Given that xy≠0, what is the value of (1/x + 1/y)/(1/x) - x/y ?

Question

Given that xy ≠ 0, what is the value of  \frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x}} - \frac{x}{y}  ?

A. 1/2
B. 1
C. 3/2
D. 2
E. 3

BrentGMATPrepNow · Answer

Replace x and y with 1 and evaluate....

\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x}} - \frac{x}{y}=\frac{\frac{1}{1} + \frac{1}{1}}{\frac{1}{1}} - \frac{1}{1}

=\frac{2}{1}-1

=2-1

= 1

Answer: B

Cheers,
Brent

prabsahi · Answer

Brent,

Can you please explain the rationale behind plugging the numbers here.

I tried another set and I got the same answer.

BrentGMATPrepNow · Answer

You bet. 
We're asked to evaluate (find the value of) the given expression. 
Since all of the answer choices are constants (just numbers), we know that the given expression must evaluate to be ONE of the answer choices FOR ALL VALUES OF X.

-------------ASIDE-----------------------------------------
We know that 2x/x = 2 for all non-zero values of x. 
For example, if x = 7, then 2x/x = (2)(7)/7 = 2
If x = 3, then 2x/x = (2)(3)/3 = 2
If x = -5, then 2x/x = (2)(-5)/(-5) = 2
-----------------------------------------------------------

So, if the given expression must evaluate to be ONE of the answer choices FOR ALL VALUES OF X, let's plug in some value of x. 
I chose a nice number (x = 1), but you can choose to plug in any x-value, and still find that the expression still evaluates to be 2.

Cheers,
Brent

prabsahi · Answer

Thanks for replying back.

I got the point that the options are numbers and so we are trying to plugin the values of numbers and alos the fact 2*x/x

But my doubt is we have variable y as well here and we dont know how the entire expression is going to reduce ..like say 2x/x or 2*xy .

BrentGMATPrepNow · Answer

The same principle applies to expressions with more than 1 variable. 
The key point here is that the original answer choices tell us that all of the variables must cancel out, leaving us with some number.

Here's a similar example:

If x ≠ y, what is the value of  \frac{3x - 3y}{x - y} ? 
A) 1
B) 2
C) 3
D) 4
E) 5

If we use an algebraic approach, we can write:  \frac{3x - 3y}{x - y}=\frac{3(x - y)}{(x - y)}= 3 
Notice that the 2 variables cancel out.
This means the values of x and y are irrelevant;  \frac{3x - 3y}{x - y}  MUST equal 3, for all values of x and y (as long as x ≠ y)

To verify this, let's assign some values of x and y

Let's try x = 2 and y = 7
We get:  \frac{3x - 3y}{x - y}=\frac{3(2) - 3(7)}{2 - 7}=\frac{-15}{-5}=3

Now let's try x = 5 and y = 1
We get:  \frac{3x - 3y}{x - y}=\frac{3(5) - 3(1)}{5 - 1}=\frac{12}{4}=3

Now let's try x = 6 and y = 0.1
We get:  \frac{3x - 3y}{x - y}=\frac{3(6) - 3(0.1)}{6 - 0.1}=\frac{17.7}{5.9}=3

And so on.

Cheers,
Brent

prabsahi · Answer

Thanks for replying back.

I got the point that the options are numbers and so we are trying to plugin the values of numbers and alos the fact 2*x/x

But my doubt is we have variable y as well here and we dont know how the entire expression is going to reduce ..like say 2x/x or 2*xy .

The same principle applies to expressions with more than 1 variable. 
The key point here is that the original answer choices tell us that all of the variables must cancel out, leaving us with some number.

Here's a similar example:

If x ≠ y, what is the value of  \frac{3x - 3y}{x - y} ? 
A) 1
B) 2
C) 3
D) 4
E) 5

If we use an algebraic approach, we can write:  \frac{3x - 3y}{x - y}=\frac{3(x - y)}{(x - y)}= 3 
Notice that the 2 variables cancel out.
This means the values of x and y are irrelevant;  \frac{3x - 3y}{x - y}  MUST equal 3, for all values of x and y (as long as x ≠ y)

To verify this, let's assign some values of x and y

Let's try x = 2 and y = 7
We get:  \frac{3x - 3y}{x - y}=\frac{3(2) - 3(7)}{2 - 7}=\frac{-15}{-5}=3

Now let's try x = 5 and y = 1
We get:  \frac{3x - 3y}{x - y}=\frac{3(5) - 3(1)}{5 - 1}=\frac{12}{4}=3

Now let's try x = 6 and y = 0.1
We get:  \frac{3x - 3y}{x - y}=\frac{3(6) - 3(0.1)}{6 - 0.1}=\frac{17.7}{5.9}=3

And so on.

Cheers,
Brent

I am sorry Brent.I understand that  2. x/x or 3(x-y)/(x-y) or 5.(x-y-z)/(x-y-z)  .. Here essentially the expressions are cancelling out in numerator and denominator 
since they are same.

But My doubt remains same say we have 2 x/y or 2. xy or 2.x^2/y(just quoting it can be even different combination)..so plugin the values might change .We cant say that num/deno is cancelling here.

I guess I am missing out something very basic/elementary that you are trying to explain and I am waiting for that takeaway.

BrentGMATPrepNow · Answer

The answer choices tell us that the variables must cancel out. 
Also, it can't be the case that the expression can have more than 1 value, since only one answer choice can be correct.

Have you tried plugging in any values for x and y to see what happens?

prabsahi · Answer

---Yes this is the first thing I did and it gave the same result . Thanks a lot Brent for your patience and time to help me figure out what I was missing

ScottTargetTestPrep · Answer

We can simplify the first fraction by multiplying it by xy/xy, and we have:

(y + x)/y

So, now we have:

(y + x)/y - x/y = y/y = 1

Answer: B

Kinshook · Answer

Given that xy ≠ 0, what is the value of  \frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x}} - \frac{x}{y}  ?

\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x}} - \frac{x}{y}

\frac{(x+y)x}{xy} - \frac{x}{y}

\frac{x}{y }+ 1 - \frac{x}{y} = 1

IMO B

Given that xy≠0, what is the value of (1/x + 1/y)/(1/x) - x/y ?

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Given that xy≠0, what is the value of (1/x + 1/y)/(1/x) - x/y ?

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