X and Y are positive integers. If 1/x + 1/y < 2, which of the followin

Answer is B:

1/X + 1/Y < 2

The maximum value of 1/X is 1 because if X equals any other number greater than one it will be a fraction. The same is true with 1/Y.

So 1/X and 1/Y will always be less than 2 as long as both X and Y are not both equal to one at the same time.

Another way of putting it is:

X*Y>1

My take on this :

From the equation :
x + y < 2xy


=> xy > (x+y)/2

So if x and y are two different positive integers, taking the two least values as 1 and 2, we have x > 1.5 at least. Hence xy > 1.

Answer B.

Let X=1,
1+1/Y<2
1/Y<1
1<Y

Y>1 when X=1,
A --> yes and no
B --> yes
C--> yes and no
D--> yes and no

Answer: B

Guys, can you guide me how D is not true? coz last time i checked, square of any number is greater than 0. Even if x is less than y, still, it's square would me more than 0..unless, x = y...

Given: 1/X + 1/Y < 2
Say X = 2, Y = 2
These values satisfy the inequality: 1/2 + 1/2 < 2

But they do not satisfy (D)
(X-Y)^2>0
(2-2)^2 = 0, not greater than 0
Hence (D) must not be true for all values. There are values that satisfy the inequality but does not satisfy (D)

Karishma, now i need a confirmation on GMAT questions...
lets say that if two unknowns are given (like X and Y ), can we assume that these two are equals? I thought if we say x and y, they are implicitly different numbers..

Until and unless they mention 'distinct numbers' or 'X not equal to Y', X and Y can be equal. The equality can be a deal breaker/maker sometimes so you have to make sure that you have analyzed its effects too.

Thanks Karishma for the information.

Let x=2
Let y=2

\(\frac{1}{2} + \frac{1}{2}< 2\)

A) X+Y=4 OUT!
B) 2*2 > 1 HOLD!
C) 2/2 + 2/2 = 2 < 1 OUT!
D) (2-2)^2 = 0 OUT!

Answer: B

My Solution:

Given : x and y are positive integers

Stem: 1/x+1/y<2---x+y<2xy (as x and y are positive we can cross multiply)

So A) x+y>4 Try x=1 & y = 2 (Not true)

B) xy>1 Try x=1 and y = 2 (Always true) ) [Note (x=1 and y =1 is not possible values because with these values stem doesn't holds true]

This is our answer as not more then one correct answer choice is possible but we can try all choices for more clarity:

C) x/y+y/x<1 Try x=1 and y = 2 (Not true)

D) (x-y)^2 Try x=2 and y=2 (Not true)

E) Can never be true

Answer is B

I thought it is some kind of trap here..
we can rewrite the original as:
x+y<2xy

A - x=2, y=2 -> x+y is not greater than 4, yet 1/2 + 1/2 < 2. so A is out.
B - if x and y are both positive integers, xy>1 all the times - looks good.
C - x/y +y/x <1 or x^2 + y^2 < xy - which will never be true, if x and y are positive integers.
D - x^2 + y^2 > 2xy - suppose x=2 and y=2. 4+4 = 8. 2*2*2=8. 8=8, it's not an inequality.
E - since B works, e is out.

\(\frac{1}{x} + \frac{1}{y} < 2\)

Let x = 1 and y = 2

\(\frac{1}{1} + \frac{1}{2} = \frac{3}{2}\)

\(\frac{3}{2} < 2\) -- Those numbers for x and y work

MUST be true?

A. X+Y>4 - NO
1 + 2 = 3, which is not greater than 4

B. XY>1 YES
Because x and y are positive integers, there is only one way XY would NOT be greater than 1: if both x and y = 1. Then XY = 1.
But x = y = 1 violates the prompt: their reciprocals summed must be less than 2; in that case, they equal 2. This choice must be true.

C. X/Y + Y/X < 1 - NO
\(\frac{1}{2} + \frac{2}{1}=\frac{5}{2}\)
\(\frac{5}{2}\) is not less than 1

D. (x-y)^2 > 0 NO
For this option, let x=y=2.
\((2-2)^2 = 0^2 = 0\)
0 is not greater than 0

E. none - NO - One of the answers, B, must be true.

Answer B

Official Solution:

If \(x\) and \(y\) are positive integers and \(\frac{1}{x}+\frac{1}{y} \lt 2\), which of the following must be true?

A. \(x + y \gt 4\)
B. \(xy \gt 1\)
C. \(\frac{x}{y} + \frac{y}{x} \lt 1\)
D. \((x - y)^2 \gt 0\)
E. none of the above


Given that \(x\) and \(y\) are positive integers, it follows that they can take on the values \(1, 2, 3, \ldots\). However, if \(x=y=1\), then \(\frac{1}{x}+\frac{1}{y}=2\), which violates the condition that \(\frac{1}{x}+\frac{1}{y}1\). This means that option B is always true.

We can demonstrate that options A, C, and D are not always true by providing counterexamples. For instance, if we set \(x=y=2\), then we find that none of the options holds true.


Answer: B