Is the value of x less than 0?

Is x<0?

(1) The product of x and y is negative.
xy is negative. So one of x or y is negative and other positive.
If y>0, x<0, but if y<0, x>0.

(2) The difference between x and its reciprocal is positive.
\(x-\frac{1}{x}>0...\frac{x^2-1}{x}>0\)
If \(x>0..x^2-1>0..x>1\)....\(x=2....2-\frac{1}{2}=\frac{3}{2}\)....x>0
If \(x<0..x^2-1<0....x<0\)...\(x=-\frac{1}{2}...-\frac{1}{2}-(-2)=\frac{3}{2}\)....x<0

Combined..
x<0 or x>1
Insuff

E

Is the value of x less than 0?

(1) The product of x and y is negative.

it means x and y are of opposite sign. but since no other info is given, it is NOT SUFFICIENT.

(2) The difference between x and its reciprocal is positive.

The difference between any 2 numbers is always non-negative.
It is zero if the numbers are equal. Otherwise always positive.

The difference between two numbers is defined as the distance between the two numbers on the number line.

Statement 2 states that the x is not equal to 1/x or x is not equal to 1 or -1.
No other info, hence NOT SUFFICIENT.

Even after combining 2 statements, we are unable to conclude anything.
Hence it is E
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We are to determine if x is less than 0.

Statement 1: xy<0
Clearly insufficient. This is because x can be either positive or negative depending on whether y is positive or negative. When y is positive, x is negative, and when y is negative, x is positive. The answer is, therefore, yes and no.

Statement 2: The difference between x and it's reciprocal is positive.
Statement 2 is not sufficient. This is because x can be 51, implying its reciprocal is 15. 51-15=36, leading to answer No to the question posed. X can also be -15, and its reciprocal is -51. -15--51=36 satisfying the given condition and this now leads to the answer Yes to the question posed.

1+2
Still insufficient. Both statements do not restrict the value of x to less than 0, equal to 0, or greater than 0. When y is negative, x=51 satisfies statement 1 and statement 2, and leads to the answer No to the question asked.
On the other hand, when y>0 x can equal -15, which satisfies both statements and leads to the answer Yes.

The right answer in my view is E.

(1) The product of x and y is negative.....Clearly insufff.....Either of x,y can be +ve,-ve
(2) The difference between x and its reciprocal is positive.....x-1/x can be positive if x is positive(1,2...) also for x=-1/2,-4/3....So instuff



OA:E

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(1) The product of x and y is negative.
--> x & y have opposite signs (If x > 0, y < 0 or of x <0, y > 0) --> Insufficient

(2) The difference between x and its reciprocal is positive.
--> x - 1/x > 0
--> (x^2 - 1)/x > 0

Formula: If A/B > 0 --> Either both A>0 & B>0 or both A<0 & B<0
Similarly, if (x^2 - 1)/x > 0

Case 1: x^2 - 1 > 0 & x > 0
--> x > 1 (x is positive)

Case 2: x^2 - 1 < 0 & x < 0
--> -1 < x < 0 (x is negative)
--> Insufficient

Combining (1) & (2),

Also, x can be either positive or negative --> Insufficient

IMO Option E

Is the value of x less than 0?
x < 0 ?

(1) The product of x and y is negative.
\(x.y < 0\)
(+).(-), x < 0 NO
OR
(-).(+), x < 0 YES

INSUFFICIENT.

(2) The difference between x and its reciprocal is positive
\(x - \frac{1}{x}\) > 0
It's possible only when x > 0 since x < 0 results in a negative difference.

Let x = 2 then \(2 - \frac{1}{2} = \frac{3}{2}\) > 0
x = -2 then \(-2 - \frac{1}{(-2)} = \frac{-3}{2}\) < 0
Hence x has to be positive. x < 0 NO

SUFFICIENT.

IMO Answer B.
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Is the value of x less than 0?

Is x < 0?

(1) The product of x and y is negative.
xy < 0
x < 0 or y < 0
insufficient

(2) The difference between x and its reciprocal is positive.

x - 1/x > 0
x2 -1 > 0
(x+1)(x-1) >0
x< -1 or x> 1 - we get that x can be both positive or negative

insufficient

taken (1) + (2) together, x can still be both positive and negative - insufficient
Thus, E is the correct answer

#1
The product of x and y is negative.
either of x&y is -ve insufficient
#2

The difference between x and its reciprocal is positive.
x=2 ; reciprocal 1/2
∆ ; 2-1/2 ; 3/2
x=-1/2 ; reciprocal ; -2
∆ -1/2+2 ; 3/2
insufficient
from 1 &2
nothing conclusive
IMO E

Is the value of x less than 0?

(1) The product of x and y is negative.
(2) The difference between x and its reciprocal is positive.

We have to find whether or not x<0.

(1) xy<0
=> Either of x and y is negative and the other is positive.
So, we do not have sufficient information to find out whether or not x<0.
Thus, insufficient.

(2) x-\(\frac{1}{x}\)<0
=> \(\frac{(x^2-1)}{x}\)<0
=> x>1 or -1<x<0
So, x can be positive or negative.
Thus, insufficient.

From (1) and (2) together, we do not have sufficient information to find out whether or not x<0.
Thus, insufficient.

Therefore, the correct answer is option E.
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Asked: Is the value of x less than 0?

(1) The product of x and y is negative.
xy < 0
x<0;y>0 or y<0;x>0
NOT SUFFICIENT

(2) The difference between x and its reciprocal is positive.
x - 1/x > 0
(x^2 - 1)/x > 0
(x-1)(x+1)/x >0
x> 1 or -1<x<0
NOT SUFFICIENT

(1) + (2)
(1) The product of x and y is negative.
xy < 0
x<0;y>0 or y<0;x>0
(2) The difference between x and its reciprocal is positive.
x - 1/x > 0
(x^2 - 1)/x > 0
(x-1)(x+1)/x >0
x> 1 or -1<x<0
x may or may not be <0
NOT SUFFICIENT

IMO E
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Imo. E

Is the value of x less than 0?
Is x = -ve?

(1) The product of x and y is negative.
xy<0, x can or can not be -ve. Insufficient.

(2) The difference between x and its reciprocal is positive.
x-1/x = +ve
x can be 2 or -1/2, So x can be or can not be -ve. Insufficient.

1 + 2)
X can be or can not be -ve. Insufficient.

Q : x<0 ?
Q type: Y/N

a) xy<0 => x and y have opposite signs. Either of them can be negative. Insufficient

b) The difference between x and its reciprocal is positive. => (x - 1/x) > 0 => (x^2 -1) > 0
or, x^2 >1
x can be positive or negative. (e.g. If x^2 =4, the x can have two values +2 or -2) Insufficient

a) + b) = x can be positive or negative.

Correct answer: E

(1) The product of x and y is negative. insufic.

\(xy<0:[x>0,y<0]…or…[x<0,y>0]\)

(2) The difference between x and its reciprocal is positive. insufic.

\(x-\frac{1}{x}>0…\frac{x^2-1}{x}>0…\frac{(x-1)(x+1}{x}>0…[x<-1,x>1]\)

(1 & 2): insufic.

Answer (E)

x< 0 ???

(Statement1): x*y< 0
—> either x or y could be positive or negative to get this inequality correct.
Insufficient

(Statement2): \(x—\frac{1}{x}\)>0

\(\frac{(x^{2}—1)}{x}\)>0

\(\frac{(x+1)(x—1)}{x}\)> 0
—> —1<x <0 and x>1
Clearly insufficient

Taken together 1&2,
—> it depends on statement1
If x< 0, then Yes
If x>0, then No

Insufficient

The answer is E.

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Is the value of x less than 0?

is X negative?

(1) The product of x and y is negative.
meaning either x or y is negative.
we don't know which one.
Therefore, insuffucient

(2) The difference between x and its reciprocal is positive.
x- (1/x) is positive
(1/x)-x is positive
there is no such number that will give those both statement true
insufficient

together, we know y doesn't give anything and with those statement in b doesn't add any value.

Therefore, E

1. xy<0 means either x or y is negetive. so it is not enough

2. the reciprocal of x is 1/x
so x-1/x is positive. here this is positive result as x is larger then 1/x. but if the value of x is negetive then the result will be negetive. so this is also wrong

so answer is E

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The trap here is to blindly assume value of X to be an integer.

Statement 1 is clearly not sufficient as X can be greater than 0 or less than 0. Not sufficient.

Statement 2-.. tricky part.

Let's see , X -1/X >0
(X^2-1)/X > 0, at first one might think for this expression to be true, X has to be positive.. and if one thinks that way, he/she has made a wrong assumption of X to be a positive integer..

Take X = 4
(16-1)/4 > 0, I get a no
Take X = -0.5
(0.25-1)/-0.5 > 0, but X is less than 0 and I get a yes . This statement is not sufficient as well .

Combining both statements is of no use because X can be positive or X can be negative.. eliminate C and E is the right answer..

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