Average value > Median

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crackthegmat2011: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sat Oct 30, 2010 6:09 am; Thanked: 1 times

by crackthegmat2011 » Thu Jan 06, 2011 9:05 am

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.

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Source: Beat The GMAT — Data Sufficiency | Thu Jan 06, 2011 9:05 am

prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Thu Jan 06, 2011 10:45 am

IMO : A

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Thu Jan 06, 2011 12:40 pm

hi all
for me it is terrible question, and on real test i will not even try to solve, just click, to me the answer is B
i will try to prove that 1 st insufficient, and if the answer is right my rationale about 2 will follow if not it does not make sence
(1) strange st, we are already given that largest number is 3 more than the median from the problem but anyway
1,2,3,4,6, median-3, and largest 3+3=6
mean =(1+2+3+4+6)/5=16/5 >3 (median) the answer is yes

another set, as we have no restrictions on the values
-4,-2,1,2,4 median=1, largest 1+3=4, all different
mean=(-4-2+1+2+4)/5=1/5<1 the answer is no
so to me 1 st insufficient

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Thu Jan 06, 2011 6:18 pm

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.

Numbers of the set (in increasing order) : a, b, c, d, e [c -> median]
e = c+3

question : a+b+c+d+e > 5c
=> a+b+2c+d+3 > 5c
=> a+b+d > 3c -3 ? ---> question stem

Statement 1:
a,b,c,d,e -> different -> Insuffcient

Statement 2:
c = a+10
Put in question stem => a+b+d > 3(a+10) -3 ?

=> b+d > 2a+27 ?

Check for minimum value :
Minimum value of b = a
Minimum value of d = c = a+10

So question stem becomes : a+a+10 > 2a+27 ? (which is NO)

Check for maximum value :
Maximum value of b = a+10
Maximum value of d = e = c+3 = a+13 (As e is c+3 and e is the largest number of the set)

So question stem becomes : a+10+a+13 > 2a+27 ? (which is NO again)

Hence; Sufficient. B

Thanks
Anshu

(Every mistake is a lesson learned )

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Thu Jan 06, 2011 11:37 pm

anshumishra 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.
Numbers of the set (in increasing order) : a, b, c, d, e [c -> median]
e = c+3

question : a+b+c+d+e > 5c
=> a+b+2c+d+3 > 5c
=> a+b+d > 3c -3 ? ---> question stem

Statement 1:
a,b,c,d,e -> different -> Insuffcient

Statement 2:
c = a+10
Put in question stem => a+b+d > 3(a+10) -3 ?

=> b+d > 2a+27 ?

Check for minimum value :
Minimum value of b = a
Minimum value of d = c = a+10

So question stem becomes : a+a+10 > 2a+27 ? (which is NO)

Check for maximum value :
Maximum value of b = a+10
Maximum value of d = e = c+3 = a+13 (As e is c+3 and e is the largest number of the set)

So question stem becomes : a+10+a+13 > 2a+27 ? (which is NO again)

Hence; Sufficient. B

anshumishra,can you please give a set of values satisfying a+b+d > 3c -3.Thanks!

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Fri Jan 07, 2011 4:57 am

prachich1987 wrote: anshumishra,can you please give a set of values satisfying a+b+d > 3c -3.Thanks!

As per statement 1 , a+b+d > 3c -3 ? is sometimes true, sometimes false.
Choose : a=1,b=2,c=3,d=4
a+b+d = 7; 3c-3 = 6 ; So a+b+d>3c-3
Choose : a=-4,b=-2,c=1,d=2
a+b+d = -4; 3c-3 = 0; So a+b+d < 3c -3 -> That is why statement 1 was insufficient

For statement 2 (As it has been proved), you will always get the value of a+b+d to be less than 3c-3.
One such example can be :
a=1,b=2,c=11,d=12,e=14
a+b+d = 15; 3c-3 = 30 => a+b+d < 3c-3

Other one is :
a=1, b=11,c=11,d=14,e=14
a+b+d = 26; 3c-3 =30 => a+b+d < 3c-3 So Sufficient.

Thanks
Anshu

(Every mistake is a lesson learned )

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aleph777: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Jun 18, 2010 10:19 am; Location: New York, NY; Thanked: 10 times

by aleph777 » Fri Jan 07, 2011 7:20 am

Great problem. Is it an official question?

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Fri Jan 07, 2011 7:23 am

my approach to the st 2 is almost the same as that of anshumishra
here out set, median-x, largest-(x+3) smallest (x-10) and two others say a,b
x-10, a, x, b, x+3
mean of the set=(x-10+a+x+b+x+3)/5=(3x-7+a+b)/5. and we need to estimate if
(3x-7+a+b)/5>x( median) or simplifying, is a+b-7>2x?
the largest values for the a=x, b=x+3.
x+x+3-7=2x-4 is 2x-4>2x obviously not we can cancel 2x and left with -4<0 other values for a and b also lead to the same result
say a=x-10, b=x, 2x-10-7 VS 2x again the answer is no
B suff

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crackthegmat2011: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sat Oct 30, 2010 6:09 am; Thanked: 1 times

by crackthegmat2011 » Fri Jan 07, 2011 7:25 am

Answer is B.

Thanks anshumishra, for providing detailed explanation.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Fri Jan 07, 2011 7:28 am

crackthegmat2011 wrote:Answer is B.

Thanks anshumishra, for providing detailed explanation.

You are welcome!

Thanks
Anshu

(Every mistake is a lesson learned )

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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Wed Jan 26, 2011 12:24 am

very hard

A is not sufficient

consider B

pick number, minimum is 0, median is 10, maximum is 13

call x, y are two other number,x<y.
make x, y max we have x=10, y=13,

we have total is 46, average is 46/5< Median

when we make x, y max, Average< Median, when x,y are less of course we have Average<Median

B is sufficien

it take me too long to do this,

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

by GMATGuruNY » Wed Jan 26, 2011 4:12 am

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.

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chendawg: Master | Next Rank: 500 Posts; Posts: 163; Joined: Tue Nov 24, 2009 7:56 amFollowed by: 4 members; Location: Philadelphia; Thanked: 13 times; GMAT Score:660

by chendawg » Wed Jan 26, 2011 4:48 pm

The way that I approached this problem was this:

First we want to rephrase. We can call the numbers A B C D E. Is (A+B+C+D+C+3)/5>C? This boils down to A+B+D>3C-3?

Statement 1: No way to know. We know the LARGEST number in the set is 3 more than median. I just picked an easy number for the median, 10, to visualize. Largest number would be 13 in the set. Since we are told nothing about the smallest numbers, they could be anything. Median could be larger or smaller or equal to the median. Insufficient.

Statement 2: Now they tell us the median is 10 more than the smallest number. Again, we can pic numbers to visualize. I picked 10 again. We now want to minimize the values to see if we can get a yes no. If the median is 10, then the largest # is 13 and the smallest is 0. Is (0+0+10+10+13)/5 larger than 10? No. Now we want to maximize the values. Is (0+10+10+13+13)/5 larger than 10? No. Thus, this statement is sufficient. We want to pick B.

For me personally it's easier to visualize with numbers plugged in than with variables. I ended up not using the rephrase, just cuz it was just as easy not using it for me personally.

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Zerks87: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sat Oct 16, 2010 8:15 am; Thanked: 9 times

by Zerks87 » Sun Feb 06, 2011 10:59 am

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

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tgou008: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Sat Jan 15, 2011 11:26 am; Thanked: 9 times

by tgou008 » Tue May 10, 2011 7:50 am

I agree with Zerks; I don't think you need to go to the time and effort of solving using algebra or plugging numbers. I would only do so if you were confused / struggling.

I got to B within 2 minutes and my apporach was as follows

{_,_,x,_,x+3} is average of set > median of set?
It is crucial to realize at this pt that given we are dealing with an odd set of #s (5), then the median and average will be the same if the set is evenly spaced. So if we are told either that the set is evenly spaced then the lowest number will be x-3 and therefore that the median = mean.
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

Therefore, question can be rephrased to - is the set of numbers evenly spaced? AND IF NOT, what is the distribution of numbers to the LHS of the median (i.e., median - min #)

STMT 1: Each # (1-5) is different. This doesn't help very much at all - the lowest number could be anything e.g., x-3, or it could be x-2, or x-4 etc.
INSUFFICIENT - Cross off A and D

STMT 2: Median = smallest# + 10. This tells us that the range is greater on the LHS of the median (i.e., median - min value = 10) than on the RHS of the median (i.e., max value - median = 3. Therefore the mean will be more heavily skewed to the LHS / smaller numbers. Therefore, the mean will be smaller than the median.
SUFFICIENT

ANSWER = B

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