M15-21

Official Solution:

If \(x = 80.0127^2\), which of the following must be true?

A. \(6400 \lt x \lt 6400.5\)
B. \(6400.5 \lt x \lt 6401\)
C. \(6401 \lt x \lt 6401.5\)
D. \(6401.5 \lt x \lt 6402\)
E. \(6402 \lt x \lt 6402.5\)


\(80.0127^2\) will be slightly greater than \(80.0125^2\). To calculate \(80.0125^2\), we proceed as follows:

\(80.0125^2=\)

\(= (80 + 0.0125)^2 = \)

\(=80^2 + 2*80*0.0125 + 0.0125^2 = \)

\(=6400 + 2 + 0.0125^2 =\)

\(=6402 + 0.0125^2\).

Given the options provided, only in choice E is \(x\) greater than 6402, while in all other options it's less than 6402. Therefore, choice E must be the correct one.


Answer: E
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Thanks for the solution .... this one was a bit tricky ....
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Good question. Thanks for the solution... Much faster than doing long multiplication

I think this is a high-quality question and I agree with explanation.

Conceptual issues tested and corrected.Suppose I have a number 10.5..and now I want find the square of this number..now what i was thinking is 100.25..but hold on..

see actual is 110.25 so its a big big difference...

Good quality question!!

I solved it in bit different manner...
80 + 0.0127

What happens that after seeing answer choices so close, I felt rounding off will make things difficult.
for second part - 2*80*0.0127
2*80*127/10000
16 * 127/1000
1270 + 127*6 = 2032
which will give 2 hence option E

I think this is a high-quality question and I agree with explanation. Amazing question!

80.0127 is nearly equal to \(80+(\frac{125}{10000})\)=\(80+\frac{1}{80}\)

\((x+\frac{1}{x})^2= x^2+\frac{1}{x^2}+2\)
\((80+\frac{1}{80})^2>6402\)

E

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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