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
Divisibility Problem
This topic has expert replies
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
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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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.
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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- Ozlemg
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wouldnt the red coloured equation be ](x+y)^2 - 2xygmatfeel 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
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Hi,Ozlemg wrote:wouldnt the red coloured equation be ](x+y)^2 - 2xygmatfeel 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
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
Cheers!
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Could you please elaborate how did you get (x-y)^2 >= 0 in the picture?Frankenstein wrote:Hi,Ozlemg wrote:wouldnt the red coloured equation be ](x+y)^2 - 2xygmatfeel 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
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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Hi,blackjack wrote: Could you please elaborate how did you get (x-y)^2 >= 0 in the picture?
It is the square of a number (x-y). It is always non-negative(>=0) irrespective of whether (x-y) is negative or positive.
Cheers!
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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.
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.
Becky
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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.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.
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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