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4^p × (1/3) ^q < 1

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4^p × (1/3) ^q < 1

by sanju09 » Fri May 21, 2010 1:31 am
Is p and q are positive integers, is 4^p × (1/3) ^q < 1?

(1) q = 2 p.

(2) q = 4.
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by liferocks » Fri May 21, 2010 1:41 am
if 4^p X (1/3) ^q < 1
we get (4/3)^p *(1/3)^(q-p)<1
since 4/3 >1 and p is positive integer,(4/3)^p>1..hence (1/3)^(q-p)<1 or q-p>0 or q>p

From 1
q>p..sufficient

From 2
No info about p...not sufficient

Ans option A
Last edited by liferocks on Fri May 21, 2010 4:10 am, edited 2 times in total.
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by sanju09 » Fri May 21, 2010 2:06 am
liferocks wrote:if 4^p X (1/3) ^q < 1
we get (4/3)^p *(1/3)^(q-p)<1
since 4/3 >1 and p is positive (only positive or integer too?),(4/3)^p>1..hence (1/3)^(q-p)<1 or q-p>0 or q>p

From 1
q>p..sufficient

From 2
No info about p...nit sufficient

Ans option A
great work though
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by liferocks » Fri May 21, 2010 3:43 am
sanju09 wrote:
liferocks wrote:if 4^p X (1/3) ^q < 1
we get (4/3)^p *(1/3)^(q-p)<1
since 4/3 >1 and p is positive (only positive or integer too?),(4/3)^p>1..hence (1/3)^(q-p)<1 or q-p>0 or q>p

From 1
q>p..sufficient

From 2
No info about p...nit sufficient

Ans option A
great work though
edited.
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by sanju09 » Fri May 21, 2010 3:58 am
liferocks wrote:if 4^p X (1/3) ^q < 1
we get (4/3)^p *(1/3)^(q-p)<1
since 4/3 >1 and p is positive integer,(4/3)^p>1..hence (1/3)^(q-p)<1 or q-p>0 or q>p

From 1
q>p..sufficient

From 2
No info about p...nit sufficient

Ans option A
Want to edit more? What is "nit"?
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by liferocks » Fri May 21, 2010 4:15 am
sanju09 wrote:
Want to edit more? What is "nit"?
Again edited!!..hope nothing else is left to edit.
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