If the operation Ø is defined for all positive integers x and w by x Ø

Question

If the operation Ø is defined for all positive integers x and w by  x Ø w = \frac{2^x}{2^w} then  (3 Ø 1) Ø 3 =  ?

(A) 2
(B) 4
(C) 8
(D) 16
(E) 32

hgp2k · Answer

Step 1) 
3 Ø 1 = 2^3/2^1 = 4

Step 2)
(3 Ø 1) Ø 3 = 4 Ø 3
4 Ø 3 = 2^4/2^3 = 2

asterixmatrix · Answer

Ans 2

(3 Ø 1) = (2^3)/(2^1) = 4
 then we have 4 Ø 3 = (2^4)/(2^3) = 2

PS: this should be in problem solving section

spyguy · Answer

Thanks for the explanation. It seems so very blatant and easy once it is explained. Does anyone have any help tips or tricks when you come across these on the gmat? I find that I lose all sense of rationalization and can't begin to take a step forward. It's like I am staring at a foreign language I know nothing about while stuck in cement that has dried.

KarishmaB · Answer

Question: (3 Ø 1) Ø 3 = ?

Always start with the innermost bracket. Forget the rest of the expression while you evaluate it.
(3 Ø 1) = ?
Given x Ø w =  (2^x)/(2^w) 
Here, x = 3 and w = 1
So (3 Ø 1) =  (2^3)/(2^1) = (2^2) = 4

So we got the value of (3 Ø 1) = 4. Let us substitute it and move ahead.
(3 Ø 1) Ø 3 = 4 Ø 3 
Now, x = 4 and w = 3
So So (4 Ø 3) = (2^4)/(2^3) = (2^1) = 2

A suggestion: Go through exponents and roots theory. That will give you a lot of confidence.

pike · Answer

I think you just need to get past the notation. I am sure you are capable of solving  2^4 / 2^2  – that's just a simple exponent rule.

So its probably the symbols and notation in this question that throws you.

When I see this, I just instantly think about this in functional notation. Get rid of a symbol you have never seen before (Ø), and put into something familiar. My thought process was this:

Ø is defined for all positive integers x and w by x Ø w  = (2^x)/(2^w)

I see this as:  f(x,w) = (2^x)/(2^w)

(3 Ø 1) Ø 3

Is now:  f(f(3,1),3) = f((2^3)/(2^1),3) = f(4,3) = 2^4/2^3 = 2^1 = 2

Maybe this approach will help.

spyguy · Answer

Thank you again Karishma and Pike that helps a great deal when looking at these types of problems. I feel comfortable doing exponents but when a problem such as this one arises I tend to forget all logic. The structured approach you both mentioned helps a great deal. +1 to each and much kudos.

KarishmaB · Answer

Responding to a pm:

Ø is not a standard operator. The question defines it for you: Ø is defined for all positive integers x and w by  x Ø w = \frac{2^x}{2^w}

So it tells you how to operate on terms.

x Ø w = \frac{2^x}{2^w}

So  1 Ø 2 = \frac{2^1}{2^2}

5 Ø 2 = \frac{2^5}{2^2}

and so on...

Now check the solution above.

Also, another question may define Ø in another way such as

Ø is defined for all x and y by  x Ø y = (x + y) * (x - y)

In that case, 
 1 Ø 2 = (1 + 2) * (1 - 2)

Hope this clarifies user defined operators.

arjunvadla · Answer

Thank You Karishma for clarifying!

Sarmadk5 · Answer

Solution:- this is easy one.

As operation will be applied twice ... First within bracket then again with outer term.

1st bracket operation= 2^3/2^1 = 2^2=4

Again applying operation.f(4,3)

2^4 / 2^3 = 2^(1) = 2

Ans :A

Posted from my mobile device

bumpbot · Answer

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If the operation Ø is defined for all positive integers x and w by x Ø

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If the operation Ø is defined for all positive integers x and w by x Ø

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