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805+ Level|   Graphs|   Math Related|            
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ruis
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Interesting question it took 5mins to solve this question, I got it right once I understood the first part. Some calculations using the calculator.....the second part is easy if you take the slope and the compare the angle of the two lines which will indicate that rate of change from 31-41 degrees celsius is greater.

It took me longer than average to solve this problem as I felt a little overwhelmed with the amount of information that was thrown at me.
­Hi,
Can you elaborate step by step how you went about this question?

MartyMurray GMATCoachBen KarishmaB chetan2u can you please help out with a structured approach to solving this qs?
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Took around 5 min to solve the question. the second part was easier since i just compared the slopes but the first part took some time and i had to use the calculator to solve it. Is there any other way (other than directly calculation the percentage change) to solve it?
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Can someone explain the first part of the question
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hrishi84 - The first part of the question stem is tricky, but if you read it carefully, you can figure it out.

I'll try to break it down:
a. At 37C, "the weight of water vapor in 1m^3 of air at its saturation point"- The term "saturation point" can also be interpreted as Maximum Absolute Humidity. Hence, we need to locate the "Absolute Humidity" at 37C on the graph, which equates to 41 g/m^3

b. At 37C, "the weight of 1m^3 of air containing no water vapor"- This is where you need to apply the formula given in the description part of the question stem. Apply the Formula for T=37, you will get 10^5/(287)(37+273)= 10^5/287*310 kg/m^3

To obtain the desired %, convert kg to g or g to kg as the units differ for both values.

41*100/(10^8/287*310)= 3.64%­
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­1.)

At 37°C the weight of water vapor in \(1m^3\) is around \(40\)g of absolute humidity.

At 37°C the weight of air containing no water vapor is \(\frac{10^5}{287(37+273)}\) kg.

As the weight of water vapor is in grams and the weight of air containing no water vapor is in kg, one needs to alter one so that they are in the same unit of measure. \(40\)g is equal to \(\frac{40}{1000}\)kg

To find the what percent the weight of water vapor in 1\(m^3\) of air at its saturation point is of the weight of 1\(m^3\) of air containing no water vapor one will go:

\(\frac{\frac{40}{1000}}{\frac{10^5}{287(37+273)}}*100\)

\(\frac{40}{1000}*\frac{287(310)}{10^5}*100\) [Let 287*310 ≈ 90 000]

\(4 * \frac{9}{10}\)

\(3.6\)%





 ­
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What's the significance of the term '1m3' in the question? That confused me a bit
Nidzo
­1.)

At 37°C the weight of water vapor in \(1m^3\) is around \(40\)g of absolute humidity.

At 37°C the weight of air containing no water vapor is \(\frac{10^5}{287(37+273)}\) kg.

As the weight of water vapor is in grams and the weight of air containing no water vapor is in kg, one needs to alter one so that they are in the same unit of measure. \(40\)g is equal to \(\frac{40}{1000}\)kg

To find the what percent the weight of water vapor in 1\(m^3\) of air at its saturation point is of the weight of 1\(m^3\) of air containing no water vapor one will go:

\(\frac{\frac{40}{1000}}{\frac{10^5}{287(37+273)}}*100\)

\(\frac{40}{1000}*\frac{287(310)}{10^5}*100\) [Let 287*310 ≈ 90 000]

\(4 * \frac{9}{10}\)

\(3.6\)%





­
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