If M is the set of all consecutive multiples of 9 between 100 and 500,

Question

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

CrackverbalGMAT · Accepted Answer

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

Palak74 · Answer

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

MathRevolution · Answer

Last multiple of 9 should be 9 * 55 = 495

Thanks

Palak74 · Answer

Oh correct

I made a mistake

So that changes the answer to 44 digits between 55 and 12 i.e the 34th digit in the range and so the answer is  306 option E

Is the approach correct?

priyachandank · Answer

first multiple= 108; last multiple= 495.
Avg= 108+495/2= 301.5

Or terms= 495-108/9= 43. So the number would be the avg of 21st and 22nd terms i.e. 297 and 306 which is 301.5

Pyro27 · Answer

Question mentions multiples of 9 b/w 100 & 500
Hence, first term of the set=108, last term of the set= 495

No of terms (n) in the set ---> 495=108+(n-1)9--->n=44

Median of 44 (even) nos in the set =  22nd+23rd term/2  
i.e. a+21d+a+22d/2= 2a+43d/2--->2*108+43*9/2--->603/2

Median =  301.5 
Option-C

shivanks · Answer

in the question inclusive isn't mention... so in that sinario should we assume it....? because if they said find the median of muliple of 9 between 100- 500 exclusive then the middle turm is 22 so in that case ans must must mulitiple of 9... and when they say inclusive then ans is sum of 22nd term and 23rd term devided by 2.. can you please explain it...

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If M is the set of all consecutive multiples of 9 between 100 and 500,

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