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Is the range of the set equal to a^2–a^5 ?

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Source: — Data Sufficiency |
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by ygdrasil24 » Mon Jun 03, 2013 9:51 pm
himu wrote:A set contains the following 5 numbers:
a^2,a^5,a,a/2, and a/5

Is the range of the set equal to a^2-a^5 ?

a is negative


a^5<a
I. Since a is <0, a2 will be max and a5 will be least hence range is a2-a5. SO A is sufficient
II. a5<a , implies a < 0 , so like statement I, B is sufficient.

Hence Answer is C
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by himu » Mon Jun 03, 2013 9:56 pm
U mean both are individually SUFFICIENT ?
U MEAN D ??

NO its not D.

I'll put the OA tomorrow after few more replies.


Cheers,
~Himu.
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by J N » Mon Jun 03, 2013 11:18 pm
Statemnet 1 ok

Statement 2 no

a= 1/2
a^2= 1/4
a/2= 1/4
a/5= 1/10
a^5 = 1/32

largest- smallest = a-a^5
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by apoorva.rattan » Tue Jun 04, 2013 1:27 am
IMO A.
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by Atekihcan » Tue Jun 04, 2013 1:39 am
The question is simply asking whether a² and a� are the largest and smallest among the five numbers, respectively.

Now, whenever comparison of powers of a variable is involved always consider positive and negative values as well as values greater than and smaller than 1 and -1.

Statement 1: a < 0
If a = -2.0, a² is largest and a� is smallest ---> YES
If a = -1/2, a² is largest and a is smallest ---> NO

So, statement 1 is not sufficient

Statement 2: a� < a
So either a < -1 or 0 < a < 1
If a = -2.0, a² is largest and a� is smallest ---> YES
If a = 1/2, a is largest and a� is smallest ---> NO

So, statement 2 is not sufficient

Both statements together: a < -1
So, a² is largest and a� is smallest ---> YES

So, both statements together is sufficient

Answer : C
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by ygdrasil24 » Tue Jun 04, 2013 10:37 pm
Ok. I made a mess out there
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by burningman » Fri Jun 07, 2013 12:44 pm
@ Atekihcan

Statement 1: a < 0
If a = -1/2, a² is largest and a is smallest ---> NO

Was it not supposed to be a^5 in the above case ?
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by Atekihcan » Tue Jun 11, 2013 12:13 am
burningman wrote:If a = -1/2, a² is largest and a is smallest ---> NO

Was it not supposed to be a^5 in the above case ?
If a = -1/2, a^5 = -1/32 > -1/2
So, a is the smallest not a^5
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