If the operation ∆ is one of the four arithmetic operations addition

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. # can be either multiplication or addition. In BOTH cases (6#2)#4 = 6#(2#4) is true: (6*2)*4=6*(2*4)=48 and (6+2)+4=6+(2+4)=12. Sufficient.

(2) 3#1 = 3. # can be either multiplication or division. If it's division the the answer to the question is No and if it's multiplication answer to the question is YES. Two different answers. Not sufficient.

Answer: A.
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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

Oh!! Anyways thanks Bunuel..I just saw two operations and so felt like i cant solve !!Now I gt it..

Same explanation like Bunnel. Dnt worry I used to make more silly mistakes than what you have done now...And it is all these silly mistakes that afect our score in a huge way....So just relax and solve the problems.....
I am understanding that Panic=failure during exams

You're given that the operation satisfies the associate property. The associative property is true of addition [(6+2) + 4 = 6 + (2+4)] and multiplication [(6*2) * 4 = 6 * (2*4)], but not of subtraction or division. If we can determine whether the operation is multiplication or addition, or if it's subtraction or division, then we have enough information.

(1) A quick test of each operation will show you that the triangle is either multiplication or addition. Regardless of which one it is, we know that it will satisfy the associative property. Sufficient.

(2) Again, a quick test will show you that the triangle is either multiplication or division. Since multiplication satisfies the property, but division does not, we can't determine the answer. Insufficient.

(A)

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At the risk of sounding stupid, the GMAT Quant thus far has taught me to consider all possible conditions when solving a DS problem and choose only that answer which holds true universally in all cases. Hence my question in this case is:

(1) says the operation can either be Addition (+) or Multiplication (X)

The question asks if LHS = RHS using the above operations in place of the symbol
My observation :
(a) if you use multipication in place of the symbol then LHS = RHS
(b) if you use addition in place of the symbol then LHS = RHS
(c) But if you use a combination of multiplication and addition in the equation then LHS not equal to RHS.

Shouldn't the above (C) be a possibility too, since all we know from (1) is that it can be any operation (+ or X). No where is it specified that the same operation is to be performed on both sides of the equation

Similarly,

(2) says the operation can either be Multiplication (X) or Division (/)

Just as in the above reasoning, if you use a combination, then LHS not equal to RHS

THus using (1) and (2) together, only Multiplication holds good to satisfy both conditions

Therefore shouldn't the answer be (C).

Sorry if this is a stupid question.

Experts Please assist....

If the operation # is ONE of the four arithmetic operations: addition, subtraction, multiplication and division. Is (6#2)#4 = 6#(2#4) ?

The above means that # in the expression, (6#2)#4 = 6#(2#4), is addition OR subtraction OR multiplication OR division.

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Hope it helps.
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Nice way to test the Assocative Law!

We learn this in primary school, but we see custom characters in the GMAT and tend to freak out...

Target question: Is (6#2)#4 = 6#(2#4)?
This is a good candidate for rephrasing the target question.

Under what circumstances does (6#2)#4 = 6#(2#4) ?
Let's test each possible operation:

ADDITION: If # represents addition, we get: (6+2)+4 = 6+(2+4)
Simplify to get: 12 = 12. WORKS!
So, (6#2)#4 = 6#(2#4) when # represents addition

SUBTRACTION: If # represents subtraction, we get: (6-2)-4 = 6-(2-4)
Simplify to get: 4-4 = 6-(-2).
Doesn't work.

MULTIPLICATION: If # represents multiplication, we get: (6x2)x4 = 6x(2x4)
Simplify to get: 12x4 = 6x8. WORKS!
So, (6#2)#4 = 6#(2#4) when # represents multiplication

DIVISION: If # represents division, we get: (6÷2)÷4 = 6÷(2÷4)
Simplify to get: 3÷4 = 6÷(1/2)
Evaluate to get: 3/4 = 12
Doesn't work.

So, (6#2)#4 = 6#(2#4) when # represents EITHER addition OR multiplication
REPHRASED target question: Does # represent EITHER addition OR multiplication?

Aside: Here’s a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100

Statement 1: 3#2 > 3
This inequality holds true when # represents addition or multiplication
Case a: If # represents addition, then the answer to the REPHRASED target question is YES, # DOES represent either addition or multiplication
Case b: If # represents multiplication, then the answer to the REPHRASED target question is YES, # DOES represent either addition or multiplication
In both cases, we get the SAME answer to the REPHRASED target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 3#1 = 3
The above equation holds true when # represents division or multiplication
Case a: If # represents division, then the answer to the REPHRASED target question is NO, # does NOT represent either addition or multiplication
Case b: If # represents multiplication, then the answer to the REPHRASED target question is YES, # DOES represent either addition or multiplication
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

It's worth noting that we don't necessarily need to know which of the four operations ∆ stands for. The question is essentially, "WHICHEVER operation ∆ stands for, is the left hand side of the equation equal to the right hand side of the equation?"

(1) ∆ could stand for either additon or multiplication. Since both of those operations will give you an equation in which the left and right hand sides of the equation are equal to each other, this statement gives us sufficient information to answer the question (and the answer to the question happens to be, "Yes. The two sides are equal to each other.") We don't need to know whether ∆ is addition or multiplication.

(2) ∆ could stand for division or multiplication. We already know that multiplication will give us an equation in which the left hand side equals the right hand side, but you'll find that the same is not true if we use division. Statement (2) is therefore insufficient.

Answer A.

A video explanation can be found here:
https://www.youtube.com/watch?v=MmdwbUAljBw

from given info (6#2)#4 = 6#(2#4)

# means + or *
#1
3#2 > 3
sufficient when #= + or * ; sufficient
#2
3#1=3
possible when # = divide not sufficient
IMO A

Solution:

Since the operations addition and multiplication are associative (but subtraction and division are not), we see that if # is either addition or multiplication, then we have (6#2)#4 = 6#(2#4).

Statement One Only:

3#2 > 3

We see that the inequality holds if # is either addition or multiplication. Since # is either addition or multiplication, we have (6#2)#4 = 6#(2#4).

Statement one alone is sufficient.

Statement Two Only:

3#1 = 3

We see that the equality holds if # is either multiplication or division. If # is multiplication, then we have (6#2)#4 = 6#(2#4). However, if # is division, then we DO NOT have (6#2)#4 = 6#(2#4).

Statement two alone is not sufficient.

Answer: A
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(1) 3#2 > 3
it could be either addition or multiplication both satisfies the equation
Clearly sufficient

(2) 3#1 = 3
it could be multiplication or divison
Former is satisfied however later is not therefore out


Therefore IMO A

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