Lesson Notes By Weeks and Term - Senior Secondary School 1

REVISION OF STANDARD FORM AND APPROXIMATION

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  4. REVISION OF STANDARD FORM AND APPROXIMATION

SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 1st TERM

REFERENCE BOOKS

  • New General Mathematics SSS 1     M.F. Macrae et al 

 

 

 
WEEK FIVE

TOPIC: REVISION OF STANDARD FORM AND APPROXIMATION

CONTENT

  • Revision of Standard Form
  • Rounding off of Numbers, Decimal Places and Significant Figures

REVISION OF STANDARD FORM

A number written in the form of A X 10n, such that A is a number between 1 and 10 (1 ≤     A ≤10) and n is a whole number (integer) is said to be in standard form.

Examples:  2 X 106, 7 X 10-3, 2.5 X 104, 8.6 X10-9e.t.c

 

Work Example: 

  1. Express the following numbers in standard form:
  1. 300000 (b) 55 (c) 2,300,000 (d)720,000,000 (e)9,400,000,000

Solution

  1. 300,000 = 3.0 X 100,000

        =3.0 X 105

  1. 55            =5.5 X 10

        =5.5 X 101

  1. 2,300,000 =2.3 X 1,000,000

            =2.3 X 106

  1. 720,000,000 = 7.2 X 100,000,000

    =7.2 X 108

  1. 9,400,000,000 = 9.4 X 1,000,000,000

     =9.4 X109

  1. Change the following from standard form to ordinary form:
  1. 5.1 X 107 = 5.1 X 10,000,000

          =51,000,000

  1. 2.5 X 106 = 2.5 X 1,000,000

          =2,500,000

  1. 3.4 X 101 = 3.4 X 10 =34
  2. 9.8 X 105 =9.8 X 100,000    =    980,000
  3. 6 X 108 = 6 X 100,000,0000    =     600,000,000

Since decimal fraction can be expressed as power of 10,they can also be expressed in standard form as shown in the example below:

 

  1. Express the following fractions in standard form
  1. 0.0015  (b) 0.000026 (c) 0.000000067 (d) 0.3

Solution

  1. 0.0015  =    15             

                 10,000

              = 1.5 X10

                   10,000

              = 1.5 

                103

             = 1.5 X 10-3

(Since from the 4th law of indices 1/xa = x-a)

  1. 0.000026 =     26

                     1,000,000

                    =2.6 X 10

                      1,000,000

                   = 2.6

                     100,000

                 =   2.6

                      105

                   = 2.6 X 10-5

  1. 0.000000067 =          67

                            1,000,000,000

                        =     6.7 X 10

                           1,000,000,000

                      =   6.7 X           1

                                     1,000,000,000

                         = 6.7   X      1

                                            108

                         =6.7 X 10-8

  1. 0.3  =   3

            10

        = 3 X 1

                 101

      =3 X 10-1

 

  1.     Express the following as decimal fractions

(a )9.4 X10-5 (b)8.8 x 10-3 (c) 1.8 x 10-1 (d) 2x10-7

Solutions

  1. 9.4  X 10-5 

9.4 X 1    

  105 

( by using the 4th law of indices as explained in the example 3 above)

    =     9.4

        100,000

    = 0.000094

  1. 8.8  x 10-3  = 8.8 x 1

    103

                   = 8.8

                      1000

                  =   0.0088

  1. 1.8 X 10-1  = 1.8 X 1

                       101

                  = 1.8 

                     10

                 =0.18

  1. 2 X10-7  = 2 X 1

                   107

               =      2 X 1 

                  10,000,000

              = 0.0000002

Note that for decimal fraction, n is a negative integer



EVALUATION

  1. Change (a) 9.18 X 105 (b)6.75 x 10-8 to ordinary number
  2. Express the following  in standard form (a) 0.0000058 (b) 458000

 

ROUNDING OFF NUMBERS

When rounding off number digits 1,2,3,4 are rounded down and digits 5, 6, 7, 8, 9 are rounded up.

Examples:

Round off the following to the nearest 

  1. Thousand 
  2. Hundred 
  3. Ten
  1. 4517
  2. 30,637

Solutions

  1. 4517≈ 500 to the nearest thousand
  2. 4517≈4500 to the nearest hundred 
  3. 4517≈4520 to the nearest tens
  1. (a) 30,637≈31,000 to the nearest thousand
  1. 30,637≈30,600 to the nearest hundred 
  2. 30,637≈30,640 to the nearest ten

 

SIGNIFICANT FIGURES

The significant figures begin from the first non-zero digit at the left hand side of a number. As before, digits 1,2,3,4, are rounded down and digits 5,6,7,8,9 are rounded up. Digits should be written with their correct place value.

Note that zero in between non-zero digits in a number are significant. E.g the zero in 8.0296 is a significant while zero in 0.0000925 are not significant.

Examples:

Round off the following to;

  1. 1 significant figure.
  2. 2 significant figures.
  3. 3 significant figures.
  1. 26,002
  2. 2.00567
  3. 0.006307

Solution

  1. (a) 26,002≈30,000 to 1 significant figure.

(b) 26,002≈26,000 to 2 significant figures.

(c) 26,002≈26,000 to 3 significant figures.

  1.   (a)2.00567≈2 to 1 significant figure.

(b)2.00567≈2.0 to 2 significant figures.

(c)2.00567≈2.01 to 3  significant figures.

  1.   (a)0.006307≈0.006 to 1 significant figure.

(b)0.006307≈0.0063 to 2 significant figures.

(c)0.006307≈0.00631 to 3 significant figures.

 

DECIMAL PLACES

Decimal places are counted from the decimal point. Zero after the point is significant and also counted.  Digits are rounded up and down as before. Place value must be kept.

Examples:

Round off the following to:

  1. 1 d.p (b) 2 d.p (c) 3 d.p
  1. 0.0089
  2. 0.9002
  3. 1.9875

Solutions

  1. (a) 0.0089≈0.0 to 1 decimal place.

(b)0.0089≈0.01 to 2 decimal places.

(c)0.089≈0.009 to 3 decimal places.

  1. (a)0.9002≈0.9 to 1 decimal place.

        (b)0.9002≈0.90 to 2 decimal places.

(c)0.9002≈0.900 to 3 decimal places.

  1. (a)   1.9875≈2.0 to 1 decimal place.
  1. 1.9875≈1.99 to 2 decimal places.
  2. 1.9875≈1.988 to 3 decimal places.

 

EVALUATION

  1. Express the following in standard form
  1. 3,500,000    (b) 28       (c) 0.47    (d) 0.0000003

 

  1. In the following statements round each number to two significant figures then write it out in full:
  1. It will cost #3.28 billion to renovate state’s classrooms (b) The area of Ghana is 23.9 million hectares
  2. It was estimated that the population of Lagos was about 9.44 millions in 2000

 

READING ASSIIGNMENT

NGM SSS1, pages 6-7, review test 3 and 4.  

 

GENERAL EVALUATION

  1. Express the following in standard form   (a) 0.000423        (b) 628500
  2. (a)  Change  4.23 X 107 to ordinary form  (b) Change 3.4 X 10-6 to decimal fraction
  3. Approximate 72899 to the nearest   (a)Ten    (b) hundred    (c) thousand
  4. approximate 0.0065734 to    (a) 1 s.f    (b) 2 s.f    (c) 3 s.f
  5. Approximate 99.99054 to      (a)1 d.p    (b) 2 d.p   (c) 3 d.p

 

WEEKEND ASSIGNMENT

  1. Round off 0.004365 correct to 2 significant figures (a) 0.04        (b) 0.0044 (c) 0.00437 (d) 0.0043   (e) 0.44
  2. What is 0.002568 correct to 3 decimal places          (a) 0.00        (b) 0.002     (c) 0.003    (d) 0.00256 (e)0.00257
  3. Write 0.0000549 in standard form     (a) 5.49 X 10-5 (b) 5.49 X 10-4 (c) 5.49 X 10-3 (d)5.49 X 104 (e) 5.49 X 105
  4. Write 5.48 X 10-4 as decimal fraction (a) 0.0548 (b) 0.00548 (c) 0.000548 (d) 0.0000548 (e) 0.00000548
  5. If 0.00725 Is written as 7.25 X 10n, the value of n is _______ (a) -4 (b) -3 (c) -2 (d)-1 (e)3

 

THEORY   

  1. The Page of a book are numbered 1 to 300
  1. How many thicknesses of the paper make 300 pages
  2. If the thickness of the book is 15mm, calculate the thickness of one leaf. Give your answer in meters in standard form
  1.   A length of wire is given as 6.8cm correct to 2 significant figures. What is the least possible length of the wire

         (a)Give the number 29,542 to the nearest ten   (b) Write 0.07258 to 3 significant figures

















































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