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

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

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New post 08 Jan 2015, 00:42
2
12
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

76% (01:21) correct 24% (02:00) wrong based on 280 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)  [#permalink]

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New post 08 Jan 2015, 01:52
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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
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 08 Jan 2015, 13:38
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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.

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 "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

Final Answer:

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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 08 Jan 2015, 14:19
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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.
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 08 Jan 2015, 21:46
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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

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)  [#permalink]

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New post 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

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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 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 = (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.
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 06 Aug 2017, 16:09
PareshGmat wrote:
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:
\(= \sqrt[4]{\frac{2 * 100^2 -1^2 + 1^2}{2} - 1}\)
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((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post Updated on: 30 Sep 2017, 19:31
PareshGmat wrote:
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

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.
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 01 Oct 2017, 02:57
gmatcracker2017 wrote:
PareshGmat wrote:
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


\((100-1)^2 + (100+1)^2=100^2 - 2*100+1 + 100^2+2*100+1=2*100^2+2\)
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 01 Oct 2017, 07:46
Bunuel wrote:
gmatcracker2017 wrote:
PareshGmat wrote:
Answer = D = 10

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

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

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

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

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

= 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-)
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 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
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Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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New post 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, (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
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Re: ((99^2 + 101^2)/2 - 1)^(1/4) &nbs [#permalink] 11 Jan 2019, 09:56
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