gud prob

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gud prob

by vivek.kapoor83 » Tue Oct 07, 2008 1:01 am
9. If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

2) z^2 = x^2+y^2

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by stop@800 » Tue Oct 07, 2008 1:19 am

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by vivek.kapoor83 » Tue Oct 07, 2008 1:25 am
how cm A

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by vivek.kapoor83 » Tue Oct 07, 2008 1:27 am
OA is A. Pls explain..X> y+z
and this holds true if no. are +ve
wt abt -ve no.

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by stop@800 » Tue Oct 07, 2008 1:31 am
vivek.kapoor83 wrote:OA is A. Pls explain..X> y+z
and this holds true if no. are +ve
wt abt -ve no.
You are given in the question X, Y and Z are positive integers, :)

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by namitrajiv » Tue Oct 07, 2008 1:45 am
hI stop@800,
Can you kindly prove, hows b insufficient ,

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by stop@800 » Tue Oct 07, 2008 1:53 am
namitrajiv wrote:hI stop@800,
Can you kindly prove, hows b insufficient ,
x – y – z > 0

x > z + y

y is positive
so if x is greater than z+y, it will certainly be greater than z-y as we are subtracting something [2y] positive.

so
B is sufficient

Example:

100 > 50

100 will always be greater than 50-x
where x is positive integer

Hope this helps!!!

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by namitrajiv » Tue Oct 07, 2008 1:56 am
In case B is sufficient then answer must be D

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by stop@800 » Tue Oct 07, 2008 2:12 am
namitrajiv wrote:In case B is sufficient then answer must be D
Buddy, I justified A
and A is the answer.

I assumed that you wanted me to explain A and you wrote B by mistake.

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by namitrajiv » Tue Oct 07, 2008 2:23 am
no actually I wanted to knw hw B is insufficient,

i.e z^2 = x^2+y^2 ,
it seems to me this is sufficient to prove that
x>z-y

e.g z= 13, y = 12, x =5
z= 13 , y = 12, x = 5

I cant find a case where the condition is insufficient

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by stop@800 » Tue Oct 07, 2008 2:44 am

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by 4meonly » Tue Oct 07, 2008 9:26 am
stop@800 wrote:36, 77, 85
z^2 = x^2+y^2
85^2 = 36^2 + 77^2
36>85-77
77>85-36
x>z-y
So B suff with D as answer

I agree with A but your example made me unsure

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by Ian Stewart » Tue Oct 07, 2008 10:01 am
stop@800 explained well above why the first statement is sufficient.

The second statement is also sufficient. If we know that x, y and z are positive, and that the Pythagorean relationship holds:

x^2 + y^2 = z^2

then we know that we can make a right angled triangle with sides x, y and z, where z is the hypotenuse. And in any triangle, the sum of two sides is always larger than the third side:

x + y > z
x > z - y

So the answer should be D.
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by cubicle_bound_misfit » Tue Oct 07, 2008 11:12 am
Hi Ian,

in GMATLAND, ain't 0 a positive integer?

Also, the question stem does not say X Y Z are all different.

how can phythagoras be used then?
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by Ian Stewart » Tue Oct 07, 2008 2:56 pm
cubicle_bound_misfit wrote:Hi Ian,

in GMATLAND, ain't 0 a positive integer?

Also, the question stem does not say X Y Z are all different.

how can phythagoras be used then?
No, 0 is not a positive integer. It is an integer, but it isn't positive, and it isn't negative.

If you have two lines of length x and y, and you connect them at right angles, the hypotenuse z will automatically satisfy

x^2 + y^2 = z^2

That's Pythagoras. It doesn't matter if x and y are different or equal. So you can make a triangle with sides x, y and z, and x+y must be greater than z.

(and while it's not especially important, in this question, x actually must be different from y if the second statement is true, because we know x, y and z are all integers -- if x were an integer and equal to y, then z would need to be equal to root(2)*x, which wouldn't be an integer).
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