[email protected] wrote:What is the average of eleven consecutive integers?
1) The average of first nine integers is 7
2) The average of last 9 integers is 9
The answer given was D. i.e each statement is suffecient.
Can someone please explain me the approach towards the solution of this problem.
regards,
Sagar
the solution above is correct, but you don't actually need counting to solve this problem - if you have the requisite INTUITION.
namely:
you are restricted to ELEVEN CONSECUTIVE INTEGERS.
this means that
you have absolutely no freedom in the placement of these integers. there are eleven of them in a row. this means that
if you can place even one number in the series at a specific location, then you've got the whole series.
for instance, if i tell you that the seventh number is 19, then i've got the whole series in place: there are six numbers right before that, and four right after it. no freedom.
(notice that it's not good enough, of course, for me to just tell you something like "one of the integers is 19".)
--
(1)
because these are consecutive integers, they're evenly spaced. therefore, the average of the first nine is equal to the number that's directly in the middle of those nine, which is the fifth one.
therefore, the fifth integer is 7, which means we can pinpoint the whole list.
sufficient.
(2)
because these are consecutive integers, they're evenly spaced. therefore, the average of integers #3 through #11 (i.e., the last nine) is equal to the number that's directly in the middle of those nine, which is the seventh one.
therefore, the seventh integer is 9, which means we can pinpoint the whole list.
sufficient.
ans (d)
Ron has been teaching various standardized tests for 20 years.
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