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

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A certain kennel will house 24 dogs for 7 days. Each dog [#permalink] New post 26 Feb 2007, 11:43
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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 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?
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Re: PS [#permalink] New post 26 Feb 2007, 13:27
OK, so let's figure out how many ounces all 24 dogs will need per day.
24x10 = 240
Then, let's multiply by number of days the dogs need to be fed.
240x7=1680
Now, we know that each case of dog food contains 30 cans, and each can contains 8 ounces of food. Let's calculate how many ounces each case contains:
30*8= 240
To find out how many cases of dog food a kennel need, all we need to do is to divide 1680 by 240.
1680/240=7

Makes sence?
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 [#permalink] New post 06 Mar 2007, 13:47
Although this question is an easy one, it may still cause some difficulties when one computes the numbers. Therefore I do the following: I just represent the numbers as factors. For example:
24 Dogs x 10 ounces a dog is simply 24x10 ("x" meaning "times")
then we know that we have 7 dogs->total number of ounces 7x24x10
represent the number of ounces needed in the same way (most of the numbers cancel each other out) and you will get the same answer much faster and easier!
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 [#permalink] New post 07 Mar 2007, 09:35
Point to be noted here.

Though 1680 is divisible by 24 but if you find that total oz of needed food is, for ex. 1690, which is not divisible by 24, number cases would be the next integer number, which, in this example, is 8.
  [#permalink] 07 Mar 2007, 09:35
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A certain kennel will house 24 dogs for 7 days. Each dog

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