Normally, I'm not arrogant enough to think that a book is wrong, but I think your source is wrong. The way to set this probelm up is

[N(First term + Last term)] / 2

where N equal number of terms, in this case 11 (there are 11 even numbers from 39- 41).

First term is 40

Last Term is 60

The answer for x and y would be 550. If you were to add x and y then the answer would be 1100. If you doubt the equation, all you have to do is get a calculator and add 40 +42 + 44 etc. and you will see that the sum is 550.

msch14 wrote:X is the sum of even integers from 40 to 60 inclusive and y is the number of even integers from 40 to 60 inclusive, what is x+y?

A. 550
B. 551
C. 560
D. 561
E. 572

OA is D. How do you set up the problem?

Q: (Sum of even numbers from 40 to 60) + (# of even terms from 40 to 60) = ?

Two different things to calculate, one is much easier than the other. When in doubt, start with the easy part of the question.

The number of even integers from 40 to 60 inclusive is 11 ((difference/2) + 1 = 10 + 1 = 11... or you can just brute force count them out).

Q: (sum of even numbers from 40 to 60) + 11 = ?

First thing to note: the sum of even numbers will always be even. An even number + 11 will be odd: eliminate (a), (c) and (e).

Next, we can calculate the sum of the set.

Sum of a set of consecutive numbers = Average of the set * # of terms.

To find the average of a set of consecutive numbers, we can simply calculate (first term + last term)/2.

Here, we get (40+60)/2 = 100/2 = 50

Sum of the set = 50 * 11 = 550

Final answer: 550 + 11 = 561... choose (D).

Thank you Stuart!