If x is equal to the sum of the even integers from 40 to 60 inclusive,

This is solved by using AP series.

40,42,44,46..........................60
This is a AP series with common difference of 2

Number of terms = A + (n-1)d
A = first term (40)
n = number of terms (need to be calculated)
d = common difference (2 in this case)

60 = 40 + (n-1)2
or n = 11

Sum of series = [2A + (n-1)d ] * n/2
Sum = 550

So ans = 550 + 11 = 561

Try using this formula instead:

quick-tips-for-adding-numbers-x-to-y-79541.html#p598233

For me,

step1: find y 40, 42, 44, ... 60 therefore y=11

step2: since consecutive numbers/even/odd, find the mean (40+60)/2=50

step3: x = 50(11)=550

step4: 550+11=561

sum of even integers = even number ( x is even)

number of even integers =11 ( y is odd )
so x+y = odd

A, C, E out ( all even)

Left with B and D :551, 561

If you have ever added even numbers you see that the pattern is 0,2,4,6,8 and 2+4+6+8 =20

{ there are 11 integers 5 in the 40's , 5 in the 50's and 60 , so when u add the u get 200+250+20+20+60 = 550}

hence the sum is 550 ( x=550) or x+y cannot be 551 since x =550

hence answer is D

IMO D



Solution:

The number of even integers from 40 to 60 inclusive = 11 (40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60)
Sum of integers = 550

Thus, total = 550 + 11 = 561

Sum can be calculated using Arithmetic Progression

\(Sum = (n/2)(a+(n-1)*d)\)

In this case a(first term) = 40, d(difference) = 2(since nos are even)

\(n = ((60-40)/2)+1\) = 11

Thus sum = 550 (substituting the values)

and the number of terms have already been calculated as 11

Thus x + y = 550+11 = 561

Hope it helps,
meshtrap

Another approach ('sum of pairs'):
Step 1: 11 numbers
Step 2: 40+60 = 42+58 = 100 (total 5 pairs, with exception of number 55 that does not have a pair)
Step 3: 500 + 55 (the middle number with no pair)+ 11 = 561

Advantages: you don't need to know formulas nor you can make mistake in formulas

x = (60 + 40)/2 * y

60 = 40 + (y-1)*2

=> y = 20/2 + 1 = 11

so 50 * 11 + 11 = 561

Answer - D

I used to take a long time solving these kinds of problem until I learned about this formula:

Average = Sum of integers / number of terms
Average = (first term + last term) / 2 ==> this works for both consecutive and even integers

SOLUTION:

x (SUM) = Average x number of terms

Average = 40 +60 /2 = 50
number of terms = ((60 -40)/2)+1 = 11
x (SUM) = 50 x 11 = 550

Therefore,
x + y = 550 + 11 = 561

x = 40+42+....60 = (mean).N = (mean)y

=> x+y = = (51).11 = 561

Answer D.

This is a perfect example of why you should not use formulas without understanding them properly. If you understand them, you will not make a mistake and will save time.
The formula quoted by the original poster: n(n+1) is absolutely fine. But one needs to understand that n is the number of even terms starting from the first even term. (I discuss why this is so here:
sum-of-even-numbers-68732.html#p849905)

Sum of even numbers from 40 to 60 using this formula will be:
30*31 - 19*20 = 10(3*31 - 19*2) = 550
Since number of terms is 11, required sum is 561

But, I would not use this formula for this question and would do it the way many of you have done:
Average = 50 (it is the middle number), Number of terms = 11 (No formula again. Any 10 consecutive integers have 5 even and 5 odd numbers. 41 to 60 will have 10 even integers and 40 is the 11th one)
Sum = 50*11 + 11 = 561

I agree with the answer responses above. I'd avoid fancy formulas and sequences if you're not familiar with them. Just step back and ask yourself " what is the total ("the sum"). Total is your average times your count. In this case, list out all the even numbers. Average is 50. There's 11 even integers (your count). 50 X 11 = 550. Add the 11. Boom. 561. I like this way too; list it out and split out the the numbers and do the math. Example: 40 + 0, 40 + 2, 40 + 4...and so forth. Count the number of 40's, which is 5, so 40 x 5 = 200, plus 2 + 4 + 6 + 8 = 20, totals 220. Do the same for the 50s. Remember to add the y. 561 is your total. Forced method is time consuming and causes errors.

Similar questions to practice:
if-x-is-equal-to-the-sum-of-the-integers-from-30-to-127276.html
if-m-equals-the-sum-of-the-even-integers-from-2-to-128426.html

Using the properties of an evenly spaced set=>
Here y=60-40/2+1=11
x=11/2[100]=50*11

x+y=11(50+1)=11*51 = 561


Hence D

Even integers from 40 to 60 inclusive = { 40 , 42 , 44 , 46 , 48 ........56 , 58 , 60 }

Sum will be 40 + 42 + 46 + .....56 + 58 + 60 = 2 ( 20 + 21 + 22..... 30 )

So, Sum will be 40 + 42 + 46 + .....56 + 58 + 60 = 2 *275 = 550


Number of even integers from 40 to 60 inclusive is ( 30 - 20 ) + 1 = 11

So, Value of x + y = 561

Answer will hence be (D) 561

x = sum of even integers from 40 to 60 inclusive
y = no. of even integers from 40 to 60 inclusive
x+y = ?

x = 40+42+44+46+.......+60

We need to find the no. of terms between 40 and 60 (both inclusive) which are even in nature.
As per arithmetic progression, the \(n^{th}\) term of a sequence whose first term is 'a', common difference between 2 consecutive terms is 'd' and the no. of terms in sequence is 'n' is given by:

\(n^{th}\) term = a+(n-1)d
in the context of this question, we have:

a = 40
n = need to determine
d = 2
\(n^{th}\) term = 60

60 = 40+(n-1)2

n = 11 = y

now to calculate x, we need to apply the formula for sum of the terms of sequence which is in arithmetic progression.
The formula is given by:
Sum = \(\frac{n}{2}\)(a+last term)

Sum = \(\frac{11}{2}\)(40+60)
x = 550

x+y = 550+11 = 561

y= (60-40)/2 + 1= 11
x= 40+42+....+58+60= 11/2 * (100)
x+y = 11 (50 + 1) = 561