Average value > Median

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Tue May 10, 2011 8:37 am

Zerks87 wrote:For this problem you dont need math at all, you just need to know the properties of sets

Stem: There is a five number set where the largest number is greater than the median by 3. They ask, basically whether the mean of the set is greater than the median.

Remember for odd number sets that in any set with an odd number of items that are evenly spaced, the mean will equal the median. There for we are looking for whether this is an evenly spaced set or not.

(1) All numbers are different - fills one of the requirements but does not tell us whether the set is evenly spaced, N.S.

(2) Media is ten more than the smallest number, therefore we have a set that look like this {x-10,....,x,....,x+3}. Clearly this is not an evenly spaced set or x+3 would be x+10 or vice versa. Therefore we know whether the mean is greater than the median. The mean is less than the median as it would be weighted more towards x-10 than x+3. SUFF

Pick B

Be careful.
It is true that if a set of values is evenly spaced, then the average equals the median.
But if a set of values is not evenly spaced, we cannot conclude that the average does not equal the median.

For example: {7,9,10,12,12}.
Average = (7+9+10+12+12)/5 = 50/5 = 10.
Median = 10.
The set is not evenly spaced, but the average = median.

If statement 2 had said that the median is 5 more than the smallest value, then the correct answer would not be B.
The values could be {5,10,10,12,13}.
Average = (5+10+10+12+13)/5 = 50/5 = 10.
Median = 10.
Average = Median.

The values could be {5,10,10,13,13}.
Average = {5+10+10+13+13} = 51/5 = 10.2.
Median = 10.
Average > Median.
Insufficient.

Even though the smallest value is 5 away from the median and the largest value is only 3 away from the median, the average in the second case is greater than the median.

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by Ian Stewart » Tue May 10, 2011 8:39 am

tgou008 wrote:
Similarly, if we are told that the set is not evenly spaced e.g., the smallest number is less than x-3 then the median will be greater than the mean. Alternatively, if the smallest number is greater than x-3 then the mean will be greater than the median

This isn't correct. It is certainly true that when a set is evenly spaced, the mean and median are equal. But if a set is not evenly spaced, anything can happen: the mean and median can still be equal, or one might be greater than the other. For example, the set {0, 3, 4, 6, 7} is not equally spaced, but its mean and median are both 4. By lowering or raising the value of the smallest element you can then get examples where the mean is smaller or larger than the median.

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sushantgupta: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Sun Mar 27, 2011 9:54 am; Thanked: 2 times

by sushantgupta » Sun Jul 03, 2011 9:46 pm

lets say we have 5 no = a,b,c,d,e

for quetsion we have e= c+3

avg = (a+b+c+d+c+3)/5 > c
=> a+b+d+3 > 3c (1

Statement 1 : All no. are different
so d can be c+1 or c+2

hence (1 => a+b+4 > 2c (2
or a+b+5 > 2c (3

if a = c-10 and b = c- 9

(2 => -13 > 0
(3 => -14 > 0

if a = c-2 and b = c-1
2) => 1 >0
3) => 2 >0

So statement 1 is insufficient.

Statement 2: c = a+10

(1 => a+b+d+3 > 3c = a+b+d+3 > 3a+30 = b+d > 2a+27 (4
max value of b =c = a+10
max value of d = e = a+13
so (4 => 2a+23 > 2a+27 = 0 > 4 (5

min value of b = a
min value of d = c = a+10

so (4 => 2a+10 > 2a+27 = 0 > 17 (6

from 5 and 6 we can deduce that median is greater than avg. Hence statement 2 is sufficeint

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amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 am; Location: pune; Thanked: 36 times; Followed by:3 members

by amit2k9 » Mon Jul 04, 2011 1:12 am

using weighted average. difference between median and smallest > difference between largest and median.
thus average will be between smallest and median.

B is clean.

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prashant misra: Master | Next Rank: 500 Posts; Posts: 101; Joined: Thu Aug 25, 2011 9:39 pm

by prashant misra » Tue Nov 22, 2011 5:11 am

i was not able to properly visualize this question ended up choosing wrong option.i tried picking numbers first statement is not sufficient but for 2nd statement we can also have numbers 1 10 11 13 20 so the median would 10 more the the smallest digit but the mean and median would be equal so if we pick other numbers this can be prove untrue.can anyone explain me this.

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 am; Location: Ahmedabad; Thanked: 31 times; Followed by:10 members

by ronnie1985 » Thu Mar 29, 2012 7:52 am

It is not specifically mentioned in the question whether the numbers are real or they are integers.

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moussaobeid: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Feb 25, 2012 4:38 am

by moussaobeid » Fri May 18, 2012 4:06 pm

GMATGuruNY wrote:
crackthegmat2011 wrote:In a set of 5 numbers if the largest number is 3 more than the median, Is the average value of set > median of set ?

1. All the numbers are different.

2. The median is 10 more than the smallest number in the set.
Let x = median. Largest number = x+3.

Statement 1: All the numbers are different.
No way to compare the average to the median.
Insufficient.

Statement 2: Smallest number = x-10
Largest possible set of values: x-10, x, x, x+3, x+3
Average = (x-10 + x + x + x+3 + x+3)/5 = (5x - 4)/5 = x - 4/5
x - 4/5 < x.
Using the largest values possible, average < median.
Sufficient.

The correct answer is B.

Thanks for your great explanations, always awsome, straight to the point and most concise

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Lifetron: Master | Next Rank: 500 Posts; Posts: 308; Joined: Thu Mar 29, 2012 12:51 am; Thanked: 16 times; Followed by:3 members

by Lifetron » Wed Aug 29, 2012 7:34 am

Good 1 !

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rajeshsinghgmat: Master | Next Rank: 500 Posts; Posts: 171; Joined: Tue Jan 08, 2013 7:24 am; Thanked: 1 times

by rajeshsinghgmat » Sat Apr 06, 2013 2:04 am

Let it B.

STATEMENT I:

{1,2,3,3,6}: Here Median = Average = 3

{1,2,3,5,6}: Here Median < Average

STATEMENT II:

{x, a, (x+10), b, (x+13)}

Here Median = x+10

Average = (a+b+3x+23)/5

if Median = Average, then, (a+b+3x+23)=5x+50

a+b=2x+27

maximum value of b = x+13

when b is maximum, a = x+14

which shows that a > Median which is not possible.

Hence we can definitely conclude whether the answer is yes or no.

Now lets consider Average > Median

i.e. a+b+3x+23> 5x+50

i.e. a+b>2x+27

when b is maximum, b = x +13
i.e. (a + x+ 13)>2x+27

i.e. a > x+14 which is impossible.

We can conclude Median > Average