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805+ Level|   Graphs|   Math Related|            
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bb
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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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the average rate of change is not matching in calculation. 31-41 = 50-30/30 = 2/3 = 0.666 and 11-21 is 16-9/9 = 0.77 what am i missing out here?
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your equation doesn't make any sense. Just looking at the graph should be enough to solve it but if you wanted to calculate the exact difference for some reason you would just take the difference of the weight between the given temperature intervals and see which one's greater.
shisingh
the average rate of change is not matching in calculation. 31-41 = 50-30/30 = 2/3 = 0.666 and 11-21 is 16-9/9 = 0.77 what am i missing out here?
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Could u make a video on this please?! @E-Gmat
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Part 2 can be solved qualitatively
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RenB

­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?

Here is the solution to this problem at 00:25:50: https://www.youtube.com/watch?v=Y-eGp95nuog
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I did the same mistake, key thing to note here is when the questions says "Maximum Absolute Humidity", the absolute humidity refers to the vaccuum formula, in which case we can omit the calculations and take a quantitative call based on the values given (9-->16 & 30-->50)
shisingh
the average rate of change is not matching in calculation. 31-41 = 50-30/30 = 2/3 = 0.666 and 11-21 is 16-9/9 = 0.77 what am i missing out here?
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bb
The units don’t play much of a role in this question. The units are the same and there’s no conversion needed, you only have percentage conversion.

I solved this on the back of the napkin without a calculator or the napkin. Of course I was casually looking on my phone without timing myself for the time pressure of the question but it is possible to solve it without the calculator. You should make a challenge for yourself to figure out how to solve every DI question without a calculator. I don’t mean to force yourself to do calculations in your head but figure out if there is a shortcut that would allow you to solve the question without doing calculations.

1. If you look at this scary equation you will notice that at the top you have 100,000 in the numerator and denominator is basically going to be 300×300. I don’t think you need to have a more precise calculation there.

This means the weight of air going to be 1.1 kg approx. Since it’s over 1 , and water vapor Weighs 40 g, it’s not going to be 4% it’s going to be approx 10% less than 4% but not quite as little as 3.2% which is 20% less.

I hesitated a little because it was slightly over 40 g but 40 to 41 It’s only a 2 1/2% change whereas from one to 1.1 is a 10% change. So the final answer would likely be slightly higher than 3.6. Maybe it’s 3.7 but still less than 4% and closer to 3.6%

The second question is fairly easy because you just look at the chart and you see the rate of change and it’s clearly greater because it’s exponential or at least it looks to me like that and that was the correct answer 😇

The key to this question was a good understanding between the numerator and denominator so you understand what happens when you divide a number by 1.1 or 0.9. Key for mental math.

Hi bb,

Great approach - thanks for sharing... can you please explain the yellow highlighted part ? What logic did you use ?
Couldn't understand it clearly.

Thank you! :)
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Hi. the yellow part comes from the graph/chart and the knowledge of denominator/numerator. Look at the chart at 37 degrees and it is 40 grams. Since we have to divide it by 1.1, the answer is 3.6.

That's all you need to solve the first drop down - one line. Do not solve this question in this way: https://gmatclub.com/forum/the-graph-sh ... l#p3411248 - this is like eating soup with chopsticks.

The second drop down is just looking at the slopes and that can be answered in 10 seconds. The slope is increasing, so it is greater but you need to understand the concept of a slope.


rushimehta


Hi bb,

Great approach - thanks for sharing... can you please explain the yellow highlighted part ? What logic did you use ?
Couldn't understand it clearly.

Thank you! :)
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