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Re: Modulus question [#permalink]
05 Oct 2011, 05:40

Question is basically asking if X is a fraction or 0.

A is insufficient because we can see by testing numbers that it can be either a negative fraction or a positive integer and be true. For example: If X = -1/2 that gives us -1/32 > -1/8 which is true. If X = 2, the equation is 32 > 8 which is also true.

B is insufficient because it can be any fraction or negative number and still fulfill this prompt. For example:

If X = -1/2 then we get 1/4 > -1/8 which is true If X = 1/2 then we get 1/4 > 1/8 which is true If X = -2 then we get 4 > -8 which is true

The answer is C because together we know that X must be a negative fraction.

Re: Modulus question [#permalink]
05 Oct 2011, 06:43

Good question.

If we draw the graph, we can figure out that individually both the option doesn't satisfy but if we join together, the value of x satisfy between the region - (-1< x < 0)

Hence, C is the answer.

Cheers!
_________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

Re: Modulus question [#permalink]
05 Oct 2011, 06:55

1

This post received KUDOS

Ajay369 wrote:

Good question.

If we draw the graph, we can figure out that individually both the option doesn't satisfy but if we join together, the value of x satisfy between the region - (-1< x < 0)

Hence, C is the answer.

Cheers!

Good thing to draw a number line to solve this type of questions.

In range (1), the expression is negative too so it is part of the range In range (2), the expression is negative also so it is part of the range In range (3), the expression is positive, so it does not belong to the range

Now even though I got x<1 for the range the same as fluke I'm wondering if I did the key points correctly because we should have +,-,+,- in that order for the ranges. I am guessing it has something to do with the fact that x^2 will always yield a positive number

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.

Attachment:

MSP824621h8d31791668agg000062b1e415iai4e4i2.gif

Sufficient.

Answer: C.

How do you do it (x+1)x(x-1)>0 --> [m]-1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link

One statement says "x must be either greater than 1 OR between -1 and 0" Another says "x must be less than 1"

So what values can you give x such that both statements are satisfied? You cannot have x greater than 1 since second statement says x must be less than 1. So x must be between -1 and 0. This satisfies both the statements. -1 < x < 0
_________________

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.

Attachment:

MSP824621h8d31791668agg000062b1e415iai4e4i2.gif

Sufficient.

Answer: C.

How do you do it (x+1)x(x-1)>0 --> [m]-1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link

One statement says "x must be either greater than 1 OR between -1 and 0" Another says "x must be less than 1"

So what values can you give x such that both statements are satisfied? You cannot have x greater than 1 since second statement says x must be less than 1. So x must be between -1 and 0. This satisfies both the statements. -1 < x < 0

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.

Attachment:

MSP824621h8d31791668agg000062b1e415iai4e4i2.gif

Sufficient.

Answer: C.

How do you do it (x+1)x(x-1)>0 --> -1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link

I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1; -1<x<0; 0<x<1; x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

(1) x^5 > x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(2) x^2 > x^3. Reduce by x^2: 1>x (x\neq{0}). Not sufficient.

(1)+(2) Intersection of the ranges is -1<x<0.

Attachment:

MSP824621h8d31791668agg000062b1e415iai4e4i2.gif

Sufficient.

Answer: C.

How do you do it (x+1)x(x-1)>0 --> -1<x<0

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link

I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1; -1<x<0; 0<x<1; x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

Yes i was asking for the first sentence; i wonder about the inequality x(x+1) (x-1) >0 x>0 x+1>0; x>-1 x-1>0; x>+1 To satisfy the Q Is [X] <1 ; so the above range will be displayed as two options -1<x<0 / 0<x<+1.

Is that what you have suggested in previous equations ?

But i am still skeptical about the Inequality. By the way thanks for help; Appreciate .

I mean from above x > 1 or x>-1 or x>0 hw that -1<x<0 ?? Please clarify / link

I guess you ask about the first statement:

(1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0. The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges:

x<-1; -1<x<0; 0<x<1; x>1.

Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive.

Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - +. So, the ranges when the expression is positive are: -1<x<0 (2nd rabge) and x>1 (4th range).

Yes i was asking for the first sentence; i wonder about the inequality x(x+1) (x-1) >0 x>0 x+1>0; x>-1 x-1>0; x>+1 To satisfy the Q Is [X] <1 ; so the above range will be displayed as two options -1<x<0 / 0<x<+1.

Is that what you have suggested in previous equations ?

But i am still skeptical about the Inequality. By the way thanks for help; Appreciate .

Please follow the links provided in my previous post.
_________________

For second option I tried to solve it with handy trick. x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

I am not sure where I am making mistake with this trick.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos

For second option I tried to solve it with handy trick. x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

Ohh Thanks Bunuel I understood and can relate it to this trick.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos

For second option I tried to solve it with handy trick. x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

I am not sure where I am making mistake with this trick.

The squared term, x^2, should be omitted because it's always positive and thus doesn't affect the sign of the left side.

Ahhh now I get it, that's awesome thanks. Still I don't quite know what's wrong with including x^2 in the key points, I know it will give you a different valid range but I would like to understando the logic behind it.