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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

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.

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