# Eight Days In The Dust Bowl

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Eight Months in the Dust Bowl One group of ninth graders was put to the task of surviving one winter, 240 days, in the dust bowl with limited food and water. During this eight months the group of four, two males and two females, had only one cow, one bull, 500 bushels of wheat, and 500 gallons of drinkable water. This group decided that the best way to survive would be that every person would get 2.6 gallons of water to last them 5 days and after that five days pass each person would get an additional 2.6 gallons. Furthermore, the best way for the food would be to kill the bull and the cow straight away and eat 1.5 servings of meat every two days and three bushels of wheat for the first 120 days, then to eat up to 6 servings of wheat per day…show more content…
Firstly, they would drink 2.6 gallons every 5 days for 240 days. Secondly, 240 days overall divided by every 5 days would be 48 days of distributing drinking water (240/5=48). Thirdly, you would have 48 distributing days multiplied by 2.6 which is the amount of water distributed this would give them 124.8 gallons distributed in the 240 days to a single person (48 x 2.6= 124.8). In the final analysis, they have multiplied 124.8 by 4 since that is how many people are in their group which gave them 499.2 gallons of water used (124.8 x 4= 499.2). Since only 41% of the cow and bull’s weight are consumable the group has a total of 409147.2 grams, or 902 servings, of consumable meat divided by four (902/4=225.5) then into 1.5 servings, or 680.4 grams, per 2 days and 3 bushels of wheat that is plenty for the first 120 days. This group has 500 pounds of wheat, which transfers into 13,608,000 grams of wheat. One serving of wheat is 100 grams, and that is 339 calories and each person gets up to 6 servings of wheat a day which is 2,034 calories. A woman has to have at least 1200 calories to survive and a male has to have at least 1800 calories. This meaning each person gets plenty of food and water all 240