OG 13, operator problem

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sohailmbaprep: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Thu Feb 23, 2012 11:18 pm

OG 13, operator problem

by sohailmbaprep » Mon Sep 17, 2012 5:26 am

Q) If @ represents one of the operators +,- and * , is k@(l+m)=(k@l)+(k@m) for all numbers k,l and m ?
1) k@1 is not equal to 1@k for some number k.
2) @ represents substraction.

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Source: Beat The GMAT — Data Sufficiency | Mon Sep 17, 2012 5:26 am

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 am; Location: Utrecht, The Netherlands; Thanked: 318 times; Followed by:52 members

by neelgandham » Mon Sep 17, 2012 6:09 am

Q) If @ represents one of the operators +,- and * , is k@(l+m)=(k@l)+(k@m) for all numbers k,l and m ?

1)k@1 is not equal to 1@k for some number k.

Let @ be *, then k*1 = 1*k. So, @ can't be *.
Let @ be +, then k+1 = 1+k. So, @ can't be +.
Let @ be -, then k-1 != 1-k(not always).
So, @ is '-'. and the value of k-(l+m) is not equal to (k-l)+(k-m), for all k,l,m.
Statement I is sufficient to answer the question.

2)@ represents substraction.

@ is '-'. and the value of k-(l+m) is not equal to (k-l)+(k-m), for all k,l,m.
Statement II is sufficient to answer the question.

IMO D

Anil Gandham
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everything's eventual: Master | Next Rank: 500 Posts; Posts: 101; Joined: Sun Jun 03, 2012 10:10 pm; Thanked: 10 times; Followed by:1 members

by everything's eventual » Mon Sep 17, 2012 5:56 pm

Hello Anil, if k = 1, then k-1 = 1-k so IMO we cannot def conclude that @ = '-'.

IMO answer is [spoiler]B)[/spoiler]

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kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 am; Location: madrid; Thanked: 171 times; Followed by:64 members; GMAT Score:790

by kevincanspain » Tue Sep 18, 2012 4:24 am

Note that (1) tells us that k@1 is not equal to 1@k for some number k. In other words, there is at least one value of k for which k@1 is not equal to 1@k. Since 1-0 is not equal to 0-1, @ being subtraction is consistent with (1). Thus @ could be subtraction.

However, since 1 x k = k x 1 for all values of k, @ being x is not consistent with (1). Thus @ cannot be x (or + for that matter).

Clearly, then, according to (1), @ must be subtraction. It is the only one of the three operations that does not contradict (1).

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