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11 Oct 2015, 08:17
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
43% (00:26) correct
57%
(01:19)
wrong
based on 7
sessions
History
Date
Time
Result
Not Attempted Yet
N, a set of natural numbers is partitioned into subsets S1 = {1}, S2 = {2,3}, S3 = {4,5,6}, S4 = {7,8,9,10} and so on. The sum of the elements of the subset S50 is
(a) 61250
(b) 65525
(c) 42455
(d) 62525
(e) 55425
Source - quant4gmat
(a) 61250
(b) 65525
(c) 42455
(d) 62525
(e) 55425
Source - quant4gmat
This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
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11 Oct 2015, 12:28
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HI excelingmat,
The question is based heavily on pattern-matching skills, and is far more 'layered' than what you would likely see on the Official GMAT on this subject. You'll likely find it easiest to define the patterns one-at-a-time, so that you can see how they impact the question.
To start, we're given information on the sub-sets of numbers:
-The 1st set has 1 number
-The 2nd set has 2 numbers
-The 3rd set has 3 numbers
Etc.
So the 50th set will have 50 numbers in it.
Since the numbers in the subsets are consecutive, we can figure out ALL of the numbers that show up BEFORE we get to the 50th set by using "bunching"...
The 1st set has 1 number and the 49th set has 49 numbers....1 + 49 = 50 numbers
The 2nd set has 2 numbers and the 48th set has 48 numbers...2 + 48 = 50 numbers
Etc.
This pattern will continue all the way down to...
The 24th set has 24 numbers and the 26th set has 26 numbers...24 + 26 = 50 numbers
Thus, we have 24 'pairs' of 50 numbers and one set (the 25th set with 25 numbers) that doesn't have a 'pair-match'...
So 24(50) + 25 = 1225 consecutive numbers exist in the 49 sets BEFORE the 50th set - that's the numbers 1 to 1225, inclusive...
Thus, the 50th set starts with 1226 and ends with 1275. We can use 'bunching' again to figure out the sum of this group (although there's no 'middle term' this time)...
25(1226+1275) =
25(2501) =
62525
Final Answer:
GMAT assassins aren't born, they're made,
Rich
The question is based heavily on pattern-matching skills, and is far more 'layered' than what you would likely see on the Official GMAT on this subject. You'll likely find it easiest to define the patterns one-at-a-time, so that you can see how they impact the question.
To start, we're given information on the sub-sets of numbers:
-The 1st set has 1 number
-The 2nd set has 2 numbers
-The 3rd set has 3 numbers
Etc.
So the 50th set will have 50 numbers in it.
Since the numbers in the subsets are consecutive, we can figure out ALL of the numbers that show up BEFORE we get to the 50th set by using "bunching"...
The 1st set has 1 number and the 49th set has 49 numbers....1 + 49 = 50 numbers
The 2nd set has 2 numbers and the 48th set has 48 numbers...2 + 48 = 50 numbers
Etc.
This pattern will continue all the way down to...
The 24th set has 24 numbers and the 26th set has 26 numbers...24 + 26 = 50 numbers
Thus, we have 24 'pairs' of 50 numbers and one set (the 25th set with 25 numbers) that doesn't have a 'pair-match'...
So 24(50) + 25 = 1225 consecutive numbers exist in the 49 sets BEFORE the 50th set - that's the numbers 1 to 1225, inclusive...
Thus, the 50th set starts with 1226 and ends with 1275. We can use 'bunching' again to figure out the sum of this group (although there's no 'middle term' this time)...
25(1226+1275) =
25(2501) =
62525
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
Looking for better-quality questions? Check out the 'Similar Questions' block below
for a list of similar but high-quality questions.
Want to join other relevant Problem Solving discussions? Visit our Problem Solving (PS) Forum
for the most recent and top-quality discussions.
Thank you for understanding, and happy exploring!
