Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 21 Jul 2019, 03:56

### 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

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

# ((99^2 + 101^2)/2 - 1)^(1/4)

Author Message
TAGS:

### Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

08 Jan 2015, 01:42
2
16
00:00

Difficulty:

35% (medium)

Question Stats:

72% (01:52) correct 28% (02:16) wrong based on 313 sessions

### HideShow timer Statistics

$$\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =$$

A: 15

B: 12

C: 11

D: 10

E: 9

_________________
Kindly press "+1 Kudos" to appreciate
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

08 Jan 2015, 02:52
5
1
2

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

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

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

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

$$= \sqrt[4]{10^4}$$

= 10
_________________
Kindly press "+1 Kudos" to appreciate
##### General Discussion
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14603
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

08 Jan 2015, 14:38
2
1
HI All,

In complex-looking calculations, if you find that you don't see an immediate pattern that will help you to 'break down' the math, sometimes you just have to think about what the numbers represent in real simple terms.

99^2 is like saying "ninety-nine 99s" --> imagine a big row of 99s that you have to add up....but don't add them up yet....
101^2 is like saying "one hundred one 101s" --> it's the same idea...a big row of 101s that you have to add up....

Now, take ONE 99 and add it to ONE 101 and you get 200. How many of those 200s do you have here?

You have ninety-nine 200s with two extra 101s left over....This gives us....

99(200) + 2(101) = 19,800 + 202 = 20,002

Next, we're asked to divide this sum by 2 and then subtract 1...

20,002/2 = 10,001

10,001 - 1 = 10,000

Finally, we're asked to take the fourth root (or quad-root) of 10,000....

since 10^4 = 10,000.....the fourth root of 10,0000 = 10

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Manager
Joined: 01 May 2013
Posts: 61
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

08 Jan 2015, 15:19
3
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

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9449
Location: Pune, India
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

08 Jan 2015, 22:46
1
2
CCMBA wrote:
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

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.

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

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

$$\sqrt[4]{10000}$$

$$10$$
_________________
Karishma
Veritas Prep GMAT Instructor

Current Student
Joined: 12 Aug 2015
Posts: 2609
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

29 Nov 2016, 14:30
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

_________________
Board of Directors
Joined: 17 Jul 2014
Posts: 2539
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

02 Dec 2016, 08:55
PareshGmat wrote:
$$\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =$$

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.

_________________
Intern
Joined: 16 Jul 2017
Posts: 11
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

06 Aug 2017, 17:09
PareshGmat wrote:

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

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

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

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

$$= \sqrt[4]{10^4}$$

= 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:
$$= \sqrt[4]{\frac{2 * 100^2 -1^2 + 1^2}{2} - 1}$$
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 274
((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

Updated on: 30 Sep 2017, 20:31
PareshGmat wrote:

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

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

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

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

$$= \sqrt[4]{10^4}$$

= 10

hi

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

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

Originally posted by testcracker on 30 Sep 2017, 17:38.
Last edited by testcracker on 30 Sep 2017, 20:31, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 56306
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

01 Oct 2017, 03:57
gmatcracker2017 wrote:
PareshGmat wrote:

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

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

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

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

$$= \sqrt[4]{10^4}$$

= 10

hi

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

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

$$(100-1)^2 + (100+1)^2=100^2 - 2*100+1 + 100^2+2*100+1=2*100^2+2$$
_________________
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 274
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

01 Oct 2017, 08:46
Bunuel wrote:
gmatcracker2017 wrote:
PareshGmat wrote:

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

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

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

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

$$= \sqrt[4]{10^4}$$

= 10

hi

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

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

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

thanks man
you are so great ....
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6967
Location: United States (CA)
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

04 Oct 2017, 16:48
PareshGmat wrote:
$$\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =$$

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

VP
Joined: 09 Mar 2018
Posts: 1002
Location: India
Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

### Show Tags

11 Jan 2019, 10:56
PareshGmat wrote:
$$\sqrt[4]{\frac{99^2 + 101^2}{2} - 1} =$$

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)

_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Re: ((99^2 + 101^2)/2 - 1)^(1/4)   [#permalink] 11 Jan 2019, 10:56
Display posts from previous: Sort by