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17 Apr 2015, 13:45
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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:
45%
(medium)
Question Stats:
68% (02:13) correct
32%
(02:14)
wrong
based on 287
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The sum of 99 consecutive integers is 9999. What is the greatest integer in the set?
1. 100
2. 120
3. 149
4. 150
5. 151
1. 100
2. 120
3. 149
4. 150
5. 151
17 Apr 2015, 16:00
Kudos
Bookmarks
Hi Turkish,
This question is based on a few Number Properties (and you probably already know what they are, you're just not used to thinking about then when dealing with a large group of numbers).
To start, we're told that a group of 99 CONSECUTIVE integers has a sum of 9999. This tells us a couple of things:
1) The AVERAGE of the group will be the "middle" number (meaning the 50th number) in the sequence.
2) The AVERAGE IS 9999/99 = 101
We're asked for the LARGEST number in the sequence. Since we know.....
the 50th number is 101.....
and there are 49 consecutive numbers "below it" and 49 consecutive numbers "above it"...
The LARGEST number must be 101 + 49 = 150
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This question is based on a few Number Properties (and you probably already know what they are, you're just not used to thinking about then when dealing with a large group of numbers).
To start, we're told that a group of 99 CONSECUTIVE integers has a sum of 9999. This tells us a couple of things:
1) The AVERAGE of the group will be the "middle" number (meaning the 50th number) in the sequence.
2) The AVERAGE IS 9999/99 = 101
We're asked for the LARGEST number in the sequence. Since we know.....
the 50th number is 101.....
and there are 49 consecutive numbers "below it" and 49 consecutive numbers "above it"...
The LARGEST number must be 101 + 49 = 150
Final Answer:
GMAT assassins aren't born, they're made,
Rich
General Discussion
30 Jul 2015, 09:20
Kudos
Bookmarks
another way : uses sequence:
sum of n numbers = (n(2a+(n-1)d))/2 where n = no of terms a = 1st term , d - difference between two terms (ie d= a2-a1)
here sum = 9999 , n= 99 d = 1 (since consecutive numbers)
9999= (99/2) * (2a+(99-1)1)
from this 'a' (ie the 1st term) = 52
nth term in a sequence : nth term = a + (n-1) d
here a = 52 , n = 99 , d = 1
So nth term = 52 + (99-1)*1
therefore Nth term = 150
sum of n numbers = (n(2a+(n-1)d))/2 where n = no of terms a = 1st term , d - difference between two terms (ie d= a2-a1)
here sum = 9999 , n= 99 d = 1 (since consecutive numbers)
9999= (99/2) * (2a+(99-1)1)
from this 'a' (ie the 1st term) = 52
nth term in a sequence : nth term = a + (n-1) d
here a = 52 , n = 99 , d = 1
So nth term = 52 + (99-1)*1
therefore Nth term = 150
