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
DS-Questions from Gmat Prep
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Source: Beat The GMAT — Data Sufficiency |
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sujaysolanki
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sujaysolanki
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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
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
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sujaysolanki
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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
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
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sujaysolanki
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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
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
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sujaysolanki
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I have no idea about the last one ..the one with the slopes..maybe someone cud throw some light 
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jayhawk2001
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eqn of line t : y = m1*x + c1sujaysolanki wrote:I have no idea about the last one ..the one with the slopes..maybe someone cud throw some light
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...
