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GMAT Club Olympics 2024 (Day 1): Humidity percentage is the water

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08 Jul 2024, 10:11

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200 tons with x% and 100 tons with y%
using allegation-
(x - 62)*2 = 62 - y
2x + y = 186
2*64 + 58 = 186

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08 Jul 2024, 10:11

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First two containers -59% Third container-64%

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Solution: When we add all three containers with 100 tons of compressed air each into one new container, the new container has 300 tons of compressed air. This implies that the water vapor mass is $ 62\% \times 300 = 186$ tons. This means that the sum of vapor mass in each of the three containers must add up to 186 tons.

Then, we continue by multiply each of the entry in the third column with 2, and then subtract that number from 186 to see if the result matches any of the five entries in the third column. By trial and error, we arrive at the answer that the first container and the second container each has 64% of humidity percentage while the last one has 58% of humidity percentage.

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08 Jul 2024, 10:16

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The first two containers have the same humidity percentage. Let = x
The third container has = y
(2x + y ) / 300 * 100 = 62
x= (186 - y ) / 2

If y = 58 , x = 64 , only these two options fits in amongst the all the given options.

Hence , First two containers = 64 and third container = 58 Ans.

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08 Jul 2024, 10:18

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To solve this problem, let's denote the humidity percentage of the first two containers as H1 and
the humidity percentage of the third container as H2.
We know the following:

Each of the first two containers has 100 tons of compressed air with the same humidity percentage.
The third container also has 100 tons of compressed air.
When the air from all three containers is mixed, the overall humidity percentage is 62%.
Given:
H is the humidity percentage of the first two containers.
X is the humidity percentage of the third container.
The combined humidity percentage is 62%.
The total mass of air is 100+100+100 = 300 tons.
The equation for the combined humidity percentage is: [(2*100)*H + 100*X]/300 = 62

Simplify the equation:
2*H + X = 186

We need to find pairs (H , X) from the provided values that satisfy this equation.

Let's test each pair from the given options:
By Taking ( H=64 & X=58 ) the eq. 2H + x = 186 gets satisfied.

Hence, the consistent pair of humidity percentages is:
First two containers: 64%
Third container: 58%

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08 Jul 2024, 10:24

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Let the first two containers have x% humidity each and y% humidity in the third.

Since we are mixing all the three air containers completely and all have same capacity of 100 ton, the resulting humidity % would be the arithmetic mean of the % in individual containers.

Therefore, 2x + y = 62*3 = 186. From here on, it's hit and trial with the given options.

For x= 58%,y= 70% but not present in any options.
For x= 59%, y= 68% not present in options
For x=60%, y= 66% not present
For x= 64% y= 58% which is present. Therefore, our answers respectively.

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08 Jul 2024, 10:30

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First container Second container third container New container that is a mixture of first, second and third containers

Weight of air 100 tons 100 tons 100 tons 300 tons
Humdity percentage x% x%. y% 62%

Then the equation to set up would be

$\frac{x}{100}* 100 + \frac{x}{100}* 100 + \frac{y}{100}* 100 = \frac{62}{100}* 300$

$2x + y = 186$

Substitute x value as 58, then you get y as 70
Substitute x value as 59, then you get y as 68
Substitute x value as 60, then you get y as 66
Substitute x value as 64, then you get y as 58 (this is available, so mark these values)

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08 Jul 2024, 10:35

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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 % from 2 containers + y% from third container/300 = 62/100
2x + y = 186
Subtituting X= 64, we get y =58

Answer : D and A

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08 Jul 2024, 10:38

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First two containers E 65%
Third container B 59%

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08 Jul 2024, 10:38

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3 containers of same volume are mixes to from content of fourth container.

Hence a%+a%+b%/3=62%

2a+b=186

Now we can do trail and error with options.

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08 Jul 2024, 10:42

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If we consider that each container contains 100 tons of compressed air and x is the part of water vapor mass contained within the total mass of air inside container 1 and 2 and y is the part of water vapor mass contained within the total mass of air inside container 3.
then Humidity percentage in containers 1 and 2 is equal to x/100 and Humidity percentage in containers 3 is equal to y/100.

If the air from all three containers was mixed into a container that had practically no air before the experiment then :
the part of water vapor mass contained within the total mass of air inside the mixed container is (x+x+y)/(100+100+100) = (2x+y)/300 (*)
If we use directly the values from the table in the formula (*) :

the only way to fave a humidity percentage of 62% from a total of 300 is 186. (0.62=186/300)
From the values proposed, the only combination is : 64, 64, 58
So the first two containers are at : 64% and the third container is at 58%

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08 Jul 2024, 10:53

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Using weighted avg:

(2H1+H2)/(2+1)=62
=> 2H1+H2=183

Putting H1=58, 59,60,64,65
We get H2=70,68,66,58,56

We have (H1,H2)=(64,58) as jointly consistent

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08 Jul 2024, 10:56

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Humidity percentage of combined air is 62%. (i.e., 186 of the 300 ton).
Now, we need to find the humidity percentage of first two containers (which have equal H% and the third one container.

Let's assume H% of first two containers as a, of 3rd as b. Therefore, 2a+b=186.
Using the options, if we use 60 as a, b comes out as 66 (which is not in options.)
Trying 64 as a, we get b as 58.

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Bunuel

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This can be solved using allegation method, let X be the humidity of the first two container, Y be the humidity of third container

X Y
\ /
\ /
62
/ \
/ \
2 1

X-62/62-Y = 1/2 or 62-X/Y-62 =1/2
Looking at the values 58 and 64 will satisfy the above eq

64- 62 / 62- 58 = 1/2

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08 Jul 2024, 11:24

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Humidity% = $\frac{(Water Vapour Mass)}{(Total Mass of Air)}*100$ inside a specified volume

Let the three containers be - A, B and C
Mass of air in each container = 100 Tons

Humidity% of A = Humidity% of B
Water Vapour Mass of A = Water Vapour Mass of B, since mass of air same in both A and B

Let Water Vapour Mass of A = Water Vapour Mass of B = $x$
Water Vapour Mass of C = $y$

Once the air of three containers are mixed:
Humidity% = $\frac{((Water Vapour Mass of A) + (Water Vapour Mass of B) + (Water Vapour Mass of C))}{300}*100$

Humidity% = $\frac{(x+x+y)}{300}*100$
$\frac{(2x+y)}{3} = 62$
$2x+y = 186$

The equation is satisfied when $x = 64%$ and $y = 58%$

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08 Jul 2024, 11:41

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Humidity is a function of water vapour in the air from studies.

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08 Jul 2024, 12:01

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Bunuel

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let Humdity percentage (HP) of two same container may be x and other container may be y

x% of 100 + x% of 100 + y%100 = 62% *300

2x+ y= 186

2*64+58 = 186.

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08 Jul 2024, 12:27

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Given, the combined air was 62% humid.

Combined air quantity: 100+100+100 = 300 tonnes.

Amount of water vapour in the combined air = 62%of300 = 186 tonnes.

So, if container A and B had x tonnes each and C had y tonnes, 2x+y=186.

Now, instead of putting all values everywhere, we can see that y=186-2x, which is always even.

So we can put y=58,60 and 64 and get values for x. Upon doing so, we get x=64 for y=58. Thus we found the solution.

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08 Jul 2024, 12:56

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The final mix will have 300 tonnes of Air and hence with 62% humidity, total amount of water vapour mass will be 300*62% = 186.

We need to find values x (humidity of first 2) and y (humidity of 3rd) such that 2x + y = 186
Looking at the answer choices, correct asnwers are 2* 64 + 58 = 186; menaing humidity of first two conatainers as 64 and third one as 58 (all are in %, removed it for the sake of easy typing :p)

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08 Jul 2024, 13:00

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Bunuel

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Using the weighted average formula for humidity percentages, where the masses are proportional to their contributions:

200x+100y/300 =62

Here, x and y represent the humidity percentages of the first two and third containers, respectively. Solving this equation:
200x+100y=62×300
200x+100y=18600

Testing the options:
If x = 60%
200×60+100y=18600
12000+100y=18600
y=66%