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# If @ represents one of the operations +, -, and x, is k@(l+m

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If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  14 Dec 2012, 06:13
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If @ represents one of the operations +, -, and x, 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 numbers k.
(2) @ represents subtraction.
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  14 Dec 2012, 06:27
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If @ represents one of the operations +, -, and x, 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 numbers k. @ is neither addition (as $$k+1=1+k$$) nor multiplication (as $$k*1=1*k$$), thus @ represents subtraction. Knowing that we can determine whether $$k-(l+m)=(k-l)+(k-m)$$ for all numbers k, l,and m. Sufficient.

(2) @ represents subtraction. The same here. Sufficient.

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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  16 Dec 2012, 11:04
Bunuel wrote:
If @ represents one of the operations +, -, and x, 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 numbers k. @ is neither addition (as $$k+1=1+k$$) nor multiplication (as $$k*1=1*k$$), thus @ represents subtraction. Knowing that we can determine whether $$k-(l+m)=(k-l)+(k-m)$$ for all numbers k, l,and m. Sufficient.

(2) @ represents subtraction. The same here. Sufficient.

Dear Bunnel,
I would like to understand the above question first..
If we take the @ as subtraction from statement 1 and 2 then the equation stands as $$k-l-m=2k-l-m$$, which is not equal in both the side.

I was wondering whether the question asks about the operation of the @ sign, which makes the equation of k@(l+m)=(k@l)+(k@m) okay from both end.

Thanks
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  16 Dec 2012, 23:07
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Drik wrote:
Bunuel wrote:
If @ represents one of the operations +, -, and x, 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 numbers k. @ is neither addition (as $$k+1=1+k$$) nor multiplication (as $$k*1=1*k$$), thus @ represents subtraction. Knowing that we can determine whether $$k-(l+m)=(k-l)+(k-m)$$ for all numbers k, l,and m. Sufficient.

(2) @ represents subtraction. The same here. Sufficient.

Dear Bunnel,
I would like to understand the above question first..
If we take the @ as subtraction from statement 1 and 2 then the equation stands as $$k-l-m=2k-l-m$$, which is not equal in both the side.

I was wondering whether the question asks about the operation of the @ sign, which makes the equation of k@(l+m)=(k@l)+(k@m) okay from both end.

Thanks

No, the question asks: "is k@(l+m)=(k@l)+(k@m) for ALL numbers k, l,and m", where @ represents one of the operations +, -, and x.
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  01 Aug 2013, 09:31
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If we take k=l=m=0 & k=1, l=2, m=3, from statement 1 we will get both "yes" or "no". Similarly Statement 2 also gives the same result.

The question doesn't specify anything about k,l,m

Could you explain?
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  09 Aug 2013, 01:35
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Qoofi wrote:

If we take k=l=m=0 & k=1, l=2, m=3, from statement 1 we will get both "yes" or "no". Similarly Statement 2 also gives the same result.

The question doesn't specify anything about k,l,m

Could you explain?

From (1) we got that @ is subtraction. So, the question becomes: is k-(l+m)=(k-l)+(k-m) for ALL NUMBERS k, l,and m? This equation holds if k=0. Therefore the equation does NOT hold true for ALL NUMBERS (it holds if k=0).

The same applies to the second statement.

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Hope it helps.
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  12 Oct 2013, 11:20
The way I approached this question was basically "is @ multiplication"?

That is the only symbol that will make the equation in the question stem equal.

The first statement tells us indeed that @ is not multiplication or even addition. The only other option is subtraction...so we have our answer and it is not multiplication. Sufficient.

The second statement tells us @ is subtraction. Ok so we know it is not multiplication. Sufficient.
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  16 Nov 2013, 20:37
The way i see the question,
k o (l +m) = (k o l) + (k o m) is only true where o is x(multiplication)
for o = + and o = -, it's not true.

1. k o 1 not equal to 1 o k. This statement is true only when o is subtraction (-). But we know that the above statement is valid only for multiplication. So this option is SUFFICIENT.
2. o represents subtraction . This statement is SUFFICIENT , as we know that the question is valid only for multiplication.
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  16 Nov 2013, 22:24
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If @ represents one of the operations +, -, and x, 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 numbers k.
(2) @ represents subtraction.

The answer to this question could be a yes or a no. If we can somehow say for sure - yes or no, then we know the option is sufficient.
2) clearly says @ is subtractn. Therefore, the equation in the question is NOT true for all nos. k,l,m. SUFFICIENT.

1)k@1 != 1@k implies that @ s not x . This could be + since if k is neg, -k+1 is not equal to 1-(-k)
This could be - since k-1 != 1-k.
substituting in the question, for +: is k+(l+m)=(k+l) + (k+m). NO.
for - : is k-(l-m)= (k-l) + (k-m) . NO.

There the equation is NOT true for all nos. SUFFICIENT.

D it is!
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Re: If @ represents one of the operations +, -, and x, is k@(l+m [#permalink]  27 Nov 2014, 00:08
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Re: If represents one of the operations +, -, and x, is k(l+m [#permalink]  13 Jul 2015, 08:07
St1 : @ is substraction

if we take K=5, l=3, m=2
Then 0 is not equal to 5....as per question stem answer is NO.

If we take K=0,l=1,m= 2

Then -3= -3....so answer is YES

Since we get yes and no both....shouldn't this statement be insufficient ?

Where am I doing the mistake ?
Re: If represents one of the operations +, -, and x, is k(l+m   [#permalink] 13 Jul 2015, 08:07
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