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Question Number 71696 by TawaTawa last updated on 18/Oct/19

The side of a square is measured to be 12cm long cofrect to the nearest cm. Find the maximum absolute error and the maximum percentage error for (a) The length of the square (Answer: 0.5cm, 4.17%) (b) The area of the square. (Answer: 12.25cm, 8.5%)

$$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$

Answered by MJS last updated on 19/Oct/19

“to the nearest cm” means ±.5cm because you will round 11.5≈12.0 and 12.49^(.) ≈12.0 ⇒ absolute error of length is ±.5cm ⇒ percentage is ((.5)/(12))≈4.17% 11.5≤x≤12.5 ⇔ 132.25≤x≤156.25 x=12±.5 ⇔ x^2 =144_(−11.75) ^(+12.25) ⇒ absolute error of area is 12.25 ⇒ percentage is ((12.25)/(144))≈8.51%

$$``\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{cm}''\:\mathrm{means}\:\pm.\mathrm{5cm}\:\mathrm{because} \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{round}\:\mathrm{11}.\mathrm{5}\approx\mathrm{12}.\mathrm{0}\:\mathrm{and}\:\mathrm{12}.\mathrm{4}\overset{.} {\mathrm{9}}\approx\mathrm{12}.\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{absolute}\:\mathrm{error}\:\mathrm{of}\:\mathrm{length}\:\mathrm{is}\:\pm.\mathrm{5cm} \\ $$$$\Rightarrow\:\mathrm{percentage}\:\mathrm{is}\:\frac{.\mathrm{5}}{\mathrm{12}}\approx\mathrm{4}.\mathrm{17\%} \\ $$$$\mathrm{11}.\mathrm{5}\leqslant{x}\leqslant\mathrm{12}.\mathrm{5}\:\Leftrightarrow\:\mathrm{132}.\mathrm{25}\leqslant{x}\leqslant\mathrm{156}.\mathrm{25} \\ $$$${x}=\mathrm{12}\pm.\mathrm{5}\:\Leftrightarrow\:{x}^{\mathrm{2}} =\mathrm{144}_{−\mathrm{11}.\mathrm{75}} ^{+\mathrm{12}.\mathrm{25}} \\ $$$$\Rightarrow\:\mathrm{absolute}\:\mathrm{error}\:\mathrm{of}\:\mathrm{area}\:\mathrm{is}\:\mathrm{12}.\mathrm{25} \\ $$$$\Rightarrow\:\mathrm{percentage}\:\mathrm{is}\:\frac{\mathrm{12}.\mathrm{25}}{\mathrm{144}}\approx\mathrm{8}.\mathrm{51\%} \\ $$

Commented by TawaTawa last updated on 19/Oct/19

I appreciate your time sir. God bless you sir

$$\mathrm{I}\:\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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