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OG 2016 #87 PS

Expert replies
by Azizakaria » Fri Oct 09, 2015 9:41 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

87. The sum of all the integers k such that −26 < k < 24 is
(A) 0
(B) −2
(C) −25
(D) −49
(E) −51

i don't understand why the answer is D
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Source: — Problem Solving |

by [email protected] » Fri Oct 09, 2015 9:47 am
Hi Azizakaria,

We're limited to the INTEGERS that fall in the range -26 < K < 24....so that would be....

-25, -24, -23.....0.....22, 23

Each of the positive integers form 1 to 23 (inclusive) would be 'cancelled out' by the negative integers from -1 to -23 (inclusive). The sum of ALL of those terms would be 0. But then you have to consider the other remaining integers: -24 and -25. Adding those two numbers gets us -49.

Final Answer: D

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by Brent@GMATPrepNow » Fri Oct 09, 2015 9:51 am
Azizakaria wrote:87. The sum of all the integers k such that −26 < k < 24 is
(A) 0
(B) −2
(C) −25
(D) −49
(E) −51

i don't understand why the answer is D
(-25) + (-24) + (-23) + (-22) + (-21) + . . . . .+ 21 + 22 + 23

Notice that the blue values all cancel out, because for each positive value, there's a negative value with the same magnitude (e.g., -23 + 23 = 0, and -22 + 22 = 0, etc)

So, we're left with (-25) + (-24), which equals -49
Brent Hanneson - Creator of GMATPrepNow.com
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by Azizakaria » Sat Oct 10, 2015 2:44 am
now I understand, thanks for your help, the explanation in the OG is so unclear.
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