Consec int+avg

[sogmat]

Set A contains some even consecutive positive integers, and set B contains some odd consecutive integers. IS the avg of numbers in B less than Median number in A

1- Sum of nos in A is greater than sum of nos in B
2- There are same no of integers in A&B


OA is c but i think its e

[srn]

A and B alone are insufficient
consider both together
few examples
A = 2,4,6,8,10 Sum 30 Median 6
B = 1,3,5,7,9 Sum 25 Avg 5

A = 2,4,6,8 Median 5
B = 1,3,5,7 Avg 4

The starting number of B must be less than the starting number of A, else sum of number's in B would be greater than A

Hence sufficient. C

[vittalgmat]

I echo srn's reply. One more vote for C.

Some more observations:

Both A and B have consecutive ints, A having consec even ints while B having consec Odd ints.

Stmt 1 says that the starting number in set A is greater than that of B.
Since this stmt does not say anything of the sizes of the two sets A and B we cannot figure out exactly the solution. hence insufficient.

Stmt 2 tells us the sizes are same. Now we are comparing apple with apples.

Now we are guaranteed that Avg of B < median of A.

One trick is to choose the sizes of sets A and B to be Odd. That way, the middle element can be easily compared, eliminating one small math operation to find the median.

Hope I havent missed anything.
-V

[sanju09]

I am starting after deleting A/B/D.

With an equal number of elements in both sets, and with &#8721;A > &#8721;B, let's take there be an odd number of elements in each set; following are few possibilities with A and B that fit in the scene here:

A {2, 4, 6} and B {1, 3, 5} or A {4, 6, 8} and B {3, 5, 7} or A {2, 4, 6, 8, 10} and B {1, 3, 5, 7, 9} or A {2, 4, 6} and B {-1, 1, 3}.

Even if we take there be an even number of elements in each set; following are few possibilities with A and B that fit in the scene here:

A {2, 4, 6, 8} and B {1, 3, 5, 7} or A {2, 4, 6, 8} and B {-1, 1, 3, 5} or A {2, 4, 6, 8, 10, 12} and B {1, 3, 5, 7, 9, 11} or A {2, 4} and B {-1, 1}.

In all the above cases, we can see that the average of numbers in B is less than the median number in A.

hence C

[vpr]

For this question there is no need to pick number. You waste a lot of time picking numbers for this type of question.

First thing to remember is , for equally spaced set, mean = median (= first term+last term/2). So question essentially asks whether mean of set B > mean of set A.

mean = sum of entries/no. of entries. In this question odd and even doesnt matter. It is mentionned to confuse you. You need both (1) and (2) to make sure mean of B > mean of A.

[vpr]

For this question there is no need to pick number. You waste a lot of time picking numbers for this type of question.

First thing to remember is , for equally spaced set, mean = median (= first term+last term/2). So question essentially asks whether mean of set B > mean of set A.

mean = sum of entries/no. of entries. In this question odd and even doesnt matter. It is mentionned to confuse you. You need both (1) and (2) to make sure mean of B > mean of A.

[cramya]

Elegant indeed, VPR.

Well done.

Regards,
CR