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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2206:30 AM PST
-08:30 AM PST
Let’s dive deep into advanced CR to ace GMAT Focus! Join this webinar to unlock the secrets to conquering Boldface and Paradox questions with expert insights and strategies. Elevate your skills and boost your GMAT Verbal Score now!
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
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.
12 Nov 2024, 03:43
Kudos
Bookmarks
Maximum range: 30.5
Minimum range: 21.5
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:
34% (02:57) correct
66%
(03:11)
wrong
based on 38
sessions
History
Date
Time
Result
Not Attempted Yet
In a class of 20 students, the average weight of the students is 40 kg and the range is 10. However, the average weight increases to 41 when a new student joins the class in the new session.
Based on the above information, select for Maximum range the maximum possible range of the weight of the new class and for Minimum range the minimum possible range of the weight of the new class.
Based on the above information, select for Maximum range the maximum possible range of the weight of the new class and for Minimum range the minimum possible range of the weight of the new class.
| Maximum range | Minimum range | |
| 10.5 | ||
| 21 | ||
| 21.5 | ||
| 30 | ||
| 30.5 | ||
| 31 |
ShowHide Answer
Official Answer
Maximum range: 30.5
Minimum range: 21.5
12 Nov 2024, 03:43
Kudos
Bookmarks
Official Solution:
• The average weight of 20 students being 40 kg implies their total weight is 20 * 40 = 800 kg.
• After adding a new student, the average weight increasing to 41 kg implies the total weight became 21 * 41 = 861 kg.
• Thus, the weight of the new student is 861 - 800 = 61 kg.
We also know that the range of weights for the 20 students was 10 kg.
To maximize the new range, we need to minimize the weight of the lightest student in the class. This can be achieved by assuming one student weighs \(x\) kg, and the remaining 19 students each weigh \(x + 10\) kg. We get \(x + 19(x + 10) = 800\), which gives \(x = 30.5\). Therefore, the maximum possible range is 61 - 30.5 = 30.5.
To minimize the new range, we need to maximize the weight of the lightest student in the class. This can be achieved by assuming 19 students each weigh \(x\) kg, and the remaining student weighs \(x + 10\) kg. We get \(19x + (x + 10) = 800\), which gives to \(x = 39.5\). Therefore, the minimum possible range is 61 - 39.5 = 21.5.
Correct answer:
Maximum range "30.5"
Minimum range "21.5"
Bunuel
• The average weight of 20 students being 40 kg implies their total weight is 20 * 40 = 800 kg.
• After adding a new student, the average weight increasing to 41 kg implies the total weight became 21 * 41 = 861 kg.
• Thus, the weight of the new student is 861 - 800 = 61 kg.
We also know that the range of weights for the 20 students was 10 kg.
To maximize the new range, we need to minimize the weight of the lightest student in the class. This can be achieved by assuming one student weighs \(x\) kg, and the remaining 19 students each weigh \(x + 10\) kg. We get \(x + 19(x + 10) = 800\), which gives \(x = 30.5\). Therefore, the maximum possible range is 61 - 30.5 = 30.5.
To minimize the new range, we need to maximize the weight of the lightest student in the class. This can be achieved by assuming 19 students each weigh \(x\) kg, and the remaining student weighs \(x + 10\) kg. We get \(19x + (x + 10) = 800\), which gives to \(x = 39.5\). Therefore, the minimum possible range is 61 - 39.5 = 21.5.
Correct answer:
Maximum range "30.5"
Minimum range "21.5"
26 Nov 2024, 22:28
Kudos
Bookmarks
TetyanaRokobit
This question uses the concept of maximizing and minimizing values in case of a given mean. You can check out it from my content over this weekend (details in my signature).
20 students with an average of 40 kg and range of 10 could be placed such that 1 person is at 30.5 kg and 19 are at 40.5 kg. Minimum value of C1.
They could also be placed such that 19 people are at a weight of 39.5 and 1 person is at 49.5 kg. Maximum value of C1.
Since the new person raises the average of 21 people's (including himself) mean by 1, it means he had 21 more than the mean i.e. his weight was 61 kg.
This is discussed here: https://youtu.be/W-qhIZ29UIs
Hence
Maximum range of new set = 61 - 30.5 = 30.5 kg
minimum range of new set = 61 - 39.5 = 21.5 kg
Select 30.5 and 21.5
