If x and y are positive integers and yx , what is the

Question

If x and y are positive integers, and y > x, what is the median of the set of consecutive integers from x to y?

(1) The range of the set of integers from x to y inclusive is 31. 
(2) The mean of the set of integers from x to y inclusive is 16.

jamifahad · Answer

what is the median of set of consecutive integers from x to y?

Stmt1:Range of set of integers from x to y inclusive is 31.
Set can be S={1,2,3,4.....,32} with median = (1+32)/2=16.5
Set can be S={10,11,12,.....41} with median=(10+41)/2=25.5. 
Not sufficient.

Stmt2:Mean of set of integers from x to y inclusive is 16.
Median of set of consecutive integers = Mean of set of consecutive integers
Median = 16. Sufficient.

OA B.

Zarrolou · Answer

If x & y are positive integers and y > x, what is the median of the set of consecutive integers from x to y?

(i) the range of the set of integers from x to y inclusive is 31 
Since we do not know where the set "starts" we are not able to determine its median.

(ii) the mean of the set of integers from x to y inclusive is 16 
In a set of consetutive integers median=mean, so the median is  16 .
IMO B

karjan07 · Answer

"In a set of consecutive integers median=mean, so the median is 16"

Will this hold true if total no. of consecutive integers is even??

Zarrolou · Answer

Yes it will hold true.

Example  3,4,5,6

median = 4.5, mean = 4.5

Generally speaking the rule  mean=median  holds true of sets of  evenly spaced integers .

sankarraman · Answer

Mean, Median, Mode coincides in case of normal data or distribution. Hence Median can be mean rather than range

ugo2602 · Answer

I agree, but where do we know that integers from x to y are consecutives integers?

Bunuel · Answer

The set of integers from x to y inclusive means all integers from x to y inclusive, so this set must be a set of consecutive integers. For example, the set of integers from 1 to 5 inclusive is {1, 2, 3, 4, 5}.

Hope it's clear.

sayansarkar · Answer

@ugo2602

look at the numbers. it is said that mean of x to y (inclusive) is 16. in case of evenly spread numbers average = {(first no)+(last no)}/2 or {(second no)+(second last no)}/2 and so on. They are paired averages.

Hence (x+y)/2=16 ie. x+y=32
so consider the following
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, 16 ,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31 holds true
median is 16=mean
again 
take out one number from the beginning and the end
2,3,4,5,6,7,8,9,10,11,12,13,14,15, 16 ,17,18,19,20,21,22,23,24,25,26,27,28,29,30
here again the mean =16 =median even though the number of terms has been reduced to 29

if you keep on reducing pairs of numbers from the outer ring you will end up with 16 as the last number

Crytiocanalyst · Answer

(1) The range of the set of integers from x to y inclusive is 31. 
it can be from 1 to 31 or could be from 31 to 62 both the possibilities exist therefore we cannot conclusively decide on the same

(2) The mean of the set of integers from x to y inclusive is 16.
Since x and y are consecutive positive numbers these median from being we can eleminate all other poddibilities other than starting from 1 to 16 we can progressively arrive at a conclusion 
Therefore IMOB

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If x and y are positive integers and y>x , what is the

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