gmatprep2..avg
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let first nos be x so series is
x , x+1 ,......x+10
stmt 1 : add x ...X+6 /9 = 7 SUFF
stmt 2: add the 3rd nos onwards th i.e X+2 ... X=10/9 = 9 SUFF
D
x , x+1 ,......x+10
stmt 1 : add x ...X+6 /9 = 7 SUFF
stmt 2: add the 3rd nos onwards th i.e X+2 ... X=10/9 = 9 SUFF
D
Regards
Samir
Samir
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I agree with D for this one. The average of an odd number of consecutive integers will just be the middle number in the list. (for example, average of 5, 6, 7 = 6). So in a list of 11 consecutive integers, the 6th number in the list will be the average. Similarly, in a list of 9 consecutive integers, the 5th number will be the average.
Statement (1) tells us that the average of the first 9 consecutive integers in the list is 7. This means that 7 is the 5th integer in the list. Therefore, 8 is the sixth number in the list and is the average of the larger list of 11. In other words the list of 11 must be 3,4,5,6,7,8,9,10,11,12,13.
Statement (2) tells us that the average of the last 9 consecutive integers in the list is 9. This means that 9 is the 5th number from the end of the list (or the 7th from the beginning) and we can construct the entire list from this and determine the overall average to be 8.
Statement (1) tells us that the average of the first 9 consecutive integers in the list is 7. This means that 7 is the 5th integer in the list. Therefore, 8 is the sixth number in the list and is the average of the larger list of 11. In other words the list of 11 must be 3,4,5,6,7,8,9,10,11,12,13.
Statement (2) tells us that the average of the last 9 consecutive integers in the list is 9. This means that 9 is the 5th number from the end of the list (or the 7th from the beginning) and we can construct the entire list from this and determine the overall average to be 8.