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

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16 00:00

Difficulty:   35% (medium)

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

### HideShow timer Statistics $$\sqrt{\frac{99^2 + 101^2}{2} - 1} =$$

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

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

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

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

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

= 10
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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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.

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

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

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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

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

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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

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.

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

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

$$\sqrt{10000}$$

$$10$$
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GRE 1: Q169 V154 Re: ((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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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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PareshGmat wrote:
$$\sqrt{\frac{99^2 + 101^2}{2} - 1} =$$

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.

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

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PareshGmat wrote:

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

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

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

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

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

= 10

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

Wouldn't the 1's cancel out? for example:
$$= \sqrt{\frac{2 * 100^2 -1^2 + 1^2}{2} - 1}$$
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((99^2 + 101^2)/2 - 1)^(1/4)  [#permalink]

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PareshGmat wrote:

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

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

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

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

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

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

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gmatcracker2017 wrote:
PareshGmat wrote:

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

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

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

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

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

= 10

hi

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

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

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

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Bunuel wrote:
gmatcracker2017 wrote:
PareshGmat wrote:

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

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

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

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

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

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

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PareshGmat wrote:
$$\sqrt{\frac{99^2 + 101^2}{2} - 1} =$$

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.

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

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PareshGmat wrote:
$$\sqrt{\frac{99^2 + 101^2}{2} - 1} =$$

A: 15

B: 12

C: 11

D: 10

E: 9

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

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)

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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
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