[siddharthaduggirala]

Originally Posted by guy with guts

For the first one,compare the lines with the standard equation of y=mx+c and if m positive slope positive.
for first statement-
m=(5+7)/(2+1)=4
second statement,m=4
so either is suffeciet-D

2.answer is C.
from the point of intersection, the two equations will be-
k=>12=16m'+5c'
l=>12=16m"+5c"

from first statement m'*m"=-1
and from second statement 5C'=0,m'=3/4
so m"=-4/3.

[/LEFT]

Hey guywithguts,
U need not actually calculate the values to answer this particular question.

[nuttyvarun]

Originally Posted by siddharthaduggirala

Hey guywithguts,
U need not actually calculate the values to answer this particular question.

@siddharthaduggirala: I guess that was to help dumbos like me.. as I pointed out that I am bad @ CG..

@gutsy: thanks

[siddharthaduggirala]

Originally Posted by nuttyvarun

@siddharthaduggirala: I guess that was to help dumbos like me.. as I pointed out that I am bad @ CG..

@gutsy: thanks

No one is a dumb buddy..

[amsey1382]

Time to try some tough one...
IS X a negative number?

1) X^2 is a positive number
2) x. |Y| is not a positive number

If a & b are integers, and |a| > |b|, is a.|b|< a-b ?
1) a < 0
b) ab >= 0.

[getneonow]

Originally Posted by amsey1382

Time to try some tough one...
IS X a negative number?

1) X^2 is a positive number
2) x. |Y| is not a positive number

1)=> X #0 but X>0 or X<0

2)=> since |Y| is always >= 0 => x<=0

from 1 and 2 => x <0

Hence ans C) Answered using both but not alone..

If a & b are integers, and |a| > |b|, is a.|b|< a-b ?
1) a < 0
b) ab >= 0.

too drowsy to ans this now..on a rough mental calc is it A??

[srijack]

Originally Posted by amsey1382

Time to try some tough one...
IS X a negative number?

1) X^2 is a positive number
2) x. |Y| is not a positive number

From 1.

X not equal to 0.

x can take both +ve and -ve. so cannot ans from 1.

From 2:

x.|y| is zero or -ve.

a.if |y| is not 0 , x has to be -ve . => not +ve
b.If y=0 then x can take any number. => not +ve
c. y any number ( 0 , -ve,+ve) and x is 0. => not +ve

so, from 2 we cannot conclude whether x is -ve.

from 1 & 2 :

x is not equal to 0.

cases a, b.
nothing says |Y| is is not zero

so 1 and 2 are not sufficient to ans the q

If a & b are integers, and |a| > |b|, is a.|b|< a-b ?
1) a < 0
b) ab >= 0.

from 1.
|a| > |b| and a<0

b can be 0,+ve ( but < a),b -ve (but mag of b < mag of a )
so range of b can be (-a,a)

a.|b|< a-b satisfies when b is -ve but not when b is (0,+ve)

so 1 is not sufficient.

from 2.

|a| > |b|
implies |a|-|b| > 0

a cannot be 0. but b can be 0

for ab>=0
If a > 0 , b (0,+ve)
if a < 0 , b (0,-ve)

is a.|b|< a-b , satisfies when a and b are -ve
fails when any a,b are +ve

so 2 alone is not sufficient.

from 1 & 2

a is <0 implies b is <0

so .can be ans using both.

...............

OAs pls

[deepakraam]

For the below q

S X a negative number?

1) X^2 is a positive number
2) x. |Y| is not a positive number


It cannot be answered by using both the statements together.

[deepakraam]

For the below q

If a & b are integers, and |a| > |b|, is a.|b|< a-b ?
1) a < 0
b) ab >= 0.


Statement 1 alone is not sufficient as b=0 case is not clear from statement 1

From statement 2 :it is also not clear :a<0, b<0 or a>0 ,b>0 or a<0 ,b=0 or a>0 , b=0.

Even combining both the statements
a<0 and ab>=0.so b can be -ve or b=0.

Even combining both the statenebts will not give solution.

[nuttyvarun]

Originally Posted by amsey1382

IS X a negative number?

1) X^2 is a positive number
2) x. |Y| is not a positive number

Option 1: X^2 = positive number
X can be +ve or -ve.
NOT SUFF

Option 2: X.|Y| <> +ve
now, |Y| is always +ve
but <> +ve can mean X.|Y| = -ve or X.|Y| = 0
NOT SUFF

Now taking option 1 and option 2 together, we find that X is -ve.. thus correct answer must be option C

[bhavin422]

Originally Posted by nuttyvarun

Option 1: X^2 = positive number
X can be +ve or -ve.
NOT SUFF

Option 2: X.|Y| <> +ve
now, |Y| is always +ve
but <> +ve can mean X.|Y| = -ve or X.|Y| = 0
NOT SUFF

Now taking option 1 and option 2 together, we find that X is -ve.. thus correct answer must be option C

hey varun ...not necessary that x wud be -ve.
St 2 has 2 alternatives. y could be zero and x could be any real no OR x is -ve ...
So combines we cannot sat that x is necessarily negative ...

Ans E