sum of the even integers from 40 to 60 inclusive

IMO D

x=sum
y=no of integers=11

x=[(40+60)/2]*11
x=50*11=550

thus x+y=550+11=561

it's 561... 50*11+11

uptowngirl92 wrote:if x is equal to the sum of the even integers from 40 to 60 inclusive and y is the number of even integers from 40 to 60 inclusive,what is the value of x+y?
550
551
560
561
572

Guys I applied the formula for "sum of consecutive evn nos." but i am going wrong somewhere.Pease help.
y=11
x=sum of consecutive even integers=n(n+1)
where n= 1st even+last even/2 -1

Therefore,here n=40+60/2-1=50-1=49
So,x=49 x 50 =2450
Hence, x+y=2450+11=2461??!??!

For consecutive integer problems or any evenly spaced problems use this formula:

Number of terms = {[last term - first term] / increment } + 1

Here, increment = 2 since it is mentioned they are even consecutive numbers. Last term = 60 first term = 40

so Number of terms = {[60 - 40] / 2} + 1 = 11

Sum = Average * number of terms

Average of the even integers from 40 to 60 is simply average of the first and the last terms which gives us [40+60]/2 = 50

Sum = 50 * 11 = 550 = x

Number of terms = y = 11

x + y = 550 + 11 = 561

thanks guys i got the answer but I still dont know where im going wrong!!am i following the formulae wrong??please refer to link below from where i got the formula:

https://www.beatthegmat.com/formula-for- ... 17241.html