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the positions on

by sanju09 » Fri Jul 23, 2010 3:58 am
Considering the positions on the number line shown, which of the following could be a value for x?
A. 5/3
B. 3/5
C. -2/5
D. -5/2
E. none

[spoiler]Source: majortests.com[/spoiler]
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by kvcpk » Fri Jul 23, 2010 4:09 am
x^3 <x^2
x(x^2-1)<0
x<0 or x^2<1

x has to be between -1 and 1 or x has to be negative
A is out

x<x^2
x(1-x)<0
x<0 or x>1

B is out

x<x^3
x(1-x^2)<0
x<0 or x does not lie between -1 and 1
C is out

lets test for D
-5/2 = -2.5
25/4 = 6.25
-125/8 = -15.6
x>x^3
D is also out

pick E
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by sanju09 » Fri Jul 23, 2010 4:15 am
kvcpk wrote:x^3 <x^2
x(x^2-1)<0
x<0 or x^2<1

x has to be between -1 and 1 or x has to be negative
A is out

x<x^2
x(1-x)<0
x<0 or x>1

B is out

x<x^3
x(1-x^2)<0
x<0 or x does not lie between -1 and 1
C is out

lets test for D
-5/2 = -2.5
25/4 = 6.25
-125/8 = -15.6
x>x^3
D is also out

pick E
I didn't get how you threw C out.
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by pathaniaus » Fri Jul 23, 2010 4:23 am
im a little confused by kvcpk's reasoning.

my thinking is the following: (-2/5) < (-8/125) < (4/25); which would mean that x<x^3<x^2
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by kvcpk » Fri Jul 23, 2010 4:24 am
sanju09 wrote: I didn't get how you threw C out.
Yeah.. I think I missed something there.. C should be the answer.
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by mj78ind » Fri Jul 23, 2010 5:33 am
Actually C and D both can be true........

C as explained earlier, D we are not told that x>0 or all numbers are to the right of zero

Hence when x = -0.4, x^3 = - 0.064, x^2 = 0.16

which also satisfies the diagram...........
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by sanju09 » Fri Jul 23, 2010 5:49 am
mj78ind wrote:Actually C and D both can be true........

C as explained earlier, D we are not told that x>0 or all numbers are to the right of zero

Hence when x = -0.4, x^3 = - 0.064, x^2 = 0.16

which also satisfies the diagram...........
How is D true, my friend?
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by mj78ind » Fri Jul 23, 2010 6:20 am
sanju09 wrote:
mj78ind wrote:Actually C and D both can be true........

C as explained earlier, D we are not told that x>0 or all numbers are to the right of zero

Hence when x = -0.4, x^3 = - 0.064, x^2 = 0.16

which also satisfies the diagram...........
How is D true, my friend?
My bad it is B and C both can be true.......

B says x = 0.6, x^3 = 0.216, x^2 = 0.36 .... hence works

C says x = -0.4 x^3 = -0.064 x^2 = 0.16 ..... hence works
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by sanju09 » Fri Jul 23, 2010 6:22 am
mj78ind wrote:
sanju09 wrote:
mj78ind wrote:Actually C and D both can be true........

C as explained earlier, D we are not told that x>0 or all numbers are to the right of zero

Hence when x = -0.4, x^3 = - 0.064, x^2 = 0.16

which also satisfies the diagram...........
How is D true, my friend?
My bad it is B and C both can be true.......

B says x = 0.6, x^3 = 0.216, x^2 = 0.36 .... hence works

C says x = -0.4 x^3 = -0.064 x^2 = 0.16 ..... hence works
Why do you feel that B works?
The mind is everything. What you think you become. -Lord Buddha



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by mj78ind » Fri Jul 23, 2010 6:31 am
sanju09 wrote:
mj78ind wrote:
sanju09 wrote:
mj78ind wrote:Actually C and D both can be true........

C as explained earlier, D we are not told that x>0 or all numbers are to the right of zero

Hence when x = -0.4, x^3 = - 0.064, x^2 = 0.16

which also satisfies the diagram...........
How is D true, my friend?
My bad it is B and C both can be true.......

B says x = 0.6, x^3 = 0.216, x^2 = 0.36 .... hence works

C says x = -0.4 x^3 = -0.064 x^2 = 0.16 ..... hence works
Why do you feel that B works?
Good ques ......... I do not know .... It should be C (unless i changed a few quant rules) ;-)

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by selango » Fri Jul 23, 2010 8:34 am
If x>0

x>x^3>x^2

x=a/b,x^3=a^3/b^3,x^2=a^2/b^2

x>x^3

a/b>a^3/b^3

a.b^3>a^3.b-->Eqn1

x^3>x^2

a^3.b^2>a^2.b^3-->Eqn2

If x<0

x^2>x^3>x

x=a/b,x^3=a^3/b^3,x^2=a^2/b^2

x^2>x^3 -->Eqn3

x^3>x--Eqn 4

A.5/3

5.27<125.3,Eqn 1 Not satisifed

B.3/5

3.125>27.5,Eqn 1 satisfied

27.25<9.125,Eqn2 not satisfied

C.-2/5=-0.4

0.16>-0.4,Eqn3 satisfied

-0.064>-0.4,Eqn 4 satisfied

-2/5 possible

D.-5/2=-2.5

6.25>-15.625,Eqn 3 satisifed

-15.625<-2.5,Eqn 4 not satisifed

Only C works for both values of x
--Anand--
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