If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x-1

[chitrangada.maitra]

Source: mgmat question bank
Equations and inequalities, #16
If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x-1)(y-1)]^-1 ?

(1) x = 2y
(2) x + y > 0

OA: A

Explanation:
(1) SUFFICIENT: If we plug x = 2y into our simplified question we get the following:

Is 2y^2/3y > 2y^2 ?
Since 2y^2 must be positive we can divide both sides of the inequality by 2y^2 which leaves us with the following:

Is 1/3y> 1 ?
If we investigate this carefully, we find that if y is an nonzero integer, 1/3y is never greater than 1. Try y = 2 and y = -2, In both cases 1/3y is less than 1.

My question is:
If we plug the fraction 1/9 for Y, we get, 1/3y >1, which renders the first statement insufficient.

Alternatively, if we simplify and rephrase the question like this:
Is y/3 > y^2, and plug 1/10 (or 1/ 9) for Y, we get, Y/3 > y^2. However, for integar values of Y, we get, Y^2 > y/3
Therefore, statement 1 is insufficient.

Please let me know where am I going wrong.
Thanks,

[gokul_nair1984]

chitrangada:

Is 1/3y> 1 ?
If we investigate this carefully, we find that if y is an nonzero integer, 1/3y is never greater than 1. Try y = 2 and y = -2, In both cases 1/3y is less than 1.

I absolutely agree..

My question is:
If we plug the fraction 1/9 for Y, we get, 1/3y >1, which renders the first statement insufficient.

However,how can you plug a fractional value here, when the question stem specifies x and y to be non zero integers?

Alternatively, if we simplify and rephrase the question like this:
Is y/3 > y^2, and plug 1/10 (or 1/ 9) for Y, we get, Y/3 > y^2. However, for integar values of Y, we get, Y^2 > y/3
Therefore, statement 1 is insufficient.

How did you arrive at this?You can rephrase the stem as:
xy/x+y>xy
Using Statement 1(x = 2y), we can further reduce it to
2(y^2)/3y>2(y^2) =>Is 1/3y>1? Whatever integer value you substitute will result in a No as an answer. Hence "A" is sufficient.
I Hope you are clear with this...

[chitrangada.maitra]

oops!

Careless mistake.. one of many.

Thanks

[tim]

:)

[wxandr]

If x and y are nonzero integers, is 
(x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?
We rewrote this question: 
Is 1/(1/x + 1/y) > (1/x * 1/y)^-1
Is 1/(y/xy + x/xy) > (1/xy)^-1
Is 1/((y + x)/xy) > xy
Is xy/x+y > xy?

(1) x = 2y 
(2) x + y > 0

(1) x = 2y.  

Substituting 2y for x in the rephrased
Question:
Is 2y^2/3y > 2y^2? 

Since 2y^2 is positive we can divide both sides of the inequality by 2y^2. rephrased:

Is 1/3y > 1? 

The prompt states that y is an integer
that is not zero. The integer property of y eliminates 0 < y < 1
and therefore the possibility that 1/3y > 1. 
The answer is definitely "no."

(2) x + y > 0.
Since (x + y) is positive, multiply both sides of
xy/x+y > xy by (x + y) to get:
Is xy > xy(x + y)

x and y are positive or negative integers.
(x + y) is positive.

x positive, y negative (or vice versa), x > -y , xy is negative
no. if |x| - |y| = 1.
yes. if |x| - |y| > 1.

x positive, y positive. xy < xy(x + y) for all integers 1 or greater.

There are several possibilities for both the signs and range of
x and y, leading to variable answers.

[nakul.maheshwari000]

Guys,
I have a question please.
I tried substituting numbers rather than solving the eq 1st.

1) x = 2y
If you substitute y = 1, then x = 2.
If you solve the inequality, you get 2/3 > 0, which will give you a "yes"
If you substitue y = -2, then x = -4
Now if you solve the inequality, you get -4/3 < 1/15 and hence the answer is "no"

Should this make choice "A" insufficient?

-Nakul

[jnelson0612]

Guys,
I have a question please.
I tried substituting numbers rather than solving the eq 1st.

1) x = 2y
If you substitute y = 1, then x = 2.
If you solve the inequality, you get 2/3 > 0, which will give you a "yes"
If you substitue y = -2, then x = -4
Now if you solve the inequality, you get -4/3 < 1/15 and hence the answer is "no"

Should this make choice "A" insufficient?

-Nakul

I'm confused, because -4/3 IS less than 1/15, which would be YES. Can you advise?

[maferchouza]

I have a question about this:

If we investigate this carefully, we find that if y is an nonzero integer, 1/3y is never greater than 1.

What if Y=9, then 1/3Y= 3 which is bigger than 1?

[RonPurewal]

in this discussion, "1/3y" means 1/(3y).

it doesn't mean (1/3)y, which is presumably how you're interpreting it. for that, you'd write (1/3)y, or, preferably, just y/3.