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Bunuel
If x = 80.0127^2, which of the following is true about x ?


A. \(6400.0 \lt x \lt 6400.5\)

B. \(6400.5 \lt x \lt 6401.0\)

C. \(6401.0 \lt x \lt 6401.5\)

D. \(6401.5 \lt x \lt 6402.0\)

E. \(6402.0 \lt x \lt 6402.5\)


M15-21

\(x = (80.0127)^2 = (80 + 0.0127)^2 = 80^2 + 2*80*0.0127 + 0.0127^2\)

Now note 80 * 0.0125 = 1, so 80*0.0127 > 1. The 2nd term above would be slightly greater than 2.

Then \((80.0127)^2 = \) 6400 + >2 + >0 = ">6402". The only option above is E.

Ans: E
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We shall split the number into \(80+0.0127\) just to see are we greater than 6401 or 6402.

We know that:\( (a+b)^2 = a^2 + 2ab + b^2\)

= \(80^2 + 2(80)(0.0127) + (0.0127)^2\)

Notice that the 2nd term can be approximately taken as \(2*80*0.0125 = 2\), hence we can assume that multiplying with 0.0127 would make it >2
Thus, we have the final expression =\( 6400 + (>2)\)

Which gives us the answer that x is greater than 6402 ----> Option E
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This is tagged into combination for some reason.
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This is tagged into combination for some reason.
­Fixed the tags. Thank you.
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