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If x and y are positive, is xy > x + y?  [#permalink]

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

(1) x < y
(2) 2 < x

OG Q 2017 New Question(Book Question: 199)

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Re: If x and y are positive, is xy > x + y?  [#permalink]

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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
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Re: If x and y are positive, is xy > x + y?  [#permalink]

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

(1) x<y
(2) 2<x

OG Q 2017 New Question(Book Question: 199)

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.

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Re: If x and y are positive, is xy > x + y?  [#permalink]

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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 ?
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Re: If x and y are positive, is xy > x + y?  [#permalink]

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sagargoyal wrote:
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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If x and y are positive, is xy > x + y?  [#permalink]

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1
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AbdurRakib wrote:
If x and y are positive, is xy > x + y?

(1) x < y
(2) 2 < x

OG Q 2017 New Question(Book Question: 199)

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.
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GMAT 1: 710 Q49 V38 Re: If x and y are positive, is xy > x + y?  [#permalink]

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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!
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Re: If x and y are positive, is xy > x + y?  [#permalink]

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sdlife wrote:
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.
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If x and y are positive, is xy > x + y?  [#permalink]

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MathRevolution wrote:
AbdurRakib wrote:
If x and y are positive, is xy > x + y?

(1) x < y
(2) 2 < x

OG Q 2017 New Question(Book Question: 199)

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.

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)

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Re: If x and y are positive, is xy > x + y?  [#permalink]

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

(1) x < y
(2) 2 < x

OG Q 2017 New Question(Book Question: 199)

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 ;
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