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Math Tricky Questions

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Math Tricky Questions

by Fab » Mon Jul 21, 2008 10:17 pm
I'll appreciate if you could help me with this ones:

1. If x is an integer, is x*lxl<2^x

(1) x<0
(2) x=-10

I don't get this one.
___________________________________



2. Is 1/p > r/(r^2)+2

(1) p=r
(2)r>0

I thougt answer was A, however is C.

___________________________________

3. If x is diferent from -y, is (x-y/x+y)>1

(1) x>0
(2) y<0

This one is very tricky..

Thanks for your help.
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by pepeprepa » Tue Jul 22, 2008 1:44 am
Post one question per post please.

1. If x is an integer, is x*lxl<2^x

(1) x<0
(2) x=-10


1) x<0 so x*lxl<0 and 0<2^x
So x*lxl<2^x

2) x<0 so x*lxl<0 and 0<2^x
So x*lxl<2^x

C) Each statement alone is sufficient



___________________________________



2. Is 1/p > r/(r^2)+2

(1) p=r
(2)r>0

I thougt answer was A, however is C.


Instinctivly I would say A like you !
2) r>0 then we have 0<r/((r^2)+2)<1
The problem is that 1/p could be <0 or >1 or between 0 and 1
So we can't say it is true or not.
That is strange.

___________________________________

3. If x is diferent from -y, is (x-y/x+y)>1

(1) x>0
(2) y<0

I would say E but I do not have time.
I just tried with 1) and 2)
x>0 and y<0
take x=3 and y=-2
and after x=3 and y=-6
It depends on x<y or x>y
So i would say E, not sure.
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by Fab » Tue Jul 22, 2008 1:28 pm
THANKS.

1. Is clear.

2. Not clear, this is an 11th OG GMAT question (DS # 145). The explanation in the book is not clear at all.
I think it's A because of simple substitution (p=r). Choice 2 doesn't make any sense since r>0 doesn't tell you anything about P.

3. Correct, answer is E. However this is what i did:

PROBLEM: (x-y/x+y) > 1
then: x-y > x+y
then -y>y
so, this is true for choice B

is this way is wrong?

THANKS AGAIN
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by sibbineni » Sun Aug 03, 2008 10:52 am
Is 1/p > r/(r^2)+2

(1) p=r

1/r>r/(r^2)+2

if r=1
1>1/3
then 1/r>r/(r^2)+2 holds true
if r=-1
-1>-1/3 (wrong)
then
then 1/r>r/(r^2)+2 holds false
so Insuff

(2)r>0

p can be >0 or <0 .... insuff

together (1) and (2) makes suff

IMO C
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by richlittl » Mon Aug 04, 2008 10:11 am
Hi Fab, here should be the reasons:

1. If x is an integer, is x*lxl<2^x

(1) x<0
(2) x=-10

The reasoning behind this is you have one side = x*|x| and another side is 2^x

please keep in mind that 2^x will ALWAYS be positive no matter what number x is. However, x*|x| will be negative when x = negative. Argument is valid and correct answer is C

2. Is 1/p > r/(r^2)+2

(1) p=r
(2)r>0

I thougt answer was A, however is C.

One thing to keep in mind whenever you see an equation like this(greater or lesser sign in there), is you always have to be careful to test whether the variable will be positive or negative. FYI: positive and negative will change the <> sign

Indeed if p,r >0 and p=r, 1/p > r/[(r^2)+2] is valid , since the right side of the function will always have greater neumerator value, however, this situation will change if r,p <0
___________________________________

3. If x is diferent from -y, is (x-y/x+y)>1

(1) x>0
(2) y<0


To simplify the function, under circumstance x does not equal -y ( this assumption must be valid otherwide we cannot even do this question due to neumerator will be 0)

x-y>x+y further down, we have 0>2y, then 0>y (if this is your solution, you are wrong!!!) please note in Q2, we stated the concern on positive and negative, right here, since there is no way we can tell if x+y is positive, the function will have two results

if x+y > 0, then we have 0>2y
if x+y < 0, then we have x-y<x+y, 0<2y, which gives us conflicting results. Thus E is correct

Hope this helps :D
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by *arch* » Tue Aug 05, 2008 7:43 pm
If x is an integer, is x*lxl<2^x

(1) x<0
(2) x=-10

.. In the question above, shouldn't the answer be (D)?

As the equation x*|x| will always be lesser than 2^x for negative values of x. I'm thinking that statements (1) and (2) should work because they both say that x<0, and are therefore independently sufficient to arrive at x*|x|<2^x.
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by india_supports » Fri Aug 08, 2008 6:45 am
richlittl wrote: ___________________________________

3. If x is diferent from -y, is (x-y/x+y)>1

(1) x>0
(2) y<0

since there is no way we can tell if x+y is positive, the function will have two results

if x+y > 0, then we have 0>2y
if x+y < 0, then we have x-y<x+y, 0<2y, which gives us conflicting results. Thus E is correct

Hope this helps :D
Hey richlittl

I like the approach, one question: can we conclude E, just coz the two possible equations yield conflicting results?

cheers
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