Power and Root Questions

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cvj0873: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Jun 20, 2012 8:26 am

by cvj0873 » Thu Oct 10, 2013 12:11 am

Hello community members of Beat the GMAT!

Currently, I'm tackling Power and Roots (my weakest Quant topic) and I need some help with a few questions. These questions are from the OG GMAT Review (13e) and OG for GMAT Quantitative Review (2e). Not only did I get these questions incorrect, I don't understand the answers provided as well. Thank you very much for all your help!

[OG GMAT Review 13e]

157. If n is a positive integer and k = 5.1 X 10^n, what is the value of k?

(1) 6,000 <k< 500,000
(2) k^2 = 2.601 X 10^9

Answer: D

[OG for GMAT Quant Review 2e]

31. If x is a positive integer, is square root X an integer?

(1) Square root of 4x is an integer.
(2) Square root of 3x is not an integer.

Answer: A

51. If n is positive, is square root n > 100?

(1) Square root of (n-1) > 99
(2) Square root of (n+1) > 101

Answer: B

Thank you all so much!!!

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Source: Beat The GMAT — Data Sufficiency | Thu Oct 10, 2013 12:11 am

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Oct 10, 2013 12:32 am

Q157.

n-> +ve
k = 5.1 x 10^n

Statement 1:
6000 < k < 5,00,000
we need "n" to be minimum "4" to make figure more than 6000
n = 4
Also, if we take n = 5, then figure becomes 510000>500000
So, n = 4
We can calculate "k"
SUFFICIENT

Statement 2:
k^2 = 2.601 x 10^9
We also know
k = 5.1 X 10^n
replace the values
(5.1 X 10^n)^2 = 2.601 x 10^9
We don;t have to solve for value.. we can see that we can reach to some value of "n"
so, if we get "n" .. we can get "k"
SUFFICIENT

Answer [spoiler]{D}[/spoiler]

R A H U L

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Oct 10, 2013 12:39 am

Q31.

X - > +ve INT
To Find: square root X an integer

Statement 1:
sqrt(4x) = INT
sqrt(4) * sqrt(x) = INT
2 * sqrt(x) = INT
sqrt(x) = INT/2= INT
SUFFICIENT

Statement 2:
sqrt(3x) = NOT-INT
sqrt(3) * sqrt(x) = NOT-INT
srt(x) = NOT-INT/NOT-INT
here, we have two different cases..
CASE1: maybe the numberator "NOT-INT" is an exact multiple of denominator.. in that case sqrt(x) is INT
CASE2: maybe the numberator "NOT-INT" is NOT an exact multiple of denominator .. in that case sqrt(x) is NOT-INT
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]

R A H U L

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Oct 10, 2013 12:45 am

Q51.
N-> +ve
TO find: square root n > 100
or, n > (100)^2

Statement 1:
Square root of (n-1) > 99
square both sides
n-1 > (99)^2
n > (99)^2 + 1
Since, (99)^2 + 1 will definitely be less than (100)^2.. we cannot say whether the value of "n" will definitely be above (100)^2
INSUFFICIENT

Statement 2:
sqrt(n+1) > 101
square both sides
n+1 > (101)^2
n > (101)^2 - 1
Since, (101)^2 - 1 will be definitely larger than (100)^2 .. we can definitely say that n>(100)^2..
SUFFICIENT

Answer [spoiler]{B}[/spoiler]

R A H U L

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 10, 2013 6:04 am

Hi cvj0873,

In the future, please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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cvj0873: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Jun 20, 2012 8:26 am

by cvj0873 » Thu Oct 10, 2013 7:38 pm

Brent@GMATPrepNow wrote:Hi cvj0873,

In the future, please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions.

Cheers,
Brent

Hi Brent,

Thanks for the heads up! Will keep this in mind when I post future questions.

Thanks!

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