BTGmoderatorDC 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
OA A
Source: GMAT Prep
Let's take each statement one by one.
(1) 3#2 > 3
Say # is:
1. Addition: Thus, 3#2 > 3 => 3+2 > 3 => 5 > 3. Thus, # can be addition.
2. Subtraction: Thus, 3#2 > 3 => 3-2 > 3 => 1 < 3. Thus, # cannot be subtraction.
3. Multiplication: Thus, 3#2 > 3 => 3 x 2 > 3 => 6 > 3. Thus, # can be multiplication.
4. Division: Thus, 3#2 > 3 => 3÷2 > 3 => 1/5 < 3. Thus, # cannot be subtraction.
Thus, # is either addition or multiplication.
Let's find out the value of (6#2)#4 = 6#(2#4)
Case 1: Say # is addition.
=> (6#2)#4 = 6#(2#4) => (6+2)+4 = 6+(2+4). It's a valid result. The answer is Yes.
Case 2: Say # is multiplication.
=> (6#2)#4 = 6#(2#4) => (6x2)x4 = 6x(2x4). It's a valid result. The answer is Yes.
Unique answer. Sufficient.
(2) 3#1 = 3
1. Addition: Thus, 3#1 = 3 => 3+1 = 3 => 4 ≠3. Thus, # cannot be addition.
2. Subtraction: Thus, 3#1 = 3 => 3 - 1 = 3 => 2 ≠3. Thus, # cannot be subtraction.
3. Multiplication: Thus, 3#1 = 3 => 3x1 = 3 => 3 = 3. Thus, # can be multiplication.
4. Division: Thus, 3#1 = 3 => 3÷1 = 3 => 3 = 3. Thus, # can be division.
Thus, # is either division or multiplication.
In Statement 1, we have already seen that # can be multiplication and the answer to the question is Yes. Let's check whether # can also work as division.
=> (6#2)#4 = 6#(2#4) => (6÷2)÷4 = 6÷(2÷4) => 3÷4 ? 6÷0.5 => 0.75 ≠12. The answer is No.
There is no unique answer. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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