k + i = 10 + 3 = 13 :)

thanks. i love this approach

:-D
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

SVP
SVP
avatar
Joined: 04 May 2006
Posts: 1647
Schools: CBS, Kellogg
Premium Member
Re: Re:  [#permalink]

Show Tags

New post 14 Feb 2008, 19:45
bmwhype2 wrote:
Fig wrote:
Let set some vaiables
o C : the number of cats
o D : the number of dogs

C + D = 55

In this problem, we do not care which one from dogs or cats is in a greater number. So, arbitrarily, we say that the cats are in a bigger number.

To maximise the number of sold pents, we have to maximise C and to respect:
o C/4 = integer
o D/5 = integer

Here, we can turn it to those equations
o C = 4*k
o D = 5*i

4*k + 5*i = 55
<=> k = 5*(11-i)/4

Therefore, (11-i)/4 must be a positive integer and the biggest possible. It's i = 3.

Then, k + i = 10 + 3 = 13 :)

thanks. i love this approach

:-D


Me too, I like it. This problem wastes me more than 5 minutes but finally it did not work for me! many thanks
_________________

GMAT Club Premium Membership - big benefits and savings

SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1820
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re:  [#permalink]

Show Tags

New post 05 Mar 2014, 21:45
Fig wrote:
Let set some vaiables
o C : the number of cats
o D : the number of dogs

C + D = 55

In this problem, we do not care which one from dogs or cats is in a greater number. So, arbitrarily, we say that the cats are in a bigger number.

To maximise the number of sold pents, we have to maximise C and to respect:
o C/4 = integer
o D/5 = integer

Here, we can turn it to those equations
o C = 4*k
o D = 5*i

4*k + 5*i = 55
<=> k = 5*(11-i)/4

Therefore, (11-i)/4 must be a positive integer and the biggest possible. It's i = 3.

Then, k + i = 10 + 3 = 13 :)



I took a single variable; Let dogs = x so Cats = 55-x & solved it

However now I see that taking 2 variables just reduces the time taken to solve (in this type of problems) where we need not have to individually count the no of dogs / cats
_________________

Kindly press "+1 Kudos" to appreciate :)

Current Student
avatar
B
Joined: 23 May 2013
Posts: 186
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
GMAT ToolKit User
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 16 Apr 2014, 08:31
1
1
bmwhype2 wrote:
On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday.

A. 11
B. 12
C. 13
D. 14
E. 20


Here's my approach:

\(C+D = 55\)

\(Adopted = \frac{C}{5} + \frac{D}{4}\) <-- This is what we want to maximize.

Letting C = 55-D, we have:

\(Adopted = \frac{55-D}{5} + \frac{D}{4}\)
\(Adopted = 11 - \frac{D}{5} + \frac{D}{4}\)
\(Adopted = 11 + \frac{D}{20}\)

Therefore, for the number of adopted to be an integer, D must be a multiple of 20, which means D could either be 0, 20, or 40, corresponding to Adopted = 11,12, and 13.

13 is then the maximum.

Answer: C
Intern
Intern
avatar
B
Joined: 21 Jun 2013
Posts: 26
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 24 Jun 2017, 23:06
I found this much simpler way of doing this question!
55 cats and dogs - Cats should be divisible by 5 and dogs by 4. Only two breakups are possible :
35 cats and 20 dogs or
15 cats and 40 dogs
we have to consider either 20 dogs or 40 dogs because if we take any other multiple of 4, it will lead to certain number of cats which are not divisible by 5.
So, out of these two, we take the 2nd possibility of 15 cats and 40 dogs as that gives us the greatest possible number of pets that could be adopted.
Manager
Manager
User avatar
B
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 65
On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 07 Feb 2019, 06:58
Do we necessarily need to account for the divisibility of dogs/cats as mentioned? I understood the min/max element related to fractions, but just picked 50 dogs and 5 cats and rounded down 50/4 to a whole number. Using 50 dogs, 10 cats also yield 13 this way as does 45 dogs and 15 cats. Is it a fluke of the question or is it generally fine to round down to a whole number in Min/Max cases?
Intern
Intern
avatar
B
Joined: 27 Oct 2017
Posts: 6
Location: United States
GMAT 1: 710 Q45 V42
GPA: 3.59
CAT Tests
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 07 Feb 2019, 07:33
bmwhype2 wrote:
On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday.

A. 11
B. 12
C. 13
D. 14
E. 20


I tackled this problem by trying to maximize the number of dogs. All else being equal, there will be more dogs given away than cats if their numbers are equal (ex. 1/4*x > 1/5*x).

Given this info, we want the largest number of dogs (and fewest number of cats) possible, so both should be an integer. Cats have a nice 5 number, so they are easier to work with:

5c + 50d --> d not an integer
10c + 45d --> d not an integer
15c + 40d --> d IS an intger

(1/3*15) + (1/4*40) = 3+10 = 13

Hope this helps!
SVP
SVP
User avatar
G
Joined: 18 Aug 2017
Posts: 1926
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member CAT Tests
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 07 Feb 2019, 09:31
bmwhype2 wrote:
On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday.

A. 11
B. 12
C. 13
D. 14
E. 20


given
c+d=55
anf
c/5+d/4 were adopted
or say
c=55-d
55-d/55+ d/4
solve
11+d/20

now d has to be a multiple of 20 ; d can be 20,40 max
so
11+40/20 = 13
IMO C
_________________

If you liked my solution then please give Kudos. Kudos encourage active discussions.

Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4955
Location: United States (CA)
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

Show Tags

New post 10 Feb 2019, 07:33
bmwhype2 wrote:
On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday.

A. 11
B. 12
C. 13
D. 14
E. 20


We see that we have to “split” the number 55 into two positive numbers such that one of them is divisible by 5 and the other by 4. Since 55 is already a multiple of 5, then one of the numbers must be both a multiple of 4 and 5 so that the other is always a multiple of 5. Therefore, one of the numbers must be either 20 or 40.

If there are 20 dogs and 35 cats, then 1/4 x 20 + 1/5 x 35 = 5 dogs + 7 cats = 12 animals are adopted.

If there are 40 dogs and 15 cats, then 1/4 x 40 + 1/5 x 15 = 10 dogs + 3 cats = 13 animals are adopted.

Answer: C
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

GMAT Club Bot
Re: On Monday, a certain animal shelter housed 55 cats and dogs.   [#permalink] 10 Feb 2019, 07:33
Display posts from previous: Sort by

On Monday, a certain animal shelter housed 55 cats and dogs.

  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®.