BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Inequalities Question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Oct 08, 2008 11:29 am
Location: Toronto, Ontario
Thanked: 5 times
by anayeri » Fri Nov 21, 2008 4:06 pm
If x<0 and 0 < (x/y) + 1 < 1, which of the following must be true?

I. y > 0

II. x/y > -1

III. (1/x) + (1/y) < 0

A. I
B. I+II
C. I + III
D. II + III
E. I + II + III

OA is E. I get why I and II work, but I'm having a lot of difficulty seeing why III works.

Here's my rationale:

simplifying the original equation, we get:

0.1 < (x/y) + 1 < 0.9
so, -0.9 < x/y < -0.1

Going to III)

(1/x) + (1/y) < 0
= (1/x) < -(1/y)
= 1 < - (x/y)
= -1 > (x/y), which is not true, given that -0.9 < x/y < -0.1

so wouldn't that make III impossible?

Clearly I've missed something. Also, what's the fastest way to answer this type of question?

thanks.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 168
Joined: Thu Nov 13, 2008 4:34 am
Location: Pittsburgh
Thanked: 9 times

by Tryingmybest » Fri Nov 21, 2008 4:15 pm
(1/x) + (1/y) < 0
or we need to prove (x+y)/xy <0

X + Y is +ve as X is negative and Y is positive
(also the magnitude of Y must be greater than X from 0 < (x/y) + 1 < 1)
XY is -ve

so (x+y)/xy <0 =>(1/x) + (1/y) < 0

Hence E
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Oct 08, 2008 11:29 am
Location: Toronto, Ontario
Thanked: 5 times

by anayeri » Fri Nov 21, 2008 4:29 pm
Nice!!! thanks for the quick reply.

Question: where was I flawed in my rationale of the inequality? Why doesn't that work?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 248
Joined: Mon Aug 11, 2008 9:51 am
Thanked: 13 times

by earth@work » Fri Nov 21, 2008 4:29 pm
x<0 and 0 < (x/y) + 1 < 1
I. y > 0
0 < (x/y) + 1 < 1 = 0 < (x+y) < y= 0<y true
II. x/y > -1
subtract -1 from inequality gives: -1<x/y<0 true
III. (1/x) + (1/y) < 0
0<(x+y)/y<1 divide by x = 0>(x+y)/yx>1/x (note the equality sign is reversed here as x is negative)
0>(1/x) - (1/y) true
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 248
Joined: Mon Aug 11, 2008 9:51 am
Thanked: 13 times

Re: Inequalities Question

by earth@work » Fri Nov 21, 2008 4:37 pm
anayeri wrote:
simplifying the original equation, we get:

0.1 < (x/y) + 1 < 0.9
so, -0.9 < x/y < -0.1

Going to III)

(1/x) + (1/y) < 0
= (1/x) < -(1/y)
= 1 < - (x/y)
= -1 > (x/y), which is not true, given that -0.9 < x/y < -0.1

so wouldn't that make III impossible?

Clearly I've missed something. Also, what's the fastest way to answer this type of question?

thanks.
hi , when we simplify
(1/x) + (1/y) < 0
= (1/x) < -(1/y)
till here it's fine, but i guess since x is negative so the equality is reversed
when multiplied by x
= 1 > - (x/y)
= -1 < (x/y)
Also solving 0 < (x/y) + 1 < 1
we get -1<x/y<0 ...therefore III is true[/i]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Fri Nov 21, 2008 5:14 pm
Question: where was I flawed in my rationale of the inequality? Why doesn't that work?
Look at it this way if it helps:

xy is always going to be negative (x is negative given y has to be positive for 0 < (x/y) + 1 < 1 to be true also given)

to prove 1/x+1/y<0

i.e y+x/xy<0
1/x * x+y/y <0

We know x+y/y > 0 (given 0 < (x/y) + 1 < 1 i.e 0<x+y/y<1)

x being negative 1/x*x+y/y is always < 0

Hope I have not missed something here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Sat Sep 20, 2008 10:06 am
Location: JAPAN

by ravikirancheni » Sat Nov 22, 2008 6:53 am
Cramya,

I have been going through your answers and posts for long time now and i really appreciate them !!..

well with the case above i have a slight spinoff ..

If

0 < (x/y) + 1 < 1 is to be satisfied then for sure
(x/y) must be <0 , a fractional value and absolute value of fraction should be less than 1.

1.so as X <0 , Y must be > 1 - OK

2.But second reason is a bit flawed becoz ..if (x/y) > -1 does not make sense as it has to be 0> (x/y)> -1

3.(1/x) <0 (always) as x<0 and (1/y) <0 (always) as y >0 and y>1.

Hope iam clear.....pls help if iam wrong.

I go for I + III
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 168
Joined: Thu Nov 13, 2008 4:34 am
Location: Pittsburgh
Thanked: 9 times

by Tryingmybest » Sat Nov 22, 2008 7:40 am
0 < (x/y) + 1 < 1 will be true only if x/y > -1

let us take x/y = -1 then 0 < (x/y) + 1 < 1 will not hold good

let us take x/y < -1 for example X = -2 then 0 < (x/y) + 1 < 1 will not hold good

So the condition x/y > -1 must be established for 0 < (x/y) + 1 < 1

So the answer will be E :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Sat Nov 22, 2008 7:50 am
0 < (x/y) + 1 < 1 is to be satisfied then for sure
(x/y) must be <0 , a fractional value and absolute value of fraction should be less than 1.

1.so as X <0 , Y must be > 1 - OK

2.But second reason is a bit flawed becoz ..if (x/y) > -1 does not make sense as it has to be 0> (x/y)> -1

3.(1/x) <0 (always) as x<0 and (1/y) <0 (always) as y >0 and y>1.

Hope iam clear.....pls help if iam wrong.

Ravi,
I dint quite catch the whole thing above. Tell me if this helps

0 < (x/y) + 1 < 1

If x/y >-1 then that would result in 0 < negative number which is not true

Therefore x/y has to be negative fraction > -1 i.e between -1 to 0 so that when uts added with 1 its greater than 0 but less than 1

Hope this hleps! Let me know if thsi is not what u asked.

Good luck!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1