amanlalwani wrote:
Bunuel wrote:
SOLUTION
List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?
(A) 2
(B) 7
(C) 8
(D) 12
(E) 22
For any evenly spaced set median = mean = the average of the first and the last terms.
So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term;
The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;
The difference will be (x + 9) - (x - 3) = 12.
Answer: D.
Hi Bunel,
I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term:
mean = (x - 7) + 2*2 = x - 3;"
Hi,
there are 10 consecutive odd numbers , means each number is 2 more than the previous number...
if the least number here is x, the next number will be x+2, third will be x+2*2...
and so on till 10th term= x+9*2..
also we can find this through arithmetic progression..
Nth term = first term + (N-1)d, d is the constant difference between two consecutive numbers..
2ND part..
"the mean of T will simply be the median or the third term:
mean = (x - 7) + 2*2 = x - 3in the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here..
the least integer in s is 7 less than T, so it will become x-7...
the third term here will be (x-7) + 2*2..
same as nthterm above