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

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SVP
Joined: 21 Jan 2007
Posts: 2262
Location: New York City
On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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16 Feb 2007, 05:03
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Difficulty:

55% (hard)

Question Stats:

64% (02:22) correct 36% (02:23) wrong based on 306 sessions

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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
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16 Feb 2007, 05:22
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
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Joined: 04 Apr 2006
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16 Feb 2007, 05:28
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
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14 Feb 2008, 10:32
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

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14 Feb 2008, 20:45
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

Me too, I like it. This problem wastes me more than 5 minutes but finally it did not work for me! many thanks
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05 Mar 2014, 22:45
1
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
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Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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16 Apr 2014, 09:31
1
2
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.

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Joined: 21 Jun 2013
Posts: 25
Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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25 Jun 2017, 00:06
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.
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On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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07 Feb 2019, 07:58
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?
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GMAT 1: 710 Q45 V42
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Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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07 Feb 2019, 08:33
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!
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Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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07 Feb 2019, 10:31
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
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
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Re: On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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10 Feb 2019, 08:33
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.

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On Monday, a certain animal shelter housed 55 cats and dogs.  [#permalink]

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16 Apr 2019, 10:36
1st eq --> c + d = 55
2nd eq --> c/5 + d/4 = ?

what do we do with ? plug the different answers starting by (b)

? = 12

c + d = 55
c/5 + d/4 = 12
Result : d=20 ; c=35

In order to make sure that 12 is the maximum number of adopted pets, let's try now (d):

c + d = 55
c/5 + d/4 = 14
Result : d=60 ; c=-5

hmm... seems impossible to have -5 cats, so 14 adopted pets seem too big number. It seems reasonable to go for answer (c) as it would still make sense and would maximize number of adopted pets more than answer (b).
On Monday, a certain animal shelter housed 55 cats and dogs.   [#permalink] 16 Apr 2019, 10:36
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