Lesson Notes By Weeks and Term - Senior Secondary School 2

SUBJECT: MATHEMATICS

CLASS:  SS 2

DATE:

TERM: 1st TERM

REFERENCE BOOKS

• New General Mathematics SSS2 by M.F. Macrae etal.
• Essential Mathematics SSS2 by A.J.S. Oluwasanmi. 

 
WEEK TWO

TOPIC: PERCENTAGE ERROR

CONTENT

• Definition of percentage error
• Calculation of percentage error
• Percentage error (range of values via approximations)
• Calculations on percentage error in relation to approximation

Definition of Percentage Error

No measurement, however, carefully made is exact (accurate) i.e if the length of a classroom is measured as 2.8m to 2 s.f the actual length may be between 2.75 and 2.85, the error of this measurement is 2.75 – 2.8 or 2.85 – 2.8 = + 0.05.

          2.75                  2.85    2.8           2.9   

Percentage error  =            error                X   100

                Actual measurement                    1   

error  =+  0.05

actual measurement  2.8

% error  =0.05   x  100

            2.8            1

                =1.785% = 1.79%

Example 2

Suppose the length of the same room is measured to the nearest cm ,280cm i.e. (280cm) calculate the percentage error.

Measurement = 280cm.

The range of measurement will be between 279.5cm or 280cm 

Error  =  280 – 279.5   =  0.5cm

% error  =       error          x  100

            Measurement        1

% error  = 0.5    x   100   =  0.178%

        280         1 

                  =   0.18% (2sf)

Example 3

The length of a field is measured as 500m; find the percentage error of the length if the room is measured to

1. nearest metre      ii. nearest 10m         iii.    one significant figure.

Solutions

1. To the  nearest  metre

    Measurement = 500m

    Actual measurement = between 499.5 - 500.5

    Error =  + 0.5m

    % error =            error        x 100

        measurement

    0.5  x  100         =   0.10%

    500    1

            =    0.10%

1. Nearest 10m

    Measurement = 500m,

    range= 495m – 505m

    error  =+ 5m

    error  =   5    x   100    =   1%

        500      1

iii.    To 1 s.f.

    measurement   =  500m

    range   =   450 – 550

    error   =   +  50

    % error  =  50   x    100    =  10%

                                500          1

Evaluation

1.         The  length  of  each  side  of  a  square  is  3.6 cm to  2s.f. (a) Write down the  smallest  and  the  largest  of  each  side.  (b) Calculate the smallest and the largest values for the perimeter.

             (c)  Find the possible values of the area.

Percentage Error (range of values via approximations)

1. Range of values measured to the nearest whole number i.e. nearest tens, hundreds etc. e.g.

Find the range of values of N6000 to:

5999. nearest naira     =    N5999.50    -    6000.50
6000. nearest N10     =    N5995        -    6005

iii.    nearest N100        =    N5950        -    6050

1. nearest N1000             =    N5500        -    6,500

2. Range of values measured to a given significant figure. E.g. find the range of value of 6000 to

    1 sf    =    5500  -  6500

    2 sf    =    5950  -  6050

    3 sf    =    5995  -  6005

    5 sf    =    5999.95- 6000.05

3. Range of values measured to a given decimal places e.g. 39.8 to a 1d.p = 39.75 – 39.85.

    Note:   if it is 1 d.p, the range of values will be in 2 d.p, if 2 d.p, the range will be in 3 d.p etc. (i.e the range = d.p + 1). The same rule is also applicable to range of values to given significant figure. 

Evaluation

Orally: From New General Mathematics SS 2 by J. B. Channon and Co 3rd edition exercise 46 no. 1a – f.

Calculations on percentage error:

Example:

Calculate the percentage error if

1. The capacity of a bucket is 7.5 litres to 1 d.p.
2. The mass of a student is 62kg to 2 s.f.

Solutions

1. Measurement     =   7.5litres  ( 1d.p)

    Range of values = 7.45  - 7.55

    Error                   =   7.5 – 7.45 = 0.05

    % error =      error            x    100

        measurement         1 

            0.05   x   100

            7.51

                        =   0.67%

2. Measurement = 62kg (2 s.f)

    Range of values    = 61.5kg to 62.5kg

    error                  =  6.2  -  61.5   =  0.5kg

    % error   =          error               x    100

        measurement          1

                0.5    x    100    =   0.81%

                62            1

EVALUATION

1. Calculate correct to 2 s.f. the percentage error in approximately 0.375 to 0.4.

GENERAL EVALUATION / REVISION QUESTION

1. A metal rod   was   measured as 9.20 m. If the real length is 9.43 m, calculate the percentage error to 3 s.f

2.A  student  measures  the  radius  of  a  circle  as  1.46 cm  instead  of   1.38 cm. Calculate  the  percentage  error.

3.The  weight  of  sugar  was  recorded  as  8.0 g  instead  of  8.2 g.  What is the percentage error?

4.A  student  mistakenly  approximated  0.03671  to  2 d.p  instead  of   2 s.f. What is the percentage error correct to 2  s.f

5.A  man’s  weight  was  measured  as  81.5 kg  instead  of  80 kg. Find the percentage error in the measurement.

WEEKEND ASSIGNMENT

What is the error in the following measurement

1. The distance between two towns is 60km to the nearest km.  (a) 5km     (b) 0.5km     (c) 8.3km     (d) 0.83km
2. The area of a classroom is 400m2 to 2 s.f.          (a) 50m2   (b) 1.25m2(c) 2.5m2     (d)  5m2
3. A sales girl gave a girl a balance of N1.15 to a customer instead of N1.25, calculate the % error.
4. A student measured the length of a room and obtained the measurement of 3.99m, if the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room?(a) 3.78m   (b) 3.80m   (c) 4.18m   (d) 4.20m
5. A man is 1.5m tall to the nearest cm, calculate his percentage error.

(a)  0.05cm    (b) 0.33%   (c) 0.033%   (d)  0.05cm

THEORY

1. A classroom is 10m by 10m; a student measured a side as 9.5m and the other side as 10m and uses his measurement to calculate the area of the classroom. Find the percentage error in a. the length of one of the sidesb. the area of the room
2. Instead of recording the number 1.23cm for the radius of a tube, a student recorded 1.32cm, find the percentage error correct to 1 d.p.

Reading Assignment

Essential Mathematics for SSS2, pages 13-22, Exercise 2.4