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Divisibility Problem

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Source: — Data Sufficiency |
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by gmatfeel » Wed Jul 13, 2011 9:38 am
From STMT (1)
IF x=y=1, then x+y=2, xy=1
IF x=y=2, then x+y=4, xy=4
IF x=y=3, then x+y=6, xy=9

NOT SUFFICIENT

FROM STMT (2)
x+y > x^2+y^2
x+y > (x+y)^2 + 2xy
x+y > 2xy
so we can infer that x+y > xy

SUFFICIENT

Hence the answer is b
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by amit2k9 » Wed Jul 13, 2011 8:26 pm
a for non negative 0 to comes into picture.

hence x=y=0 and x=y=1 give different solutions. not sufficient.

b x+y > x^2+y^2 meaning 0<x,y<1.
hence,

checking for x=0.2 and y=0.4 gives x+y > xy always.

sufficient. C it is.
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by Ozlemg » Thu Jul 14, 2011 6:21 am
gmatfeel wrote:From STMT (1)
IF x=y=1, then x+y=2, xy=1
IF x=y=2, then x+y=4, xy=4
IF x=y=3, then x+y=6, xy=9

NOT SUFFICIENT

FROM STMT (2)
x+y > x^2+y^2
x+y > (x+y)^2 + 2xyx+y > 2xy
so we can infer that x+y > xy

SUFFICIENT

Hence the answer is b
wouldnt the red coloured equation be ](x+y)^2 - 2xy
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by Frankenstein » Thu Jul 14, 2011 6:59 am
Ozlemg wrote:
gmatfeel wrote:From STMT (1)
IF x=y=1, then x+y=2, xy=1
IF x=y=2, then x+y=4, xy=4
IF x=y=3, then x+y=6, xy=9

NOT SUFFICIENT

FROM STMT (2)
x+y > x^2+y^2
x+y > (x+y)^2 + 2xyx+y > 2xy
so we can infer that x+y > xy

SUFFICIENT

Hence the answer is b
wouldnt the red coloured equation be ](x+y)^2 - 2xy
Hi,
I guess that was a typo.
x+y > x^2+y^2
=> x+y > (x-y)^2 + 2xy
As (x-y)^2 >= 0, (x-y)^2 + 2xy >= 2xy which in turn is twice xy and hence at least as much as xy.
So, x+y > xy

Hence, B
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by blackjack » Thu Jul 14, 2011 7:24 am
Frankenstein wrote:
Ozlemg wrote:
gmatfeel wrote:From STMT (1)
IF x=y=1, then x+y=2, xy=1
IF x=y=2, then x+y=4, xy=4
IF x=y=3, then x+y=6, xy=9

NOT SUFFICIENT

FROM STMT (2)
x+y > x^2+y^2
x+y > (x+y)^2 + 2xyx+y > 2xy
so we can infer that x+y > xy

SUFFICIENT

Hence the answer is b
wouldnt the red coloured equation be ](x+y)^2 - 2xy
Hi,
I guess that was a typo.
x+y > x^2+y^2
=> x+y > (x-y)^2 + 2xy
As (x-y)^2 >= 0, (x-y)^2 + 2xy >= 2xy which in turn is twice xy and hence at least as much as xy.
So, x+y > xy

Hence, B
Could you please elaborate how did you get (x-y)^2 >= 0 in the picture?
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by Frankenstein » Thu Jul 14, 2011 7:48 am
blackjack wrote: Could you please elaborate how did you get (x-y)^2 >= 0 in the picture?
Hi,
It is the square of a number (x-y). It is always non-negative(>=0) irrespective of whether (x-y) is negative or positive.
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by tpr-becky » Thu Jul 14, 2011 10:14 am
If X and Y are non-negative,is (X+Y) greater then XY?
1. X=Y
2. X+y is greater than X^2+Y^2

For the first statement you can simply plug in numbers - keeping in mind that the test is probably testing 0 or 1.

x = 0, y=0 then 0 = 0
x=1, y = 1 then 2 > 1

So insufficient.

Then look at statement 2 - the others have done the algebra but look at the concept. The only way for x + y to be greater than x(x) + y(y) is if x and y are fractions. They can't be negatives because the square will always be greater, they can't be positive integers becuase the square will always be greater and they can't be 0 or 1 because the square will always be equal, not less - thus the only way for the squares to be less than the original number is when the original number is a fraction. when you add positive fractions then sum is greater, when you multiply positive fractions less than one the result is smaller - thus if x and y must be fractions then x + y must be greater than x(y)

B is the answer.
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by Ian Stewart » Thu Jul 14, 2011 10:55 am
tpr-becky wrote:
Then look at statement 2 - the others have done the algebra but look at the concept. The only way for x + y to be greater than x(x) + y(y) is if x and y are fractions.
That's not quite right. If x + y is greater than x^2 + y^2, then it's certainly true that *one of* x or y is between 0 and 1. They do not both need to be between 0 and 1; it might be that x = 1/2, and y = 11/10, for example.

Still, as long as we know that one of our two letters is between 0 and 1, we can be certain that xy is less than x + y (since if, say, 0 < x < 1, then xy < y < x + y must be true), so a conceptual approach works well here.
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