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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 ec=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 unintuitive, 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...
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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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14 Nov 2012, 23:40
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and ec=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 ec = 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
A<M (10, 10, 0, 0, 14), A>M (10, 0, 0, 14, 14)
Combining both cases
a=5, c=15, e =19
M =15 min values A = 5+5+15+15+19/5 < M
Max Values A = 5+15+15+19+19/5 < M
Hence C is the answer.
Hope it clarifies



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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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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 ec=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 unintuitive, 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 pluggingin 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 ec=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 ec =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 \(ec=4\) to find out e=19, and c=15, but nothing else. Not sufficient. Statement 2: \(c =a+10\) or \(a = c10\). Now notice, b is a number between a and c and it can be written as \(c10<= b <=c\) similarly d is a number between c and e or \(c <= d <=c+4\) If we add these two: \(2c10 <=b+d <= 2c+4\)To find out the average, we need sum. lets just take a look at sum \(Sum = a+b+c+d+e\) or \(Sum = c10 + b + c + d +c+4\) =>\(Sum= 3c6 + b +d\) using b+d \(3c6 + 2c10 <= Sum <= 3c6 +2c+4\) \(5c 16<=Sum <=5c2\) Hence maximum limit of sum is 5c2, 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. Ans B it is!Hope it helps.
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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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15 Nov 2012, 01:46
as i explained in case B, there are 2 situations M>A M<A
hence B alone is not sufficient.



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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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suryanshg wrote: as i explained in case B, there are 2 situations M>A M<A
hence B alone is not sufficient. Look at the highlighted portion. 1st mistake c=a+10 e=c+4 => e=a+14considering a = 10 c = 0 e = 142nd mistake A<M (10, 10, 0, 0, 14), A>M (10, 0, 0, 14, 14) A=14/5, M=0 how is A<M ? Probably should give u some idea.
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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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15 Nov 2012, 01:58
oops! silly error! definitely would have cost me 20 GMAT points. thanks



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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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15 Nov 2012, 02:01
suryanshg wrote: oops! silly error! definitely would have cost me 20 GMAT points. thanks 20 GMAT points and just thanks? where is 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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anon1 wrote: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and ec=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 unintuitive, 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=c34 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 ec=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. HTH Jyothi
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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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anon1 wrote: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and ec=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 ec = 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. B.
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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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Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e [#permalink]
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anon1 wrote: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and ec=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 unintuitive, 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... Economist: You have 2 equations with 2 unknowns  you can find the values of c and e: > ec = 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), > ec=4 > c=a+10 > from the first equation: > c=e4 Plug this into the second equation: > e4 = a+10 > ea = 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.
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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.....................>>



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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, 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, c10≤b≤c≤d≤c+4
Average is: {3c6+b+d}/5
Question was A>M and in our situation is wether b+d>5c3c6=> 2c6?
we can go back to the formula and it max it out were b+d: 2c10≤b+d≤2c+4 as such we can conclude that its not possible for A>M



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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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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 Apr 2017, 10:04
it may take more than 3 min to solve the problem, but if you want to guess, then just go for B because we know how a, c , e relate to each other. While in statement 1, a,b,d can be any number.




Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e
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