How do you solve this!

a bit long, but we can solve this algebraically...

T= x, x+2, x+4, x+6, x+8

sum of all elements of T
= x + x+2 + x+4 + x+6 + x+8
= 5x + 2(1+2+3+4)
= 5x + 20
A.M (T) = (5x+20)/5 = x+4



S = x+7,x+9,x+11,x+13 .... , x+25 ( since the least element of S is 7 more than x )

sum of all elements in S
(x+7)+ (x+9) + (x+11) + .... + (x+25)
7*10 + {x + (x+2) + (x+4) + (x+6) .... + (x+18)}
70 + 10x + 2 ( 1+2+3 .... +9)
70 + 10x + 2*45 = 10x + 160

A.M (S) = (10x+160)/10 = x+16


x+16 - (x+4) = 12

Another easy way is as follows

in Arithmetic Progression, the median is always equal to the Mean i.e. Average

Therefore if least of 5 even integer is x then least of 10 odd integers is x+7
Therefore mean of 5 even integer is x+4 (x, x+2, x+4,...) then mean of 10 odd integers is x+7+9 [average of (x+7)+8 and (x+7)+10]

The difference is = (x+16) - (x+4) = 12 ANSWER

AkiB wrote: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

We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.

Since the least integer in S is 7 more than the least integer in T, x + 7 = the least integer in S, and so S has the following integers: x + 7, x + 9, x + 11, x + 13, x + 15, x + 17, x + 19, x + 21, x + 23, and x + 25.

Since each list is an evenly spaced set, the average of each list is the respective median. Since the median of the integers in T is x + 4, and the median of integers in S is [(x +15) + (x + 17)]/2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively.

Therefore, the average of list S is (x + 16) - (x + 4) = 12 more than the average of list T.

Answer: D