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.
Jul 0911:00 AM PDT
-12:30 PM PDT
Welcome to the M7 MBA Spotlight Series, featuring MIT Sloan School of Management! This video is part of GMAT Club’s exclusive MBA Spotlight event, where top business schools meet prospective applicants like you.- Jun 30
08:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates! - Jul 10
11:00 AM PDT
-11:00 AM PDT
In our conversation, Kishore talks about how strategically he used data analytics to identify high impact areas to focus on as well as his early morning study routine (3 AM prep!) that optimized his GMAT preparation. 
Jul 1211:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Updated on: 15 Oct 2024, 09:24
Originally posted by Sajjad1994 on 15 Mar 2020, 07:27.
Last edited by Bismuth83 on 15 Oct 2024, 09:24, edited 5 times in total.
Last edited by Bismuth83 on 15 Oct 2024, 09:24, edited 5 times in total.
Moved to new forum
Kudos
Bookmarks
In calculating the final score for the contest, the weight loss for a fitness trainer's third client (Client 3) had equal weighting as the weight loss on Client 2.: Yes
The median final score for all fitness trainers was 27.40: Yes
In the data set for "Client 1" clients who worked with a fitness trainer having four years of experience, the range was 8.: No
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:
36% (03:10) correct
64%
(03:10)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
TA.jpg [ 56.11 KiB | Viewed 4509 times ]
(Sort ↕ the table by clicking on the headers)
| Years of Experience | Fitness Trainer | Client 1 (100- 125kg) | Client 2 (125- 150kg) | Client 3 (150- 250kg) | Final Score |
|---|---|---|---|---|---|
| 1 | Susan | 10 | 20 | 35 | 20.5 |
| 3 | Megan | 14 | 22 | 45 | 25.7 |
| 4 | Tom | 25 | 35 | 70 | 41.5 |
| 4 | Brad | 20 | 25 | 66 | 35.3 |
| 2 | Peter | 22 | 28 | 49 | 31.9 |
| 1 | Melissa | 25 | 33 | 62 | 38.5 |
| 3 | Nick | 16 | 18 | 52 | 27.4 |
| 2 | Russel | 14 | 15 | 39 | 21.8 |
| 2 | Patty | 8 | 12 | 22 | 13.4 |
For each of the following statements, select Yes if the statement is true based on the information provided; otherwise select No.
| Yes | No | |
| In calculating the final score for the contest, the weight loss for a fitness trainer's third client (Client 3) had equal weighting as the weight loss on Client 2. | ||
| The median final score for all fitness trainers was 27.40 | ||
| In the data set for "Client 1" clients who worked with a fitness trainer having four years of experience, the range was 8. |
ShowHide Answer
Official Answer
In calculating the final score for the contest, the weight loss for a fitness trainer's third client (Client 3) had equal weighting as the weight loss on Client 2.: Yes
The median final score for all fitness trainers was 27.40: Yes
In the data set for "Client 1" clients who worked with a fitness trainer having four years of experience, the range was 8.: No
17 Mar 2020, 06:58
Kudos
Bookmarks
a)Let the weights for Client1 -x,Client 2-y,Client3-z,
Now per the option it is asking whether y=z
So let's take y=z and then recheck it with the data in the table
So let's form two equations from the table
25X+105Y=41.5-Tom
X= x/(x+2y),Y=y/(x+2y)
25X+95Y= 38.5-Melissa
So solving for X&Y, so we get Y=0.3, so it's 30%,X=0.4-> 40%
So now let's check these value for let's suppose Megan so = 14*.4 + 67*
.3= 25.7, so our calculation is right. And hence answer is Yes
b)Yes the median is 27.40(since it's the 5th value among the 9 values)
c)No,the range comes around 5(25-20)
Posted from my mobile device
Now per the option it is asking whether y=z
So let's take y=z and then recheck it with the data in the table
So let's form two equations from the table
25X+105Y=41.5-Tom
X= x/(x+2y),Y=y/(x+2y)
25X+95Y= 38.5-Melissa
So solving for X&Y, so we get Y=0.3, so it's 30%,X=0.4-> 40%
So now let's check these value for let's suppose Megan so = 14*.4 + 67*
.3= 25.7, so our calculation is right. And hence answer is Yes
b)Yes the median is 27.40(since it's the 5th value among the 9 values)
c)No,the range comes around 5(25-20)
Posted from my mobile device
General Discussion
03 Mar 2025, 10:15
Kudos
Bookmarks
what is this formula X= x/(x+2y),Y=y/(x+2y)?
And Why the mean is varying all over the table? (It's not following the formula: \(\frac{(x+y+z)}{3}\) for some reason)
And Why the mean is varying all over the table? (It's not following the formula: \(\frac{(x+y+z)}{3}\) for some reason)
Apt0810
a)Let the weights for Client1 -x,Client 2-y,Client3-z,
Now per the option it is asking whether y=z
So let's take y=z and then recheck it with the data in the table
So let's form two equations from the table
25X+105Y=41.5-Tom
X= x/(x+2y),Y=y/(x+2y)
25X+95Y= 38.5-Melissa
So solving for X&Y, so we get Y=0.3, so it's 30%,X=0.4-> 40%
So now let's check these value for let's suppose Megan so = 14*.4 + 67*
.3= 25.7, so our calculation is right. And hence answer is Yes
b)Yes the median is 27.40(since it's the 5th value among the 9 values)
c)No,the range comes around 5(25-20)
Posted from my mobile device
Now per the option it is asking whether y=z
So let's take y=z and then recheck it with the data in the table
So let's form two equations from the table
25X+105Y=41.5-Tom
X= x/(x+2y),Y=y/(x+2y)
25X+95Y= 38.5-Melissa
So solving for X&Y, so we get Y=0.3, so it's 30%,X=0.4-> 40%
So now let's check these value for let's suppose Megan so = 14*.4 + 67*
.3= 25.7, so our calculation is right. And hence answer is Yes
b)Yes the median is 27.40(since it's the 5th value among the 9 values)
c)No,the range comes around 5(25-20)
Posted from my mobile device
