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06 Apr 2021, 11:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:28) correct
31%
(02:29)
wrong
based on 300
sessions
History
Date
Time
Result
Not Attempted Yet
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?
A. 300
B. 301
C. 301.5
D. 302.5
E. 306
A. 300
B. 301
C. 301.5
D. 302.5
E. 306
08 Apr 2021, 03:21
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Solution:
Ask yourself- What's the first and last number in the set?
12 x 9 =108 and should be the first number in the set(a multiple of 9 and greater than 100)
500 / 9 = 55 .55
55 x 9 = 495
=> The last number in our set is therefore 495
We know that
(a)Median of a sequence of numbers = Average
(b)Average in a sequence of numbers can be obtained by adding the first and last terms of the sequence and dividing by 2
108 + 495 = 603
& 603 / 2 = 301.5
=> 301.5 is the median of Set M (option c)
Devmitra Sen
GMAT SME
Ask yourself- What's the first and last number in the set?
12 x 9 =108 and should be the first number in the set(a multiple of 9 and greater than 100)
500 / 9 = 55 .55
55 x 9 = 495
=> The last number in our set is therefore 495
We know that
(a)Median of a sequence of numbers = Average
(b)Average in a sequence of numbers can be obtained by adding the first and last terms of the sequence and dividing by 2
108 + 495 = 603
& 603 / 2 = 301.5
=> 301.5 is the median of Set M (option c)
Devmitra Sen
GMAT SME
General Discussion
07 Apr 2021, 23:16
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I think the answer is Option C 301.5
Between 100 and 500 the first and last multiple of 9 are
9*12=108 and 9*54=486
Now the number of digits between 12 and 54 including both are 43
since 43 is an odd number
the median will lie between 21 st and 22nd multiples of 9 in the above range
since our first number starts from 12th multiple in the range above
we add 21 and 22 to 12
ie. 33rd and 34th multiple of 9
297 and 306
Taking the average for the above number gives us 301.5
Between 100 and 500 the first and last multiple of 9 are
9*12=108 and 9*54=486
Now the number of digits between 12 and 54 including both are 43
since 43 is an odd number
the median will lie between 21 st and 22nd multiples of 9 in the above range
since our first number starts from 12th multiple in the range above
we add 21 and 22 to 12
ie. 33rd and 34th multiple of 9
297 and 306
Taking the average for the above number gives us 301.5
