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

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On Monday, a certain animal shelter housed 55 cats and dogs. [#permalink] New post 16 Feb 2007, 04:03
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61% (02:50) correct 38% (01:39) wrong based on 83 sessions
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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 [#permalink] New post 16 Feb 2007, 04:22
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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 :)
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 [#permalink] New post 16 Feb 2007, 04:28
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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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Re: [#permalink] 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
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Re: Re: [#permalink] 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
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Re: On Monday, a certain animal shelter housed 55 cats and dogs. [#permalink] New post 01 Mar 2014, 09:52
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Re: [#permalink] 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
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Re: On Monday, a certain animal shelter housed 55 cats and dogs. [#permalink] New post 16 Apr 2014, 08:31
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


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
Re: On Monday, a certain animal shelter housed 55 cats and dogs.   [#permalink] 16 Apr 2014, 08:31
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