Question

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?

Answer 1

Answer:

7 cases will be needed by kennel to feed all of the dogs for 7 days.

Explanation:

Amount of dog food required by a dog = 10 ounces

Number of dogs in kennel = 24

Number of days dog  housed in kennel = 7

Total amount of dog food = T

$T=10\times 24\times 7 ounces=1,680 ounces$

Number of cans in single case = 30

Amount of dog food in a can = 8 ounces

Total amount of dog food in a case = 30 × 8 ounces = 240 ounces

Let the X cases of dog are needed for 7 days:

$X\times 240 ounces=1,680 ounces$

$X=\frac{1,680 ounces}{240 ounces}=7$

7 cases will be needed by kennel to feed all of the dogs for 7 days.