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12 Jul 2019, 00:04
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A sequence consists of the consecutive integers from 1 through m. Does m equal to 49?
(1) The average of the integers in the sequence is not an integer.
(2) The median of the integers in the sequence is not an integer.
(1) The average of the integers in the sequence is not an integer.
(2) The median of the integers in the sequence is not an integer.
Sum of 49 terms/ 49 is an integer hence m can't be 49. Sufficient
B. Median is not an integer, sonumber of terms must be even.hence m can't be 49
Hence answer is D
Posted from my mobile device
B. Median is not an integer, sonumber of terms must be even.hence m can't be 49
Hence answer is D
Posted from my mobile device
12 Jul 2019, 00:21
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IMO : D
A sequence consists of the consecutive integers from 1 through m. Does m equal to 49?
(1) The average of the integers in the sequence is not an integer.
(2) The median of the integers in the sequence is not an integer.
The avg of the sequence that consist of consecutive integers= (first+last)/2.
1) so (1+49)/2=50/2=25 ; which is an integer, that means m is not equal to 49.
Sufficient.
2) the median of the sequence from 1 to 49 =25, which is again a integer, so sufficient m is not =49.
D
A sequence consists of the consecutive integers from 1 through m. Does m equal to 49?
(1) The average of the integers in the sequence is not an integer.
(2) The median of the integers in the sequence is not an integer.
The avg of the sequence that consist of consecutive integers= (first+last)/2.
1) so (1+49)/2=50/2=25 ; which is an integer, that means m is not equal to 49.
Sufficient.
2) the median of the sequence from 1 to 49 =25, which is again a integer, so sufficient m is not =49.
D
