# Approximation and Errors - Mathematics Form 3 Notes

## 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 = 1350
30

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.

## 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

### 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 - 12/3| = 1/15

### Propagation of Errors

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

The same can also be found by taking half the interval between the maximum area and the minimum area
1/213.6425-12.8625) = 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

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 )

Alternatively

Relative 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

## Past KCSE Questions on the Topic.

1.
1. Work out the exact value of R  =                1
0.003146 - 0.003130
2. An approximate value of R may be obtained by first correcting each of the decimal in the denominator to 5 decimal places
1. The approximate value
2. The error introduced by the approximation
2. The radius of circle is given as 2.8 cm to 2 significant figures
1. If C is the circumference of the circle, determine the limits between which C/π lies
2. By taking ∏ to be 3.142, find, to 4 significant figures the line between which the circumference lies.
3. The length and breadth of a rectangular floor were measured and found to be 4.1 m and 2.2 m respectively. If possible error of 0.01 m was made in each of the measurements, find the:
1. Maximum and minimum possible area of the floor
2. Maximum possible wastage in carpet ordered to cover the whole floor
4. In this question Mathematical Tables should not be used
The base and perpendicular height of a triangle measured to the nearest centimeter are 6 cm and 4 cm respectively.
Find
1. The absolute error in calculating the area of the triangle
2. The percentage error in the area, giving the answer to 1 decimal place
5. By correcting each number to one significant figure, approximate the value of 788 x 0.006. Hence calculate the percentage error arising from this approximation.
6. A rectangular block has a square base whose side is exactly 8 cm. Its height measured to the nearest millimeter is 3.1 cm
Find in cubic centimeters, the greatest possible error in calculating its volume.
7. Find the limits within the area of a parallegram whose base is 8cm and height is 5 cm lies. Hence find the relative error in the area
8. Find the minimum possible perimeter of a regular pentagon whose side is 1 5.0cm.
9. Given the number 0.237
1. Round off to two significant figures and find the round off error
2. Truncate to two significant figures and find the truncation error
10. The measurements a = 6.3, b= 1 5.8, c= 1 4.2 and d= 0.001 73 have maximum possible errors of 1%, 2%, 3% and 4% respectively. Find the maximum possible percentage error in ad/bc correct to 1 sf.

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