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If x is the sum of the first 250 positive even integers and y is the sum of first 250 positive odd integers, which of the following is equal to x-y?
(A) 25
(B) 50
(C) 200
(D) 250
(E) 500
(A) 25
(B) 50
(C) 200
(D) 250
(E) 500
28 Oct 2023, 22:14
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DanTheGMATMan
If x is the sum of the first 250 positive even integers and y is the sum of first 250 positive odd integers, which of the following is equal to x-y?
(A) 25
(B) 50
(C) 200
(D) 250
(E) 500
(A) 25
(B) 50
(C) 200
(D) 250
(E) 500
The sum of first n even integers is given by = x = \(n(n+1)\)
The sum of first n odd integers is given by = y = \(n^2\)
\( x - y = n^2 + n - n^2 = n\)
In this question, n = 250
Option D
23 Nov 2024, 13:09
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ignore the formulas here.
just find the pattern.
since there are 250 of each even and odd, our range is 1-500
list x: 2,4,6,8,10
list y: 1,3,5,7,9
sum x =30
sum y =25
delta = 5
how many differences of 5 are there? well we found that there is delta for every group of 10. we know that we go up to 500, so there must be 50 groups of 10, split in half. or just, 25, since there were 250 of each even and odd.
5*25=250
(to see how this shortcut works, you could also do (30*50)-(25*50)=250. Where 50 represents the
just find the pattern.
since there are 250 of each even and odd, our range is 1-500
list x: 2,4,6,8,10
list y: 1,3,5,7,9
sum x =30
sum y =25
delta = 5
how many differences of 5 are there? well we found that there is delta for every group of 10. we know that we go up to 500, so there must be 50 groups of 10, split in half. or just, 25, since there were 250 of each even and odd.
5*25=250
(to see how this shortcut works, you could also do (30*50)-(25*50)=250. Where 50 represents the
