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zisis
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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zisis
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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Okay.. I read the question wrong...
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(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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zisis
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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zisis
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

Could someone edit this question so it will be an actual question? There is no proper question stem.
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reto
zisis
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

Could someone edit this question so it will be an actual question? There is no proper question stem.

What do you mean? Everything is correct there. Similar questions: Operations/functions defining algebraic/arithmetic expressions

Check functions related questions in our Special Questions Directory.

Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions
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reto
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Bunuel
reto
zisis
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

Could someone edit this question so it will be an actual question? There is no proper question stem.

What do you mean? Everything is correct there. Similar questions: Operations/functions defining algebraic/arithmetic expressions

Check functions related questions in our Special Questions Directory.

Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions


The question mark is missing in the question stem.
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Bunuel
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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.

I don't understand why the answer is not E. if a = 0, then yes, if a is not equal to 0, then no. So not sufficient, as nothing can be said conclusively ...
anybody out there to help me ....?

thanks
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Bunuel
BalakumaranP
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.

I don't understand why the answer is not E. if a = 0, then yes, if a is not equal to 0, then no. So not sufficient, as nothing can be said conclusively ...
anybody out there to help me ....?

thanks

The question asks: is \(a#(b-c)=(a#b)-(a#c)\) for all numbers \(a\), \(b\) and \(c\)? From each statement we got a deinite NO answer - NO the equation doe NOT hold true for all numbers, it's true if a = 0 but not true if a is not 0.
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Bunuel
BalakumaranP
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.

I don't understand why the answer is not E. if a = 0, then yes, if a is not equal to 0, then no. So not sufficient, as nothing can be said conclusively ...
anybody out there to help me ....?

thanks

Thanks Bunuel, you are soooo... great .... :clap:
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Key word in A --> 'not equal to' -- thus the only time a#1 not 1#a is when # = minus
Really need to read questions properly!
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Key word in A --> 'not equal to' -- thus the only time a#1 not 1#a is when # = minus
Really need to read questions properly!
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Bunuel GMATPrepNow VeritasKarishma EMPOWERgmatRichC

If a = zero, LHS = RHS , and ans to main Q stem is YES
If a\(\neq{0}\) zero, LHS\(\neq{0}\) RHS, ans to main q stem is NO.
How is St 1 suff?
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Bunuel GMATPrepNow VeritasKarishma EMPOWERgmatRichC

If a = zero, LHS = RHS , and ans to main Q stem is YES
If a\(\neq{0}\) zero, LHS\(\neq{0}\) RHS, ans to main q stem is NO.
How is St 1 suff?

Answered here: https://gmatclub.com/forum/if-represent ... l#p1909860

From (1) we get that the equation is NOT true for ALL numbers. It's true only if a = 0 (so not for ALL numbers).
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zisis
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

Given: # represents one of the operations +,- and *

Asked: 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
a+1 = 1+a
a*1 = 1*a
\(a-1 \neq 1-a\)
# = -
a#(b-c) = a-b+c
(a#b) – (a#c) = (a-b) - (a-c) = c -b
a # (b-c)\(\neq\) (a#b) – (a#c)
SUFFICIENT

(2) # represents subtraction
# = -
a#(b-c) = a-b+c
(a#b) – (a#c) = (a-b) - (a-c) = c -b
a # (b-c)\(\neq\) (a#b) – (a#c)
SUFFICIENT

IMO D
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Bunuel GMATPrepNow VeritasKarishma EMPOWERgmatRichC

If a = zero, LHS = RHS , and ans to main Q stem is YES
If a\(\neq{0}\) zero, LHS\(\neq{0}\) RHS, ans to main q stem is NO.
How is St 1 suff?

Hi adkikani,

This question actually has a built-in 'design flaw' which makes the information in the two Facts irrelevant. The way that the question is phrased, there are only 3 possibilities and they ALL lead to a "Sufficient" answer. There are a lot of details in the wording that make the prompt 'complicated looking' - so instead, here's a much simpler question that is based on the same design flaw:

"Is this lightbulb always on?"

IF... the lightbulb is ALWAYS on... then we have a definitive/Sufficient answer to the question that is asked (Sufficient YES).
IF... the lightbulb is NEVER on... then we also have a definitive/Sufficient answer to the question that is asked (Sufficient NO).
IF... the lightbulb is SOMETIMES on... then we still have a definitive/Sufficient answer to the question that is asked (Sufficient NO).

From the date on this post, we can clearly see that this is an old question (and it's likely that whoever wrote it didn't even recognize the flaw in it). While you'll likely see at least one Symbolism question on Test Day - and you'll certainly have to work through lots of little Arithmetic steps during the Exam - it's incredibly unlikely that you will face any flawed questions on the Official GMAT. This is all meant to say that you should ignore this question.

GMAT assassins aren't born, they're made,
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