If x is equal to the sum of the integers from 30 to 50, incl

If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of EVEN integers from 30 to 50, inclusive,
what is the value of x+y ?

a) 810
b) 811
c) 830
d)850
e)851

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I started solving like this .....

Sum of integers = (Mean x No of Terms)

So for X - {30 to 50} , No. of terms = 21, Mean = 50+30 / 2 = 40
So Sum of X = No of terms x Mean = 21 x 40 = 840
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Now for Y - {30, 32,34,36,38,40,42,44,46,48,50} Now since these have already been added before.....what am i supposed to do?......i am lost here..

Sum = x + y = ????

The sum of the integers from 30 to 50 is \(x=\frac{first \ term+ last \ term}{2}*# \ of \ terms=\frac{30+50}{2}*21=40*21=840\);

The number of even integers from 30 to 50 is \(y=\frac{50-30}{2}+1=11\) (check this: totally-basic-94862.html);

x+y=840+11=851.

Answer: E.

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Hi Bunuel how do you calculate no. of terms(i.e 21). By counting or is there any method. Counting no. of terms in given range or no. of hours have always been a problem for me. i always count which consumes a lot of time.

I started solving like this .....

Sum of integers = (Mean x No of Terms)

So for X - {30 to 50} , No. of terms = 21, Mean = 50+30 / 2 = 40
So Sum of X = No of terms x Mean = 21 x 40 = 840
----------------------------------------------------------------------------------------------------------------
Now for Y - {30, 32,34,36,38,40,42,44,46,48,50} Now since these have already been added before.....what am i supposed to do?......i am lost here..

Sum = x + y = ????