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Using alligations: Ratio of qty of first two containers and third container= 200/100=2/1

Two cont-----------------------3rd cont
--x%-----------------------------y%--
-----------------62%-----------------
--2------------------------------1---

We need to find the two values such that 62% is closer to the value of x% since the qty of two containers is more and the difference between the two values is a multiple of 3. On scanning the answer choices we see 58% and 64% have a difference of 6 and this difference is split in the ratio 1:2 such that 62% and 64% have a gap of 2 units and 62% and 58% have a gap of 4 units
Thus x=64% and y=58%
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Bunuel
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.


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­Let 100 tons of each container combines, then out of the 300 ton of air, 62% is humidity i.e., 186 ton. So check in the options of the humidity content that fits in the below equation:
2* ( Humidity content % of First two containers) + (humidity percentage for the Third container) = 186

By this, 

Humidity content % of First two containers = 68% and 
humidity percentage for the Third container = 58%.­
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Bunuel
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.


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­Let mass of water vapur in 1st container = x tons, 2nd = x tons (as first two containers have same humidity), and 3rd = y tons out of 100 tons of air.

After mixing, total volume of compressed air = 300 tons
Total mass of water vapur = 2x+y

Humidity = \( (2x+y)/300 = 62% = 62/100  => 2x+y = 186\)

Iterating the values from options. we get x = 64 and y = 58
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­First 2 containers: 64% 
Third container: 58%

The total humidity in the final container would be 62% * 300 (total qty. of air in 3 containers) = 186, so we need to find 2 percentages that give us 186 and we know that the 2 containers combined will give us a higher % of humidity and would be a closer % to 62% to skew the average upwards and the third container will have a % that is below 62% and would be a further away from 62% than the % of the first 2 containers

First 2 containers
64%: 64% * 200 = 128 

, then we have 58 left (186 - 128), so the third container is 58% * 100 = 58

128+58 = 186
 
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2 x + y = 3* 62
Solve for x & y
At x= 64, y= 58 satisfies the condition
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if x tons of water is in first two containers and y tons on water in the last container then new % humidity is
% humidity = (2x + y) / 3 = 62

we know from above equation that y has to be even so we only need to check for remaining 3 options and select the option that is present.

x = 64, y = 58
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[quote="Bunuel"]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%.

A=64% B = 58%

Explanation:

Humidity of air in containers Aand B - x%
humidity in container C-y%

(100*x+100*x+100*y)/300= 62
2x+y=62*3
2x+y= 186

x= 64 y= 58 satisfies this equation

Posted from my mobile device
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Bunuel
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.


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­x+x+y=62 product 3 i.e. 186. So,  64+64+58=186 fits the answer
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First & second Container - D
Third container - A
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­Let x be the percentage of air mass in the first two containers, and y be the percentage in the third one.

It is given that -> (2x/100*100 + y/100*100)/300 = 62/100

Upon simplifying-> 2x + y = 186

x = 58, and y = 64 are consistent with this equation.
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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%.

Let h be humidity % in first and second container, and h' be humidity % in third container.
from given info, we find that 2h + h' = 62*3 = 186

and from the given values in options, if h = 64%, we find h' = 58%.


Hence, the humidity percentage for the First two containers is 64% and the humidity percentage for the Third container is 58%.
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The fact that humidity percentage is defined as the water vapor MASS contained with the total MASS of air inside a speicified volume of air is very important for this question. This means that the container shape or size is not important because we are dealing with the mass of water vapor against the mass of air. Different container sizes will not change the mass of air or water inside. Moreover, we also know that the three containers have the exact same MASS of compressed air: 100 tons. Furthermore, we also know that the humidity percentage of the first two containers are the same; and since we know the mass of compressed air for these two containers are equal, we can conclude that the mass of water vapor in the first two containers must also be the same. Therefore the only unknown is the mass of water vapor in the third container. What we do know, is that when we mix the contents of the three containers into a container with no air, meaning no air or water vapor, the humidity percentage of the combine air was found to be 62%. 

We know that the last container now has a total mass of air of 300 tons and a total humidity percentage of 62%. Since humidity percentage is the relationship between air and water vapor and we know the three containers had equal air masses, then we know that the water vapor mass must be directly proportionate to the humidity percentage and therefore 62% is the average of the three containers. Let x be the humidity percentage of the first container and also the second container since they are the same, and let y be the humidity percentage of the third container. We can express the average humidity percentage, 62%, as equal to (2x+y)/3. Lets simplify the equation and move the x and y to different sides. The equation can now be written as 2x=186-y. Now all you have to do is look through the table at the y values. Plug in a value for y from the table and then divide by 2 and see if that value is also in the table. When you plug in 58 from the table for y, you get 186-58 which is equal to 128. When you subtract 128 by 2 you get 64, which also happens to be in the table. Therefore the first two containers have a humidity percentage of 64 and the third container has a humidity percentage of 58. 
Bunuel
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.


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­
­
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Bunuel
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.


­
 


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­Let say for container 1 and 2 humidity x% and for container 3 humidity is y%. Now as all 3 are mixed into another container then resulting humidity is 62%.
It means, (100*x% + 100*x% + 100*y%) / 300 = 62%
         => 2x + y = 62

Now, from given options if we put y = 58 then we get x = 64.

Answers: container 1 and 2: 64%, container 3: 58%.
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Can be solved using ­Weighted average.

Two containers containing same humidity % = 2x
One container having different humidity % compared to first two = y

(2x + y)/3 = 62
2x + y = 186

Scanning the options 64% for two containers and 58% for third container fits.

So 64,58. 
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­x = humidity percentage in the first container
x = humidity percentage in the second container
y = humidity percentage in the third container

water vapor mass in the first container: x/100 * 100 = x tons
water vapor mass in the second container: x/100 * 100 = x tons
water vapor mass in the third container: y/100 * 100 = y tons

In the combined container we have 300 tons of compressed air and 2x+y of water vapor mass.

(2x+y)/300 * 100 = 62
2x+y= 186

y must be even because if it were odd, x wouldn't be an integer and we only have integers in the table for x.
if y=64 then x would be 61 that is not in the table.
if y=60 then x would be 63 that is not in the table.

if y=58 then x would be 64 that IS in the table.

IMO 64% for the 1st and 2nd container and 58% for the 3rd container.
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Bunuel
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.


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Assume 1st two container has humidity % - a
3rd container has - x

so, 

­(2a+x)/3=62
2a+x=186

a=93-(x/2)

we now know x will be even. try putting even values.

a=93-(58/2)

a=64

so, a is 64 and x is 58 (thats our answer)
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First two containers' humidity level: X

Third containers: Y

(2x+y)/3=62

(2*64+58)/3=62

for easier math calculate 2*(X-62)=62-Y­
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