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If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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Updated on: 12 Aug 2017, 08:52
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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
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Md. Abdur Rakib
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Originally posted by AbdurRakib on 06 May 2017, 01:26.
Last edited by Mahmud6 on 12 Aug 2017, 08:52, edited 2 times in total.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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06 May 2017, 10:00




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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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06 May 2017, 02:11
AbdurRakib wrote: 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 IMO, (D) should be the answer. If statement I is true, then y > 0. Let us choose y = 1. And it is given that, x<0 and 0 < \(\frac{x}{y}\) + 1<1 Putting y = 1, and taking any value of x < 0 (say, x = 1) , we find that, \(\frac{x}{y}\) + 1 = 0. Now it is NOT given that 0 <= \(\frac{x}{y}\) + 1<1, but 0 < \(\frac{x}{y}\) + 1<1. Therefore statement I cannot be true. Only option (D) is present which does not contain statement I. So, IMO (D) is the answer. If you like my post, please encourage by giving KUDOS.



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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06 May 2017, 04:40
moutikli wrote: AbdurRakib wrote: 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 IMO, (D) should be the answer. If statement I is true, then y > 0. Let us choose y = 1. And it is given that, x<0 and 0 < \(\frac{x}{y}\) + 1<1 Putting y = 1, and taking any value of x < 0 (say, x = 1) , we find that, \(\frac{x}{y}\) + 1 = 0. Now it is NOT given that 0 <= \(\frac{x}{y}\) + 1<1, but 0 < \(\frac{x}{y}\) + 1<1. Therefore statement I cannot be true. Only option (D) is present which does not contain statement I. So, IMO (D) is the answer. If you like my post, please encourage by giving KUDOS. Hello If we simplify the given equation 0<x/y +1 < 1 It equals 0<x+y < y Which implies y>0 Hence I is true Is this correct? Or am i doing something wrong Thanks Sent from my SMN9200 using GMAT Club Forum mobile app



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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06 May 2017, 06:54
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 Sent from my BLNL22 using GMAT Club Forum mobile app



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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13 May 2017, 23:57
Bunuel Please let me know whether my method of deriving III as true is valid ? From the question stem we know that x<0 and and \(\frac{x}{y}\) lies between 0 and 1 i.e. 1<\(\frac{x}{y}\)<0 Hence we necessarily have to have y>0 Multiplying Y on both sides (without changing signs because y>0) \(x>y\) result 1 Given in III is \(\frac{1}{x}+\frac{1}{y}<0\) Therefore, \(\frac{1}{x}<\frac{1}{y}\) Taking resiprocals \(\frac{x}{1}\)>\(\frac{y}{1}\) Hence \(x >y\) result 2 result 1 and result 2 are the same Hence III is correct



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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14 May 2017, 01:43



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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14 May 2017, 04:01
Bunuel wrote: KM2018AA wrote: Bunuel Please let me know whether my method of deriving III as true is valid ? From the question stem we know that x<0 and and \(\frac{x}{y}\) lies between 0 and 1 i.e. 1<\(\frac{x}{y}\)<0 Hence we necessarily have to have y>0 Multiplying Y on both sides (without changing signs because y>0) \(x>y\) result 1 Given in III is \(\frac{1}{x}+\frac{1}{y}<0\) Therefore, \(\frac{1}{x}<\frac{1}{y}\) Taking resiprocals
\(\frac{x}{1}\)>\(\frac{y}{1}\)Hence \(x >y\) result 2 result 1 and result 2 are the same Hence III is correct Your solution missing steps. In the highlighted part, do you know the sign of y? How? If not then you cannot do that. So then what is the rule with taking reciprocals in inequalities? (if any)



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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14 May 2017, 05:36



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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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18 May 2017, 20:02
AbdurRakib wrote: 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 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
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?
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16 Jun 2018, 12:31
AbdurRakib wrote: 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 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
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