Commutator

Expert replies

Sher1: Master | Next Rank: 500 Posts; Posts: 116; Joined: Fri Feb 20, 2009 9:26 am; Location: New Jersey; Thanked: 7 times; GMAT Score:660

by Sher1 » Thu Aug 06, 2009 9:20 pm

For any operation ? that acts on two numbers x and y, the commutator is defined as x?y – y?x. For which of the following operations is the commutator not equal to zero for some values of x and y?

I. x?y = xy
II. x?y = (x – y)^2
III. x?y = x^3 + 3x^2y + 3xy^2 + y^3

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Like this post

Source: Beat The GMAT — Problem Solving | Thu Aug 06, 2009 9:20 pm

real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Thu Aug 06, 2009 9:30 pm

I don't find the answer among the choices available

Like this post

aspiregmat: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sun Nov 11, 2007 9:58 pm; Thanked: 1 times

by aspiregmat » Thu Aug 06, 2009 9:31 pm

I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.

Like this post

real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Thu Aug 06, 2009 9:35 pm

aspiregmat wrote:I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.

can u pl. explain how do u get the highlighted term? I mean why it is (y-x)^3 and not (y+x)^3

Like this post

aspiregmat: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sun Nov 11, 2007 9:58 pm; Thanked: 1 times

by aspiregmat » Thu Aug 06, 2009 9:41 pm

real2008 wrote:
aspiregmat wrote:I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.
can u pl. explain how do u get the highlighted term? I mean why it is (y-x)^3 and not (y+x)^3

My mistake ...

it should be (x+y)^3 - (y+x)^3 so Ans IMO none of the above

, which is not in the ans choice....

Like this post

Sher1: Master | Next Rank: 500 Posts; Posts: 116; Joined: Fri Feb 20, 2009 9:26 am; Location: New Jersey; Thanked: 7 times; GMAT Score:660

by Sher1 » Fri Aug 07, 2009 11:50 am

I also thought the answers should be none. This is a Manhattan Gmat challenge question I found on MBA mission. Below is the exp if anyone can make sense of it please share.

https://www.mbamission.com/blog/2009/08/ ... llenge-63/

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review