1
n is divisible by 3
Example 3,6 ..

N is odd and even ...So insufficient. AD out Left with BCE

From stmt2
Lets pick numbers
n= 3 2n = 6 n is divisible by 1,3 2n div by 1,3,6
n= 4 2n = 8 n is divisible by 1,2,4 2n div by 1,2,4 8

Even after picking more numbers you will see n is always odd
hence B

Question type y/n

yes if v > 0 no if v < 0

Stem mv < pv < 0

From stmt 1 m < p
Say m = -2 p = -1 v = 1
i.e. mv < pv <0> 0

Now say m = 1 p = 2 v = -1
i.e. m < p
i.e. mv < pv < 0 does not hold
i.e. v < 0

For mv < pv <0> 0
m and v need to -ve and v >0 Hence Sufficient

Knock off BCE ..left with AD

From stmt2

m < 0

Say m = -2 p = -1 v = 1
i.e. mv < pv <0> 0

IF u pick more numbers u will see that v > 0 for the stem to hold true

Hence eliminate A ..D it is

3.
From stmt 1 and stem
say k = 2 n = 9 r = 1

let take one more no
say k=4 n=25 r=1

Hence knowck off BCE left with AD

From stmt 2

k=5 n cud be anything hence r will vary insufficient

Hence A

4. Question type

yes if Xy > 0 no if xy <0> -2

Say x = 3 y = 1
x-y > -2
Xy > 0

Now say x = 3 y =-1
x-y > -2
But xy < 0

Hence AD gone left with BCE

Stmt2 says x -2y <6> 6

Say x = 2 y = 5
2y -x > 6
xy > 0

Now say x=-2 y =5
2y -x > 6
But xy <0> -2 and 2y -x > 6

Pick x = 3 , y =2 and x = -3 y = -2 the condition holds

Hence C

I have no idea about the last one ..the one with the slopes..maybe someone cud throw some light :

sujaysolanki wrote:I have no idea about the last one ..the one with the slopes..maybe someone cud throw some light :

eqn of line t : y = m1*x + c1
eqn of line k : y = m2*x + c2

1 - insufficient. product of x intercepts = (-c1/m1) * (-c2/m2)
= c1c2 / m1m2 = +ve. We don't know if m1m2 is +ve or -ve

2 - insufficient. We just know c1c2 = -ve

Combining 1 and 2, we know m1m2 is negative. So, sufficient. Hence C.

The loose end here is the data-point (4,3) that I have not used
anywhere...

Thank you guys. It`s all clear.