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a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]

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14 Nov 2012, 18:09

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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]

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14 Nov 2012, 23:40

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]

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15 Nov 2012, 00:12

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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]

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28 May 2013, 21:48

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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]

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29 May 2013, 00:16

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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

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]

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04 Jun 2014, 01:43

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Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]

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18 Aug 2015, 13:07

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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!

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Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]

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18 Aug 2015, 19:29

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]

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18 Aug 2015, 22:21

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
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18 Aug 2015, 22:21

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