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

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

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New post 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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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

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.
Edited the question.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 06 May 2017, 10:00
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4
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


\(0 < \frac{x}{y} + 1<1\)

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.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 06 May 2017, 02:11
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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.

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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 06 May 2017, 04:40
1
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.

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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

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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 06 May 2017, 06:54
2
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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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 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 ?  [#permalink]

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New post 14 May 2017, 01:43
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.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 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 ?  [#permalink]

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New post 14 May 2017, 05:36
KM2018AA wrote:
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)


You should know the signs. Check the links below for more:

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... 09468.html
http://gmatclub.com/forum/everything-is ... 08884.html
http://gmatclub.com/forum/graphic-appro ... 68037.html

Hope this helps.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 18 May 2017, 20:02
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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 ?  [#permalink]

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New post 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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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 28 Sep 2018, 09:32
Bunuel 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


\(0 < \frac{x}{y} + 1<1\)

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.


Even though Statement II is true in this context, it is not true for values of \(\frac{x}{y}\)>0 because that contradicts the statement in the question stem after manipulating it (\(-1 < \frac{x}{y}<0\)). But then again, it is true in the context of this question. So if we get something similar in that sense, do we consider the statement to be true or not (Choice 'C' vs. Choice 'E')? How does that fit in the "must be true" VS. "could be true" context? Does anyone share the same misunderstanding in GMAT questions? I would appreciate if anyone could clarify this concept.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?  [#permalink]

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New post 01 Jan 2019, 23:38
Bunuel 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


\(0 < \frac{x}{y} + 1<1\)

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.

But for II. to be "must true" x/y must also be less than 0, which is not mentioned in II. then why is it correct? can anyone please help me out? also, i'm having the same doubt with III.
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Re: If x < 0 and 0 < x/y + 1 < 1, which of the following must be true ?   [#permalink] 01 Jan 2019, 23:38
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