Maths Question (Approximation & Errors)

1)maximum relative error = maximum absoulte error / measured value
maximum absoulte error = maximum relative error X measured value
∴the absolute error of weight of each piece of pin
= [168 X (1/84)] / 100
= 0.02g
the weight of each piece of pin = 168 / 100 = 1.68g
upper limit of weight of each piece of pin
= measured value + maximum absoulte error
= (1.68 + 0.02)g
= 1.70g
lower limit of weight of each piece of pin
= measured value﹣maximum absoulte error

= (1.68﹣0.02)g
= 1.66g
∴the possible range of weight of each piece of pin is between 1.66g and 1.70g.

2)the maximum absoulte error of the length
= (1/2)(0.5)
= 0.25 cm
the lower limit of the length
= (8.5﹣0.25)cm
= 8.25cm
∴the smallest possible volume of the box
= (8.25)3 cm3
= 561.52 cm3

3)maximum relative error = maximum absoulte error / measured value
maximum absoulte error = maximum relative error X measured value
∴the maximum absolute error of width
= (16.8)(1/840)
= 0.02cm
the upper limit of the width
= (16.8 + 0.02)cm
= 16.82cm
the lower limit of the width
= (16.8﹣0.02)cm
= 16.78cm
∴the range of the actual width is between 16.78cm and 16.82cm.
the maximum absolute error of length
= (25.58)(25/1279%)
= 0.005cm
the upper limit of the length
= (25.58 + 0.005)cm
= 25.585cm
the lower limit of the length
= (25.58﹣0.005)cm
= 25.575cm
∴the range of the actual length is between 25.575cm and 25.585cm.
the upper limit of the area
= (16.82)(25.585)cm2
= 430.3397cm2
the lower limit of the area
= (16.78)(25.575)cm2
= 429.1485cm2
∴the range of the actual area is between 429.1485cm2 and 430.3397cm2