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08 Sep 2020, 10:58
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Is \small \frac{x+1}{x-3}< 0 ?
(1) \small -1< x< 1
(2) \small x^{2}-4<0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
(1) \small -1< x< 1
(2) \small x^{2}-4<0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
08 Sep 2020, 11:36
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Hey Vibho,
The question stem is a bit unclear but if it were to be this Is x+1/x-3<0
1) -1<x<1
2) x^2 - 4<0
Then
Statement 1) x can take values between -1 and 1
for any value of x between -1<x<1, (x+1) will be positive, take x=-0.99, then we have x+1 as 0.01
and as with the denominator for any x between -1<x<1, (x-3) will be negative
Take x=0.99, we have 0.99-3= - 2.01, for x=-0.99, -3.99
So during this interval we have a positive / negative which implies that within this range the given equation is always negative, or <0
Hence (1) is sufficient
2) x^2 - 4<0
Implies that x^2<4
or -2<x<2
Now consider the numerator, x+1
For x>-1, x+1 will be positive
but for values between -2<x<-1, we have x+1 as negative
Similarly with the denominator
for all values of x between -2<x<2, x-3 will be negative
Since the numerator is hovering between positive and negative depending on values of x, we can say that (2) is sufficient
Hence the answer is A
The question stem is a bit unclear but if it were to be this Is x+1/x-3<0
1) -1<x<1
2) x^2 - 4<0
Then
Statement 1) x can take values between -1 and 1
for any value of x between -1<x<1, (x+1) will be positive, take x=-0.99, then we have x+1 as 0.01
and as with the denominator for any x between -1<x<1, (x-3) will be negative
Take x=0.99, we have 0.99-3= - 2.01, for x=-0.99, -3.99
So during this interval we have a positive / negative which implies that within this range the given equation is always negative, or <0
Hence (1) is sufficient
2) x^2 - 4<0
Implies that x^2<4
or -2<x<2
Now consider the numerator, x+1
For x>-1, x+1 will be positive
but for values between -2<x<-1, we have x+1 as negative
Similarly with the denominator
for all values of x between -2<x<2, x-3 will be negative
Since the numerator is hovering between positive and negative depending on values of x, we can say that (2) is sufficient
Hence the answer is A
08 Sep 2020, 11:51
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Asked: Is \( \frac{x+1}{x-3}< 0 ?\)
Values of x such that -1 < x < 3, will satisfy the equation
(1)\( -1< x< 1\)
SUFFICIENT
(2) \( x^{2}-4<0\)
-2<x<2
NOT SUFFICIENT
IMO A
Values of x such that -1 < x < 3, will satisfy the equation
(1)\( -1< x< 1\)
SUFFICIENT
(2) \( x^{2}-4<0\)
-2<x<2
NOT SUFFICIENT
IMO A
