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DS Problem...

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DS Problem...

by ramonsa » Thu Apr 16, 2009 3:43 pm
Can someone help me with this problem?

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

(1) k * 1 is not equal to 1 * k for some numbers k.
(2) * represent subtraction.

Thanks
Ramon
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by vittalgmat » Thu Apr 16, 2009 3:59 pm
Let me take a shot at this.


stmt 1: This says the operation * is not commutative.
Addition is commutative but not subtraction.
so * is - operator.

Now lets see if k - (l+m) = (k -l) + (k -m)
Pluging numbers will help us here

let k = 1, l= 2, m=3 (these need not be integers.. but choosing ints is fine here)
LHS
1 - (2 +3) = 1 - 5 = -4

RHS
(1 - 2) + (1 -3) = -1 + -2 = -3

LHS Not = RHS.

now lets try k =l = m=0
LHS = RHS
so 2 diff answers. Not sufficient.


Stmt 2.
* represents subtraction.

Same as stmt 1. Insufficient.

Both stmts together is not sufficient.
so
E

What is the OA ?
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by ramonsa » Thu Apr 16, 2009 4:01 pm
D is the answer on the book.
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by sands_of_time » Thu Apr 16, 2009 8:14 pm
See, here we are talking about associative law, i.e.
a x (b + c) = a x b + a x c

Now, this associative law holds good for addition and multiplication. (You can substitute numbers and check)

Now, we will be able to get an answer if we know that the symbol * means either multiplication/addition OR subtraction.

Now consider the options:
a) 1*k is not equal to k*1

This is true only if * is '-' and for subtraction, the associative law doesn't hold good(again, you can pick numbers to understand this more)

Statement Sufficient.

b) * is '-'

In accordance with the logic above, sufficient.

Hence, the answer D
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by vittalgmat » Thu Apr 16, 2009 9:12 pm
@Sands_of_time
Ur explanation is correct as long as k, l, m, n are all NOT 0.

Try k =l=m=n=0.

So I think if the Q said that the variables are not 0, the answer is D
else it is E.


Feedback appreciated.
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by sands_of_time » Thu Apr 16, 2009 9:54 pm
vittalgmat wrote:
So I think if the Q said that the variables are not 0, the answer is D
else it is E.
quote]

I think you are right here vittalgmat.

Thanks for pointing it out!
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by ramonsa » Fri Apr 17, 2009 3:24 am
Thanks!!!
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by shargaur » Thu May 14, 2009 7:36 am
@vittalgmat

if you consider l,m,k = 0 then second statement will fail...which should nt happen.

1*0=0*1 in any case.
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by PAB2706 » Fri May 15, 2009 6:37 am
hi,

no where in the question they have said that k l and m shud be positive number... if we take some positive and some negetive numbers it is obvious that we will get varying answers.

if that is the case taking the operator as - will straight away give us varying answers always.. so we wont have to plug numbers....pretty obvious..isnt it? or m i missing something.

can this be a faster process...

cheers!!!
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