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First two containers: 64%
Third container: 58%
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
79% (02:44) correct
21%
(02:55)
wrong
based on 1785
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Humidity percentage is the water vapor mass contained within the total mass of air inside a specified volume of air. An observatory has three large containers each containing 100 tons of compressed air. The first two containers have the same humidity percentage. In an experiment, the air from all three containers was mixed into a container that had practically no air before the experiment. The humidity percentage of the combined air was found to be 62%.
In the table, select a humidity percentage for the First two containers and a humidity percentage for the Third container that are jointly consistent with the given information. Make only two selections, one in each column.
In the table, select a humidity percentage for the First two containers and a humidity percentage for the Third container that are jointly consistent with the given information. Make only two selections, one in each column.
| First two containers | Third container | |
| 58% | ||
| 59% | ||
| 60% | ||
| 64% | ||
| 65% |
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Official Answer
First two containers: 64%
Third container: 58%
|
09 Jul 2024, 01:11
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Bunuel
Experts' Global Explanation:
The amount of air in each container = 100 tons.
Let the amount of water vapor in the first container be x.
Since the first two containers have the same humidity percentage, the amount of water vapor in the second container will also be x.
Let the amount of water vapor in the third container be y.
When the air from all three containers was mixed:
The total amount of air = 100 + 100 + 100 = 300 tons.
The total amount of water vapor = x + x + y = 2x + y.
Since the humidity percentage of the combined air was found to be 62%:
2x + y = 0.62(300)
2x + y = 186
If y = 58, then:
2x + 58 = 186
2x = 186 – 58
2x = 128
x = 64
General Discussion
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08 Jul 2024, 07:29
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Humidity is computed as water-vapour mass/total mass of air. The first two container have same humidity percentage, say x%. And each container has 100 tonnes of air. Hence, it can be said that each of the first two containers have -> (x/100)*100 = x tonnes of water-vapour.
Also, lets assume the third container has y tonnes of water-vapour.
Moreover, since each container contains 100 tonnes of air. So, when combined into one container, the final large container has 300 tonnes of air. And the humidity percentage of final container is 62%.
Converting the above statements into an equation -
(total mass of water-vapour of the three containers / total mass of air of the three conatiners) * 100 = 62
((2x + y)/300) * 100 = 62
2x + y = 186.
Now by filling in the values from the available answer choices, we conclude that (humidity in each of first two containers) x = 64, and (humidity of third container) y = 58
Also, lets assume the third container has y tonnes of water-vapour.
Moreover, since each container contains 100 tonnes of air. So, when combined into one container, the final large container has 300 tonnes of air. And the humidity percentage of final container is 62%.
Converting the above statements into an equation -
(total mass of water-vapour of the three containers / total mass of air of the three conatiners) * 100 = 62
((2x + y)/300) * 100 = 62
2x + y = 186.
Now by filling in the values from the available answer choices, we conclude that (humidity in each of first two containers) x = 64, and (humidity of third container) y = 58
