Last visit was: 12 Oct 2024, 18:05 It is currently 12 Oct 2024, 18:05
Decision Tracker
My Rewards
Unanswered
Active Topics
New posts
New comers' posts
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

The graph shows data for physicians who, in 2000 and 2008, were survey

Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11498
Own Kudos [?]: 36752 [107]
Given Kudos: 333
Send PM
Reviews Badge
The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   07 Mar 2024, 07:38
7
Kudos
100
Bookmarks
Expert Reply
00:00

Difficulty:

95% (hard)

Question Stats:

41% (03:14) correct 59% (03:21) wrong based on 587 sessions
Hide Show DIFFICULTY AND TIMER STATISTICS
­

The graph shows data for physicians who, in 2000 and 2008, were surveyed about the average number of times per month they used each of 3 methods (telephone, email, and video conferencing) to communicate with their patients. The bars represent the percentage of the respondents who reported using the indicated method the indicated number of times per month.

Select from each drop-down menu the option that completes the statement so that it most accurately reflects the information provided.

It must be the case that the median number of times per month respondents reported having communicated with patients byin the 2008 survey was less than that in the 2000 survey.

It must be the case that the percentage of respondents who reported having communicated with patients by both was as great or greater in the 2008 survey as it was in the 2000 survey.

Show Spoiler
ID: 700210
Submit
Click Submit to add this question to your Error log
ShowHide Answer
Official Answer
Dropdown 1: telephone Dropdown 2: telephone and email
Most Helpful Reply
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 15361
Own Kudos [?]: 68724 [9]
Given Kudos: 444
Location: Pune, India
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   30 Mar 2024, 19:26
5
Kudos
4
Bookmarks
Expert Reply
chetan2u
­
­

The graph shows data for physicians who, in 2000 and 2008, were surveyed about the average number of times per month they used each of 3 methods (telephone, email, and video conferencing) to communicate with their patients. The bars represent the percentage of the respondents who reported using the indicated method the indicated number of times per month.


Select from each drop-down menu the option that completes the statement so that it most accurately reflects the information provided.


It must be the case that the median number of times per month respondents reported having communicated with patients by[Dropdown Placeholder #1]in the 2008 survey was less than that in the 2000 survey.

It must be the case that the percentage of respondents who reported having communicated with patients by both[Dropdown Placeholder #2] was as great or greater in the 2008 survey as it was in the 2000 survey.

 ­
The tricky part of this question is the data given to us. Once we understand it, the questions are fairly simple.

Say 10 physicians were surveyed.

2000 survey would look something like this

Question: How many times in a month did you communicate with patients using telephone?
Response of the 10 physicians: 0, 0, 1, 1, 2 , 4, 4, 5, 7, 9
20% people said 0, 20% said 1, 30% said more than 1 but less than or equal to 4 and 30% said more than 4.

Same was repeated for email, video and then again in 2008 too. 


It must be the case that the median number of times per month respondents reported having communicated with patients by _____ in the 2008 survey was less than that in the 2000 survey.

For which of telephone/email/video was the median for 2008 less than the median for 2000?
The median for telephone in 2000 is 2 or 3 or 4. Black region is where 50% is. 
The median for telephone in 2008 is 1. The 50% corresponds to the yellow region.

Hence the median is lower for telephone (ANSWER). For both email and video, median is the same in both years i.e. 0.


It must be the case that the percentage of respondents who reported having communicated with patients by both ______ was as great or greater in the 2008 survey as it was in the 2000 survey.

So we are looking for the overlap of two sets. We need this overlap to be higher or same in 2008 than in 2000. 
In 2008, 80% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 40% physicians communicated via email with their patients. (all regions except white)
Hence there must be an overlap of at least 20% here. 20% physicians must have communicated via both. Think Sets.

In 2000, 83% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 18% physicians communicated via email with their patients. (all regions except white)
Is it even possible that 20% communicated via both? No. At the most only 18% could have communicated via both. 

Hence answer is telephone and email. ANSWER

Maxsparrow

Was it hard to arrive at this? Not at all. Notice the graph. The "communication regions" (yellow, black and blue in which the physician communicated at least once with the patients) for email are much higher in 2008 and the overlap is necessitated with telephone only because its communication region is very high. So you can straight away jump to "telephone and email" and evaluate that to know if it is the answer. 

Check out the discussion on another graph here:
​​​​​​​https://youtu.be/ilMxPjHNeic


­
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11498
Own Kudos [?]: 36752 [6]
Given Kudos: 333
Send PM
Reviews Badge
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   07 Mar 2024, 07:39
4
Kudos
2
Bookmarks
Expert Reply
Solution­ for statement 2 added: 

The statement is
It must be the case that the percentage of respondents who reported having communicated with patients by both A and B was as great or greater in the 2008 survey as it was in the 2000 survey.

We can see that we know the % individually but not combined. So, we have to find a pair whose combined % in 2008 would always be greater than that in 2000. For this, you will have to compare worst case scenario of 2008 with best scenario of 2000.
Worst scenario: Minimum overlap between A and B
Best scenario: Maximum overlap between A and B­

Let us take telephone and email
­Best scenario: Maximum overlap between the two in 2000. So, take all connected through email were connected through telephone, that is complete overlap. Answer: entire 19% of email.
­Worst scenario: Minimum overlap between the two in 2008. tele - 80% and email - 40%. So, take all connected through email fill up % not connected through tele, so 100-80 or 20% is filled up. But remaining 40-20 or 20% have to overlap with tele. Answer: 20% of email.

Thus, worst in 2008 is better than best in 2000. Hence, 'Telephone and email' is the answer
­
General Discussion
User avatar
G
SKDEV
Joined: 25 Feb 2024
Status:a smooth sea never made a skilled sailor
Posts: 81
Own Kudos [?]: 53 [2]
Given Kudos: 120
Send PM
CAT Tests
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   Updated on: 14 Mar 2024, 01:06
2
Kudos
Could only get the first one : Telephone
Approach:­
Median would be the black shaded area (1<n<=4), so  
Quote:
It must be the case that the median number of times per month respondents reported having communicated with patients by___?____in the 2008 survey was less than that in the 2000 survey.
It is asking us to find for which method, the black shaded region reduced, which will be for Telephone (It remained same for Video Conferencing)

Originally posted by SKDEV on 09 Mar 2024, 05:10.
Last edited by SKDEV on 14 Mar 2024, 01:06, edited 2 times in total.
User avatar
B
mickkronik
Joined: 16 Apr 2023
Posts: 1
Own Kudos [?]: 5 [0]
Given Kudos: 20
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   10 Mar 2024, 14:39
­My solution: Telephone, Email and Video Conferencing

Telephone:
As you see in the pic I put the times per month for every communication type in an ascending order so the median will fall into the 50%. 

Email and Video Conferencing:
Adding the highlighted percentages for 2000 and 2008 for each communication type that n>=1 show that:

2000
E-mail: 10+5+5= 20
Video Conf: 5

2008
E-mail: 20+10+10= 40
Video Conf: 5+5=10

Hence, Email and Video Conferencing is the correct answer.­
Attachments

Screenshot_182.png
Screenshot_182.png [ 36.84 KiB | Viewed 8034 times ]

User avatar
B
manasp35
Joined: 18 Jan 2024
Posts: 31
Own Kudos [?]: 36 [4]
Given Kudos: 29
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   11 Mar 2024, 22:48
4
Kudos
 
Quote:
­
Drop down 2 solution : say X, Y are two methods
(X & Y ) = physicians communicated with both X and Y methods.

We need to find
(X & Y) in 2008 greater or equal to (X & Y) 2000 
­ 
imply 

min of ((X & Y) in 2008) greater or equal to max of ((X & Y) 2000) ?     ------ eq (1)

if X & Y is Email & video
min of ((X & Y) in 2008) = 0  (it's possible that no one who used email also used video)
max of ((X & Y) 2000) = 4% (it's possible that all the ones using video also used email)
doesn't satisfy eq (1) 

if X & Y is Telephone& video
min of ((X & Y) in 2008) = 0  (it's possible that no one who used telephone also used video)
max of ((X & Y) 2000) = 4% (it's possible that all the ones using video also used telephone)
doesn't satisfy eq (1) 

if X & Y is Email & Telephone
min of ((X & Y) in 2008) = 20%  (atleast 20% should use both)
max of ((X & Y) 2000) = 19% (it's possible that all the ones using email also used telephone)
does satisfy eq (1) 

Ans: Email & Telephone
 
User avatar
S
Maxsparrow
Joined: 19 Aug 2023
Posts: 68
Own Kudos [?]: 38 [0]
Given Kudos: 82
Location: Indonesia
GMAT Focus 1:
695 Q85 V83 DI85
GMAT 1: 640 Q49 V27
GMAT 2: 640 Q50 V26
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   17 Mar 2024, 07:28
chetan2u
Kudos for correct solution
­I dont get the second question, would you please elaborate?
User avatar
B
Gmatguy007
Joined: 07 Feb 2024
Posts: 83
Own Kudos [?]: 9 [0]
Given Kudos: 104
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   13 Apr 2024, 02:27
KarishmaB
chetan2u
­
­

The graph shows data for physicians who, in 2000 and 2008, were surveyed about the average number of times per month they used each of 3 methods (telephone, email, and video conferencing) to communicate with their patients. The bars represent the percentage of the respondents who reported using the indicated method the indicated number of times per month.


Select from each drop-down menu the option that completes the statement so that it most accurately reflects the information provided.


It must be the case that the median number of times per month respondents reported having communicated with patients by[Dropdown Placeholder #1]in the 2008 survey was less than that in the 2000 survey.

It must be the case that the percentage of respondents who reported having communicated with patients by both[Dropdown Placeholder #2] was as great or greater in the 2008 survey as it was in the 2000 survey.

 ­
The tricky part of this question is the data given to us. Once we understand it, the questions are fairly simple.

Say 10 physicians were surveyed.

2000 survey would look something like this

Question: How many times in a month did you communicate with patients using telephone?
Response of the 10 physicians: 0, 0, 1, 1, 2 , 4, 4, 5, 7, 9
20% people said 0, 20% said 1, 30% said more than 1 but less than or equal to 4 and 30% said more than 4.

Same was repeated for email, video and then again in 2008 too. 


It must be the case that the median number of times per month respondents reported having communicated with patients by _____ in the 2008 survey was less than that in the 2000 survey.

For which of telephone/email/video was the median for 2008 less than the median for 2000?
The median for telephone in 2000 is 2 or 3 or 4. Black region is where 50% is. 
The median for telephone in 2008 is 1. The 50% corresponds to the yellow region.

Hence the median is lower for telephone (ANSWER). For both email and video, median is the same in both years i.e. 0.


It must be the case that the percentage of respondents who reported having communicated with patients by both ______ was as great or greater in the 2008 survey as it was in the 2000 survey.

So we are looking for the overlap of two sets. We need this overlap to be higher or same in 2008 than in 2000. 
In 2008, 80% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 40% physicians communicated via email with their patients. (all regions except white)
Hence there must be an overlap of at least 20% here. 20% physicians must have communicated via both. Think Sets.

In 2000, 83% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 18% physicians communicated via email with their patients. (all regions except white)
Is it even possible that 20% communicated via both? No. At the most only 18% could have communicated via both. 

Hence answer is telephone and email. ANSWER

Maxsparrow

Was it hard to arrive at this? Not at all. Notice the graph. The "communication regions" (yellow, black and blue in which the physician communicated at least once with the patients) for email are much higher in 2008 and the overlap is necessitated with telephone only because its communication region is very high. So you can straight away jump to "telephone and email" and evaluate that to know if it is the answer. 

Check out the discussion on another graph here:
​​​​​​​https://youtu.be/ilMxPjHNeic


­
­KarishmaB firstly many kudos for your solution and the way you broke down the question !! In the second question, I supposed that the overlap was the extra part beyond 100% but in the email it seems that I misunderstood it since in telephone is 80% + 40% - 100% = 20% but in email is 83% + 18% - 100% = 1%. I understood the rest of the solution, I just got confused in this part.

Could you help me?
User avatar
B
cookiemonsterpsy
Joined: 29 Oct 2023
Posts: 18
Own Kudos [?]: 4 [0]
Given Kudos: 31
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   13 Apr 2024, 19:37
mickkronik
­My solution: Telephone, Email and Video Conferencing

Telephone:
As you see in the pic I put the times per month for every communication type in an ascending order so the median will fall into the 50%. 

Email and Video Conferencing:
Adding the highlighted percentages for 2000 and 2008 for each communication type that n>=1 show that:

2000
E-mail: 10+5+5= 20
Video Conf: 5

2008
E-mail: 20+10+10= 40
Video Conf: 5+5=10

Hence, Email and Video Conferencing is the correct answer.­
­this was my intitial answer too :S 2nd qn is confusing
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 15361
Own Kudos [?]: 68724 [0]
Given Kudos: 444
Location: Pune, India
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   14 Apr 2024, 22:42
Expert Reply
Gmatguy007
KarishmaB
chetan2u
­
­

The graph shows data for physicians who, in 2000 and 2008, were surveyed about the average number of times per month they used each of 3 methods (telephone, email, and video conferencing) to communicate with their patients. The bars represent the percentage of the respondents who reported using the indicated method the indicated number of times per month.


Select from each drop-down menu the option that completes the statement so that it most accurately reflects the information provided.


It must be the case that the median number of times per month respondents reported having communicated with patients by[Dropdown Placeholder #1]in the 2008 survey was less than that in the 2000 survey.

It must be the case that the percentage of respondents who reported having communicated with patients by both[Dropdown Placeholder #2] was as great or greater in the 2008 survey as it was in the 2000 survey.

 ­
The tricky part of this question is the data given to us. Once we understand it, the questions are fairly simple.

Say 10 physicians were surveyed.

2000 survey would look something like this

Question: How many times in a month did you communicate with patients using telephone?
Response of the 10 physicians: 0, 0, 1, 1, 2 , 4, 4, 5, 7, 9
20% people said 0, 20% said 1, 30% said more than 1 but less than or equal to 4 and 30% said more than 4.

Same was repeated for email, video and then again in 2008 too. 


It must be the case that the median number of times per month respondents reported having communicated with patients by _____ in the 2008 survey was less than that in the 2000 survey.

For which of telephone/email/video was the median for 2008 less than the median for 2000?
The median for telephone in 2000 is 2 or 3 or 4. Black region is where 50% is. 
The median for telephone in 2008 is 1. The 50% corresponds to the yellow region.

Hence the median is lower for telephone (ANSWER). For both email and video, median is the same in both years i.e. 0.


It must be the case that the percentage of respondents who reported having communicated with patients by both ______ was as great or greater in the 2008 survey as it was in the 2000 survey.

So we are looking for the overlap of two sets. We need this overlap to be higher or same in 2008 than in 2000. 
In 2008, 80% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 40% physicians communicated via email with their patients. (all regions except white)
Hence there must be an overlap of at least 20% here. 20% physicians must have communicated via both. Think Sets.

In 2000, 83% physicians communicated via telephone with their patients. (all regions except white)
In 2008, 18% physicians communicated via email with their patients. (all regions except white)
Is it even possible that 20% communicated via both? No. At the most only 18% could have communicated via both. 

Hence answer is telephone and email. ANSWER

Maxsparrow

Was it hard to arrive at this? Not at all. Notice the graph. The "communication regions" (yellow, black and blue in which the physician communicated at least once with the patients) for email are much higher in 2008 and the overlap is necessitated with telephone only because its communication region is very high. So you can straight away jump to "telephone and email" and evaluate that to know if it is the answer. 

Check out the discussion on another graph here:
https://youtu.be/ilMxPjHNeic


­
­KarishmaB firstly many kudos for your solution and the way you broke down the question !! In the second question, I supposed that the overlap was the extra part beyond 100% but in the email it seems that I misunderstood it since in telephone is 80% + 40% - 100% = 20% but in email is 83% + 18% - 100% = 1%. I understood the rest of the solution, I just got confused in this part.

Could you help me?
­
You haven't misunderstood anything. 1% is the minimum overlap in 2000 and 18% is the maximum possible overlap (notice the highlighted part)
In 2008, minimum possible overlap is 20% and hence the overlap of 2008 will always be higher. 
User avatar
B
Rytthhm
Joined: 21 Mar 2021
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 92
Send PM
CAT Tests
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   22 Apr 2024, 11:37
1.

Telephone -
2000
25% - 4+ times, 35% - 1 to 4 times; 20% - once

2008
20% - 4+ times; 20% - 1 to 4 times, 40% - once

2. understand the overlaps
2000 - email & telephone [maximise overlap] - say 18%

2008 - email & telephone [minimise overlap] - say 80% + 40% = 102% not possible; 20% overlap has to be there
User avatar
B
Ratikainen
Joined: 25 Apr 2024
Posts: 16
Own Kudos [?]: 5 [0]
Given Kudos: 23
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   12 Jun 2024, 05:54
With this much confusion from Q2 and time pressure of DI, I'd just guess and move on.
User avatar
G
Gemmie
Joined: 19 Dec 2021
Posts: 576
Own Kudos [?]: 203 [0]
Given Kudos: 75
Location: Viet Nam
GMAT Focus 1:
695 Q87 V84 DI83
GPA: 3.55
Send PM
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink] The graph shows data for physicians who, in 2000 and 2008, were survey   06 Aug 2024, 07:12
­It must be the case that the median number of times per month respondents reported having communicated with patients by [telephone] in the 2008 survey was less than that in the 2000 survey.

Median is the part of the bar passing the 50% tick mark

Tel:
+) 2008: 1 < median <= 4
+) 2000: 0 < median <= 1

Median for both email and video conferencing in both years is 0


It must be the case that the percentage of respondents who reported having communicated with patients by both [telephone and email] was as great or greater in the 2008 survey as it was in the 2000 survey.

Min combined percentage (of 2 methods) in 2008 must be >= Max combined percentage in 2000

Min: when the overlapping of those who reported having communicated is smallest
Max: when the overlapping of those who reported having communicated is largest

Min combined percentage (of 2 of the methods) in 2008:
+) Tel + Email: 10%
+) Email + Vid Conf: 0%
+) Tel + Vid Conf: 0%

=> The best chance is Tel + Email­
GMAT Club Bot
gmatclubot
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
06 Aug 2024, 07:12
Moderators:
Bunuel
Math Expert
96080 posts
chetan2u
RC & DI Moderator
11498 posts
Bismuth83
88 posts