Relative error #2

Classical Mechanics Level 4

You are measuring the length, width and height of a box with a ruler. You got 10 cm, 5 cm and 4 cm respectively as the length,width and height. How much relative error was occurred in your measurement?

If your answer is $x.\text{Something}\%$ , enter your answer as $x + 2$

Note: The ruler can only measure in cm.

The answer is 31.

1 solution

Guilherme Niedu
Feb 21, 2018

The volume will be $v = 10 \cdot 5 \cdot 4 = 200$ $cm^3$

Since the error is $\pm 0.5$ cm, the minimum possible volume is $9.5 \cdot 4.5 \cdot 3.5 = 149.625$ $cm^3$ and the maximum possible volume is $10.5 \cdot 5.5 \cdot 4.5 = 259.875$ $cm^3$ . So, the maximum relative errror is:

$\large \displaystyle e = max \left ( \left | \frac{ 259.875 - 200 }{200} \right |, \left | \frac{ 149.625 - 200 }{200} \right | \right )$

$\color{#20A900} \boxed{ \large \displaystyle e = 29.9375 \% }$

$\color{#3D99F6} \large \displaystyle x = 29, \boxed{ \large \displaystyle x+2 = 31}$

Why is the error $\pm 0.5$ cm

Aman thegreat - 3 years, 3 months ago

Relative error #2

The answer is 31.

1 solution

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