Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Dropdown 1: 3.6
Dropdown 2: greater than
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
55% (03:39) correct
45%
(03:37)
wrong
based on 1297
sessions
History
Date
Time
Result
Not Attempted Yet
The graph shows the maximum absolute humidity—the maximum amount of water vapor that atmospheric air at sea level can hold—in grams of water vapor per cubic meter (g/m^3), for integer-valued temperatures on the Celsius scale from 1°C through 41°C. When the air contains this maximum amount of water vapor, the air is said to be at its saturation point. In addition, at sea level and at a temperature of T °C, the density of air containing no water vapor is \(\frac{10^5}{287(T + 273)}\) \(kg/m^3 \)where 1 kg = 1 kilogram = 1,000 g.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
At 37°C, and at sea level, 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.
The average rate of change of maximum absolute humidity from 31°C to 41°C is the average rate of change of maximum absolute humidity from 11°C to 21°C.
ID: 700158
Graph.png [ 198.78 KiB | Viewed 14881 times ]
Attachment:
Graph.png [ 198.78 KiB | Viewed 14881 times ]
ShowHide Answer
Official Answer
Dropdown 1: 3.6
Dropdown 2: greater than
Trying to understand why the right answer is right?
GMAT Club Tests explanations walk through concepts, shortcuts, and common traps on many questions.
See how →
25 Oct 2024, 21:06
Kudos
Bookmarks
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.
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.
18 Jun 2024, 07:17
Kudos
Bookmarks
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]
\(4 * \frac{9}{10}\)
\(3.6\)%
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]
\(4 * \frac{9}{10}\)
\(3.6\)%
