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If # represents one of the operations +,- and *, is a # [#permalink]
19 Sep 2010, 15:49
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If # represents one of the operations +,- and *, is a # (b-c) = (a#b) – (a#c) for all numbers a, b and c. (1) a#1 is not equal to 1#a for some numbers a (2) # represents subtraction Mods, please to DS section...posted by mistake in PS - apologies
Last edited by zisis on 19 Sep 2010, 16:09, edited 1 time in total.
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Re: operations (?) [#permalink]
19 Sep 2010, 15:58
I converted question to 3 euqations where # can be +, - or * So I) When # = * ab-ac = ab-ac (true for all values) II) When # = + Is a+b-c=b-c III) When #=- Is a-b+c = -b-c? With option A we know # is not * as 1*a is always equal to a*1. So given equation is not equal With option B , equation is not equal. So answer choice D
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Re: operations (?) [#permalink]
19 Sep 2010, 16:50
zisis wrote: If # represents one of the operations +,- and *, is a # (b-c) = (a#b) – (a#c) for all numbers a, b and c.
(1) a#1 is not equal to 1#a for some numbers a
(2) # represents subtraction
Mods, please to DS section...posted by mistake in PS - apologies 1. From choice 1 it is clear that # is subtraction. coz a+1=1+1, a*1=1*a but a-1!=(not equal) 1-a. 2. Choice 2 says directly that it is subtraction. Hence answer is D
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Re: operations (?) [#permalink]
19 Sep 2010, 19:38
vigneshpandi wrote: zisis wrote: If # represents one of the operations +,- and *, is a # (b-c) = (a#b) – (a#c) for all numbers a, b and c.
(1) a#1 is not equal to 1#a for some numbers a
(2) # represents subtraction
Mods, please to DS section...posted by mistake in PS - apologies 1. From choice 1 it is clear that # is subtraction. coz a+1=1+1, a*1=1*a but a-1!=(not equal) 1-a. 2. Choice 2 says directly that it is subtraction. Hence answer is D IMO, it cannot be "not equal" always. If # is subtraction, a#(b-c)=a-(b-c)=a-b+c (a#b)-(a#c)=(a-b)-(a-c)=a-b-a+c=-b+c a-b+c=-b+c when a=0 and not equal for other values.. So both are insufficient.. Am I missing something here?
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Re: operations (?) [#permalink]
20 Sep 2010, 00:14
BalakumaranP wrote: IMO, it cannot be "not equal" always.
If # is subtraction,
a#(b-c)=a-(b-c)=a-b+c
(a#b)-(a#c)=(a-b)-(a-c)=a-b-a+c=-b+c
a-b+c=-b+c when a=0 and not equal for other values.. So both are insufficient.. Am I missing something here? If # represents one of the operations +, - and *, is a#(b-c)=(a#b)-(a#c) for all numbers a, b and c. (1) a#1 is not equal to 1#a for some numbers a. # is neither addition (as a+1=1+a) not multiplication (as a*1=1*a), so # is a subtraction. Then LHS=a#(b-c)=a-b+c and RHS=(a#b)-(a#c)=(a-b)-(a-c)=c-b, so the question becomes "is a-b+c=c-b for all numbers a, b and c?" --> "is a=0". So when a=0 (and # is a subtraction) then a#(b-c)=(a#b)-(a#c) holds true but not for other values of a, so not for all numbers a, b and c. Answer to the question is NO. Sufficient. (2) # represents subtraction --> the same as above. Sufficient. Answer: D. Hope it's clear.
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Re: operations (?) [#permalink]
20 Sep 2010, 03:09
Okay.. I read the question wrong...
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Re: operations (?) [#permalink]
20 Sep 2010, 19:10
(1) a#1 is not equal to 1#a for some numbers a
a + 1 = 1 + a for all a a*1 = 1*a for all a
a - 1 != 1 - a
Statement (1) implies # is a minus sign, so it has same meaning as (2).
Ans D
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Re: operations (?) [#permalink]
21 Sep 2010, 10:55
zisis wrote: If # represents one of the operations +,- and *, is a # (b-c) = (a#b) – (a#c) for all numbers a, b and c.
(1) a#1 is not equal to 1#a for some numbers a
(2) # represents subtraction
Mods, please to DS section...posted by mistake in PS - apologies You're told that "#" is either addition, subtraction, or multiplication, and then asked if "#" satisfies the distributive property. Of these three, distribution only holds for multiplication, so if "#" is "*", it holds, and if "#" isn't "*", then it does not hold. All we really need to know is what operation "#" really is. (1) This is only true of subtraction, so # is subtraction and the distributive property does not hold. Sufficient. (2) Same as above. Sufficient. (D)
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Re: operations (?)
[#permalink]
21 Sep 2010, 10:55
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