Chapter 1 standard form

1. Created By: Mohd Said B Tegoh

2. Required Basic Mathematical Skills
Rounding off whole
numbers to a specified
place value
Round off 1 688 to the
nearest hundred
1off 430 618 to the
700
Round
nearest thousand
431 000

3. Round off 30 106 correct to
the
nearest hundred.
0
3 0106
0<5

4. Round off 14.78
to the nearest whole number
+1
15. 7 8
4
Add I the decimals
Drop all to digit 4
Understand !!!
The first digit on the
right is greater than 5

5. Required Basic Mathematical Skills
Rounding off whole
numbers to a
specified number of
decimal places
Express 1.8523 to
three decimal places
1.852 to
Express 0.4968
two decimal places
0.50

6. Round off 5.316
to 1 decimal place
5 . 3 1 6
Do not
The first digit on the
change
Underline digit 3
right is less than 5
digit 3
st
(1 decimal place)

7. Round off 4.387
to 2 decimal places
+1
9
4.38 7
Add 1 todigit on
The first digit 8
Underline
8
the right is more
(2 nd decimal 5
than place)

8. CALC
√
sin
cos
ab/c
x2
tan
(
M+
ENG
log
ln
RCL
x -1
CONST
hyp
fdx
DEL
AC
7
0
8
.
+
9
=
EXP
(-)
^
Ans

9. Before Getting started……
 MODES
Before starting a calculation, you must
enter the correct mode as indicated in
the table below
MODE
1
MODE
2

10. MODE
3
MODE
1
MODE
2
MODE
2

11. Arithmetic Calculations
Use the MODE key to enter the COMP
when you want to perform basic
calculations.
MODE
COMP
1
1

12. FIX, SCI, RND
(Fix)
: Number of Decimal Places
1
(Sci)
2
: Number of Significant
Digits
(Norm) : Exponential of significant
Digits
3

13. Round off 5.316
to 1 decimal place
5x
MODE
Fix
1
1
Fix 0 ٨ 9 ? 1
5 . 3
= 5.3
1
6

14. Round off 5.316
to 2 decimal place
5x
MODE
Fix
1
1
Fix 0 ٨ 9 ?
5 . 3
= 5.32
5.32
1
6
2

15. Round off 4.387
to 2 decimal place
5x
MODE
Fix
1
1
Fix 0 ٨ 9 ?
4 . 3
= 4.39
4.39
8
7
2

16. Round off 4.387
to 1 decimal place
5x
MODE
Fix
1
1
Fix 0 ٨ 9 ?
4 . 3
= 4.4
4.4
8
7
1

17. Required Basic Mathematical Skills
Law of Indices
10m x 10n = 10m + n
10 ÷ 10 = 10
m
n
m -n
Simplify the following
103 x 10-5
10-2
102 ÷ 106
10-4

18. Very large and very small numbers are conveniently
rounded off to a specified number of significant figures
The concept of significant figures is another way
of stating the accuracy of a measurement
ignificant figures refer to the relevant digits in an integer
or a decimal number which has been rounded off to
a given degree of accuracy

19. ositive numbers greater than 1 can be
ounded off to a given number of significa
gures

20. The rules for determining the
number of significant figures in
a number are as follows:
All non-zero digits are significant
figures
2.73 has 3 significant figures
1346 has 4 significant figures

21. The rules for determining the
number of significant figures in
a number are as follows:
All zeros between non-zero are
significant figures
2.03 has 3 significant figures
3008 has 4 significant figures

22. The rules for determining the
number of significant figures in
a number are as follows:
In a decimal, all zeros after any
non-zero digit are significant
figures
3.60 has 3 significant figures
27.00 has 4 significant
figures

23. The rules for determining the
number of significant figures in
a number are as follows:
In a decimal, all zeros before
the first non-zero digit are
not significant
0.0032 has 2 significant figures
0.0156 has 3 significant
figures

24. The rules for determining the
number of significant figures in
a number are as follows:
All zeros after any non-zero
digit in a whole number are not
significant unless stated other
wise
1999 = 2000 ( one s.f )

25. The rules for determining the
number of significant figures in
a number are as follows:
All zeros after any non-zero
digit in a whole number are not
significant unless stated other
wise
1999 = 2000 ( two s.f )

26. The rules for determining the
number of significant figures in
a number are as follows:
All zeros after any non-zero
digit in a whole number are not
significant unless stated other
wise
1999 = 2000 ( three s.f )

27. State the number of significant figures in
each of the following
(a)
4 576
(b)
603
(c)
25 009
(d)
2.10
(a)
0.0706
(f)
0.80
4
3
5
3
3
2

28. Example 1
Express 3.15 x 105 as a single
number
3 . 1 5
= 315000
EXP
5

29. 3
.
1
5
EXP
5
5x
= 3.15 x 105
Norm 1^2 ?
MODE Norm
2 315000
3
3

30. Example 2
Express 4.23 x 10-4 as a single
number
4 . 2 3
= 0.000423
EXP
(-)
4

31. 4
.
2
3
EXP
(-)
4
5x
= 4.23 x 10-4
Norm 1^2 ?
MODE Norm
2 0.000423
3
3

32. hod of rounding off to a specified number of significant figur
Identify the digit (x) that is to be rounded off
Is the digit after x greater than or equal to 5
YES
Add 1 to x
NO
x remains unchanged
Do the digit after x lie before the decimal point?
YES (BEFORE)
Replace each digit with zero
NO (AFTER)
Drop the digits
Write the number according to the specified number of significant figures

33. Round off 30 106 correct to
three
significant figures.
0
3 0106
0<5

34. Round off 30 106 correct to
three
significant figures.
5x
MODE
3
=
Sci
2
0
2
1
Sci 0 ٨ 9 ?
0
3.01 x 10 44
3.01 x 10
30 100
30 100
6
3

35. Round off 0.05098 correct to
three
significant figures.
1 0 +1
0. 0 5098
8>5
00
0 . 0 51 9 8

36. Round off 0.05098 correct to three
significant figures.
5x
MODE
0
=
Sci
2
.
2
0
Sci 0 ٨ 9 ?
5
5.10 x 10-2
5.10 x 10-2
0.0510
0.0510
0
3
9
8

37.  To clear the Sci specification……
5X
Press
MODE
Norm
3
Norm 1 ⱱ 2 ?
3
1
 To continue the Sci specification……
ON
Press

38. Round off 0.0724789 correct to
four significant figures.
+1
8
0. 0 724789
8>5
8
0. 0 724789

39. Round off 0.0724789 correct to
four significant figures.
5x
MODE
Sci
2
2
Sci 0 ٨ 9 ?
0
.
=
7.248 x 10 -2
7.248 x 10 -2
0
7
0.07248
0.07248
2
4
4
7
8
9

40. Complete the following table (Round off to)
Number
3 sig. fig.
2 sig. fig.
1 sig. fig.
47 103
47100
47000
50 000
20 464
20 500
20 000
20 000
1 978
1 980
3.465
3.47
2 000
3.5
2
0003
70.067
70.1
70
70
4.004
4.00
4.0
4
0.04567
0.0457
0.046
0.05
0.06045
0.0605
0.060
0.06
0.0007805
0.000781
0.00078
0.0008

41. We usually use standard form for writing very large a
very small numbers
A standard form is a number that is written as the
product of a number A (between 1 and 10) and
a power of 10
A x 10n, where 1 ≤ A < 10, and n is an integer

42. Positive numbers greater than or equal
to 10 can be written in the standard form
A x 10n , where 1 ≤ A ≤ 10 and n is the
positive integer, i.e. n = 1, 2, 3,………
Example 58 000 000 = 5.8 x 107

43. Positive numbers less than or equal
to 1 can be written in the standard form
A x 10n , where 1 ≤ A ≤ 10 and n is the
negative integer, i.e. n = …..,-3, -2, -1
Example 0.000073 = 7.3 x 10-5

44. Express 431 000 in standard form
Express the number as a product of
A (1 ≤ A < 10) and a power of 10
A
Power of 10
431 000 = 4.31 x 100 000
4.31 x 105
=
431 000 = 4 3 1 0 0 0
4.31 x 105
=
5 is the number of places, the decimal point
is moved to the left

45. Express 431 000 in standard form
5x
MODE
4
=
Sci
2
3
2
1
Sci 0 ٨ 9 ?
3
0
0
4.31 x 1055
4.31 x 10
0

46. Express 0.000709 in standard form
Express the number as a product of
A (1 ≤ A < 10) and a power of 10
Power of 10
A
1
0.000709 = 7.09 x
10000
1
= 7.09 x
4
10
7.09 x 10-4
=

47. Express 0.000709 in standard form
0.000709 = 0 . 0 0 0 7 0 9
= 7.09 x 10-4
-4 is the number of places, the decimal point
is moved to the right

48. Express 0.000709 in standard form
5x
MODE Sci
2
2
Sci 0 ٨ 9 ?
0
.
=
7.09 x 10-4
7.09 x 10-4
0
0
0
3
7
0
9

49. Write the following numbers in standard
form
NUMBER
8765
32154
6900000
0.7321
0.00452
0.0000376
0.0000000183
STANDARD FORM
8.765 x 103
3.2154 x 104
6.9 x 106
7.321 x 10-1
4.52 x 10-3
3.76 x 10-5
1.83 x 10-8

50. Number in the standard form, A x 10n , can be
converted to single numbers by moving the decimal
point A
(a) n places to the right if n is positive
(b) n places to the left if n is negative

51. Express 1.205 x 104 as a single number
3.405 x 10
4
=3 . 4 0 5 0
Move the decimal point 4 places to the right
=34050

52. Express 3.405 x 104 as a single number
MODE
COMP
1
1
3 . 4 0
= 34 050
5
EXP
4

53. Express 7.53x 10-4 as a single number
7.53 x 10
-4
= 0 0 00 7.5 3
Move the decimal point 4 places to the left
= 0.000753

54. Express 7.53 x 10-4 as a single number
7 . 5 3
= 0.000753
EXP
(-)
4

55. Express the following in single numbers
STANDARD FORM
4.863 x 103
7.2051 x 104
4.31 x 106
5.164 x 10-1
1.93 x 10-3
2.04 x 10-5
9.16 x 10-8
NUMBER
4863
72051
4310000
0.5164
0.00193
0.0000204
0.0000000916

56. 3.25 X 105 = 325000
-5
7.14 X 10 = 0.0000714
4537000 = 4.537 X 106
0.0000006398 = 6.398 X 10-7

57. 325 X 105
32.5 X 106
=
3.25 X 107
=
0.325 X 108
=

58. 431 X 10-8
43.1 X 10-7
=
4.31 X 10-6
=
0.431 X 10-5
=

59. Two numbers in standard form can be added
or subtracted if both numbers have the same
index

60. s
MA R T
a x 10 + b x 10
m
= (a + b) x 10
m
m
a x 10 - b x 10
m
= (a - b) x 10
m
m

61. 5.3 x 105 + 3.8 x 105
(5.3 + 3.8 ) x 105
=
9.1 x 105
=
7.8 x 10-2 - 3.5 x 10-2
(7.8 - 3.5 ) x 10-2
=
4.3 x 10-2
=

62. Two numbers in standard form with
difference indices can only be added or
subtracted if the differing indices are made
equal

63. 4.6 x 106 + 5 x 105
4.6 x 106 + 0.5 x 106
=
(4.6 + 0.5 ) x 106
=
5.1 x 106
=

64. 6.4 x 10-4 - 8 x 10-5
6.4 x 10-4 - 0.8 x 10-4
=
(6.4 - 0.8) x 10-4
=
5.6 x 10-4
=

65. Calculate 3.2 x 10 4 – 6.7 x 10 3 . Stating
your answer in standard form.
3.2 x10 − 0.67 x10
4
= (3.2 − 0.67) x10
= 2.53x10
4
4
4

66. Calculate 3.2 x 104 – 6.7 x 103. Stating your
answer in standard form.
5x
MODE Sci
2
3
6
.
.
2
Sci 0 ٨ 9 ?
2
EXP
7
EXP
2. 53 x 1044
2. 53 x 10
4
3
3
=

67. 0.0000398_
3.98x10
−5
2.9 x10
− 0.29 x10
= (3.98 − 0.29) x10
= 3.69 x10
−5
−5
−5
−6

68. 0.0000398_
2.9 x10
−6
5x
MODE
Sci
2
2
0
.
0
0
0
0
3
9
8
-
2
.
6
=
3.69 x 10-5
3.69 x 10-5
EXP (-)
Sci 0 ٨ 9 ?
3
9

69. When two numbers in standard form are
multiplied or divided, the ordinary numbers
are multiplied or divided with each other
While their indices are added or subtracted

70. s
MA R T
a x 10 x b x 10
m+n
= (a x b) x 10
m
n
a x 10 ÷ b x 10
m-n
= (a ÷ b) x 10
m
n

71. 9.5 x 103 x 2.2 x 102
(9.5 x 2.2) x (103 x 102)
=
20.9 x 103+2
=
20.9 x 105
=
2.09 x 106
=

72. 7.2 x10
−2
6 x10
7 .2
5 −( −2 )
=
x10
6
7
= 1.2 x10
5

73. Calculate 1.17 x 10-2 . Stating your answer in
3 x 106
standard form.
1.17
−2 −6
x10
3
−8
= 0.39 x10
= 3.9 x10
−9

74. Calculate 1.17 x 10-2 . Stating your answer in
3 x 106
standard form.
5x
MODE Sci
2
1
.
3
EXP
2
3
Sci 0 ٨
9?
1
7
6
=
EXP
(-)
2
3. 90 x 10 -9
÷

75. −3 2
Calculate (9.24 ×10 ) , expressing the answer
6 ×10 −2
in standard form.
(9.24) 2 x10 −3 x 2
6 x10 −2
85.4
=
x10 −6−( −2 )
6
= 14.2 x10 −4
= 1.42 x10 −3

76. −3 2
Calculate (9.24 ×10 ) , expressing the answer
6 ×10 −2
in standard form.
5x
MODE Sci
2
2
Sci 0 ٨ 9 ? 3
(
9
.
2
4
EXP
(-)
3
)
x2
÷
(
6
EXP
(-)
2
)
=
1. 42 x 10-3
1. 42 x 10-3

77. 1 km2 = (1000 x 1000) m2
= (103 x 103) m2
= 106 m2

78. The area of a piece of rectangular land is
6.4 km2. If the width of the land is 1600 m,
calculate the length, in m, of the land
Length of the land = Area
Width
6.4 x10 6
=
1.6x10 3
6.4
=
x103
1.6
3
= 4 x10 m

79. Round off 0.05098 correct to
three
significant figures.
+1
1 0
A
B
C
D
0.051
0.0500
0.0509
0.0510
0. 0 5098
8>5
00
0 . 0 51 9 8

80. Round off 0.05098 correct to three
significant figures.
5x
MODE
0
=
Sci
2
.
2
0
Sci 0 ٨ 9 ?
5
5.10 x 10 -2
5.10 x 10 -2
0.0510
0.0510
0
3
9
8

81. Round off 0.08305 correct to three significant
figures.
A
B
C
D
0.083
0.084
0.0830
0.0831
1 +1
0. 08305
5=5
0
0 . 0 8 31 5

82. Round off 0.08305 correct to three
significant figures.
5x
MODE
0
=
Sci
2
.
2
0
Sci 0 ٨ 9 ?
8
8.31 x 10-2
8.31 x 10-2
0.0831
0.0831
3
3
0
5

83. Round off 30 106 correct to
three
significant figures.
A
B
C
D
30 000
30 100
30 110
30 200
0
3 0106
0<5

84. Round off 30 106 correct to
three
significant figures.
A
B
C
D
30 000
30 100
30 110
30 200
5x
MODE
3
=
Sci
2
0
2
1
Sci 0 ٨ 9 ?
0
3.01 x 1044
3.01 x 10
30 100
30 100
6
3

85. Express 1.205 x 104 as a single number
A
B
C
D
1 205
12 050
1 205 000
12 050 000
MODE
COMP
1
1 2 0 50
1
1 . 2 0
= 12 050
5
EXP
4

86. Express 4.23 x 10-4 as a single number
A
B
C
D
0. 423
0. 0423
0. 00423
0. 000423
4 . 2 3
= 0.000423
EXP
(-)
4

87. Express 52 700 in standard form.
A
B
C
D
5.27 × 102
5.27 × 104
5.27 × 10−2
5.27 x 10-4
52 700
5x
MODE
5
=
Sci
2
2
2
7
Sci 0 ٨ 9 ?
0
5.27 x 1044
5.27 x 10
0
3

88. 3.2 x10 + 6900 =
4
A
B
C
D
3.89 x10 8
3.89 x10
4
1.01x10
8
1.01x10
4
3.2 x10 + 6900
4
= 3.2 x10 + 0.69 x10
4
= (3.2 + 0.69) x10
= 3.89 x10
4
4
4

89. 3.2 x10 + 6900 =
4
5x
MODE
Sci
2
2
3
.
2 EXP 4
6
9
0
Sci 0 ٨ 9 ?
0
=
3
+
3.89 x 1044
3.89 x 10

90. 8.15x10
A
B
C
D
−6
−1.8x10
−7
=
6.35x10 −6
6.35x10
−7
7.97 x10
−6
7.97 x10
8.15x10
−7
−6
− 0.18x10
= (8.15 − 0.18) x10
= 7.97 x10
−6
−6
−6

91. 8.15x10
−6
−1.8x10
−7
=
5x
MODE Sci
2
8
.
2
1
.
Sci 0 ٨ 9 ?
5
EXP
8
EXP
-
1
=
7. 97 x 10-6
7. 97 x 10-6
3
(-)
(-)
6
7

92. 2.96 x10
=
−4 2
(4 x10 )
−3
A
B
C
D
7.24 x10 5
7.24 x10
4
1.85x10 5
1.85x10
4
2.96 x10
−3
16x10
2.96
-3(-8)
=
x10
16
5
= 0.185x10
−8
= 1.85x10
4

93. 2.96 x10
=
−4 2
(4 x10 )
−3
5x
MODE Sci
2
2
Sci 0 ٨ 9 ?
3
2
.
9
6
EXP
(-)
3
÷
(
4
EXP
(-)
4
)
x2
=
1. 85 x 1044
1. 85 x 10