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by BTGmoderatorRO » Sun Oct 15, 2017 9:29 am
E: 9 $$\sqrt[4]{\frac{\left(99^2+101^2\right)}{2}-1}$$
A: 15
B: 12
C: 11
D: 10
OA is

d. Can some experts help me with this? How do i come up with the correct approach to solve this question?
Thanks
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by DavidG@VeritasPrep » Sun Oct 15, 2017 11:46 am
Roland2rule wrote:E: 9 $$\sqrt[4]{\frac{\left(99^2+101^2\right)}{2}-1}$$
A: 15
B: 12
C: 11
D: 10
OA is

d. Can some experts help me with this? How do i come up with the correct approach to solve this question?
Thanks
99^2 = (100-1)^2 = 100^2 - 200 + 1 = 10^4 -200 + 1
101^2 = (100+1)^2 = 100^2 + 200 + 1 = 10^4 + 200 + 1

So 99^2 + 100^2 = 10^4 - 200 + 1 + 10^4 + 200 + 1 = 2*10^4 + 2
And [(99^2 + 100^2 )/2] - 1 = [(2*10^4 + 2)/2] - 1 =( 10^4 + 1) -1 = 10^4

All that's left to do is to take the fourth root of 10^4, which is just 10. The answer is D
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by Brent@GMATPrepNow » Sun Oct 15, 2017 12:28 pm
Roland2rule wrote:$$\sqrt[4]{\frac{\left(99^2+101^2\right)}{2}-1}$$
A: 15
B: 12
C: 11
D: 10
E: 9
We might also apply some approximation.
Notice that (99² + 101²)/2 represents the AVERAGE of 99² and 101²
The average of 99² and 101² is going to be approximately equal to 100²
So, the fourth root of (99² + 101²)/2 - 1 ≈the fourth root of 100² - 1
Next, since 100² is equal to 10,000, subtracting 1 from has little consequence. So, let's ignore it

So, $$\sqrt[4]{\frac{\left(99^2+101^2\right)}{2}-1}$$ ≈ the fourth root of (100² - 1)
≈ fourth root of (100² - 1)
≈ fourth root of (100²)
≈ fourth root of (10,000)
≈ 10

Answer: D

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by [email protected] » Sun Oct 15, 2017 1:48 pm
Hi Roland2rule,

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: D

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by Scott@TargetTestPrep » Sun Nov 24, 2019 5:23 pm
BTGmoderatorRO wrote:E: 9 $$\sqrt[4]{\frac{\left(99^2+101^2\right)}{2}-1}$$
A: 15
B: 12
C: 11
D: 10
OA is

d. Can some experts help me with this? How do i come up with the correct approach to solve this question?
Thanks

First, we have: 99^2 + 101^2 = 9,801 + 10,201 = 20,002.

Next, we have: 20,002/2 - 1 = 10,001 - 1 = 10,000.

Finally, take the 4th root of 10,000; we have: 4^√10,000 = 10.

Alternate Solution:

Notice that 99 = 100 - 1 and 101 = 100 + 1. Let's substitute these into the expression within the parentheses:

99^2 + 101^2

(100 - 1)^2 + (100 + 1)^2

100^2 - 200 + 1 + 100^2 + 200 + 1

2*(100^2) + 2

Thus:

4^√[(99^2 + 101^2)/2 - 1] = 4^√[(2*(100^2) + 2)/2 - 1] = 4^√[100^2 + 1 - 1] = 4^√(100^2) = 10

Answer: D

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