Carcass wrote:
If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?
(A) \(x > y + z\)
(B) y and z are positive.
(C) \(1 > x(y + z)\)
(D) \(1 < xy + xz\)
(E) \(\frac{1}{x} > z – y\)
Given that \(\frac{1}{x}>(y+z)\), where x<0
Multiplying 'x' both sides, (inequality sign to be flipped since x is negative)
\(x*\frac{1}{x}>x(y+z)\)
Or, \(1 < xy+xz\)
Ans. (D)QS: can we write the reciprocal as -\(\frac{1}{x}\) and then proceed. please correct me if my thought process is wrong.
YES, IT'S WRONG. Suppose x = -5, reciprocal of x is \(\frac{1}{x}\) = \(\frac{1}{(-5)}\) = -\(\frac{1}{5}\); If you say, reciprocal of x is -\(\frac{1}{x}\), then -\(\frac{1}{x}\) = \(\frac{-1}{-5}\) = \(\frac{ 1}{5}\).