a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e : GMAT Data Sufficiency (DS)
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a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e

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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, 17:09
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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...
[Reveal] Spoiler: OA
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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, 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

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

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

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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14 Nov 2012, 23:12
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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
$$c <= d <=c+4$$

$$2c-10 <=b+d <= 2c+4$$

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.

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, 00:46
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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15 Nov 2012, 00:49
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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+14
considering
a = -10
c = 0
e = 14

2nd 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, 00:58
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, 01:01
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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28 May 2013, 20:48
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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...

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.

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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28 May 2013, 23:16
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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

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.

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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04 Jun 2014, 00:43
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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, 12:07
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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...

Economist:

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.
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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, 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.....................>>
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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, 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
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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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21 Aug 2016, 10:18
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Re: a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e   [#permalink] 21 Aug 2016, 10:18
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