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bmwhype2
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at a park, people were walking their dogs and cats. There 09 Nov 2007, 14:09
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at a park, people were walking their dogs and cats. There were more dogs than people. Dogs and people had 100 feet and heads altogether. If the combined number of people and dogs were three times the number of cats, how many cats were in the park?

5
8
10
14
17



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at a park, people were walking their dogs and cats. There 09 Nov 2007, 14:29
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I got B. 8 cats.

People + Dogs = 3*8 = 24

10 people * 3 = 30

14 dogs * 5 = 70

30 + 70 = 100 heads and feet combined.
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at a park, people were walking their dogs and cats. There 09 Nov 2007, 17:28
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bmwhype2
at a park, people were walking their dogs and cats. There were more dogs than people. Dogs and people had 100 feet and heads altogether. If the combined number of people and dogs were three times the number of cats, how many cats were in the park?

5
8
10
14
17
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Answer B

P - 2 feet and 1 head
D - 4 feet and 1 head

3P + 5D = 100

D = 20 - 3P/5

possible solutions
for P = 5, D = 17 But 17+5 = 22 not divisible by 3 ( since cats are 1/3)

for P = 10 , D = 14 Total 24 . So 8 cats are there .

for P = 15 , D becomes less than 15 which is not the case as D > P
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at a park, people were walking their dogs and cats. There 10 Nov 2007, 00:09
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bmwhype2
at a park, people were walking their dogs and cats. There were more dogs than people. Dogs and people had 100 feet and heads altogether. If the combined number of people and dogs were three times the number of cats, how many cats were in the park?

5
8
10
14
17
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Getting B.
8 Cats, 14 dogs and 10 people.
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at a park, people were walking their dogs and cats. There 10 Nov 2007, 06:52
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how can this be solved just by using equations, and not backsolving by subbing answers in ?

i start with 3p+5d=100 and p+d = 3c, but this is two equations and 3 unknowns, can it be solved ?



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Hi there,
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