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Fleas

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Fleas

by MBALA2009 » Sun Jul 20, 2008 6:32 pm
A boarding kennel boards 60 animals (only dogs and cats), and is experiencing a problem with fleas. It is determined that 1/5 of the cats and 1/3 of the dogs are afflicted with fleas. What is the smallest number of animals in the kennel that could have fleas?

A) 9
B) 11
C) 13
D) 14
E) 15

ANS: D
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by Saule » Sun Jul 20, 2008 8:20 pm
Not sure if there's any formula to solve this, but the intuitive way to it is to think that since you need to find the smallest number of animals with fleas, then you need to have as many cats as possible and as little dogs as possible within the given constraints. The constraints are that the number of cats should be divisable by 5 and the number of dogs by 3.
So, say, 55 cats vs.5 dogs won't work, since 5 can't be divided by 3.
50 vs. 10 is wrong for the same reason.
But 45 cats and 15 dogs works just fine.
So, the answer is 45/5 + 15/3 = 14
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by vishubn » Sun Jul 20, 2008 8:24 pm
Hi

Maximizing and minimizing problems this is the method i employ always :

Hope it helps


N=60
cats=1/5
dogs=1/3

N=d+g where d is the multiple of 3 and c is the multiple of 5

In order to get the minimum number we need to to maximise the numebr of cats and dogs

plug in values:

the biggest number which satisfies the multiples of 3 and 5 condition
60=cat(highest multiple of 5) + dog(highest multiple of 3)
60=45+15

now, number of animals afflicted with fleas :
--> 1/5(45)=9 cats
--> 1/3(15)=5 dogs

answer =14


Vishu
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by TRANGZON » Sun Jul 20, 2008 8:34 pm
the animals are afflicted with fleas: 1/3x + 1/5y

the smallest animals afflicted --> y > x
x=3 --> y=57 not divisable 5
x=6 --> y=54 not divisable 5
x=9 --> y=51 not divisable 5
x=12 --> y=48 not divisable 5
x=15 --> y=45 divisable 5 --> ACCEPTED

--> the animals are afflicted with fleas: 1/3x15 + 1/5x45 = 5+9= 14
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