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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
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zonebp wrote:
Is there any short cut method for these kind of problems. I have took 3 minutes 15 sec to solve the problem. Is this the avg time required to solve the problem please...


Note that the function is symmetric (if you switch one variable with another, it doesn't change) except for the sign:

a ⊗ b = a/b – b/a

b ⊗ a = b/a – a/b = - (a/b - b/a)

Look for options which are symmetric too.

I. x ⊗ xy = x(1 ⊗ y)

This is not symmetric. You have x and xy so x is different from y. I will ignore this right now.

II. x ⊗ y = -(y ⊗ x)

This we already know is true.

III. 1/x ⊗ 1/y = y ⊗ x

Just check for this.

Answer (E).

In such questions, looking for symmetry often helps.
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The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
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ajit257 wrote:
The operation ⊗ is defined for all nonzero numbers a and b by a ⊗ b = a/b – b/a. If x and y are nonzero numbers, which of the following statements must be true?

I. x ⊗ xy = x(1 ⊗ y)
II. x ⊗ y = -(y ⊗ x)
III. 1/x ⊗ 1/y = y ⊗ x

A. I only
B. II only
C. III only
D. I and II
E. II and III


a ⊗ b = a/b – b/a

Lets check I option :
LHS = x ⊗ xy = x/xy -xy/x = 1/y - y
RHS = x(1 ⊗ y) = x(1/y - y)............... LHS =/= RHS

Lets check II option :
LHS = x ⊗ y = x/y - y/x
RHS = -(y ⊗ x) = -(y/x - x/y) = x/y - y/x................... LHS = RHS

Lest check III option :
LHS = 1/x ⊗ 1/y = [m]/frac{1/x}{1/y} - /frac{1/y}{1/x} = y/x - x/y
RHS = y ⊗ x = y/x - x/y ............... LHS = RHS

Answer E. II and III
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
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If you pay close attention, III is basically the operation backwards so you don't have to waste time trying it. Also, if you look at the answer choices, if III is correct then just test II to choose between C or E.
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
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norovers wrote:
For statement 3 I cross multiplied the LHS and was left with: y ⊗ x = y ⊗ x

Is this operation allowed?


What do you mean by "cross multiplied the LHS"?

There are no operations you can do on 1/x ⊗ 1/y without substituting in the expression for ⊗ since you do not know how this symbol relates the two terms without the defining expression.
If say, ⊗ = *, then (1/x)*(1/y) = 1/xy
If instead ⊗ = /, then (1/x)/(1/y) = y/x
and so on.

You only know how to handle the standard operations. ⊗ is a specially defined operation and you need to convert it to standard operations to simplify it.
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
Is there any short cut method for these kind of problems. I have took 3 minutes 15 sec to solve the problem. Is this the avg time required to solve the problem please...
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
zonebp wrote:
Is there any short cut method for these kind of problems. I have took 3 minutes 15 sec to solve the problem. Is this the avg time required to solve the problem please...


No shortcut bro.. But you can solve it in 1:30 minutes if u solve a lots of similar problems..
I solved in 1:39 minutes...

3 minutes 15 sec is a bit high time.. So you can solve in less time if you practice more....
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
shashankism wrote:
zonebp wrote:
Is there any short cut method for these kind of problems. I have took 3 minutes 15 sec to solve the problem. Is this the avg time required to solve the problem please...


No shortcut bro.. But you can solve it in 1:30 minutes if u solve a lots of similar problems..
I solved in 1:39 minutes...

3 minutes 15 sec is a bit high time.. So you can solve in less time if you practice more....


Thanks a lot mate. Im struggling to pace and just worried a lot due to the pacing issue.
Just running out of time and sometimes spending far too much time on a specific question. Slowly I started getting into countdown phobia.
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
For statement 3 I cross multiplied the LHS and was left with: y ⊗ x = y ⊗ x

Is this operation allowed?
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]
Is there a quick way to answer this question? I just dumped it in the practice test as I could see it would take up too much time
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The operation is defined for all nonzero numbers a and b [#permalink]
Bunuel wrote:
ajit257 wrote:
The operation ⊗ is defined for all nonzero numbers a and b by a ⊗ b = a/b – b/a. If x and y are nonzero numbers, which of the following statements must be true?
I. x ⊗ xy = x(1 ⊗ y)
II. x ⊗ y = -(y ⊗ x)
III. 1/x ⊗ 1/y = y ⊗ x

A. I only
B. II only
C. III only
D. I and II
E. II and III

Please can someone confirm the ans. I am not sure about the ans


Given: \(a@b=\frac{a}{b}-\frac{b}{a}\), for all nonzero numbers \(a\) and \(b\).

I. \(x@(xy) = x(1@y)\): \(LHS=x@(xy)=\frac{x}{xy}-\frac{xy}{x}=\frac{1}{y}-y\) and \(RHSx(1@y)=x(\frac{1}{y}-y)\) as you see LHS doen't equal to RHS;

II. \(x@y = -(y@x)\): \(LHS=x@y=\frac{x}{y}-\frac{y}{x}\) and \(RHS=-(y@x)=-(\frac{y}{x}-\frac{x}{y})=\frac{x}{y}-\frac{y}{x}\) --> LHS=RHS;

III. \((\frac{1}{x})@(\frac{1}{y}) = y@x\): \(LHS=(\frac{1}{x})@(\frac{1}{y})=\frac{y}{x}-\frac{x}{y}\) and \(RHS=y@x=\frac{y}{x}-\frac{x}{y}\) --> LHS=RHS.

Answer: E (II and III).

Hope it's clear.


Experts, can you use smart numbers to solve this? I set a = 2 and b=2 and then believed the unidentified operation was subtraction based on what made the left-hand side = right-hand side. However, then choices II and III did not work based on this. Are there any other ways to solve this then if you found the method above time consuming? Do you have to work from the left-hand side to right-hand side?

Also, for I. when you factor out the x on the left-hand side to arrive at (1/y - y), why does it just disappear? I did x(1/y - y)=x(1/y -y) and thought they were equal, but the expert above seems to do (1/y-y)=x(1/y - y).
Thank you in advance for your time. RonTargetTestPrep
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