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23 May 2018, 07:26
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
93% (01:31) correct
7%
(02:05)
wrong
based on 460
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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
A. 5
B. 6
C. 7
D. 8
E. 9
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23 May 2018, 07:51
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adkikani
Solve the question by breaking it into parts
Quote:
Total food requirement = 24*7*10 dog days ounce
Quote:
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)
08 Feb 2020, 13:21
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adkikani
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
