If x

Question

If x<0 and 0 < \frac{x}{y} + 1<1 , which of the following must be true ? I. y > 0 II. \frac{x}{y}>-1 III. \frac{1}{x}+\frac{1}{y}<0 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III

Bunuel · Accepted Answer

0 < \frac{x}{y} + 1<1 Subtract 1 from all 3 sides: -1 < \frac{x}{y}<0 . We see that II is true. Since x < 0, then for \frac{x}{y}<0 to be true y must be positive. I must be true. Next, divide 0 < \frac{x}{y} + 1<1 by x and flip the signs since we know that x is negative: 0 >\frac{1}{y} + \frac{1}{x}>\frac{1}{x} . III must be true. Answer: E. Hope it's clear.

Saha6812 · Answer

IMO the answer is E

0<x/y +1<1
-1<x/y<0

As x<0 and x/y <0
Y must be positive

As mod value of x/y lesser than 1, mod value of y is greater than mod value of x

So mod value of 1/x is greater than mod value of 1/y

x < 0, so 1/x < 0
As |1/x |>|1/y|
1/x+1/y<0

So all the conditions are correct
Ans E

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ScottTargetTestPrep · Answer

We can simplify the given inequality: 0 < x/y + 1 < 1 -1 < x/y < 0 Since x is negative, y must be positive. Let’s now analyze our Roman numeral answer choices: I. y > 0 Since we’ve mentioned y must be positive, Roman numeral I is correct. II. x/y > -1 Since -1 < x/y < 0 also means that x/y > -1, Roman numeral II is correct. III. 1/x + 1/y < 0 We can multiply both sides of the inequality by x to obtain: 1 + x/y > 0 Notice that we switch the inequality sign since x is negative. Now let’s subtract 1 from both sides: x/y > -1 Roman numeral III is correct also. Answer: E

GyMrAT · Answer

Given x<0 .............(i) & 0 < \frac{x}{y} + 1<1 ......(ii) can be simplified as, -1 < \frac{x}{y} < 0 ................(iii) I. y > 0 - from (iii), we can say since x<0 , y>0 is true II. \frac{x}{y}>-1 - from (iii), we can say this is true III. \frac{1}{x}+\frac{1}{y}<0 - multiply (ii) with \frac{1}{x} & since x<0 , we need to flip the signs. Hence III is also true. Answer E. Thanks, GyM

Kinshook · Answer

If x<0 and 0 < \frac{x}{y} + 1<1 , which of the following must be true ? -1 < \frac{x}{y} < 0 Since x<0 & x/y<0; y > 0 \frac{x}{y} > - 1 x/y + 1 > 0 ; dividing the equation by x<0 1/y + 1/x < 0 I. y > 0 : MUST BE TRUE II. \frac{x}{y}>-1 : MUST BE TRUE III. \frac{1}{x}+\frac{1}{y}<0 : MUST BE TRUE IMO E

AvaneeshT · Answer

Can someone help me explain the language of the question present here.

If y > 0 must be true, and x < 0 is already stated, then the equation fails at x = -999 and y = 111, even though this follows the condition mentioned in the question and the options. Since the question states that all options must be true, all values for y > 0 should be true, which is clearly impossible here.

What is the fault in my thinking? Can someone explain?

Bunuel · Answer

Your example (x = -999, y = 111) doesn't satisfy the condition -1 < x/y < 0, so it's invalid. For these values, x/y + 1 = -8, which is not greater than -1.

You should understand the methodical approach shown above, and this will avoid you from testing individual values.

If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?

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