If x and y are positive, is xy > x + y?

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

xy > x + y
⇔ xy - x - y > 0
⇔ xy - x - y + 1 > 1
⇔ (x-1)(y-1) > 1
⇔ x > 1, y > 1 or 0 < x < 1, 0 < y < 1?

Since we have 2 variables (x and y) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
Since x > 2, y > x > 2 or y > 2, both conditions together are sufficient.

Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
x = 2, y = 3: Yes
x = 1/2, y = 1: No
The condition 1) is not sufficient.

Condition 2)
x = 3, y = 4: Yes
x = 3, y = 1/2: No
The condition 2) is not sufficient.

Therefore, the answer is C.

given x and y positive

if we take x,y<1 then statement doesn't hold if x,y>1 then holds.

1) not suff for above explanation
2)no info about y

we know x>2 and so is y so 1+2 is suff

IS xy>x+y


Fact 1) x<y

Put x=1/2, Y=2 1>2.5 Answer is NO
Put x= 3, Y=4 12>7 Answer is YES
Insuff


Fact 2) X>2 Clearly Insuff


Combining 1 & 2

Take example above when x= 3, Y=4
Another example x=5/2 , Y=4
Suff.

Answer is C

Why should the answer be not B because, if we take x = 3 and y = 4, the statement is sufficient (if first condition is taken into consideration) and if we take x = 5 and y = 2 then also the statement is sufficient (if first taken not taken into consideration) so having condition 1 is not necessary.

Can anyone respond to this ?

Two examples, are not enough to prove sufficiency. What if x = 1 and y = 1? For (2), there are infinitely many pairs which give a NO answer to the question as well as there are infinitely many pairs which give an YES answer to the question.
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Hi Experts,

Can someone please provide a solution for this question? If I use number plugging, I am not able to prove that C will be sufficient. Can there be another approach (algebraic) to solving this?

Thanks for your help!

Hi

Not an expert but I will give you my perspective. I think here algebra would be difficult, at least I cannot think of that.

We are given that both x/y are positive numbers, and after we combine the two statements we get that both x/y are greater than 2. We have to determine whether their product is greater than their sum.

Now, '2' is a key number here. Upto '2', we cannot be sure whether product of two positive numbers will be greater than their sum or not.
Eg, if we take two numbers as 1,2 - then their product is actually lesser than their sum
If we take two numbers as 1,1.5 - then also their product is lesser than their sum
If we take two numbers as 0.5, 4 (here one of them is less than 2 & other is greater than 2) - then also their product is lesser than their sum (here the fractional value of first number which is less than 1, has decreased the product to a large extent and made it lesser than the sum)

And if both numbers are equal to 2 (though here its given that x < y so they cannot be equal but still to explain) - then their product will be equal to their sum (4 each)

But once each number is greater than '2', the property is such that the product will always be greater than the sum
Even if we take them as 2.01, 2.02 - we will see that their product is greater than their sum
And as the numbers keep on increasing, their product will keep on increasing and become more and more larger than their sum.

Hi MathRevolution,

Could you explain me the highlighted step after the below step
(x-1)(y-1)>1
So we have
(x-1)>1
or
(y-1)>1
So we have x> 2 and or we have y>2
What did i miss?

However, this is how I approached

xy>x+y
xy-x-y>0
adding 1 on both sides
xy-x-y+z>1
(x-1)(y-1)>1
so either
x-1>1 more so x>2
or y-1>1 more so y>2

So question is asking IS x>2 and y>2. we can have answer to our question is we know about x and y.

So statement A:
(A) x < y

we know that x>o and y>0, but we dont kow if x>2 and y>2
So insufficient.

Statement B:
(B) 2 < x
We know that x>0 so x>2 but don't have any information about y
so insufficient,

Now combining A and B we have
2<x<y
so we have x>2 and y>2


However just to remember

A * B\(\geq\) A + B.
When A & B \(\leq{0}\), it's always true.

When A & B \(\geq 2\), it's always true.

When A \(\geq{1}\), B \(\leq{1}\) (or B\(\geq{1}\), A\(\leq{1}\)), it's always false.

When 0 < A < 1, B < 0 (or 0 < B < 1, A < 0), it can be either true or false.

When 1 < A < 2, B > 0 (or 1 < B < 2, A > 0), it can be either true or false.

A * B > A + B.
when A>2 and B>2
For this question if we had two statements say
Statement 1: A>2
statement 2: B>2
On combining 1 and 2 we can say that (A*B)> (A+B)

Probus
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From (1), x<y There are many possibilities smart no.s(all values>0 though) for which you will get no unique answer sometimes xy < (x+y) & sometimes xy > x+y =>NOT SUFFICIENT ;

From (2), 2<x => No clue about value of y => NOT SUFFICIENT ;

Now, its given x & y both are +ve so,

2<x => 2y < x*y .......(a)
x<y .......(b)

Adding (a) & (b), (2y+x) < (xy+y) => (y+x) < xy => SUFFICIENT => Option C ;

Solution:

Question Stem Analysis:

We need to determine whether xy > x + y, given that x and y are both positive. That is, we need to determine whether the product of two positive quantities is greater than their sum. Usually, that is the case; however, it still relies on the values of the quantities. For example, if we can show both quantities are greater than 2, then their product will be greater than their sum.

Statement One Alone:

We see that x < y. However, this does not allow us to determine whether xy > x + y. For example, if x = 2 and y = 3, then xy > x + y. However, if x = 1 and y = 2, then xy is not greater than x + y. Statement one alone is not sufficient.

Statement Two Alone:

We see that x > 2. However, this does not allow us to determine whether xy > x + y. For example, if x = 3 and y = 2, then xy > x + y. However, if x = 3 and y = 1/3, then xy is not greater than x + y. Statement two alone is not sufficient.

Statements One and Two Together:

Since x > 2 and y > x, we see that both x and y are greater than 2.

Let’s divide each side of the inequality xy > x + y by y (which can be done because we are told that y is positive):

Is x > (x + y)/y ?

Is x > x/y + y/y ?

Is x > x/y + 1 ?

Since y > x, x/y is a proper fraction and therefore less than 1. Thus, the right hand side of the inequality is less than 2. Since x > 2, x > x/y + 1. Therefore, xy > x + y. Both statements together are sufficient.

Answer: C
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How did you open this? MathRevolution
(x-1)(y-1) > 1

MathRevolution it seems like you’ve done something wrong here: x > 1, y > 1 or 0 < x < 1, 0 < y < 1—— how did you get these ranges from the previous part you solved?



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avigutman ThatDudeKnows how to solve (x-1)(y-1)>1 algebraically? How do we get range of x and y. I know there might be more ways to solve this question but I always use this so I would like to know how to proceed from this to not be stuck in exam. Pls tell the approach for such types in GMAT

If you're interested in the ranges of x and y, Elite097, why would you go from is xy > x + y ? to is (x-1)(y-1) > 1 ? ? I'm not following your strategy here. However way you phrase that inequality, you're going to have to use some number line reasoning (see below) in order to make inferences about the possible ranges.

Given that x and y are positive, the question is just asking whether the product of two positive numbers is greater than their sum. Since we know that 2*2 = 2+2, I view 2 as an inflection point: if x and y are both greater than 2, we get a YES, and if they're both less than 2, we get a NO. If one of them is greater than 2 and the other is less than 2, then it depends on the actual values (hopefully you can "see" that intuitively, but if not, you can prove it to yourself e.g. x=1/4 and y=3 leads to a NO, but x=5/4 and y=10 leads to a YES).

I haven't yet explored the question or how you got to (x-1)(y-1)>1, but since that's what you're asking about, here's a bit on thinking through things like this.

(x-1)(y-1)>1

So (x-1)(y-1) is positive. That means that either x-1 and y-1 are both positive or x-1 and y-1 are both negative.

If both positive, at least one of x-1 and y-1 is greater than 1, so at least one of x>2 or y>2 must be true.

If both negative, at least one ox x-1 and y-1 is less than -1, so at least one of x<0 or y<0 must be true.

The things you should always be looking for are odd/even, positive/zero/negative, what happens between -1 and 0, what happens between 0 and 1, what happens with large positive values, and what happens with large negative values. With practice, it gets easier to spot on which concept a given question hinges.

Thanks for addressing my doubt. Could you please tell me how did you come to the below statemnts and then jump to the ranges (pls present your working):
If both positive, at least one of x-1 and y-1 is greater than 1 and
If both negative, at least one ox x-1 and y-1 is less than -1? x-1<0 implies x<1 and similarly y < 1 so how did you get -1 here? ThatDudeKnows

What does it mean for two numbers to be multiplied together and be greater than 1? I think it's clear that either (1) they are both positive or (2) they are both negative.

But what else do we know?

Let's say they are both positive. If they are also both greater than 1, then the product will definitely be greater than 1 (If this isn't TOTally clear, try coming up with a counterexample...WHY can't you? Make sure you understand the conceptual nature of why before you keep reading.). What if they are both less than 1? Can you think of two numbers between 0 and 1 that you can multiply together that will result in a product that's greater than 1? Why not? Well, a number between 0 and 1 is a fraction with a smaller number in the numerator than it has in the denominator. If we multiply two of those together, the numerator will be smaller than the denominator, making the product less than 1. Okay, so if both numbers are greater than 1, the product is greater than 1. And if both numbers are between 0 and 1, the product is less than 1. What if one of the numbers is greater than 1 and the other is less than 1? Can you come up with examples that make the product greater than 1? How about examples that make the product less than 1?

Once you've explored that, spend a little time doing a similar analysis on the negative side. What happens if both numbers are less than -1 (the way I prefer to phrase this would be "both numbers are farther to the left on the number than -1," since that helps me "visualize" that we are ultimately talking about direction and distance from 0)? What happens if both numbers are between 0 and -1 ("both numbers are to the left of 0 and to the right of -1")? What happens if there's one of each?

This is a good question for really wrapping your head around what these concepts mean and how there are a few places on the number line that serve as great "break points" for when something goes from true to false. Positive/negative, distance from 0 (especially using -1 and 1 as thresholds) are huge concepts. I'd recommend spending as much time as you need on this question until it REALLY sinks in. And then revisit it two days from now to make sure.

And then add an analysis of what happens if one number is positive and one is negative. What if the positive is huge and the negative is -.00001? What if the positive is 1 and the negative is less than -1? What if the positive is 0.5 and the negative is -0.1? What if the positive is 0.1 and the negative is -0.5? Etc. WHY do each of those products work out the way that they do?

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