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

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

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

let's analyze the denominator. x(x+1) = x^2+x. now x^2+x will always be positive except for numbers lying between -1 and 0. now if x lies between -1 and 0, then the fraction
\(\frac{6}{x(x+1)}\) will be negative. this violates our initial given condition that \(\frac{6}{x(x+1)} >1\). hence the expression x^2+x will always be positive.

now since expression x^2+x is positive, therefore we can cross multiply. Thus we have
x^2+x<6
x^2+x-6<0
(x+3)(x-2)<0

now for all values of x which are less than -3, (x+3)(x-2) will always be positive. similarly for all values of x , which are greater than 2, (x+3)(x-2) will always be positive.

hence our desired range is between -3 and 2. i.e. -3<x<2. now out of the given options, only option b lies inside this range. hence answer must be B

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

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

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

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

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

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.

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.

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?

Thanks Engr12, plugging in the value is guess the easiest!

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

6 > x(x+1)
\(6 > x^{2} + x\)
Plug In options
B

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

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

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

\(\frac{6}{[x^2 + x]}\) > 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 = \(-2.5\)

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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