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16 Feb 2007, 05:03
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Difficulty:
55%
(hard)
Question Stats:
66% (02:19) correct
34%
(02:28)
wrong
based on 544
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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
A. 11
B. 12
C. 13
D. 14
E. 20
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
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
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
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
