japan1453 wrote:Could you explain this. Thanks

D

Whats the QA..

I am getting D

Think first about the different operands...
+ and * are commutative, meaning a+b = b+a and a*b = b*a
- is not, meaning a-b not = b-a

1) k @ 1 is not equal to 1 @ k for all numbers means that @ cannot be + or *, so it must be -. At this point, the equation is irrelevant, you have the information you need to solve it, so you know 1 is sufficient.

2) @ represents subtraction is the same as 1), so 2 is sufficient.

Answer = D.

-Carrie

Choice 1 states that K.1 is not equal to 1.K for SOME numbers of K.

Considering . to be -ve, you have two cases

Case 1: What if K is -ve (example - 2)?

K-1 => -2-1 =>-3
1-K => 1- -2 =>3
-3 not equals 3

Case 2: What if K is +ve(example 2)?
K-1 => 2-1 =>1
1-K => 1- 2 =>-1
-1 not equals 1

In both the cases, it doesn't comply. So I am not sure if case 1 would be sufficient to answer the question.

IMO B

GID09 wrote:Choice 1 states that K.1 is not equal to 1.K for SOME numbers of K.

Considering . to be -ve, you have two cases

Case 1: What if K is -ve (example - 2)?

K-1 => -2-1 =>-3
1-K => 1- -2 =>3
-3 not equals 3

Case 2: What if K is +ve(example 2)?
K-1 => 2-1 =>1
1-K => 1- 2 =>-1
-1 not equals 1

In both the cases, it doesn't comply. So I am not sure if case 1 would be sufficient to answer the question.

IMO B

IMO D

If k # 1 is not equal to 1 # k for some values of k, then # definitely cannot be a sign of addition or multilication. This makes us believe that # represents the sign of subtraction, proving the statement 1 sufficient to answer it. D is the correct answer.