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# Which of the following COULD be correct?

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Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:00
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69% (01:48) correct 31% (01:54) wrong based on 250 sessions

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Which of the following COULD be correct?

I. $$x < x^3 < x^2$$
II. $$x < x^3 < x^2 < x^4$$
III. $$x^3 < x^2 < x^4$$
IV. $$x^3 < x < x^2 < x^4$$

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

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Updated on: 25 Jul 2019, 19:53
1
In this one we just need to validate if one value or set values comply with the statement, so we can try these values: -2, -0.5, 0.5 and 2

I. x<x^3<x^2
-2: -2<-8 <4 Not ok
-0.5: -1/2 < -1/8 < 1/4 ok

II. x<x^3<x^2 <x^4

-2: -2<-8 <4<16 Not ok
-0.5: -1/2 < -1/8 < 1/4<1/16 Not ok
0.5: 1/2 < 1/8 < 1/ 4 < 1/16 Not ok
2: 2< 8 <4 < 16 Not ok

III. x^3<x^2<x^4

-2: -8<4 <16 ok

IV. x^3<x<x^2<x^4

-2: -8<-2 <4<16 ok

So, only I, III and IV could be correct, (D) is our answer

Originally posted by Mizar18 on 25 Jul 2019, 08:10.
Last edited by Mizar18 on 25 Jul 2019, 19:53, edited 1 time in total.
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Which of the following COULD be correct?  [#permalink]

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Updated on: 26 Jul 2019, 08:36
Which of the following COULD be correct?

I. x<x3<x2
II.x<x3<x2<x4
III. x3<x2<x4
IV. x3<x<x2<x4

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

let x = -1/2 ,
IMO A
Bunuel isnt it that in such type of questions we need to test all the given conditions with same value of x? with x=-1/2 only A stands out where as at x=-2 we get option D ? isnt there an anomaly in this question ? question does not specifies that x has to be an integer value ... GMATinsight sir please advise ...

Originally posted by Archit3110 on 25 Jul 2019, 08:10.
Last edited by Archit3110 on 26 Jul 2019, 08:36, edited 2 times in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:11
1
Which of the following COULD be correct?

I. x<x^3<x^2 - Correct Possible value of x = --1/2, X^3 = -1/8, x^2 = 1/4.
II. x<x^3<x^2<x^4 - Incorrect for x<x^3 -1<x<1 but x<x^2 requires x to be negative. If x is negative and greater than -1. X =-1/2 in that case x^2 cannot be less than x^4
III. x^3<x^2<x^4 - Correct eg: X = -2
IV. x^3<x<x^2<x^4 Correct. Eg X = -2

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

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Which of the following COULD be correct?  [#permalink]

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Updated on: 26 Jul 2019, 07:12
1
Check for x < -1 --> x = -2 or
-1 < x < 0 --> x = -0.5 or
0 < x < 1 --> x = 0.5 or
x > 1 --> x = 2

I. $$x<x^3<x^2$$
If x = -0.5; $$(x, x^3, x^2)$$ = (-0.5, -0.125, 0.25) --> Possible

II. $$x<x^3<x^2<x^4$$ --> Not Possible

III. $$x^3<x^2<x^4$$
If x = -2; $$(x^3, x^2, x^4)$$ = (-8, 4, 16) --> Possible

IV. $$x^3<x<x^2<x^4$$
If x = -2; $$(x^3, x, x^2, x^4)$$ = (-8, -2, 4, 16) --> Possible

IMO Option D

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Originally posted by Dillesh4096 on 25 Jul 2019, 08:17.
Last edited by Dillesh4096 on 26 Jul 2019, 07:12, edited 1 time in total.
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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 23:01
2
IMO : D

Which of the following COULD be correct?

I. x<x^3<x^2
II. x<x^3<x^2<x^4
III. x^3<x^2<x^4
IV. x^3<x<x^2<x^4

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

there are three types of variables we can use to test,
0>x
1>x>0
and x>1

Sol:
1: it is present in all choices, that means this has to be true,
2:Not possible
3:let x=-2 then x=-2, x^3=-8 x^2=4 and x^4=16 . So -8<4<16.
4:let x=-2, then x=-2, x^3=-8 x^2=4 and x^4=16 . So -8<-2<4<16.

Originally posted by abhishekdadarwal2009 on 25 Jul 2019, 08:19.
Last edited by abhishekdadarwal2009 on 25 Jul 2019, 23:01, edited 1 time in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:20
1
Here we can solve by cases by taking numbers as this is could be question and we need to just prove one example that can be true.
Also Given I is there in all options we do not need to check one:-p
I. x<x^3<x^2 this holds true for -1/2
II. x<x^3<x^2<x^4 NO such case
III.x^3<x^2<x^4 true for -2
IV.x^3<x<x^2<x^4 true for -2

So Answer: D. I, III, and IV only
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Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:20
Which of the following COULD be correct?

We just have to find a way to make them true.

D. I, III, and IV only
I
If III is true I has to be true. By checking the options.

-.1 < -.001 < A +ve number

III
Easy by putting x=-2
-8<4<16

IV
Easy by putting x=-2
-8<-2<4<16

Let's see II can be true or not.
x^2 > x^3 iff x is -ve but then x^3 can't be more than x.
Also if we consider x=-.1, x^4 can't be more than x^2

Anomaly of [-1,1]
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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 08:54
2
Quote:
Which of the following COULD be correct?

I. x<x^3<x^2
II. x<x^3<x^2<x^4
III. x^3<x^2<x^4
IV. x^3<x<x^2<x^4

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

Take x=-2
x^2=4
x^3=-8
x^4=16

This condition satisfies III and IV.
Now if we look at the options with both III and IV only D and E survives.
Lets analyze II and we can conlude
For II to be true, x<x^3<x^2<x^4
but x^3<x^2<x^4 only when x<0 and when x<0 , x^3<x
Therefore, II cannot be true

Hence D
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Originally posted by kitipriyanka on 25 Jul 2019, 08:29.
Last edited by kitipriyanka on 25 Jul 2019, 08:54, edited 1 time in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:31
1

Taking x values as -0.1, -2,0.1 and 1
I satisfies when x = -0.1
II never satisfies
III and 1V satisfiy when x = -2

Therefroe D it is
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:38
1
Which of the following COULD be correct?

I. x<x3<x2x<x3<x2
II. x<x3<x2<x4x<x3<x2<x4
III. x3<x2<x4x3<x2<x4
IV. x3<x<x2<x4x3<x<x2<x4

This is an easy question. It could be solved through plugging in numbers. The numbers I used was .5, -.5

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:41
Values of x can be either
Lesser than -1
Between -1 to 0
Between 0 to 1
Greater than 1

Among the four given cases,
First case is satisfied when x=-0.1
Second case can not be satisfied in any case
Third case is with any negative number greater than 1
Fourth case is possible with any negative number greater than 1

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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 09:37
1
Quote:
Which of the following COULD be correct?

I. x<x3<x2
II. x<x3<x2<x4
III. x3<x2<x4
IV. x3<x<x2<x4

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

x…x2…x3…x4:
2…4…8…16
-2…4…-8…16
0.2…0.04…0.008…0.00016
-0.2…0.04…-0.008…0.00016

I. x<x3<x2: case [4]
II. x<x3<x2<x4: no cases…
III. x3<x2<x4: case [2]
IV. x3<x<x2<x4: case [2]

Originally posted by exc4libur on 25 Jul 2019, 08:44.
Last edited by exc4libur on 25 Jul 2019, 09:37, edited 1 time in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:46
1
Which of the following COULD be correct?

I. $$x<x^3<x^2$$
If $$x^3>x$$, then either $$x>1$$ or $$0>x>-1$$.
However, if $$x^3<x^2$$, then $$x$$ cannot be positive. It still can be negative and be more than -1. Let us check via example:
$$-1/2 < -(1/2)^3=-1/8 < (-1/2)^2=1/4$$ Correct

II. $$x<x^3<x^2<x^4$$
If $$x<x^3$$, then either $$x>1$$ or $$0>x>-1$$.
If $$x^3<x^2$$, then $$0>x>-1$$.
However, $$x^4$$ cannot be greater than $$x^2$$ in case $$0>x>-1$$. Impossible

III. $$x^3<x^2<x^4$$
If $$x^3<x^2$$, then either $$x<0$$ or $$0<x<1$$.
If $$x^2<x^4$$, then x cannot be $$0<x<1$$, but it still can be negative. Let us check with example below:
$$(-2)^3 < (-2)^2 < (-2)^4$$
$$-8 < 4 < 16$$ Correct

IV. $$x^3<x<x^2<x^4$$
If$$x^3 < x$$, then either $$0<x<1$$ or $$x<0$$.
If $$x<x^2$$, then it cannot be $$0<x<1$$, but $$x<0$$ is possible. In this case $$x^2<x^4$$ is also applicable. Let us check:
$$(-2)^3 < -2 < (-2)^2 < (-2)^4$$
$$-8 < -2 < 4 < 16$$ Correct

Options I, III and IV are correct.

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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 08:53
1
Which of the following COULD be correct?

$$I. x<x^3<x^2$$
For example $$x=-.5=> x^3=-.125$$ and $$x^2=.25$$
-.5(x)<-.125(x^3)<.25(x^2)
$$x<x^3<x^2$$
COULD be TRUE

$$II. x<x^3<x^2<x^4$$
For $$x<x^3<x^2$$ => -1<x<0 but in this domain $$x^2>x^4$$
For example $$x=-.5=> x^3=-.125$$ and $$x^2=.25$$
$$-.5(x)<-.125(x^3)<.25(x^2)$$
$$But x^4=.0625<x^2$$
MUST NOT BE TRUE

$$III. x^3<x^2<x^4$$
For example x=-2
$$-8(x^3)<4(x^2)<16(x^4)$$
COULD BE TRUE

$$IV. x^3<x<x^2<x^4$$
For example x=-2
$$-8(x^3)<-2(x)<4(x^2)<16(x^4)$$
COULD BE TRUE

A. I only
B. I and III only
C. I, II, and IV only
D. I, III, and IV only
E. I, II, III, and IV

IMO D
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Originally posted by Kinshook on 25 Jul 2019, 08:48.
Last edited by Kinshook on 25 Jul 2019, 08:53, edited 1 time in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 08:51
1
We have to Consider here different variables
Positive integers , negative integers, fraction
Positive integer will give straight equation x^4 > x^3> X^2>x
Negative integer will give x^4>x^2 >x^3 >x

fraction : x>x^2 >x3 >x4 and in case of negative fraction it will be x^2>x^4>x^3>x

From these equalities we can derive I, II, and IV, So Answer D
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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 23:39
1
COULD be correct- indicates we need to check whether there is at least one possibility where the inequality stands true.

I. $$x < x^3 < x^2$$
Consider x = -0.5
-0.5 < -0.125 < 0.25
Correct

2. $$x < x^3 < x^2 < x^4$$
Under no value of x this stands correct.
It cannot be correct.

3. $$x^3 < x^2 < x^4$$
Consider x = -2
(-2)^3 < (-2)^2 < (-2)^4
-8 < 4 < 16
Correct

4. $$x^3 < x < x^2 < x^4$$
Consider x = -2
(-2)^3 < (-2) < (-2)^2 < (-2)^4
-8 < -2 < 4 < 16
Correct

Originally posted by Sayon on 25 Jul 2019, 09:00.
Last edited by Sayon on 25 Jul 2019, 23:39, edited 1 time in total.
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Which of the following COULD be correct?  [#permalink]

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Updated on: 25 Jul 2019, 09:07
1
Which of the following COULD be correct?

I.$$x<x^3<x^2$$
This holds true for any negative number less than 0 but greater than -1

II. $$x<x^3<x^2<x^4$$
Can't find anything to prove this

III. $$x^3<x^2<x^4$$
This holds true for any negative number less than 1

IV. $$x^3<x<x^2<x^4$$
This holds true for any negative number less than 1

IMO D

Originally posted by ancored on 25 Jul 2019, 09:06.
Last edited by ancored on 25 Jul 2019, 09:07, edited 1 time in total.
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 09:06
1
Which of the following COULD be correct?

I. x<x^3<x^2 --Correct
let x = -1/2
-1/2 < -1/8 < 1/2

II. x<x^3<x^2<x^4--incorrect
this is not true for negative integers, positive integers , negative fractions and positive fractions

III. x^3<x^2<x^4---correct
let x = -2
-8<4<16

IV. x^3<x<x^2<x^4--Correct
let x = -2
-8<-2<4<16

D. I, III, and IV only-- is the Correct answer
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Re: Which of the following COULD be correct?  [#permalink]

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25 Jul 2019, 09:09
1
I. x<x^3<x^2 True between -1 to 0

II. x<x^3<x^2<x^4 not true

III. x^3<x^2<x^4 true between -infinity to 0

IV. x^3<x<x^2<x^4 true between -infinity to 0

D is correct
Re: Which of the following COULD be correct?   [#permalink] 25 Jul 2019, 09:09

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