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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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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
[Reveal] Spoiler: OA

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Last edited by Mahmud6 on 12 Aug 2017, 07:52, 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 ? [#permalink]

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New post 06 May 2017, 01:11
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.

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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, 03:40
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. :-D

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, 05:54
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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 06 May 2017, 09:00
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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


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

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 13 May 2017, 22:57
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, 00:43
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, 03:01
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, 04:36
Expert's post
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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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, 19:02
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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] 18 May 2017, 19:02
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