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# animal shelter holds 55 cats and dogs on Monday. On Friday,

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animal shelter holds 55 cats and dogs on Monday. On Friday, [#permalink]  16 Feb 2007, 05:03
animal shelter holds 55 cats and dogs on Monday. On Friday, exactly 1/5 of the cats and 1/4 of dogs were adopted. no new dogs or cats were taken into animal shelter during this period. what is the greatest possible number of pets that could have been adopted between monday and friday?

answer: 13

we want to find the larger ratio to maximize the amount of pets adopted so we focus on the 1/4 dogs. how do we set up this question mathematically from here without plugging in until a number fits the ratios?
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[#permalink]  16 Feb 2007, 05:22
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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]  16 Feb 2007, 05:28
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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]  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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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Re: Re: [#permalink]  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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Re: Re:   [#permalink] 14 Feb 2008, 20:45
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# animal shelter holds 55 cats and dogs on Monday. On Friday,

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