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B
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Difficulty:
5%
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Question Stats:
98% (01:03) correct
2%
(01:23)
wrong
based on 104
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If x#y = yx^y for all values x and y, then 2#(0(-2#2)=
A. -64
B. 0
C. 32
D. 64
E. 128
A. -64
B. 0
C. 32
D. 64
E. 128
Hey guys i am having a hard time understanding the solution to this question. The correct answer is B. This is the solution given by veritas prep:
This function question may look laborious, but can actually be solved quickly noting that it will require the multiplication of each of the values, and that 0 is one of the values. Accordingly, because 0 multiplied by anything is 0, the answer can only be 0. Taken step-by-step, the result of the innermost parentheses will be 2(-2)2, or 8. That term then becomes the y term of the next sequence, for which x is 0. That term will be 8(0)8, or 0. Then, that becomes the y term of the next set, which would be 0(2)0. The answer, still, is 0, and B is the correct answer.
What i am confused about is how the innermost parenthesis equal 8?
This function question may look laborious, but can actually be solved quickly noting that it will require the multiplication of each of the values, and that 0 is one of the values. Accordingly, because 0 multiplied by anything is 0, the answer can only be 0. Taken step-by-step, the result of the innermost parentheses will be 2(-2)2, or 8. That term then becomes the y term of the next sequence, for which x is 0. That term will be 8(0)8, or 0. Then, that becomes the y term of the next set, which would be 0(2)0. The answer, still, is 0, and B is the correct answer.
What i am confused about is how the innermost parenthesis equal 8?
02 Apr 2015, 16:08
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pmklings
Dear pmklings
I'm happy to help.
My friend, I believe you are confused by Order of Operations. This idea, something known as PEMDAS, specifies some quite precise ways to view calculations, and sometimes these are anti-intuitive.
Look at the algebraic expression:
yx^y
In that, what do we do first? Do we multiply y times x, and then raise that product to the power of y? Or do we raise just x to the power of y, and then multiply that power by the first y? Well, the standard Order of Operation procedure tells us that exponents take precedent over multiplication, so we have to raise just the x to the power of y, computing that, before we can even consider the first factor of y.
Thus,
(2)(-2)^2 = 2*[(-2)^2] = 2*4 = 8
Does this make sense? Does this answer your question?
You may also find this blog article helpful:
https://magoosh.com/gmat/2013/gmat-quant ... g-symbols/
Mike
23 Jan 2020, 11:39
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pmklings
First, we have; (-2#2) = 2(-2)^2 = 2(4) = 8
Next, we have: 0(8) = 0
Lastly, we have 2#0 = 0(2)^0 = 0
Answer: B
