**Question 1: **Mrs.Simpson weighed 165 lbs in her Physicians scale. But the defective
scale used by her at Home showed her weight as 160 lb. Calculate the
relative and percentage error in the measurement of her scale.

If her daughter weighs 105 lb on the defective scale, find her actual weight nearest to the tenth of a lb.

** Solution: **

The actual error in the measurement of Mr. Simpson's weight = 165 - 160 = 5 lbs

Relative error in the measurement shown by the defective scale = $\frac{5}{165}$ = $\frac{1}{33}$

Percentage error = $\frac{1}{33}$ x 100 $\approx$ 3.03%

Let the actual weight of the daughter = x lbs

As Mrs.Simpson's weighing scale shows weight less than the actual weight x > 105.

Relative error in the measurement of defective scale = $\frac{x-105}{x}$ = $\frac{1}{33}$

33(x -105) = x (Cross multiplication of the proportion)

33x - 3465 = x

32x = 3465 $\Rightarrow$ x = 108.3 lb

The actual weight of her daughter = 108.3 lb

**Question 2: **A meter scale can measure to a precision of 1 mm. A rope and a small
length of wire are measured using the scale as 57.5 cm and 4 cm.
Calculate the relative errors occurred in both the measurements read
using the scale. If it is required to measure the length of the wire to
the same accuracy level as that found in the measurement of the rope,
find the precision level required for the measuring instrument.

** Solution: **

The absolute error for the measurements taken using the scale = $\frac{1}{2}$ x precision level = $\frac{1}{2}$ x 1 mm = 0.5 mm = 0.05 cm

Relative error in the measurement of the rope = $\frac{0.05 }{57.5}$ = $\frac{1}{1150}$ $\approx$ 0.0009

Relative error in the measurement of the wire = $\frac{0.05}{4}$ = $\frac{1}{80}$ $\approx$ 0.0125

If the length of the wire is to be measured to the same accuracy
level found in the measurement of the rope, the relative error in the

measurement of the wire should also be equal to $\frac{1}{1150}$.

Let the required precision = x.

The absolute error that can occur in the measurement = $\frac{x}{2}$

Thus the relative error that can occur in the measurement of length 4 cm = $\frac{\frac{x}{2}}{4}$ = $\frac{x}{8}$

$\frac{x}{8}$ = $\frac{1}{1150}$ $\Rightarrow$ x = $\frac{8}{1150}$ = 0.007 cm

Hence the required precision level = 0.07 mm.