January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1823
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, 00:42
Question Stats:
76% (01:21) correct 24% (02:00) wrong based on 280 sessions
HideShow timer Statistics
\(\sqrt[4]{\frac{99^2 + 101^2}{2}  1} =\) A: 15 B: 12 C: 11 D: 10 E: 9
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kindly press "+1 Kudos" to appreciate




SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1823
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, 01:52
Answer = D = 10 \(\sqrt[4]{\frac{99^2 + 101^2}{2}  1}\) \(= \sqrt[4]{\frac{(1001)^2 + (100+1)^2}{2}  1}\) \(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2}  1}\) \(= \sqrt[4]{100^2 + 1  1}\) \(= \sqrt[4]{10^4}\) = 10
_________________
Kindly press "+1 Kudos" to appreciate




EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13368
Location: United States (CA)

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
08 Jan 2015, 13:38
HI All, In complexlooking 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. I'm going to start with the squared terms: 99^2 and 101^2 (since we're asked to add these numbers together). 99^2 is like saying "ninetynine 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 ninetynine 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 quadroot) of 10,000.... since 10^4 = 10,000.....the fourth root of 10,0000 = 10 Final Answer: 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
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Manager
Joined: 01 May 2013
Posts: 61

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
08 Jan 2015, 14:19
Honestly, I estimated. Solved in a matter of seconds.
99^2 ~ 100^2 ~ 101^2 > 10,000
So 2(10,000)/2 = 10,000
10,0001 ~ 10,000
10^4 = 10,000
So the 4th root is 10
The answer is D.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8804
Location: Pune, India

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
08 Jan 2015, 21:46
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
10,0001 ~ 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. \(\sqrt[4]{\frac{(99^2 + 101^2)}{2}  1}\) \(\sqrt[4]{\frac{(20000)}{2}  1}\) \(\sqrt[4]{10000}\) \(10\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Current Student
Joined: 12 Aug 2015
Posts: 2626

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
29 Nov 2016, 13:30
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
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Board of Directors
Joined: 17 Jul 2014
Posts: 2598
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
02 Dec 2016, 07:55
PareshGmat wrote: \(\sqrt[4]{\frac{99^2 + 101^2}{2}  1} =\)
A: 15
B: 12
C: 11
D: 10
E: 9 99^2 = (1001)^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.



Intern
Joined: 16 Jul 2017
Posts: 11

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
06 Aug 2017, 16:09
PareshGmat wrote: Answer = D = 10
\(\sqrt[4]{\frac{99^2 + 101^2}{2}  1}\)
\(= \sqrt[4]{\frac{(1001)^2 + (100+1)^2}{2}  1}\)
\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2}  1}\)
\(= \sqrt[4]{100^2 + 1  1}\)
\(= \sqrt[4]{10^4}\)
= 10 Hi, can you explain how you are going from \(= \sqrt[4]{\frac{(1001)^2 + (100+1)^2}{2}  1}\) 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: 273

((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
Updated on: 30 Sep 2017, 19:31
PareshGmat wrote: Answer = D = 10
\(\sqrt[4]{\frac{99^2 + 101^2}{2}  1}\)
\(= \sqrt[4]{\frac{(1001)^2 + (100+1)^2}{2}  1}\)
\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2}  1}\)
\(= \sqrt[4]{100^2 + 1  1}\)
\(= \sqrt[4]{10^4}\)
= 10 hi please let me understand the way you arrived at the following.. "2 x 100^2 + 2 x 1^2" thanks in advance
Originally posted by testcracker on 30 Sep 2017, 16:38.
Last edited by testcracker on 30 Sep 2017, 19:31, edited 2 times in total.



Math Expert
Joined: 02 Sep 2009
Posts: 52386

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
01 Oct 2017, 02:57



Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 273

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
01 Oct 2017, 07:46
Bunuel wrote: gmatcracker2017 wrote: PareshGmat wrote: Answer = D = 10
\(\sqrt[4]{\frac{99^2 + 101^2}{2}  1}\)
\(= \sqrt[4]{\frac{(1001)^2 + (100+1)^2}{2}  1}\)
\(= \sqrt[4]{\frac{2 * 100^2 + 2 * 1^2}{2}  1}\)
\(= \sqrt[4]{100^2 + 1  1}\)
\(= \sqrt[4]{10^4}\)
= 10 hi please let me understand the way you arrived at the following.. "2 x 100^2 + 2 x 1^2" thanks in advance \((1001)^2 + (100+1)^2=100^2  2*100+1 + 100^2+2*100+1=2*100^2+2\) thanks man you are so great ....



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4600
Location: United States (CA)

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
04 Oct 2017, 15:48
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. Answer: D
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Senior Manager
Joined: 09 Mar 2018
Posts: 255
Location: India

Re: ((99^2 + 101^2)/2  1)^(1/4)
[#permalink]
Show Tags
11 Jan 2019, 09:56
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, (1001)^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
_________________
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) &nbs
[#permalink]
11 Jan 2019, 09:56






