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((99^2 + 101^2)/2  1)^(1/4)
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08 Jan 2015, 01:42
Question Stats:
72% (01:52) correct 28% (02:16) wrong based on 313 sessions
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\(\sqrt[4]{\frac{99^2 + 101^2}{2}  1} =\) A: 15 B: 12 C: 11 D: 10 E: 9
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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08 Jan 2015, 02:52
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
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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08 Jan 2015, 14:38
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
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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08 Jan 2015, 15:19
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.



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Re: ((99^2 + 101^2)/2  1)^(1/4)
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08 Jan 2015, 22:46
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. \(\sqrt[4]{\frac{(99^2 + 101^2)}{2}  1}\) \(\sqrt[4]{\frac{(20000)}{2}  1}\) \(\sqrt[4]{10000}\) \(10\)
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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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
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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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 = (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.
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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06 Aug 2017, 17:09
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: \(= \sqrt[4]{\frac{2 * 100^2 1^2 + 1^2}{2}  1}\)



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((99^2 + 101^2)/2  1)^(1/4)
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Updated on: 30 Sep 2017, 20:31
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, 17:38.
Last edited by testcracker on 30 Sep 2017, 20:31, edited 2 times in total.



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Re: ((99^2 + 101^2)/2  1)^(1/4)
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01 Oct 2017, 03:57
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 \((1001)^2 + (100+1)^2=100^2  2*100+1 + 100^2+2*100+1=2*100^2+2\)
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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01 Oct 2017, 08:46
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 ....



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Re: ((99^2 + 101^2)/2  1)^(1/4)
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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. Answer: D
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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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, (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
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Re: ((99^2 + 101^2)/2  1)^(1/4)
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