Problem Solving

Can something help me solve this question:


What is the arithmatic mean of multiples of 6 that are greater than 0 and less than 1000?

a 499
b 500
c 501
d 502
e 503

The mean of all the multiples of 6 between 0 and 1000 can be easily found by simply knowing the first and last multiple of 6. In this case the 1st multiple is 6 and the last is 996. The mean of those two numbers is the same as the mean of all multiples of 6 between 0-1000. Therefore the answer is (996+6)/2 = 501

another way to solve it is to look at the pattern:

(6+12)/2=9
(6+12+18 )/3=12
(6+12+18+24)/4=15

You can see that with every multiple, the average increases by 3. There are 166 multiples of 6 from 0-1000. Subtracting the first multiple, which is 6 and you get 165. 165*3 = 495 adding the first multiple --- 495 + 6 = 501

I will go with C

Think this way, may be it will help

1st number in the set = 6 = 6 * 1
next number in the set = 12 = 6 * 2
Last number in the set = 996 = 6 * 166

so the sum can be written as 6 (1+2+3+4+5+6+.... +166)

sum of n consecutive numbers = n(n+1)/2

so sum here = (6x166x167)/2 and average = (6x166x167)/(2x166)

= 3x167 = 501