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22 Nov 2021, 06:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
81% (01:41) correct
19%
(02:26)
wrong
based on 85
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If x is equal to the sum of the even integers from m to n, inclusive, where m and n are positive even integers, which of the following represents the value of x in terms of m and n?
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2 - m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2 - m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
MK1204
Joined: 26 Jan 2020
Last visit: 17 Dec 2022
Posts: 9
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Location: India
Schools: Booth '24 Darden '24 Goizueta '24 Fuqua '24 Haas '24 ISB '23 Kelley '24 Kellogg '24 Wharton '24
GMAT 1: 700 Q49 V38
Schools: Booth '24 Darden '24 Goizueta '24 Fuqua '24 Haas '24 ISB '23 Kelley '24 Kellogg '24 Wharton '24
GMAT 1: 700 Q49 V38
Posts: 9
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22 Nov 2021, 07:04
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Let's assume m=2 n=8
2 +4+6+8 =20
We can get the ans by substituting the values
From option a we get (8+2/2) (8-2/2+1)
10/2* 3+1
5*4 = 20
IMO As we get the desired sol. from a we need not go check other options
Posted from my mobile device
2 +4+6+8 =20
We can get the ans by substituting the values
From option a we get (8+2/2) (8-2/2+1)
10/2* 3+1
5*4 = 20
IMO As we get the desired sol. from a we need not go check other options
Posted from my mobile device
22 Nov 2021, 07:19
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Bunuel
For this question, it's pretty easy to test values.
Let's say m = 2 and n = 6
So, x = 2 + 4 + 6 = 12
So, we'll plug m = 2 and n = 6 to be chances choice to see which one yields a value of 12.
A. \((\frac{2+6}{2})(\frac{6-2}{2} + 1)=(4)(3)=12\). PERFECT!
B. \(3(2+6)=24\). No good. We need an output of 12. Eliminate.
C. \(\frac{6^2 - 2^2}{2}=16\). No good. We need an output of 12. Eliminate.
D. \(6(2+6)=48\). No good. We need an output of 12. Eliminate.
E. \((\frac{2+6}{2})(\frac{6-2}{2})=(4)(2)=8\). No good. We need an output of 12. Eliminate.
