Tough one

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callmemo: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Sun Sep 28, 2008 11:22 am

by callmemo » Mon Dec 22, 2008 10:13 am

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of sqrt(288kx)?

a. 24k sqrt(3)
b. 24 sqrt(k)
c. 24 sqrt(3k)
d. 24 sqrt(6k)
e. 72 sqrt(k)

[spoiler]OA: B[/spoiler]

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Source: Beat The GMAT — Problem Solving | Mon Dec 22, 2008 10:13 am

amitabhprasad: Master | Next Rank: 500 Posts; Posts: 259; Joined: Thu Jan 18, 2007 8:30 pm; Thanked: 16 times

by amitabhprasad » Mon Dec 22, 2008 10:27 am

288 = 2^5.3^2
since x is divisible by 6 we will say x = 2*3*m
so
sqrt(288kx) = sqrt(2^5*3^2*k*2*3*m)
==>sqrt(2^6*3^2*k*3*m)
= 24sqrt(3*k*m)
from here you can easily eliminate 4 of the 5 choices and left with "B"

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callmemo: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Sun Sep 28, 2008 11:22 am

by callmemo » Mon Dec 22, 2008 10:41 am

not sure how to eliminate C & D. Can you elaborate?

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amitabhprasad: Master | Next Rank: 500 Posts; Posts: 259; Joined: Thu Jan 18, 2007 8:30 pm; Thanked: 16 times

by amitabhprasad » Mon Dec 22, 2008 11:45 am

sure
d. 24 sqrt(6k)
our equation = 24sqrt(3*k*m)
as per question k and x are +ve integers, and x is divisible by 3
this implies x = (any int)*6
we have already taken care of 6,
so to satisfy d we can pick a = 2

e. 72 sqrt(k)
following above reasoning if you pick a =3
= 24sqrt(3*k*3)
= 72sqrt(k)
c) m =1
a) m = k
b) to satisfy b we have to select m = 1/3 which will be a fraction but as per question stem our variable "m" should an integer.

Hope this helps

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umaa: Legendary Member; Posts: 727; Joined: Sun Jun 08, 2008 9:32 pmFollowed by: 1 member; Thanked: 8 times

by umaa » Mon Dec 22, 2008 9:19 pm

Hey amitabhprasad,
How did you eliminate A? K is not in the square root.

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adilka: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Thu Oct 23, 2008 9:28 am; Location: Canada; Thanked: 1 times; GMAT Score:700

by adilka » Mon Dec 22, 2008 9:44 pm

umaa wrote:Hey amitabhprasad,
How did you eliminate A? K is not in the square root.

It can be eliminated by assuming m=k, hence, sqrt(k^2)=K, we're left with 24k sqrt(3)

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umaa: Legendary Member; Posts: 727; Joined: Sun Jun 08, 2008 9:32 pmFollowed by: 1 member; Thanked: 8 times

by umaa » Mon Dec 22, 2008 9:47 pm

Right.. I thought abt it. thx..

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