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A certain kennel will house 24 dogs for 7 days. Each dog requires 10

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adkikani
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A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  23 May 2018, 07:26
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A certain kennel will house 24 dogs for 7 days. Each dog requires 10 ounces of dog food per day. If the kennel purchases dog food in cases of 30 cans each and if each can holds 8 ounces of dog food, how many cases will the kennel need to feed all of the dogs for 7 days?

A. 5
B. 6
C. 7
D. 8
E. 9
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Re: A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  23 May 2018, 07:51
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adkikani wrote:
A certain kennel will house 24 dogs for 7 days. Each dog requires 10 ounces of dog food per day. If the kennel purchases dog food in cases of 30 cans each and if each can holds 8 ounces of dog food, how many cases will the kennel need to feed all of the dogs for 7 days?

A. 5

B. 6

C. 7

D. 8

E. 9

Solve the question by breaking it into parts
Quote:
A certain kennel will house 24 dogs for 7 days. Each dog requires 10 ounces of dog food per day.

Total food requirement = 24*7*10 dog days ounce
Quote:
If the kennel purchases dog food in cases of 30 cans each and if each can holds 8 ounces of dog food

Total purchase of 30 cans = 240 ounces

So, Number of cases required for feeding the dogs for 7 days is \(\frac{24*7*10}{30*8}\) = 7 cases, Answer must be (C)
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Re: A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  08 Feb 2020, 13:21
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adkikani wrote:
A certain kennel will house 24 dogs for 7 days. Each dog requires 10 ounces of dog food per day. If the kennel purchases dog food in cases of 30 cans each and if each can holds 8 ounces of dog food, how many cases will the kennel need to feed all of the dogs for 7 days?

A. 5
B. 6
C. 7
D. 8
E. 9


Since each dog requires 10 ounces of dog food per day, 24 dogs require 10 x 24 = 240 ounces of dog food per day and 240 x 7 = 1680 ounces of dog food for the 7 days.

Since each case of dog food has 30 cans and each can has 8 ounces of dog food, a case has 30 x 8 = 240 ounces of dog food.

Therefore, 1680/240 = 7 cases of dog food are needed to feed 24 dogs for 7 days.

Answer: C
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Re: A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  09 Feb 2020, 06:14
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adkikani wrote:
A certain kennel will house 24 dogs for 7 days. Each dog requires 10 ounces of dog food per day. If the kennel purchases dog food in cases of 30 cans each and if each can holds 8 ounces of dog food, how many cases will the kennel need to feed all of the dogs for 7 days?

A. 5
B. 6
C. 7
D. 8
E. 9


Since each dog requires 10 ounces of dog food per day, 24 dogs require 10 x 24 = 240 ounces of dog food per day and 240 x 7 = 1680 ounces of dog food for the 7 days.

Since each case of dog food has 30 cans and each can has 8 ounces of dog food, a case has 30 x 8 = 240 ounces of dog food.

Therefore, 1680/240 = 7 cases of dog food are needed to feed 24 dogs for 7 days.

Answer: C
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A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  12 May 2021, 09:51
Weight of food required for 24 dogs for 7 days = 24*7*10 = 1680 ounces - (1)
Weight of food per case = 30 * 8 = 240 ounces - (2)

Therefore, number of cases required = (1) / (2) = 1680 / 240 = 7 cases.
Answer is choice (C).
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Re: A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink] New post  04 Aug 2022, 19:48
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Re: A certain kennel will house 24 dogs for 7 days. Each dog requires 10 [#permalink]
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