PROBLEMS ON ABSOLUTE ERROR RELATIVE ERROR AND PERCENTAGE ERROR

Problem 1 :

The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. find the following in calculating the area of the circular plate:

(i) Absolute error (ii) Relative error (iii) Percentage error

Solution :

(i)  Absolute error :

Absolute error  =  Actual value - Approximate value

  Actual length  =  12.5 cm approximate value  =  12.65 cm

Area of circle A(r) =  πr²

Actual change in area  =  π(12.65)² - π(12.5)²

=  π[160.0225 - 156.25]

=  π(3.7725)

=  3.7725π  ---(1)

Approximate change  =  A'(12.5) x change in radius

  =  2π(12.5) x 0.15

=  25π x 0.15

=  3.75π  ---(2)

(1) - (2)

Absolute error  =  3.7725π - 3.75π

=  0.0225π cm²

(ii)  Relative error  =  (Actual value - Approximate value)/Actual value

Relative error  =  0.0225π / 3.7725π

=  0.0059 cm²

(iii)  Percentage error  =  Relative error x 100%

=  0.0059 x 100%

=  0.6%

Problem 2 :

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9 8 . cm. Find approximations for the following :

(i) change in the volume (ii) change in the surface area

Solution :

Volume of sphere V(r)  =  (4/3)πr³

Surface area of sphere A(r) =  4πr²

(i)  Change in volume  =  Rate of change in volume x difference

V'(r)  =  (4/3)π(3r²) dr

V'(r)  =  (4πr²) dr

V'(r)  =  (4π(10)²) (0.2)

V'(r)  =  80π

Volume decreases by 80π cm³.

(ii)  Change in surface area :

A(r) =  4πr²

A'(r)  =  8πr(dr)

A'(r)  =  8π(10)(0.2)

A'(r)  =  16π cm²

Surface area decreases by 16π cm².

Problem 3 :

The time T , taken for a complete oscillation of a single pendulum with length l , is given by the equation

T  =  2π√(l/g)

where g is constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of 1.

Solution :

T  =  2π√(l/g)

T  =  2π(l/g)^1/2

log T  =  log 2π + (1/2) [log l - log g]

(1/T) dT/dl  =  0 + (1/2) [(1/l) - 0]

dT/T  =  (1/2) (1/l) dl

Percentage error  =  (1/2) (1/l) x 0.02 l x 100

=  1%

Problem 4 :

Show that the percentage error in the n^th root of a number is approximately 1/n times the percentage error in the number.

Solution :

f(x)  =  x^1/n

y  =  x^1/n

Taking log on both sides, we get

log y  =  (1/n) log x

(1/y) (dy/dx)  =  (1/n) (1/x)

dy/y  =  (1/n) (1/x) dx

Problem 5 :

Sham was given three of the grades of her classmates. These grades were the three closest values to Sham’s grade. If Sham’s grade is considered the actual value, what is the mean absolute error of her classmates’ grades?

Sham’s Grade: 96.3

Classmate 1 : 95.6

Classmate 2 : 94.8

Classmate 3 : 95.2

Solution :

The absolute error of each of the classmates is equal to,

AE = AV – MV

AE1 = 96.3 – 95.6

AE1 = 0.7

AE2 = 96.3 – 94.8

AE2 = 1.5

AE3 = 96.3 – 95.2

AE3 = 1.1

The mean or average of the absolute errors is equal to,

MAE = (0.7 + 1.5 + 1.1) / 3

MAE = 1.1

Therefore, the Mean Absolute Error is 1.1

Problem 6 :

Find the absolute and relative errors. The actual value is 125.68 mm, and the measured value is 119.66 mm.

Solution :

Absolute Error is given,

AE = AV – MV

AE = 125.68mm – 119.66mm

AE = 6.02mm

Relative Error is given,

RE = [(AV–MV)/AV] x 100

RE = [(125.68mm – 119.66mm)/125.68mm] x 100

RE = 0.0479 x 100

RE = 4.79%

Therefore, the Absolute Value is 6.02mm with a Relative Error of 4.79%.

Problem 7 :

The book’s length is 12.5cm, but Via measured only 12.4cm. Find the absolute and relative errors.

Solution :

Absolute Error is given,

AE = AV – MV

AE = 12.5cm – 12.4cm

AE = 0.1cm

Relative Error is given,

RE = [(AV – MV)/AV] x 100

RE = [(12.5cm – 12.4cm)/12.5 cm] x 100

RE = 0.008 x 100

RE = 0.8%

Therefore, the Absolute Value is 0.1cm with a Relative Error of 0.8%.

Problem 8 :

The thermometer measures up to the nearest 2 degrees. The temperature was measured at 38° C. Find the relative Error.

Solution :

The actual value is not given, but we are given 2 degrees as the values change.

That means that the value could be 37° C or 39° C. Either way, we are sure that the absolute Error is 1° C. And so the Relative Error is given,

RE = [(AV – MV)/AV] x 100

RE = (AE-AV) x 100

RE = 138 x 100

RE = 0.0263 x 100

RE = 2.63%

Therefore, the Relative Error is 2.63%.

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