How to quickly find sum of integers?

LulaBrazilia 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 ?

Define the sets -

LulaBrazilia wrote:If x is equal to the sum of the even integers from 40 to 60

X = { 40 , 42 ,44 , 46, 48, 50 , 52 , 54 , 56 , 58 , 60 }

X = 550

LulaBrazilia wrote: y is the number of even integers from 40 to 60

Y = 11 ( Count the elements in set X , it comes as 11 }

LulaBrazilia wrote:what is the value of x+y ?

So x + y = 561.

So OA is [spoiler](D)[/spoiler]

A string of numbers with constant intervals (eg. evens, or multiples of 7) is an arithmetic sequence. The sum of such a sequence is easily found by multiplying its average (the median) to the number of values. The answer is D. I go through the question in detail in the full solution below (taken from the GMATFix App).



-Patrick

For Sum use AP sum formula:

11/2 ( 2*40 + 10*2) = 11 * 50 = 550

Number of terms = 11

So, 550 + 11 = 561

[spoiler]{D}[/spoiler]

Total time = 44 seconds

LulaBrazilia 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 ?

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

Here's an approach that doesn't involve formulas.

{40, 42...58, 60}

x = SUM:
Add up PAIRS OF INTEGERS, working from the ends to the middle:
40+60 = 100
42+58 = 100
44+56 = 100
46+54 = 100
48+52 = 100
50.
x = 5*100 + 50 = 550.

y = number:
The sum above was yielded by adding 5 pairs of integers plus one more, for a total of 11 integers.
Thus, y=11.

Result:
x+y = 550+11 = 561.

The correct answer is D.

For an alternate approach -- a bit more formula-based -- check my post here:
https://www.beatthegmat.com/for-any-posi ... tml#361604

Maybe an easier way to think about this is by working with the concept of an evenly spaced set.

If you have an evenly spaced set, such as

{10, 12, 14, 16, 18, 20, 22}

You can see that each pair of terms equidistant from the median adds up to DOUBLE the median. For instance, our median is 16. 14 and 18 are one term away from the median, and 14 + 18 = 2 * 16. By the same token

12 + 20 = 2 * 16
10 + 22 = 2 * 16

etc.

So for summation purposes, it's almost like we're adding 16 + 16 + 16 + 16 + 16 + 16 + 16. Hence all you have to do is find the median (easy to do in an evenly spaced set: it's just the average of the least term and the greatest term in the set) and then multiply it by the number of terms in your set.

This ALWAYS works in evenly spaced sets -- the sets don't have to be integers -- and it comes in handy quite often on the GMAT.