Algebra

[heshamelaziry]

If x and y are integers ans xy doesn't = zero, is x-y > 0 ?

(1) x/y <1/2
(2) sqrt x^2 = x and sqrt of y^2 = y

OA is C

I don't understand the OA. There is no restricion n the value of y and i don't see a connection between x value and y value !!!?

[Talkativetree]

If x and y are integers ans xy doesn't = zero, is x-y > 0 ?

(1) x/y <1/2
(2) sqrt x^2 = x and sqrt of y^2 = y

(1) this tells us a couple of things depending on the values of x and y.

if x and y are both (+), then 0<2x<y
If x and y are both (-), then 0>x>.5y
If x is (-), but y is (+), then y>0>2x
if y is (+), but x is (+), then 2x>0>y

All this is unnecessary to write out, I only wrote it out to give a greater understanding of the problem. as you may be able to see, (1) is insufficient on it's own.

(2) sqrt x^2 = x and sqrt of y^2 = y
-looking at this, I would say insufficient as well.

So now we get either C or E. Looking at (1), lets see which options would

if x and y are both (+), then 0<2x<y
If x and y are both (-), then 0>2x>y
If x is (-), but y is (+), then y>0>2x
if y is (+), but x is (+), then 2x>0>y

Now, this is where I'm fuzzy on the exact rules of square roots that someone else could help out with, but I believe when you're taking the square root of a square, you taking it in as an absolute value. So if x was negative, then sqrt x^2 would equal a positive number. Now, with the equations in (2), we know that the solution for sqrt x^2 and sqrt y^2, are x and y respectively. because the sqrt of a square will be positive, then both x and y should both be positive, leaving us with only one possibility, 0<2x<y.

so if x =1, then y>2 (3 for example). Let's plug these in.

x-y>0
1-3>0
-2>0 is wrong.

This means that C is sufficient to answer the questions of is x-y>0.

This is a great example, because we disprove the question. Often times people get caught up in trying to answer yes, not yes or no. HOWEVER, i am not completely sure about the absolute value thing regarding the sqrt of squares rule, so take this with a few grains of salt

[sanjana]

This is how I would go about it.

Question : is x-y>0 or is x>y

Statement 1:
------------

x/y<1/2

If x and y are +ve we can say that y>x and hence this statement will be sufficient.
But we dont know the signs of x and y and hence cannot say whether x is larger or y
hence,insufficient.

Statement 2:
------------

sqrt(x^2)=x and sqrt(y^2)=y
this can be written as
|x|=x and |y|=y

Again cannot say if x is greater or y,insufficient.

Combining 1 and 2,

We have |x|/|y|<1/2
==>+ve/+ve<1/2
Hence y>x and this answers our question is x>y(NO!).
Therefore C.

[heshamelaziry]

This is how I would go about it.

Question : is x-y>0 or is x>y

Statement 1:
------------

x/y<1/2

If x and y are +ve we can say that y>x and hence this statement will be sufficient.
But we dont know the signs of x and y and hence cannot say whether x is larger or y
hence,insufficient.

Statement 2:
------------

sqrt(x^2)=x and sqrt(y^2)=y
this can be written as
|x|=x and |y|=y

Again cannot say if x is greater or y,insufficient.

Combining 1 and 2,

We have |x|/|y|<1/2
==>+ve/+ve<1/2
Hence y>x and this answers our question is x>y(NO!).
Therefore C.

So if x and y are positive you assume that y is 2 and x is 1 from statement 1.? This is how i ddid it: since x and have the same sign, i crossed multiply and got 2x < y and here the value of y is not dependant on the value of X ???? Please explain

[sanjana]

The simple logic here is when u divide if the NR and DR are Equal the result will be one
If NR>DR (BOTH +VE) - the quotient will be >1
If NR<DR (BOTH +VE) - the quotient will be <1

Here we know that the quotient of 2 +ve no:s is <1 (1/2)
hence the DR(Y) HAS TO BE GREATER THAN NR(X).

Hope this helps clear.