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If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5

Given 6/(x(x+1))>1 i.e x(x+1) is +ive implies X is positive ;
since we are dealing with all positives
6/(x(x+1))>1 ---> (x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5

Bunuel am i doing right ?
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Lucky2783
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5

Given 6/(x(x+1))>1 i.e x(x+1) is +ive implies X is positive ;
since we are dealing with all positives
6/(x(x+1))>1 ---> (x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5

Bunuel am i doing right ?

Lucky2783 ,I don't think this derivation of yours is right: x(x+1) is +ive implies X is positive. Please check!
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sytabish
Lucky2783
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5

Given 6/(x(x+1))>1 i.e x(x+1) is +ive implies X is positive ;
since we are dealing with all positives
6/(x(x+1))>1 ---> (x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5

Bunuel am i doing right ?

Lucky2783 ,I don't think this derivation of yours is right: x(x+1) is +ive implies X is positive. Please check!

thanks .
actually we do not need know the sign of X here
Given 6/(x(x+1))>1 implies (x(x+1)) is a +ive quantity so we can simply multilply both sides of inequality by (x(x+1))
6 > (x(x+1))
6> x^2 + x
x^2+x-6 < 0
(x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5
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Hi styabish,

You have to be very careful with your assumptions in the Quant section. This specific Number Property WILL show up on Test Day....

(X)(X+1) = positive

This does NOT mean that X has to be positive.

X COULD be positive....

eg
X = 1
(1)(2) = 2

X COULD be NEGATIVE though...

eg
X = -2
(-2)(-1) = 2

GMAT assassins aren't born, they're made,
Rich
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\(\frac{6}{(x(x+1))} > 1\)

\(\frac{6}{x^2 + x} > 1\)

For x = -3.5

\(\frac{6}{12.25-3.5}\) >> This would be less than 1

For x = -2.5

\(\frac{6}{6.25-2.5}\) >> This would be greater than 1

Answer = B
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viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5


isn't this what you get after factorising
x<-3 or x<2
then how is -2.5 the answer? (since -2.5 is greater than -3)
Can someone help, please?
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viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5


isn't this what you get after factorising
x<-3 or x<2
then how is -2.5 the answer? (since -2.5 is greater than -3)
Can someone help, please?

Let me ask you: what does x < -3 (x is less than -3) or x < 2 (x is less than 2) even mean?

As for the solution please see above.

Theory on Inequalities:
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html
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viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5


isn't this what you get after factorising
x<-3 or x<2
then how is -2.5 the answer? (since -2.5 is greater than -3)
Can someone help, please?

The most straightforward method for this type of question will be to use the values in the options and see which one gives you >1 . Only 1 option must satisfy this requirement.

Additionally, for algebraic solution, look below:

Given : \(\frac{6}{x(x+1)} > 1\) ----> \(\frac{6}{x(x+1)} - 1 > 0\) ---> \(\frac{6-x^2-x}{x(x+1)} > 0\) ----> \(\frac{-6+x^2+x}{x(x+1)} < 0\)

\(\frac{(x+3)(x-2)}{x(x+1)} < 0\) ----> -3<x<-1 or 0<x<2

Only -2.5 lies in this range.

Hope this helps.
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Bunuel
aggarwalpooja
viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5


isn't this what you get after factorising
x<-3 or x<2
then how is -2.5 the answer? (since -2.5 is greater than -3)
Can someone help, please?

Let me ask you: what does x < -3 (x is less than -3) or x < 2 (x is less than 2) even mean?

As for the solution please see above.

This link sorted my worries! For anyone who struggled in the last bit after factorising refer this:
https://www.khanacademy.org/math/algebr ... equalities

Thanks Bunuel! Now I know why did you ask me what didx < -3 (x is less than -3) or x < 2 (x is less than 2) even mean?
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Engr2012
aggarwalpooja
viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5


isn't this what you get after factorising
x<-3 or x<2
then how is -2.5 the answer? (since -2.5 is greater than -3)
Can someone help, please?

The most straightforward method for this type of question will be to use the values in the options and see which one gives you >1 . Only 1 option must satisfy this requirement.

Additionally, for algebraic solution, look below:

Given : \(\frac{6}{x(x+1)} > 1\) ----> \(\frac{6}{x(x+1)} - 1 > 0\) ---> \(\frac{6-x^2-x}{x(x+1)} > 0\) ----> \(\frac{-6+x^2+x}{x(x+1)} < 0\)

\(\frac{(x+3)(x-2)}{x(x+1)} < 0\) ----> -3<x<-1 or 0<x<2

Only -2.5 lies in this range.

Hope this helps.

Thanks Engr12, plugging in the value is guess the easiest!
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\(\frac{6}{(x(x+1))} > 1\)

6 > x(x+1)
\(6 > x^{2} + x\)
Plug In options
B
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Lucky2783
sytabish
Lucky2783
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5

Given 6/(x(x+1))>1 i.e x(x+1) is +ive implies X is positive ;
since we are dealing with all positives
6/(x(x+1))>1 ---> (x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5

Bunuel am i doing right ?

Lucky2783 ,I don't think this derivation of yours is right: x(x+1) is +ive implies X is positive. Please check!

thanks .
actually we do not need know the sign of X here
Given 6/(x(x+1))>1 implies (x(x+1)) is a +ive quantity so we can simply multilply both sides of inequality by (x(x+1))
6 > (x(x+1))
6> x^2 + x
x^2+x-6 < 0
(x+3)(x-2)<0
++++++(-3)----(2)++++++
so anything between -3 and 2 satisfies the inequality .
B. -2.5

hello there! :)
How did you manage to draw the line based on this (x+3)(x-2)<0 is there rule to transform it into line ? :?
thank you :)
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Hi dave13,

Let me try to respond to your query.

If \(\frac{6}{x(x+1)}\)>1, which of the following could the value of x?

note x and (x+1) are two consecutive numbers, then their product is always positive . Why
if x is negative then (x+1) is also negative and their product is always positive
if x is positive then (x+1) is also positive and their product is always positive
But x cannot be -1 and 0 , because the exp will be undefined for these values.

So
we can write \(\frac{6}{x(x+1)}\)>1 as \(\frac{6}{x(x+1)}\)-1>0

So \(\frac{(6-x(x+1)}{x(x+1)}\)>0

\(\frac{(6-x^2-x)}{x(x+1)}\)>0

\(\frac{-((x+3)(x-2))}{x(x+1)}\)>0

When we multiply by -1 on both sides we change the sign of inequality.

\(\frac{((x+3)(x-2))}{x(x+1)}\)<0

Now if you draw the number line and have positive and negative regions this is how it would look

++++++++++++(-3)---------------------(-1)++++++++++++(0)--------------------(2)++++++++++

Now the region where the inequality holds is
-3<x<-1 and 0<x<2

Now options B, C, D are greater than 2 so discard.
Option A is less than -3 so discard
Option B lies between -3<x<-1 so this could be possible value of for which the inequality will hold.


Probus
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viktorija
If 6/(x(x+1))>1, which of the following could the value of x?

A.-3.5
B.-2.5
C.2.5
D.3.5
E.4.5

\(\frac{6}{[x (x + 1)]}\)\(> 1\)

\(\frac{6}{[x^2 + x]}\) > 1

We know that \(x^2 + x\) can never be negative irrespective of the value of \("x"\), therefore

\(6 > x^2 + x\)

Or \(x^2 + x - 6 < 0\)

\((x + 3) (x - 2) < 0\)

\((x - 2) < 0\) or \((x + 3) > 0\)

\(x < 2\) or \(x > -3\)

\(-3 < x < 2\)

Answer : B = \(-2.5\)
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Solution:

6/x(x+1) > 1 and this is not possible for x(x+1)<0

=> x(x+1) >0 and in other words it implies x(x+1) is positive.

=>6 and 1 are positive values too and hence we can now confidently write

6/x(x+1) >1 => 6>x(x+1)

=>x(x+1)<6

=>x^2 + x <6

=>x^2 + x -6 < 0 (Add -6 on both sides)

=>x^2 +3x - 2x -6 < 0

=> x(x+3) - 2(x+3) <6

=> (x+3)(x-2) < 6

Use the wavy line approach now to reach the critical points as -3 and 2,plot them on the number line and consider the negative interval values only
=> +.................(-3)................. - ...................(2)...............+...

NEGATIVE INTERVAL To be considered AS "<" sign in inequality

=> -3<x<2 and the only option satisfying the range is -2.5 (option b)

Devmitra Sen
GMAT SME

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