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In a set of 5 numbers, if the largest number is 3 more than 25 Apr 2013, 03:19
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In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different.
(2) The median is 10 more than the smallest number in the set.
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B
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Bunuel
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In a set of 5 numbers, if the largest number is 3 more than 25 Apr 2013, 03:29
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In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different. If the set is {-100, -10, 0, 1, 3}, then (median)=0>(average)=(some negative number) but if the set is {-2, -1, 0, 2, 3}, then (median)=0<(average)=(some positive number). Not sufficient.

(2) The median is 10 more than the smallest number in the set. The average will be maximized for the following set: {x-10, x, x, x+3, x+3} --> (average)=(x-10+x+x+ x+3+x+3)/5=x-4/5 and (median)=x --> x-4/5<x. Sufficient.

Answer: B.
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In a set of 5 numbers, if the largest number is 3 more than 25 Apr 2013, 12:28
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See attached excel below.

B is sufficient to answer.
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mean_median.JPG [ 29.11 KiB | Viewed 5613 times ]

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In a set of 5 numbers, if the largest number is 3 more than 25 Apr 2013, 21:44
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rajatr
In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different.
(2) The median is 10 more than the smallest number in the set.
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You can think in conceptual terms too.

In a set of 5 numbers, the middle number will be the median. So say we arrange the numbers in ascending order:

___ ____ Median ____ (Median + 3)

(1) All the numbers are different.

The small numbers on the left of median could be very small making the mean smaller than median or they could be 1 less than the next such that mean is greater than median (3, 4, 5, 7, 8). Hence not sufficient.

(2) The median is 10 more than the smallest number in the set.

The smallest number is much smaller than the greatest number. Even if the fourth number is also (Median + 3), the fourth and the fifth number cannot make up the difference of 10. Hence Mean MUST be smaller than the median.

Answer (B)
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In a set of 5 numbers, if the largest number is 3 more than 22 Oct 2015, 06:35
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Bunuel
In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different. If the set is {-100, -10, 0, 1, 3}, then (median)=0>(average)=(some negative number) but if the set is {-2, -1, 0, 2, 3}, then (median)=0<(average)=(some positive number). Not sufficient.

(2) The median is 10 more than the smallest number in the set. The average will be maximized for the following set: {x-10, x, x, x+3, x+3} --> (average)=(x-10+x+x+ x+3+x+3)/5=x-4/5 and (median)=x --> x-4/5<x. Sufficient.

Answer: B.
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I just thought that B tells us that it is not an evenly spaced set hence mean can not be equal to the median.
We dont really need to do know if it is greater or lesser than the mean in a DS question.

Correct me if wrong please
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In a set of 5 numbers, if the largest number is 3 more than 22 Oct 2015, 22:12
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saadis87
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In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different. If the set is {-100, -10, 0, 1, 3}, then (median)=0>(average)=(some negative number) but if the set is {-2, -1, 0, 2, 3}, then (median)=0<(average)=(some positive number). Not sufficient.

(2) The median is 10 more than the smallest number in the set. The average will be maximized for the following set: {x-10, x, x, x+3, x+3} --> (average)=(x-10+x+x+ x+3+x+3)/5=x-4/5 and (median)=x --> x-4/5<x. Sufficient.

Answer: B.


I just thought that B tells us that it is not an evenly spaced set hence mean can not be equal to the median.
We dont really need to do know if it is greater or lesser than the mean in a DS question.

Correct me if wrong please
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A couple of things here:

If the set is evenly spaced, mean = median - correct! But does that mean that if the set is NOT evenly spaced, mean cannot be equal to median? No.
e.g. 3, 4, 7, 9, 12 - mean = median = 7
So a non-evenly spaced set DOES NOT imply mean is not equal to median.

Secondly, are we sure that it will be either greater OR less only? What if we can take numbers such that both are possible for different numbers? So under the constraints, what if we can take numbers such that mean is greater in some cases and less in other cases? So we will need to analyse the statement a little further to find that mean MUST be less than median in all cases. How much less exactly is something we don't care about as long as we know that it must be less.
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In a set of 5 numbers, if the largest number is 3 more than 23 Oct 2015, 00:39
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saadis87
Bunuel
In a set of 5 numbers, if the largest number is 3 more than the median, is the average value of the set greater than the median of the set?

(1) All the numbers are different. If the set is {-100, -10, 0, 1, 3}, then (median)=0>(average)=(some negative number) but if the set is {-2, -1, 0, 2, 3}, then (median)=0<(average)=(some positive number). Not sufficient.

(2) The median is 10 more than the smallest number in the set. The average will be maximized for the following set: {x-10, x, x, x+3, x+3} --> (average)=(x-10+x+x+ x+3+x+3)/5=x-4/5 and (median)=x --> x-4/5<x. Sufficient.

Answer: B.


I just thought that B tells us that it is not an evenly spaced set hence mean can not be equal to the median.
We dont really need to do know if it is greater or lesser than the mean in a DS question.

Correct me if wrong please

A couple of things here:

If the set is evenly spaced, mean = median - correct! But does that mean that if the set is NOT evenly spaced, mean cannot be equal to median? No.
e.g. 3, 4, 7, 9, 12 - mean = median = 7
So a non-evenly spaced set DOES NOT imply mean is not equal to median.

Secondly, are we sure that it will be either greater OR less only? What if we can take numbers such that both are possible for different numbers? So under the constraints, what if we can take numbers such that mean is greater in some cases and less in other cases? So we will need to analyse the statement a little further to find that mean MUST be less than median in all cases. How much less exactly is something we don't care about as long as we know that it must be less.
Show more


Thankyou. That makes more sense than what i said :)
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In a set of 5 numbers, if the largest number is 3 more than 14 Sep 2025, 01:12
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