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As a member of FEMA, you’re required to set up a contingency plan to supply meals to residents of a town devastated by a tornado. A breakfast ration weighs 12 ounces and the lunch and dinner rations weigh 18 ounces each. Assuming a food truck can carry 3 tons and that each resident will receive 3 meals per day, how many residents can you feed from one truck during a 10–day period?

2 Answers

5 votes

32,000 oz in a ton 96,000 oz can be held in truck

48 oz for each person per day

I'm not confident with my answer but use this to figure it out, hope I helped.

User BlueM
by
5.1k points
4 votes

Answer:

220. 4 persons approximately

Explanation:

first consider,

1 Ton = 35274 ounce

Weight of food for one person (Breakfast+lunch+dinner) for one day = 12+18+18 = 48 ounce

Weight of food for one person (Breakfast+lunch+dinner) for ten day

= (12+18+18)*10 = 480 ounce

Maximum food the truck can carry = 3 tons = 105822 ounce

Maximum persons who could be fed = 105822/480 = 220.4

Since, one person is consuming 480 ounce of food in the duration of ten days, so you can easily obtain the number of maximum people allowed by dividing total capacity of truck in ounce which is 105822 ounce to the ounces of food consumed by one person in ten days which is 480 ounces.

User Yugr
by
5.0k points
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