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06 Nov 2019, 02:33
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If sequence Q has four terms: x, 12, y, and 20, is it true that the median of sequence Q is 16?
(1) The average of sequence Q is 16
(2) x = 16
Are You Up For the Challenge: 700 Level Questions
(1) The average of sequence Q is 16
(2) x = 16
Are You Up For the Challenge: 700 Level Questions
Updated on: 06 Nov 2019, 08:12
Originally posted by CareerGeek on 06 Nov 2019, 02:46.
Last edited by CareerGeek on 06 Nov 2019, 08:12, edited 1 time in total.
Last edited by CareerGeek on 06 Nov 2019, 08:12, edited 1 time in total.
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Bunuel
Edited.
(1) The average of sequence Q is 16
—> (x + 12 + y + 20)/4 = 16
—> x + y = 32
Since the sum of lowest and highest term is 32, irrespective of any value of x and y, median = 16 always
Many values of median are possible—> Sufficient
(2) x = 16
Many values of median are possible since we do not know the value of y —> Insufficient
IMO Option A
Posted from my mobile device
Statement tells us that average of x, 12, y, and 20.
Since average of 12 and 20 is 16, average of x and y must be 16. There are 2 cases possible.
Case 1-
If one of x or y is less than 12, then other must be greater than 20. hence median of Q=\(\frac{12+20}{2}\)=16
Case 2
If one of x or y lies in the range 12-20(both inclusive), other will lie in the range 12-20 (both inclusive) too. Median would be \(\frac{x+y}{2}\)=16
Sufficient
Statement 2-
if y=16, median is 16
if y has value other than 16, median would be different from 16
Insufficient
A
Since average of 12 and 20 is 16, average of x and y must be 16. There are 2 cases possible.
Case 1-
If one of x or y is less than 12, then other must be greater than 20. hence median of Q=\(\frac{12+20}{2}\)=16
Case 2
If one of x or y lies in the range 12-20(both inclusive), other will lie in the range 12-20 (both inclusive) too. Median would be \(\frac{x+y}{2}\)=16
Sufficient
Statement 2-
if y=16, median is 16
if y has value other than 16, median would be different from 16
Insufficient
A
Bunuel
