Night reader wrote:If x is equal to the sum of the even integers from 50 to 70, inclusive, and y is the number of odd integers from 50 to 70, inclusive, what is the value of x + y ?

(A) 610
(B) 611
(C) 660
(D) 670
(E) 671

There in all 11 Even integers in between 50 to 70 ,

50 - 1 integer
50 to 60 - ( 52, 54, 56, ... 60 ) 5 integers = 56 * 5 = 280
60 to 70 - ( 62, 64, .. 70 ) 5 integers = 66 * 5 = 330

IN all 280 + 330 + 50 = 660

in all 10 odd integers 660 + 10 = 670 OPTION D

I am sure there must be a shorter way to solve this.

you have almost nailed it!

the shortest way: we are given consecutive even integers from 50 to 70. You correctly mention there are 11 (!) even integers only. You may know the consecutive number properties i.e. median/mid-point = average. Apply here mid-point = 60

60*11=660

find the number of odd integers between 50 and 70 (you correctly found this) i.e. 10

660+10 = 670

A shortcut to counting even/odd/multiples.
Number of even numbers in 50 and 70 can be found out using this-
[(Last - First)/Increment] + 1
Last and first are the starting points i.e. 50 and 70 both are even. Last and First have to be even numbers as we have to count the number of even numbers. Other case, when odd numbers have to be counted , last and first will change, first will be the first odd number (i.e. 51) and last will be the last odd in that range (here 69).
Works for multiples too, ex how many numbers between 50 and 70 are divisible by 3 , in other words multiples of 3.
1) Even numbers. F=50, L=70, Increment =2
[(70-50)/2] + 1 = 11
2) Odd numbers. F=51, L=69, Increment=2
[(69-51)/2] + 1 = 10
3) Multiples of 3, F=51, L=69, Increment=3
[(69-51)/3] + 1 = 7

Thanks everyone.

Its very helpful.