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ggarr
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At a park, people were walking their dogs and cats. It is 11 Feb 2007, 03:37
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At a park, people were walking their dogs and cats. It is known that there were more dogs than people. Altogether, dogs and people had 100 feet and heads. It is also known that there were 3 times as many dogs and people as cats. How many cats were there in the park?

6
8
9
10
15

I started like this:
100 = 4L + 2L + H +H
100 = 6L + 2H
It would be great if someone could finish what I started ... to the extent possible



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devilmirror
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At a park, people were walking their dogs and cats. It is 11 Feb 2007, 10:41
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ggarr
At a park, people were walking their dogs and cats. It is known that there were more dogs than people. Altogether, dogs and people had 100 feet and heads. It is also known that there were 3 times as many dogs and people as cats. How many cats were there in the park?

6
8
9
10
15

I started like this:
100 = 4L + 2L + H +H
100 = 6L + 2H
It would be great if someone could finish what I started ... to the extent possible
Show more


Good Question. 8-)

Assuming: P = People, D = Dog, and C = Cat

People has 1 head 2 feet. Total of people's head and feet = P + 2P = 3P
Dog has 1 head 4 feet. Total of dog's head and feet = D + 4D = 5D
From question: 3P + 5D = 100

This is the linear equation. The possible outcomes are:
(P,D)
(5, 17)
(10, 14)
(15, 11)
(20, 8)
...

However, one condition from the question told us that P < D. We are left with 2 pairs of (P,D)
(5, 17)
(10, 14)

From question (P+D)/3 = C

Only (10, 14) will make C possible.

C = (10 + 14)/3 = 24/3 = 8

There are 8 cats in the park.
:)
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At a park, people were walking their dogs and cats. It is 11 Feb 2007, 10:57
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i like your way devil....

i solved it differently.... (i hate equations) by simply back solving.

i always start with the middle answer which is 9 in this case...

if there were 9 cats, there would be 27 dogs and people. which means that there are at least 14 dogs, which add 70 to the head+feet count. together with 13 people (which add 39) this is already too much. and this was the minimum number of dogs (more dogs add even more feet)...
so 9 cats is too many.
then i tried 8. which seem feasible (24 dogs+people), then 13 dogs ad 65 and 11 people add 33... almost there... 14 dogs would do the job.

then double checked 6:
6 cats, mean 18 dogs and people, but even 18 dogs don't add enough to the feet&head count... hence cannot happen.

8 is the correct answer.

i like devil's way better, because it reduces the possible answers rather fast. you find real fast the P must be divisible by 5. then you are down to just 2 possible answers for P,D... and their sum must be a multiple of 3.... wow... fast,neat and clean....



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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