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Proportion

by manelgirona » Tue Feb 09, 2010 9:13 am
At a pet center 35 dogs of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many can of food would be needed fo 12 days if 3 of the dogs were sold?

a) 84
b) 70
c) 49
d) 25
e) 24



OA: C

Any ideas about how to solve it?
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by ajith » Tue Feb 09, 2010 9:26 am
manelgirona wrote:At a pet center 35 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many can of food would be needed fo 12 days if 3 of the dogs were sold?

a) 84
b) 70
c) 49
d) 25
e) 24



OA: C

Any ideas about how to solve it?
35 cans used in 6 days to feed 10 dogs

to feed 1 dog in a day ; we have to use 35/60 cans is used
to feed 7 dogs for 12 days; we have to use 35*7*12/60 = 49 cans
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by shashank.ism » Tue Feb 09, 2010 10:55 am
ajith wrote:
manelgirona wrote:At a pet center 35 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many can of food would be needed fo 12 days if 3 of the dogs were sold?

a) 84
b) 70
c) 49
d) 25
e) 24



OA: C

Any ideas about how to solve it?
food of dog for 1 day = 35/60
7 dogs are left now
so total cans required = 35*7*12/60 = 49 Ans C

35 cans used in 6 days to feed 10 dogs

to feed 1 dog in a day ; we have to use 35/60 cans is used
to feed 7 dogs for 12 days; we have to use 35*7*12/60 = 49 cans
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by harsh.champ » Tue Feb 09, 2010 1:22 pm
manelgirona wrote:At a pet center 35 dogscan of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many can of food would be needed fo 12 days if 3 of the dogs were sold?

a) 84
b) 70
c) 49
d) 25
e) 24



OA: C

Any ideas about how to solve it?
My problem approach will be like this:-Diet of dog per day = 35/60 cans[In this type of problems it is better to solve by unitary method like considering for per day per dog etc.]
For 12 days and 7 dogs we have 12 x 7 x 35/60 = 49 C.
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