ravitejapandiri wrote:

If the operation # is one of the four arithmetic operations addition, subtraction, multiplication and division, is (6#2)#4 = 6#(2#4)

(1) 3#2 > 3

(2) 3#1 = 3

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We basically need to answer either a YES or a NO for the equation (6#2)#4 = 6#(2#4)

And if we see the equation, then we know for sure that we can answer the question if # is :-

+ or * =>YES, Equation will hold => Sufficient

- or / => NO Equation will not hold => Sufficient

Now, looking at stetements:

(1) 3#2 > 3

# can either be + or * to hold true, which is one of our conditions as listed above, hence Sufficient

(2) 3#1 = 3

# can be either * or /, hence in this case we cannot be sure as it consists of one possibility from each condition that we had laid out earlier, and thus will result in a YES for * and a NO for /.

Thus (2) is not sufficient as we cannot come up with a definite answer.

Hence, A

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