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

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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
2
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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.
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
­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.­
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
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

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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
chetan2u wrote:
Kudos for correct solution

­I dont get the second question, would you please elaborate?
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
KarishmaB wrote:
chetan2u wrote:
­

­

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.

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?
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
mickkronik wrote:
­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
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
Gmatguy007 wrote:
KarishmaB wrote:
chetan2u wrote:
­

­

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.

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.
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
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
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
With this much confusion from Q2 and time pressure of DI, I'd just guess and move on.
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Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
­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
Re: The graph shows data for physicians who, in 2000 and 2008, were survey [#permalink]
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