What is the tenth term in a series of 89 consecutive

Thanks brstorewala for sharing this interesting math DS problem.
I wonder where are the source to get all these questions?

I already studied the Kaplan book, but wasn't feeling it too helpful.
I got most questions right in practice but I didn't do that well in real GMAT test with the time concern. Any suggestion?

With almost two minutes for each question on the math section and around 110 sec for each question on the verbal section, time does not seem to be a major concern. The problem is that the mind goes blank. I experienced it when I took my PP test at home. Especially for the RC part, I was not able to think clearly. It was like my mind was clouded and I was blank. So it is really a matter of how calm and composed you are, assuming that you have practiced sufficiently. I am a proponent of the philosophy of "practice as much as u can"......2k, 5k, 10k......any number of questions....the key is getting a feel, making sure that you don't see a different "kind" of a problem in the exam. Of course strategy is important, so I make this summary kind of a thing for each section, in which i note down the main testing points and rules. The summary would be very handy toward the end of your preparation schedule. Those are my two cents.....i am just waiting to see how my philosophy and strategy work on the "D" day.........it all boils down to those 4 bloody hours.......

Cool... A sort of philosophy... Brstorewala, have you ever tried to write a philosophic treatise? You are to be successful. No doubt.

great advice. comprehension is also my weakest part.
usually how much time people spend on preparing for the gmat?
3 months? 6 months?

There is a very good explanation of consecutive integers in MGMAT Number Properties.

If you can calculate first term of the series than you can calculate everything else (\(A_1\) and \(n\) define every other term, also average and sum)

Damn, thanks a bunch, Bunuel
I completely miss the CONSECUTIVE part... hence did not realize d=1

On a second thought, does consecutive always imply the numbers need to be increasing? Otherwise d=-1 is also possible from the given statements?

I think consecutive +ve integers are : 2, 3, 4, 5, and n but not n, ...............10, 9, 8, 7, 6 ..............1.

In that sense, answer should be D. However your thinking is also valuable if it were a +ve series only.

Wow - you guys found some really old question from 2003!!!

It is D.

And answer is 65 --- 10th term in series.

A ) Using the average, find the midpoint of series. It will be 45th term with values of 89*100/89 = 100. To get 10th term, there are 35 jumps of 1 point worth each. Hence 10th term = 100 - 35 * 1= 65.

A is sufficient.

B) Similarly here 79th term is 134. To reach 10th term, there are 69 jumps worth 1 point each. Hence 10th term = 134 - 69 = 65.

B is sufficient.

Hence correct answer choice is D.

Thanks,
Akhil M.Parekh

D

1) since these are consecutive intergers, mean will be the middle number... walk back to find the 10th

2) 79th term is know, so walk back to find the 10th term.... again, since these are consecutive integers

Consecutive integers means they are all distinct and are in AP.

In an AP with odd number of term the median and the mean are the same.

Statement 1:
100 is the mean and also the median. So 100 is the 45th terms. We just subtract 44 and get the answer. Also, it works becasue they are consecutive integer. Had it been an AP of non consecutive integers, this wouldn't have worked

Anyways, the statement is sufficient

Statement 2:
Gives one of the terms
The common difference is 1 (consecutive integers)

Sufficient to back track or use AP formulas to find any other term in the series.

Hence Each statement alone is sufficient. D

(1) The average of the integers is 100
Sufficient
(first+last term)/2=100
first + last =200
since there are total of 89 numbers and all numbers are consecutive; therefore last term = first+88
First+(First+88)=200
2*first=112
FIRST TERM=56
the first term is 56 therefore the tenth term will be 65
SUFFICIENT

(2) The 79th term in the series is 134
SUFFICIENT:-
79 term is 134, so we can reverse count the 10th term
or (79th-69th) term = (134-69)
or 10th term =65

BOTH STATEMENT GIVES US ANSWER

SO D IS THE ANSWER

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