ganesh prasath wrote:If the operation @ is one of + - * or /, is (6 @ 2) @ 4 = 6 @ (2 @ 4)
a. 3 @ 2 > 3
b. 3 @ 1 = 3


ans is A !...can anyone pls explain...Thanks..

1) 3@2>3;
now here two @ can have two possible values either * or +;

lets put * in place of @ in equation (6 @ 2) @ 4 = 6 @ (2 @ 4);
we have; (6 * 2) * 4 = 48;
6 * (2 * 4)=48;

now put + in place of @ in equation (6 @ 2) @ 4 = 6 @ (2 @ 4);
we have (6 + 2) + 4 = 12;
6 + (2 + 4)=12;

since equation (6 @ 2) @ 4 = 6 @ (2 @ 4) holds true for both +, and * hence A alone is sufficient to answer the question.

2)3 @ 1 = 3; now here two possible values of @ is * and /;

lets put * in place of @ in the equation (6 @ 2) @ 4 = 6 @ (2 @ 4);
(6 * 2) * 4 = 48;
6 * (2 * 4)=48;
now put / in place of @ in equation (6 @ 2) @ 4 = 6 @ (2 @ 4);
(6 / 2) / 4 =3/4;
6 / (2 / 4) =12;

as the equation (6 @ 2) @ 4 = 6 @ (2 @ 4) holds true for @=* and it doesn't hold true for @=/; therefore 2 alone is not sufficient to answer the question;

hence answer should be A

Hey there,

So we've got four possibilities for what @ represents: addition, subtraction, multiplication, and division.

First we see which ones of these possibilities remain viable given the info in statement (1): 3 @ 2 > 3. Well, @ could be addition, because indeed, 3 + 2 > 3, and it could be multiplication, because indeed 3 * 2 > 3. It could NOT, however, represent subtraction or division, because neither 3 - 2 nor 3 / 2 is greater than 3. So we've got some ambiguity as to what @ actually represents, but it turns out that regardless of whether it is * or +,

(6 @ 2) @ 4 = 6 @ (2 @ 4)

We can say that these two quantities will be equal even without bothering to do the computation, because we know that addition and multiplication are both associative. [If we want to prove this to ourselves with concrete numbers, though... if @ represents addition, we wind up with 8 + 4 equaling 6 + 6 (12 equaling 12); if @ represents multiplication, we wind up with 12 * 4 equaling 6 * 8 (48 equaling 48).] So from statement (1) we can answer the question with a definitive YES and say SUFFICIENT.

Now we look at Statement (2). In order for statement (2) to be true, @ could represent multiplication or division, since 3 * 1 and 3 / 1 are both 3; @ canNOT, however, equal addition or subtraction, since neither 3 + 1 nor 3 - 1 equals 3. If @ represents multiplication, the answer to the question is YES, as shown above. But if @ represents division, the answer to the question will be NO. We can anticipate this "NO" based on the fact that division is NOT associative, but we can also derive it: (6/2)/4 = 3/4, whereas 6/(2/4) = 6/(1/2) = 12. 3/4 does not equal 12, so (as anticipated) NO. Since statement (2) leaves us unsure whether to say YES or NO, it is INSUFFICIENT.

The takeaway here (with regard to statement 1) is this: the fact that there's some ambiguity *on the way* to the answer (that is, we can't determine whether @ represents * or +) does NOT necessarily mean we won't land at a definitive answer. Quite often all of several possible paths converge to one answer, and that's what we need to care about.

Hope that helps!

ganesh prasath wrote:If the operation @ is one of + - * or /, is (6 @ 2) @ 4 = 6 @ (2 @ 4)
a. 3 @ 2 > 3
b. 3 @ 1 = 3


ans is A !...can anyone pls explain...Thanks..

This question is a pain since it is basically asking you to eliminate possibilities. I am going to change DS options A and B respectively since they are traditionally called '1' and '2' in the exam.

1. 3 @ 2 > 3 can be 3+1>3 or 3*2>3
2. 3 @ 1 = 3 can be 3*1=3 or 3/1=3

With this info alone, we know nothing. Now we need to eliminate possibilities. I will call 3+1>3 '1a', 3*2>3 '1b' and so on.

(6 @ 2) @ 4 = 6 @ (2 @ 4)
1a: (6+2)+4= 6+(2+4) => 12 = 12 (Works)
1b: (6*2)*4 = 6*(6*4) = 48=96 (Does not work)

For the sake of brevity, you can find that the multiplication and division options from DS item #2 do not work. Since adding from DS item #1 is the only operation that works, the answer is A.