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Knewton test 1 DS question #15

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by sarwan » Sat Aug 07, 2010 11:10 pm
Please help me for this DS question.

If $\mathsf{p}$ is an integer greater than zero, is $\mathsf{200 < \sqrt{p}}$?

1. $\mathsf{198 < \sqrt{p-2}}$
2. $\mathsf{202 < \sqrt{p+2}}$

among a, b, c d and e which is correct for this DS ?
Pls explain.
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by selango » Sat Aug 07, 2010 11:27 pm
Can you pls check the question?Format is not clear..
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by sarwan » Sun Aug 08, 2010 4:38 am
sarwan wrote:Please help me for this DS question.

If p is an integer greater than zero, is 200 < sqrt{p}?

1. 198 < sqrt{p-2}
2. 202 < sqrt{p+2}

among a, b, c d and e which is correct for this DS ?
Pls explain.

ans is : b
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by selango » Sun Aug 08, 2010 5:27 am
p>0

sqrt(p)>200 or p>200*200

stmt1,

sqrt(p-2)>198

p-2>198^2-->p>198^2+2

p can be greater or not greater than 40000(200*200)

Insuff

stmt2,

sqrt(p+2)>202

p+2>202^2 or p>202^2-2

p>(202+sqrt(2))(202-sqrt(2))

-->p>200*200

Suff

Pick B
Last edited by selango on Sun Aug 08, 2010 5:43 am, edited 2 times in total.
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by kvcpk » Sun Aug 08, 2010 5:32 am
sarwan wrote:
sarwan wrote:Please help me for this DS question.

If p is an integer greater than zero, is 200 < sqrt{p}?

1. 198 < sqrt{p-2}
2. 202 < sqrt{p+2}

among a, b, c d and e which is correct for this DS ?
Pls explain.

ans is : b
p>0 , is 200 < sqrt{p}? impliew Is p> 200^2 ?

198 < sqrt{p-2}
198^2 <p-2
198^2 -2 <p
Implies p> 198^2 -2
P can be or cannot be greater than 200^2.
Hence INSUFF

202 < sqrt{p+2}
202^2 < p+2
p> 202^2 -2
202^2 -2 is always greater than 200^2
hence p> 200^2
SUFF

pick B

Hope this helps!!
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by kvcpk » Sun Aug 08, 2010 5:34 am
selango wrote: p+2>202^2 or p>202^2-2

p>(202+2)(202-2)
You will need to edit this Anand.
p>(202+root(2))(202-root(2))
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by selango » Sun Aug 08, 2010 5:44 am
kvcpk wrote:
selango wrote: p+2>202^2 or p>202^2-2

p>(202+2)(202-2)
You will need to edit this Anand.
p>(202+root(2))(202-root(2))
Thanks praveen.edited..
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