Last visit was: 23 Apr 2026, 14:17 It is currently 23 Apr 2026, 14:17
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
avatar
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,271
 [31]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,271
 [31]
2
Kudos
Add Kudos
29
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,271
 [15]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,271
 [15]
8
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,777
Own Kudos:
13,047
 [6]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,777
Kudos: 13,047
 [6]
3
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
avatar
CCMBA
Joined: 01 May 2013
Last visit: 03 Feb 2015
Posts: 56
Own Kudos:
104
 [3]
Given Kudos: 8
Posts: 56
Kudos: 104
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Honestly, I estimated. Solved in a matter of seconds.

99^2 ~ 100^2 ~ 101^2 --> 10,000

So 2(10,000)/2 = 10,000

10,000-1 ~ 10,000

10^4 = 10,000

So the 4th root is 10

The answer is D.
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [3]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
CCMBA
Honestly, I estimated. Solved in a matter of seconds.

99^2 ~ 100^2 ~ 101^2 --> 10,000

So 2(10,000)/2 = 10,000

10,000-1 ~ 10,000

10^4 = 10,000

So the 4th root is 10

The answer is D.

Note that you cannot always approximate 99^2 to 100^2. The difference between them is quite a bit: 99^2 = 9801, which is 199 less than 100^2. But here, it is perfectly good to approximate because you have 99^2 + 101^2 - one of these is lower than 100^2, the other is higher so both can be approximated to be 100^2 + 100^2 = 20,000 together.
The difference between this approximated value (20,000) and actual value (20,002) is quite small.

Also, a -1 won't make any difference to a fourth power of a number such as 10 or 11 etc. So you can easily ignore -1.

\(\sqrt[4]{\frac{(99^2 + 101^2)}{2} - 1}\)

\(\sqrt[4]{\frac{(20000)}{2} - 1}\)

\(\sqrt[4]{10000}\)

\(10\)
User avatar
stonecold
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,231
Own Kudos:
Given Kudos: 893
GRE 1: Q169 V154
GRE 1: Q169 V154
Posts: 2,231
Kudos: 3,643
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is a great Question testing our approximating ability
Here 99^2=100^2
and 101^2=100^2
We can do that as "We are Rounding in opposite direction so the effect will almost neutralise the error"
Also note that 1 is negligible
Hence the expression reduces to [2*100^2/2]^1/4
Hence D

So the answer here has to be D
User avatar
mvictor
User avatar
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,118
Own Kudos:
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
GMAT 1: 650 Q49 V30
Posts: 2,118
Kudos: 1,276
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =\)

A: 15

B: 12

C: 11

D: 10

E: 9

99^2 = (100-1)^2 = 10,000 +1 - 200 = 9,801.
101 ^2 = (100+1)^2 = 10,000 + 1 + 200 = 10,201
sum of them is 20,002.
divided by 2: 10,001.
subtract 1 = 10,000 or 10^4
root 4 of 10^4 = 10.

answer is D.
avatar
MPhizzle
Joined: 16 Jul 2017
Last visit: 05 May 2019
Posts: 10
Own Kudos:
Given Kudos: 139
Posts: 10
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
Answer = D = 10

\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1}\)

\(= \sqrt[4]{\frac{(100-1)^2 + (100+1)^2}{2} - 1}\)

\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2} - 1}\)

\(= \sqrt[4]{100^2 + 1 - 1}\)

\(= \sqrt[4]{10^4}\)

= 10


Hi, can you explain how you are going from \(= \sqrt[4]{\frac{(100-1)^2 + (100+1)^2}{2} - 1}\) to
\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2} - 1}\)?

Wouldn't the 1's cancel out? for example:
\(= \sqrt[4]{\frac{2 * 100^2 -1^2 + 1^2}{2} - 1}\)
User avatar
testcracker
Joined: 24 Mar 2015
Last visit: 02 Dec 2024
Posts: 199
Own Kudos:
Given Kudos: 541
Status:love the club...
Posts: 199
Kudos: 135
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
Answer = D = 10

\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1}\)

\(= \sqrt[4]{\frac{(100-1)^2 + (100+1)^2}{2} - 1}\)

\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2} - 1}\)

\(= \sqrt[4]{100^2 + 1 - 1}\)

\(= \sqrt[4]{10^4}\)

= 10

hi

please let me understand the way you arrived at the following..

"2 x 100^2 + 2 x 1^2"

thanks in advance
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,852
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatcracker2017
PareshGmat
Answer = D = 10

\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1}\)

\(= \sqrt[4]{\frac{(100-1)^2 + (100+1)^2}{2} - 1}\)

\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2} - 1}\)

\(= \sqrt[4]{100^2 + 1 - 1}\)

\(= \sqrt[4]{10^4}\)

= 10

hi

please let me understand the way you arrived at the following..

"2 x 100^2 + 2 x 1^2"

thanks in advance

\((100-1)^2 + (100+1)^2=100^2 - 2*100+1 + 100^2+2*100+1=2*100^2+2\)
User avatar
testcracker
Joined: 24 Mar 2015
Last visit: 02 Dec 2024
Posts: 199
Own Kudos:
Given Kudos: 541
Status:love the club...
Posts: 199
Kudos: 135
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
gmatcracker2017
PareshGmat
Answer = D = 10

\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1}\)

\(= \sqrt[4]{\frac{(100-1)^2 + (100+1)^2}{2} - 1}\)

\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2} - 1}\)

\(= \sqrt[4]{100^2 + 1 - 1}\)

\(= \sqrt[4]{10^4}\)

= 10

hi

please let me understand the way you arrived at the following..

"2 x 100^2 + 2 x 1^2"

thanks in advance

\((100-1)^2 + (100+1)^2=100^2 - 2*100+1 + 100^2+2*100+1=2*100^2+2\)

thanks man
you are so great .... 8-)
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 23 Apr 2026
Posts: 22,283
Own Kudos:
Given Kudos: 302
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 22,283
Kudos: 26,531
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =\)

A: 15

B: 12

C: 11

D: 10

E: 9

Let’s simplify (99^2 + 101^2)/2 first:

(99^2 + 101^2)/2 = (9,801 + 10,201)/2 = 20,002/2 = 10,001

Thus:

∜[(99^2 + 101^2)/2 - 1] = ∜(10,001 - 1) = ∜10,000 = 10

Alternate Solution:

Note that 99 = 100 - 1 and 101 = 100 + 1. Thus:

99^2 + 101^2 = (100 - 1)^2 + (100 + 1)^2 = 100^2 - 200 + 1 + 100^2 + 200 + 1 = 2*100^2 + 2

Then, (99^2 + 101^2)/2 = (2*100^2 + 2)/2 = 100^2 + 1 = 10000 + 1 = 10001.

Thus, ∜[(99^2 + 101^2)/2 - 1] = ∜(10,001 - 1) = ∜10,000 = 10.

Answer: D
User avatar
KanishkM
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 755
Own Kudos:
Given Kudos: 123
Location: India
Posts: 755
Kudos: 512
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =\)

A: 15

B: 12

C: 11

D: 10

E: 9


\(\sqrt[4]{\frac{99^2 + 101^2}{2} - 1}\)

The point you realize that, (100-1)^2 and (100+1)^2 can be substituted for 99^2 & 101^2, this question will be a breeze

Solve the equation to get 10^(4*1/4)

Answer D
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,962
Own Kudos:
Posts: 38,962
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109785 posts
Tuck School Moderator
853 posts