Inequalities Question

Question

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.

Tryingmybest · Answer

(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

anayeri · Answer

Nice!!! thanks for the quick reply. Question: where was I flawed in my rationale of the inequality? Why doesn't that work?

earth@work · Answer

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

earth@work · Answer

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]

cramya · Answer

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.

ravikirancheni · Answer

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

Tryingmybest · Answer

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

cramya · Answer

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!

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