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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
14 Nov 2012, 17:09
7
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
31% (02:55) correct
69% (01:43) wrong based on 200 sessions
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
So I actually understand the solution provided but it seemed a little un-intuitive, was hoping someone here could discover some sort of new insight about this problem. Basically, the solution suggests plugging but I wonder if there is something nicer...
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
14 Nov 2012, 22:40
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
Answer should be C
Case I e-c = 4 e+c =34 median is 15 = M e= 19, c =15 taking max values for a, b, c, d, e A = 15+15+15+19+19/5 > M Taking min values, example A = (-10)+(-10)+15+15+19/5 < M Case I = not sufficient
Case II
c=a+10 e=c+4 => e=a+14 considering a = -10 c = 0 e = 14
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
14 Nov 2012, 23:12
5
This post received KUDOS
anon1 wrote:
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
So I actually understand the solution provided but it seemed a little un-intuitive, was hoping someone here could discover some sort of new insight about this problem. Basically, the solution suggests plugging but I wonder if there is something nicer...
Problems with plugging-in are finding right numbers to plug and not knowing where to stop. A more methodogical algebric approach is sure shot way to solve DS. (but sometimes time consuming). Need to find the right balance between two.
Question says: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4
We cant assume that a,b,c,d and e are integers or positive etc. e could be 23.4 and c could be 19.4 or a, b or c could be anything negative or in decimals. Plugging in right numbers would be very hard and very time consuming. Infact- the solution given above by suryanshg 'assumes' numbers are integers.. and is therefore incorrect. -------------
Lets take a look at the problem. given is e-c =4 and a≤b≤c≤d≤e Clearly c is the median. problem is finding out avergage, A = (sum/5)
statement 1: \(e+c=34\) we can combine this with \(e-c=4\) to find out e=19, and c=15, but nothing else. Not sufficient.
Statement 2: \(c =a+10\) or \(a = c-10\). Now notice, b is a number between a and c and it can be written as \(c-10<= b <=c\) similarly d is a number between c and e or \(c <= d <=c+4\)
If we add these two: \(2c-10 <=b+d <= 2c+4\)
To find out the average, we need sum. lets just take a look at sum \(Sum = a+b+c+d+e\) or \(Sum = c-10 + b + c + d +c+4\) =>\(Sum= 3c-6 + b +d\)
using b+d \(3c-6 + 2c-10 <= Sum <= 3c-6 +2c+4\) \(5c -16<=Sum <=5c-2\)
Hence maximum limit of sum is 5c-2, therefore average A (which is Sum/5) is always going to be less than c (the median). This is exactly what we want to know. Sufficient.
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
28 May 2013, 20:48
1
This post received KUDOS
anon1 wrote:
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
So I actually understand the solution provided but it seemed a little un-intuitive, was hoping someone here could discover some sort of new insight about this problem. Basically, the solution suggests plugging but I wonder if there is something nicer...
Stmt 1 ) e+c = 34 e=c+4 e=c-34 hence solving these two gives us a value of C = 15 and e=19 . if a=15 , b=15 , c=15, d=19 and e=19 , then Mean > Median if a=1 , b=1 , c=15 , d=15 and e=19 , then Mean < Median Not Sufficient.
Stmt 2) c= a+10 .. and given that e-c=4 --> e=c+4 a ≤ b ≤ c ≤ d ≤ c+4 --------------> a ≤ b ≤ a+10 ≤ d ≤ (a+10)+4 -------------> a ≤ b ≤ a+10 ≤ d ≤ a+14
Mean > Median ? i.e (a+b+a+10+d+a+14)/5 > a+10 ? solving the above , we get - IS b+d > 2a+26 ? Lets calculate the maximum value of b+d . max value of b is a+10( since b ≤ a+10) and max d is a+14 ( since d ≤ a+14)----> hence max b+d=2a+24
Hence b+d is always ≤ 2a+24 (cannot be greater than 2a+26) Hence Mean is not greater than Median . Sufficient.
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
28 May 2013, 23:16
5
This post received KUDOS
Expert's post
anon1 wrote:
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
From the given problem, we know that the Median will always be c.
From F.S 1, we know that e+c = 34 and e-c = 4--> e=19, c = 15(Median). Now we will try to maximize the average : That can be done if a=b=c=15 and d=19--> 15,15,15,19,19, The average = 16.6, Thus, A>M.
Again, we can have a scenario where a=b=0 and c=d=15 and e=19. Thus, the average = 9.8 and A<M. Insufficient.
From F.S 2, we know that the series will be : a,b,a+10,d,a+14. Now we will find the maximum value of the average value --> That will be possible if b=c=a+10 and d=e=a+14--> \(\frac{(a+a+10+a+10+a+14+a+14)}{5}\) = \(\frac{(5a+48)}{5}\) = a+9.6. Now Median = a+10, and irrespective the value of a, A<M. Sufficient.
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
04 Jun 2014, 00:43
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
18 Aug 2015, 12:07
1
This post received KUDOS
anon1 wrote:
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?
(1) e+c=34 (2) c=a+10
So I actually understand the solution provided but it seemed a little un-intuitive, was hoping someone here could discover some sort of new insight about this problem. Basically, the solution suggests plugging but I wonder if there is something nicer...
You have 2 equations with 2 unknowns - you can find the values of c and e:
--> e-c = 4 --> e+c = 34
--> 2e = 38 --> e = 19 --> c = 15
You still do not know anything about a, b, and d. It could be that the numbers are 0,0,15,15,19, in which case the average <15 and the median is 15, and thus A<M, and the answer is 'no', or that the numbers are 15,15,15,19,19, in which case the average is >15 and the median is 15, and thus A>M, and the answer is 'yes'.You have 2 equations with 2 unknowns - you can find the values of c and e:
Stat.(1)->Maybe->IS->BCE.
According to Stat. (2),
--> e-c=4 --> c=a+10
--> from the first equation: --> c=e-4 Plug this into the second equation: --> e-4 = a+10 --> e-a = 14
So the differences between e, c and a are fixed. But you can still play around with b and d. The numbers could be 0,0,10,10,14, in which case the average is <10 and the median is 10, and thus A<M and the answer is 'no'. but is it always "No"? Plug in the other extreme, where b and d are the greatest they can be: If the numbers are 0,10,10,14,14, the average is still 48/5~9.5<10 and the median is 10, and the answer is still "No". Thus, the answer is always "No", and
Stat.(2)->No->S->B. _________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
18 Aug 2015, 18:29
ans should be B only without doubt.
Option B : a, b, c, d, e => a, b, (a+10), d, (a+14) so, median = a+10 Now, for minimum value of mean, b=a & d=a+10 => A(min) = (5a+34) / 5 = a+6.8 < median Again, For max mean, b=a+10 & d=a+14 => A(max) = (5a+48) / 5 = a+9.6 < median So, we can conclusively say that Mean can't be greater than Median.....................>>
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
18 Aug 2015, 21:21
i solved this thing in simpler and quicker way, we have: a≤b≤c≤d≤e
st1 is only good if we need to have the actual numbers and that does not seem to be the case since we are asked to solve for A>M but we could keep it in mind in case all things got kaput.
st2 is much more interesting and breaking it in give us, c-10≤b≤c≤d≤c+4
Average is: {3c-6+b+d}/5
Question was A>M and in our situation is wether b+d>5c-3c-6=> 2c-6?
we can go back to the formula and it max it out were b+d: 2c-10≤b+d≤2c+4 as such we can conclude that its not possible for A>M
gmatclubot
Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e
[#permalink]
18 Aug 2015, 21:21
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...