Approximation: Degree of accuracy and rounding up of number

Content

• Degree of Accuracy
• Rounding Up of Numbers

I.  Degree of Accuracy

Many calculations involve measurements.  The degree of accuracy of the results of the calculations depends therefore on the degree of accuracy of the measurements. It therefore means that the degree of accuracy of measurement in a calculation must be taken into consideration when determining the answer to the calculation.

Rounded –of values are sometimes used in calculations for example, pi(π) is often taken as 3.14  or 3.14 2.

II. Rounding –up of Numbers

It is not cost effective to give exact number of certain things due to the difficulty that may be encountered in the course of carrying out such task.  E.g.  Number of vehicles  plying a particular road, spectators in a stadium, population of a town etc. What is usually done is to round the number or approximate it to the nearest 10, 100,1000 and so on.

Example 1

Round the following numbers to the nearest ten

(a) 34               (b) 127            (c) 43678

Solution

• 34

:. To the nearest 10 = 30

(b) 127

 :. To the nearest 10 =130.

(c) 43678

:. To the nearest 10 = 43680.

Evaluation:

1. Round these numbers to the nearest hundred

    (a) 231         (b) 87345        (c) 567

2. The number of people at the cinema yesterday was 2576. Give this number to the nearest

  (a) 10             (b) 100            (c) 1000

Decimal Places

See the illustration below

 3.    5   7   8    6         

From  the illustration above, 3.5786 is divided into two parts by a decimal points to the right decimal to the left (whole number ).

Example 1

Give each of the following correct to 1d.p and 2 d.p

(a) 3.4567           (b) 35. 4782             (c) 4.2071

Solution

(a) 3.4567   

            i.   3.5  ( 1 d.p)

            ii. 3.46 (2d.p)

(b) 35. 4782

            i. 35.5   ( 1d.p)

ii. 35.48 ( 2d.p)

(c) 4.2071

            i.   4.2  ( 1 d.p)

            ii. 4.21 ( 2d.p)

Evaluation

Give each number correct to 2.d.p and 3d.p

(a) 5.7804                    (b) 0.007992               (c ) 16.869      (d) 28.0099.

Significant Figures

The word significant means important.  In mathematics, we need to study it in two aspects

i.  whole numbers

            3   8   0   6  9

ii.  decimal numbers

            0.   0   0  5  0  8   6

From the two illustrations above, we can conclude that zeros in the middle of a whole number are significant whilezeros at the end are not significant (insignificant)

Example 2

Give 45775 correct to (a) 1 s.f            (b)  2s.f                       (c)  3 s.f

Solution

(a) 50000         ( 1s.f)

(b)46000          ( 2 s.f)

 (c) 45 800       (3.s.f)

Example 3

Give each of the following numbers correct to 2 s.f

(a) 5.781          (b) 0.00244                 (c) 0.0507

Solution

(a) 5.781  = 5.8 ( 2 s.f)

(b) 0.00244 = 0.0024 ( 2 s.f)

(c ) 0.0507  = 0.051 ( 2 s.f)

Evaluation:

Give each number correct to 3 significant figures

(a) 57045         (b) 4540          (c )  456.56      (d) 0.5002       (e)34.0061   (f) 0.001011

Nearest Whole Number

To round a decimal number to the nearest whole number, check the number in the 1^std.p, if it is 5 or more than round the number up but if it is less than 5 do not change the number.

Example 1

Give the following correct to

i.   the nearest hundredth

ii.  the nearest thousandth

(a) 7.3425        (b) 0.00692       (c ) 7.0149     (d) 42.4739.

Solution

(a)  7.3425

 i.   7.34   (nearest hundredth)

ii. 7.343  (nearest thousandth)

(b) 0.0069

i.  0.01  (nearest hundredth)

ii. 0.007 (nearest thousandth)

(c ) 7.0149

i. 7.01 (nearest hundredth)

ii. 7.015 (nearest thousandth)

(d) 42.4739

i.   4.47  (nearest hundredth)

ii.  42.474  (nearest thousandth)

Example 2

Give each number correct to the nearest whole number

(a) 8.22            (b) 134.674

Solution

• 8.22  = 8 (nearest whole number )
• 134.674  =  135 (nearest whole number )

Evaluation:

Round off each of the following:

a.   34.8cm to the nearest cm

b.   67.1cm to the nearest cm

c. 24.6kg to the nearest kg.

Approximation in Our Everyday Activities

Approximation is a way of using rounded numbers to estimate answers to a calculation. Approximation can help us decide whether an answer to a calculation is of right size or not.  To find an approximate answer to a calculation, round the numbers to easy numbers, usually 1 s.f., or 2.s.f. or to the nearest whole number.  Then work out the approximated answer using these easy numbers.

Example 1

A boy was asked to calculate the cost of 82 oranges at N 5.80 each

Solution

Rough calculation

• = 80 and 5. 80 = 6

:. Approximated cost = 80 x 6

                  = N 480.

Actual calculation

   82 x 5. 80  

   = N 475. 60

comparing the rough calculation with the actual calculation, you will discover that the two answers

N 480  and N 475.60 are very close.

Reference material:

1. Essential mathematics for Jss I  (UBE Edition ) by AJS Oluwasanmipg 85 – 91
2. New General Mathematics for JSS I (UBE edition) by MF Macrae et al pg 178-179.

Reading Assignment

Read about quantitative reasoning and application of approximation to our everyday activities .

Weekend Assignment

1.  Give 3.9998 to 2 s.f. (a) 3.9 (b) 3.0 (c ) 4.0 (d) 4. 99

2.  Give 0.00057891  to2 s.f(a) 0.00  (b) 0.00058      (c) 0.58 (d) 0.0

3. Give 37.0567 to 2 d.p (a) 37 (b) 37.06 (c ) 37.05   (d) 37.1

4. Round 26 to the nearest ten (a) 5    (b) 20  (c) 30   (d) 40.

5. Round 7.586 to the nearest whole number (a) 8     (b) 7    (c) 6     (d) 7.6.

Theory

1.   The number of road accident in Lagos –Ibadan Expressway of Nigeria in a decade was 1294594.

Give this number to the nearest          (a) 100    (b) 1000

2.  Express each number correct to 1 d.p and 1 s.f  (a) 23.0036         (b) 6.7887.

Evaluation:

1. A farmer has N 200,000 to spend on cattle.  He wants to buy 9 calves.  Each calf costs N18500. Check, by approximation, that the farmer has enough money. Find, accurately how much change he will get after buying the calves.

2.  A bucket holds 10.5 litres. A cup holds about 320ml. Estimated the number of cups of water

that the bucket holds.

Reading Assignment

Read about Base Numbers.

i.   Essential mathematics

ii. New General Mathematics  pg 183 – 186.

Weekend Assignment

1.  Find the approximate answer to   0. 41  x 0. 92    (a) 0.6  (b)  0.36 (c ) 0.3 (d) 0.04.

2. Find the rough value of   4 ½   x 1 7/8  (a)  8         (b) 7  (c ) 10    (d) 9

3.   x = 0. 876 – 0.326. By doing a rough calculation, decide which of the following is the value of x

    (a) 0.18                    (b) 0.21           (c )0.3              (d) 0.55.

4.  A cup has a capacity of 290ml. It takes 63 cups to fill a bucket. Find the approximate capacity of the bucket in litres. (a) 9 litres (b) 10 litres          (c ) 1800 litres    (d)   18 litres,

5.  A sum of N 236000 is divided equally among 54 members of a club.Approximately  how much does each member get? (a) N 4000         (b)N2000         (c) N 20000     (d) N40000.