Secondary Notes

Approximation and Errors – Mathematics Form 3 Notes and K.C.S.E Topical Q&A

Approximation and Errors - Mathematics Form 3 Notes and K.C.S.E Topical Q&A

Approximation and Errors – Mathematics Form 3 Notes and K.C.S.E Topical Q&A

Approximation and Errors – Mathematics Form 3 Notes and K.C.S.E Topical Q&A

Approximation

Approximation involves rounding off and truncating numbers to give an estimation

Rounding Off

In rounding off the place value to which a number is to be rounded off must be stated.

The digit occupying the next lower place value is considered.

The number is rounded up if the digit is greater or equal to 5 and rounded down if it’s less than 5.

Example

Round off 395.184 to:

  1. The nearest hundreds
  2. Four significant figures
  3. The nearest whole number
  4. Two decimal places

Solution

  1. 400
  2. 395.2
  3. 395
  4. 395.18

Truncating

Truncating means cutting off numbers to the given decimal places or significant figures, ignoring the rest.

Example

Truncate 3.2465 to

  1. 3 decimal places
  2. 3 significant figures

Solution

  1. 3.246
  2. 3.24

Estimation

Estimation involves rounding off numbers in order to carry out a calculation faster to get an approximate answer .This acts as a useful check on the actual answer.

Example

Estimate the answer to 152 x 269/32
                      

Solution

The answer should be close to 150 x 270 /30= 1350

The exact answer is 1 277.75. 1 277.75 writen to 2 significant figures is 1 300 which is close to the estimated answer.

KCSE MATHEMATICS QUESTIONS AND SOLUTIONS

Accuracy and Error

Absolute Error

The absolute error of a stated measurement is half of the least unit of measurement used.

When a measurement is stated as 3.6 cm to the nearest millimeter, it lies between 3.55 cm and 3.65 cm.

The least unit of measurement is milliliter, or 0.1 cm. The greatest possible error is 3.55 – 3.6 = -0.05 or 3.65 – 3.6 = + 0.05.

To get the absolute error we ignore the sign. So the absolute error is 0.05 thus,|-0.05| =| +0.05|= 0.05.

When a measurement is stated as 2.348 cm to the nearest thousandths of a centimeters (0.001 ) then the absolute error is 1/2 x 0.001 = 0.0005.

Relative Error

Relative error =           absolute  /
actual measurements

>

Example

An error of 0.5 kg was found when measuring the mass of a bull.if the actual mass of the bull was found to be 200kg. Find th relative error

Solution

>

elative error =          absolute           = 0.5 kg = 0.0025
actual measurements    200

Percentage Error

Percentage error = relative error x 1 00%

=   absolute error      x 100%
actual measurment

>

Example

The thickness of a coin is 0.20 cm.

  1. The percentage error
  2. What would be the percentage error if the thickness was stated as 0.2 cm ?

Solution

>

The smallest unit of measurement is 0.01

Absolute error 1/2 x 0.01 = 0.005

Percentage error = 0.005 x 100% = 2.5 %
                             0.20

>

The smallest unit of measurement is 0.1
Absolute error 1/2 x 0.1 = 0.05 cm
Percentage error 0.05/0.2 x 100% = 25 %

Rounding Off Error

An error found when a number is rounded off to the desired number of decimal places or significant figures, for example when a recurring decimal 1.6 is rounded to the 2 significant figures, it becames 1.7 the rounded off error is;
1.7 – 1.6 
=17/10 – 5/1/30

Note;

>

1.6 which is a recurring decimal converted to a fraction is 5/3

KCSE MATHEMATICS QUESTIONS AND SOLUTIONS

Truncating Error

The error introduced due to truncating is called a truncation error.in the case of 1.6 truncated to 2 S.F., the truncated error is; |1.6 -1.6|= |16/10 – 12/31/15

>

Propagation of Errors

Addition and subtraction

What is the error in the sum of 4.5 cm and 6.1 cm, if each represent a measure measurement.

Solution

The limits within which the measurements lie are 4.45, i.e. ., 4.55 or 4.5 ± 0.005 and 6.05 to 6.1 5, i.e. 6.1 ±0.05.
The maximum possible sum is 4.55 + 6.15 =10.7cm
The minimum possible sum is 4.45 
+ 6.05 =10.5 cm
The working sum is 4.5 + 6.1 = 10.6
The absolute error = maximum sum – working sum
=|10.7 – 10.6|
=0.10

>

Example

What is the error in the difference between the measurements 0.72 g and 0.31 g?

Solution

>

The measurement lie within 0.72 ± 0.005 and 0.31 ± 0.005 respectively

The maximum possible difference will be obtained if we substract the minimum value of the second measurement from the maximum value of the first, i.e ;

0.725 – 0.305 cm

The minimum possible difference is 0.71 5 – 0.31 5 = 0.400.the working difference is 0.72 – 0.31 =0.41, which has an absolute error of |0.420 -0.41 | or |0.400 – 0.41 | = 0.1 0.

Since our working difference is 0.41, we give the absolute error as 0.01 (to 2 s.f)

Note:

In both addition and subtraction, the absolute error in the answer is equal to the sum of the absolute errors in the original measurements.

Multiplication

Example

A rectangular card measures 5.3 cm by 2.5 cm. find

  1. The absolute error in the rea of the card
  2. The relative error in the area of the cord

Solution

  1. The length lies within the limits 5.3 ± 0.05 cm
  2. The length lies within the limits 2.5 ± 0.05 cm

The maximum possible area is 2.55 x 5.35 =13.6425 cm2
The minimum possible area is 2.45 x 5.25 =12.8625 cm2
The working area is 5.3 x 2.5 = 13.25 cm2
Maximum area – working area = 13.6425 – 13.25 = 0.3925.
Working area – minimum area = 13.25 – 12.8625 = 0.3875
We take the absolute error as the average of the two.

Thus, absolute error = 0.3925 + 0.3875 = 0.3900
2

KCSE MATHEMATICS QUESTIONS AND SOLUTIONS


The same can also be found by taking half the interval between the maximum area and the minimum area
1/2( 13.6425-12.8625) = 0.39

The relative error in the area is :
 0.39  = 0.039 ( to 2 S.F)
13.25

Division

Given 8.6 cm ÷ 3.4 cm.Find:

  1. The absolute error in the quotient
  2. The relative error in the quotient

KCSE MATHEMATICS QUESTIONS AND SOLUTIONS

Solution

  1. 8.6 cm has limits 8.55 cm and 8.65 cm. 3.4 has limits 3.35 cm and 3.45 cm.
    The maximum possible 
    quotient will be given by the maximum possible value of the numerator and the smallest possible value of the denominator, i.e.,
    8.65 = 2.58 (to 3 s.f)
    3.35
    The minimum possible quotient will be given by the minimum possible value of the numerator and the biggest possible value of the denominator, i.e.
    8.65 = 2.48 (to 3 s.f)
    3.45
    The working quotient is;
    8.6= 2.53 (to 3 .f.)
    3.4
    The absolute error in the quotient is;
    2.53 x 2 .48 = 1/x 0.10
           2
    = 0.050 ( to 2 s.f)
  2. Relative error in the working quotient ;
    0.05 =  5
       2.53     253
    = 0.01 97
    = 0.020 (to 2 s.f )AlternativelyRelative error in the numerator is 0.05 = 0.00581
                                                     8.6
    Relative error in the denominator is 0.05 = 0.0147
                                                         3.4
    Sum of the relative errors in the numerator and denominator is
    0.00581 + 0.01 47 = 0.02051 s
    =0.021 to 2 S.F

KCSE MATHEMATICS QUESTIONS AND SOLUTIONS

FORM 1 END TERM EXAMINATIONS FREE DOWNLOAD

FORM 2 END TERM EXAMINATIONS FREE DOWNLOAD

FORM 3 END TERM EXAMINATIONS FREE DOWNLOAD

FORM 4 END TERM EXAMINATIONS FREE DOWNLOAD

PRE-MOCKS EXAMS AND MARKING SCHEMES 2022

Follow our Facebook, Instagram, Linked In, Twitter & Join our Telegram Group and Facebook Group for More Updates on WhatsApp Groups

 

Related Articles

Leave a Reply

Back to top button