Passnownow

# Classwork Series and Exercises {Mathematics- SS2}: Approximation and Errors

Approximation and Errors

Approximations are very useful when calculating. Numbers can be approximated to varying degrees in order to obtain rough estimates of calculations.

Example: If the tax deducted from a person’s salary is ₦ 93.80 per month, the amount deducted in a year is ₦ 93.80 x 1.2. A rough estimate is ₦ 90 x 10 = ₦900.

The real value of the tax deducted = ₦ 93.80 x 1.2 = ₦ 1179.6

Rough estimates are not accurate (as shown above) but they give an idea of the size or order of magnitude of the correct results of calculations. Thus they are often used to check the correctness of answer without calculators.

Rounding off Numbers

Numbers can be rounded off to different levels of accuracy such as to the nearest hundred, ten, whole numbers etc. or to a given number of decimal places (d.p) or significant figures (s.f)

Significant numbers

Numbers are sometimes rounded off to a given number of significant figures. The significance of a digit depends on its place value in the numbers. Thus in the number 146.83, the “4” with a value of 40, is more significant than the 6, the 6 more significant than the  8 whilst the 1 is the most significant of all in the set.

To round off to a certain number of significant figures, count all the numbers up to the required number of figures. Then check the next figure, if 5 or > 5, round up and add to the previous number. However if <5, round off.

Example: Approximate to 3 significant figures (i) 4567 (ii) 7654

Solution:

i.  4567; 6 is the 3rd number but 7 > 5 so we approximate and add to 6 then replace 7 by 0 = 4570

ii.7654; 5 is the 3rd number but 4< 5 so we round off, replaced by 0 = 7650

For decimal fractions, the 1st significant figure is the first non-zero digit in the fraction. In the number 0.0002408, the 1st significant figure is 2.

NOTE: The zero after 4 in 0.0002408 is significant because it is between other numbers.

Examples:  Express the numbers below

(i)      234.89 to 3 significant figures = 235

(ii)     4578 to the nearest thousand  = 5000

(iii)    4578 to the nearest hundred = 4600

(iv)    4578 to the nearest ten = 4580

(v)     0.893 to 2 significant figures = 0.89

Degree of Accuracy

In examples ii- iv, observe that the gap between the actual number 4578 and the approximate values grows wider for a higher level of approximation i.e. the gap for approximation to the nearest thousand -422, is bigger than approximation for nearest hundred- 32 which is bigger than the approximation to the nearest ten-2. Thus, it can be said that the higher the degree of approximation, the lesser the accuracy.

This shows that errors occur when numbers are approximated and the amount   of error depends on the level of approximation.

Percentage Error

No measurement however carefully made is exact. Rather they are approximate to certain degrees of accuracy. If the width of a room is measured as 2.8m to 2 s.f, the actual width of the room may lie between 2.75m and 2.85m. The error in this measurement is 2.75 – 2.8m or 2.85m – 2.75, that is -0.05 or + 0.05m. Thus the possible error in the measurement is ± 0.05.

Example: Find the possible amount of errors in the following measurements

i.  4600 if the number has been approximated to the nearest hundred

Solution:  4600 = 4550 to 4650, Error = ± 50

ii.3.80 if the number had been approximated to 2 significant figures

Solution: 3.80 = 3.75 to 3.85, Error = ±0.05

The percentage that these errors in the measurements represent when compared to the original numbers can be evaluated using the formula below:

Percentage error = Error in measurement x 100%

Actual measurement

Where Error = Difference between measured value & approximated/assumed value

In example i. above, Percentage Error = ± 50 x 100 % = ± 1.08%

4600

Examples: A student wrongly records the dimensions of a room as Length= 13m, Width = 10m. On careful measurement, his teacher finds the room to be L= 13.1m, Width= 9.90m. Find (i) The error in the measurement of the area (ii) Percentage error in measurement of area to 2 significant figures

Solution:

Wrong Area = Wrong (length x width) = (13m x 10m) = 130m2

Actual measurement = (13.1m x 9.9m) = 129.69m2

i.  Error in measurement = Wrong measurement – Actual measurement

= 130 – 129.69 = 0.31m2

ii.Percentage Error =      Error  x 100%          = 0.31 x 100% = 0.239% = 0.24% (to 2 s.f)

Actual Measurement       129.69

Tests and Exercises

1. The length of a wall is given as 20m to 1 s.f.  Calculate the possible percentage error in this measurement. (a) ± 0.25% (b) ± 2.5% (c) ± 5% (d) ±10% (e) ± 25%**

Guideline: Rounding up to 1 significant figure to get 20 could be any value from 15 to 25 since 5 can either be rounded up or down. Thus the error = 25 -20 = 5 OR 15 -20 = – 5 i.e. ± 5

Percentage Error = Error x 100% =        ± 5 x 100% = ±25%. Answer is Option E

Actual measurement        20

2. Estimate which of these has the least value (a) 0.256 (b) 0.256 (c) 2.56 (d) 2.56 (e) 25.6

48                4.8         0.48       4.8        480

Guideline: The digits in both the numerator and the denominator are the same for all the options but are of different values. To solve this, express the numerator in all the options as whole numbers BUT adjust the denominators accordingly as shown below:

a.    256    b.   256      c. 256       d. 256       e.   256

48000       4800           48          480             4800

Option A has the largest denominator so it has the least value. Answer is Option A

3. An aircraft travels 5000km (to 1 s.f) in 6 hours (to the nearest hour). Estimate its speed in km/h to 1 s.f. (a) 833.333km/hr (b) 833km/hr (c) 8km/hr (d) 800km/hr

Guideline:

Speed = Distance = 5000 =833.33 = 800km/hr to 1 sig.fig. Answer is Option D

Time            6

4. A surveyor measures a road as being 69.3km long however there is a -1% error in this measurement. What is the true length of the road to 1 decimal place? (a) 69.4km (b) 69.2km

(c) 68.6 km (d) 70.0km

Guidelines:

Let the correct (actual) measurement of the land be X

-1% error = Error x 100% =        (69.3 – X) x 100%

X                                  X

-1% = 70.0km. Answer is Option D

5.  A young child gives his age as 13 years. However his mother says he’s only 12 years and 6 months old. What is the percentage error in the boy’s approximation of his own age? (a) 0.5% (b) 2%

(c) 4%  (d) 5%

Guidelines:

Approximated Age: 13 years = (13 x 12) months = 156 months

Real Age: 12years, 6 months = {(12 x12) + 6 months} = 150 months

Error in age = 6 months

Percentage Error = Error x 100% = 6 X100 = 4%. Answer is Option C