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

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16 Feb 2007, 04:03

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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.

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

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01 Mar 2014, 09:52

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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

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, 08:31

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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.

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.

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

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14 Jul 2015, 08:45

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

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22 Aug 2016, 10:38

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

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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.