From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

netigen wrote:From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

The answer is A indeed. This question is good because it tests rarely seen concepts like mode and median. Nice job.

netigen wrote:From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

How did you select the set to be A B 5 7 7 F ... Stmt 2 says mid 2 nos are 5 and 7... the stem does not say so... ??? :

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

The bold part above explains why 5 and 7 are the 3rd and 4th numbers in the set.

Got it.. thanks

netigen wrote:Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

The bold part above explains why 5 and 7 are the 3rd and 4th numbers in the set.

Theres actually a minor error more to your median statement.

"Since the number of elements is even, the median will not be in the set" is slightly misleading. The statement should be "since the number of elements is even, the median will be the average of the 3rd & 4th number." Meaning the 3rd & 4th number could be 5 & 7, or 6 & 6. However, since we know that the mode is 7, and thus the # of occurances of 7 > # of occurances of 6, if the 3rd & 4th numbers were 6 & 6, then we would need 3 7's. Since we only have 6 elements in the set, we can't possibly have 3 7's, therefore the 3rd & 4th elements have to be 5 & 7 respectively.