Evenly Spaced Set Average Question

[ellabaro]

Hi, I took a Manhattan CAT exame and came across the following question:
Given that a is the average (arithmetic mean) of the first nine positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b?
Correct Answer: 10:13

I understand how to complete this problem the long way; however, when I tried taking a shortcut that I learned I did not come up with the correct answers. To find the average of the first nine positive multiples of six I subtracted the ninth multiple of 6 (54) from the first multiple of 6 (6) = 48 and divided 48 by 2 - 24. When I took the convential way of finding the average, my answer is 30 and leads to me to the correct answer choice.

What am I doing incorrectly in the shortcut version? I can't figure it out...

[mercytron]

The average of evenly spaced sets is (first number + last number)/2.

In this case for first 9 multiples og 6, average will be (6+54)/2 = 30.

[jnelson0612]

Thanks mercytron!

[ajithalexjacob]

The average is (6+12+18+ .... 54)/9

which is (6*1 + 6*2 + 6*3 .... 6*6)/9

therefore, the average can be written as 6*(1+2+3+4+5+...9)/9

now using sum of consecutive numbers n(n+1)/2

we can write sum of 1 to 9 as (9*10)/2

the average can be rewritten as (6*9*10)/2*9 = 30

similarly the median of the 12 multiples will be the average of the 6th and 7th term, which is 6.5*6 = 39

the ratio will be 30:39 = 10:13

[jlucero]

The average is (6+12+18+ .... 54)/9

which is (6*1 + 6*2 + 6*3 .... 6*6)/9

therefore, the average can be written as 6*(1+2+3+4+5+...9)/9

now using sum of consecutive numbers n(n+1)/2

we can write sum of 1 to 9 as (9*10)/2

the average can be rewritten as (6*9*10)/2*9 = 30

similarly the median of the 12 multiples will be the average of the 6th and 7th term, which is 6.5*6 = 39

the ratio will be 30:39 = 10:13

Your method is correct, but I'd argue that mercytron's method of finding the average (6+54)/2 = 30 is much faster.