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A typical house in the 1950's had a concrete rectangular water cistern in their basement to hold water for all the household needs. The inside dimensions were 7 feet 6 inches by 8 feet 6 inches by 8 feet. If 1 cubic foot of water weighs 74 pounds, calculate the weight of the water to the neearest TENTH of a  ton if the cistern was completely filled with water.

Question

A typical house in the 1950's had a concrete rectangular water cistern in their basement to hold water for all the household needs. The inside dimensions were 7 feet 6 inches by 8 feet 6 inches by 8 feet. If 1 cubic foot of water weighs 74 pounds, calculate the weight of the water to the neearest TENTH of a  ton if the cistern was completely filled with water.

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Solution

Sure, let's break this down step by step:

  1. First, we need to find the volume of the cistern in cubic feet. The dimensions are given in feet and inches, so we need to convert the inches to feet. There are 12 inches in a foot, so 6 inches is 0.5 feet. Therefore, the dimensions in feet are 7.5 feet by 8.5 feet by 8 feet.

  2. The volume of a rectangular prism (which is what this cistern is) is length x width x height. So, we multiply these dimensions together to get the volume: 7.5 feet * 8.5 feet * 8 feet = 510 cubic feet.

  3. We know that 1 cubic foot of water weighs 74 pounds. So, the total weight of the water is 510 cubic feet * 74 pounds/cubic foot = 37,740 pounds.

  4. We want the weight in tons, and there are 2000 pounds in a ton. So, we divide the weight in pounds by 2000 to get the weight in tons: 37,740 pounds ÷ 2000 = 18.87 tons.

So, the weight of the water to the nearest tenth of a ton is 18.9 tons.

This problem has been solved

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