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smushkas: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jul 01, 2007 5:56 pm; Thanked: 3 times

by smushkas » Tue Mar 11, 2008 2:31 pm

Hey guys,

I was surprised that there were roots of 3 and 4 like in this question. Should we really have to know them or there is some kind of shortcut or trick how to compute them quickly?

Thanks in advance!

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Last edited by smushkas on Mon Mar 17, 2008 3:59 pm, edited 2 times in total.

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Source: Beat The GMAT — Problem Solving | Tue Mar 11, 2008 2:31 pm

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Mar 11, 2008 2:43 pm

You don't actually have to know the values, you just have to apply a bit of common sense.

We know that sqrt(4) = 2.

We also know that both the cube root and quardic root of ANY number greater than 1 will themselves be greater than 1.

So, we have:

2 + (more than 1) + (more than 1) = more than 4: choose (e).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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smushkas: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jul 01, 2007 5:56 pm; Thanked: 3 times

by smushkas » Tue Mar 11, 2008 2:58 pm

Thanks Stuart!
Actually, I did answer by the same logic as you described.
Thanks again!

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musicdaemon: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 13, 2008 10:34 am; Location: India; Thanked: 3 times

by musicdaemon » Mon Mar 17, 2008 10:13 am

Stuart Kovinsky wrote:You don't actually have to know the values, you just have to apply a bit of common sense.

We know that sqrt(4) = 2.

We also know that both the cube root and quardic root of ANY number greater than 1 will themselves be greater than 1.

So, we have:

2 + (more than 1) + (more than 1) = more than 4: choose (e).

Dear Stuart,
What is your take at -> sqrt(4) = +2 or -2

if sqrt(4)=-2 then the answer is ambiguous

what do you think?

Let the Game begin

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smushkas: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jul 01, 2007 5:56 pm; Thanked: 3 times

by smushkas » Mon Mar 17, 2008 3:57 pm

Musicdaemon,
As Amitava wrote it "Actually musicdaemon, sqrt(x) is always positive and it can not be negative."

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