Download - Chapter 1 standard form
Created By: Mohd Said B Tegoh
Required Basic Mathematical Skills
Rounding off whole numbers to a specified place value
Round off 1 688 to the nearest hundred
Round off 430 618 to the nearest thousand
1 700
431 000
3 10 0 60 < 5
0 0
Round off 30 106 correct to thenearest hundred.
Round off 14.78 to the nearest whole number
4. 7 8
The first digit on the right is greater than 5
+1
Add I to digit 4
15Drop all the decimalsUnderstand !!
!
Required Basic Mathematical Skills
Rounding off whole numbers to a specified number of decimal places
Express 1.8523 to three decimal places
Express 0.4968 to two decimal places
1.852
0.50
Round off 5.316 to 1 decimal place
5 . 3 1 6
Underline digit 3
(1st decimal place)
The first digit on the right is less than 5
Do not change digit 3
Round off 4.387 to 2 decimal places
4 . 3 8 7
Underline digit 8
(2nd decimal place)
The first digit on the right is more
than 5
Add 1 to 8
9
+1
(
+
^
EXP (-)
log ln
hyp fdx
.Ans
tan
CALC √
0
ENGM+
sin cos
ab/c x2
RCL x-1 CONST
DEL AC
7 8 9 =
Before Getting started…… MODES
Before starting a calculation, you mustenter the correct mode as indicated in the table below
MODE 1
MODE 2
MODE 3
MODE 1
MODE 2
MODE 2
Arithmetic CalculationsArithmetic Calculations
Use the key to enter the COMPwhen you want to perform basiccalculations.
MODE
MODECOMP 1 1
FIX, SCI, RNDFIX, SCI, RND
(Fix) : Number of Decimal Places(Fix) : Number of Decimal Places
(Sci) : Number of Significant (Sci) : Number of Significant DigitsDigits
(Norm) : Exponential of significant(Norm) : Exponential of significant DigitsDigits
1
2
3
1
MODE5x
Fix 1
Fix 0 ٨ 9 ?
1
Round off 5.316 to 1 decimal place
5 . 3 6=
1
5.3
1MODE5x
Fix 1
Fix 0 ٨ 9 ?
2
Round off 5.316 to 2 decimal place
5 . 3 6=
1
5.325.32
1MODE5x
Fix 1
Fix 0 ٨ 9 ?
2
Round off 4.387 to 2 decimal place
4 . 3 7=
8
4.394.39
1MODE5x
Fix 1
Fix 0 ٨ 9 ?
1
Round off 4.387 to 1 decimal place
4 . 3 7=
8
4.44.4
Required Basic Mathematical Skills
Law of Indices
10m x 10n = 10m + n
10m ÷ 10n = 10m - n
Simplify the following103 x 10-5
102 ÷ 106
10-2
10-4
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
Significant figures refer to the relevant digits in an integer or a decimal number which has been rounded off to a given degree of accuracy
Positive numbers greater than 1 can be rounded off to a given number of significant figures
The rules for determining the number of significant figures in a number are as follows:
All non-zero digits are significantfigures
2.73 has 3 significant figures
1346 has 4 significant figures
The rules for determining the number of significant figures in a number are as follows:
All zeros between non-zero aresignificant figures
2.03 has 3 significant figures
3008 has 4 significant figures
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
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 arenot significant
0.0032 has 2 significant figures
0.0156 has 3 significant figures
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 )
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 )
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 )
State the number of significant figures ineach of the following
(a) 4 576
(b) 603
(c) 25 009
(d) 2.10
(a) 0.0706
(f) 0.80
4353
32
Example 1Example 1Express 3.15 x 105 as a singlenumber
3 . 1 EXP 5
=5
315000
3 . 1 EXP 5
=
5
3.15 x 105MODE5x
Norm 3 3
Norm 1^2 ? 2 315000
Example 2Example 2Express 4.23 x 10-4 as a singlenumber
4 . 2 EXP (-) 4
=3
0.000423
4 . 2 EXP (-) 4
=
3
4.23 x 10-4MODE5x
Norm 3 3
Norm 1^2 ? 2 0.000423
Method of rounding off to a specified number of significant figures
Identify the digit (x) that is to be rounded off
Is the digit after x greater than or equal to 5
Add 1 to x x remains unchanged
Do the digit after x lie before the decimal point?
Replace each digit with zero Drop the digits
Write the number according to the specified number of significant figures
YES NO
YES (BEFORE) NO (AFTER)
3 10 0 60 < 5
0 0
Round off 30 106 correct to three
significant figures.
MODESci2 2 Sci 0 ٨ 9 ?
1
3
3.01 x 1043.01 x 104
3
=
5x
0 0 6
30 10030 100
Round off 30 106 correct to three
significant figures.
Round off 0.05098 correct to three
significant figures.
0 5. 0 0 988 > 5
+11 0
0 5. 0 0 981 0
Round off 0.05098 correct to three significant figures.
MODE
Sci2 2 Sci 0 ٨ 9 ?
0
3
5.10 x 10-25.10 x 10-2
0
=
5x
. 5 0 9 8
0.05100.0510
To clear the Sci specification……
MODENorm 3 3Press
5X
Norm 1 ⱱ 2 ? 1
To continue the Sci specification……
PressON
Round off 0.0724789 correct to four significant figures.
0 7. 0 2 478 98 > 5
+18
0 7. 0 2 478 98
Round off 0.0724789 correct to four significant figures.
MODESci2 2 Sci 0 ٨ 9 ?
0
4
7.248 x 10-27.248 x 10-2
0
=
5x
. 97 2 4 7 8
0.072480.07248
Complete the following table (Round off to)
Number 3 sig. fig. 2 sig. fig. 1 sig. fig.
47 103
20 464
1 978
3.465
70.067
4.004
0.04567
0.06045
0.0007805
47100 47000 50 000
20 500 20 000 20 000
1 980 2 000 2 0003.47 3.5 3
70.1 70 70
4.00 4.0 4
0.0457 0.046 0.05
0.0605 0.060 0.06
0.000781 0.00078 0.0008
We usually use standard form for writing very large and 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
Positive numbers greater than or equal to 10 can be written in the standard formA 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
Positive numbers less than or equal to 1 can be written in the standard formA 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
Express 431 000 in standard form
431 000
Express the number as a product ofA (1 ≤ A < 10) and a power of 10
= 4.31
A
x 100 000
Power of 10
= 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
Express 431 000 in standard form
MODESci2 2 Sci 0 ٨ 9 ?
1
3
4.31 x 1054.31 x 105
4
=
5x
3 0 0 0
0.000709
Express the number as a product ofA (1 ≤ A < 10) and a power of 10
= 7.09
A
x
Power of 10
= 7.09 x 10-4
Express 0.000709 in standard form
10000
1
= 7.09 x410
1
0.000709 =
Express 0.000709 in standard form
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
Express 0.000709 in standard form
MODE Sci
2 2 Sci 0 ٨ 9 ?
. 0 0 0
3
7.09 x 10-47.09 x 10-4
0
=
7 0 9
5x
Write the following numbers in standard form
NUMBER STANDARD FORM
8765
32154
6900000
0.7321
0.00452
0.0000376
0.0000000183
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
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
Express 1.205 x 104 as a single number
3.405 x 104
= 3 . 4 0 5 0
3 4 0 5 0=Move the decimal point 4 places to the right
3 . 4 EXP 4
=0
34 050
5
MODECOMP 1 1
Express 3.405 x 104 as a single number
Express 7.53x 10-4 as a single number
7.53 x 10-4
= 7 . 5 3 0 0 0 0
= 0.000753 Move the decimal point 4 places to the left
7 . 5 EXP (-) 4
=3
0.000753
Express 7.53 x 10-4 as a single number
Express the following in single numbers
STANDARD FORM NUMBER
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
4863
72051
43100000.5164
0.00193
0.00002040.0000000916
3.25 X 105 = 325000
7.14 X 10-5 = 0.0000714
4537000 = 4.537 X 106
0.0000006398 = 6.398 X 10-7
325 X 105
32.5 X 106
3.25 X 107
0.325 X 108
===
431 X 10-8
43.1 X 10-7
4.31 X 10-6
0.431 X 10-5
===
Two numbers in standard form can be added or subtracted if both numbers have the same index
s M A R T
a x 10m + b x 10m =(a + b) x 10m
a x 10m - b x 10m =(a - b) x 10m
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
Two numbers in standard form with difference indices can only be added or subtracted if the differing indices are made equal
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
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
Calculate 3.2 x 104 – 6.7 x 103. Stating your answer in standard form.
4
4
44
1053.2
10)67.02.3(
1067.0102.3
x
x
x
x
MODE Sci2 2 Sci 0 ٨ 9 ?
. 2 EXP
3
2. 53 x 1042. 53 x 104
3
6
4 -
5x
EXP7. 3 =
Calculate 3.2 x 104 – 6.7 x 103. Stating your answer in standard form.
0000398.0 6109.2 x_
5
5
55
1069.3
10)29.098.3(
1029.01098.3
x
x
xx
MODESci2 2 Sci 0 ٨ 9 ?
0
3
0
5x
. 0 0 0
3 9 8 - 2 . 9
EXP (-) 6 = 3.69 x 10-53.69 x 10-5
0000398.0 6109.2 x_
When two numbers in standard form are multiplied or divided, the ordinary numbers are multiplied or divided with each otherWhile their indices are added or subtracted
s M A R T
a x 10m x b x 10n = (a x b) x 10m + n
a x 10m ÷ b x 10n
= (a ÷ b) x 10m - n
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
7
)2(5
2
5
102.1
106
2.7106
102.7
x
x
xx
Calculate 1.17 x 10-2 . Stating your answer in 3 x 106
standard form.
9
8
62
109.3
1039.0
103
17.1
x
x
x
Calculate 1.17 x 10-2 . Stating your answer in 3 x 106
standard form.
MODE Sci2 Sci 0 ٨
9 ?. 1 7 EXP
3
3. 90 x 10-9
1
3
(-) 2 ÷
5x
EXP 6 =
2
2
23
106
)1024.9(
Calculate , expressing the answer
in standard form.
3
4
)2(6
2
232
1042.1
102.14
10
106
10)24.9(
x
x
x6
85.4x
x x
2
23
106
)1024.9(
MODE Sci2 2 Sci 0 ٨ 9 ?
. 2 EXP
3
1. 42 x 10-31. 42 x 10-3
9
)
(-) 3
5x
EXP6x2
=
(
4
÷ ( (-)
2 )
Calculate , expressing the answer
in standard form.
1 km2 = (1000 x 1000) m2
= (103 x 103) m2
= 106 m2
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
m3
3
6
104
106.1
4.6
104.6
x
x
1.6x10x
3
Round off 0.05098 correct to three
significant figures.
A 0.051B 0.0500C 0.0509D 0.0510
0 5. 0 0 988 > 5
+11 0
0 5. 0 0 981 0
Round off 0.05098 correct to three significant figures.
MODE
Sci2 2 Sci 0 ٨ 9 ?
0
3
5.10 x 10-25.10 x 10-2
0
=
5x
. 5 0 9 8
0.05100.0510
Round off 0.08305 correct to three significant figures.
A 0.083B 0.084C 0.0830D 0.0831
0 8. 0 3 055 = 5
+11
0 8. 0 3 051
Round off 0.08305 correct to three significant figures.
MODE
Sci2 2 Sci 0 ٨ 9 ?
0
3
8.31 x 10-28.31 x 10-2
0
=
5x
. 8 3 0 5
0.08310.0831
3 10 0 60 < 5
0 0
Round off 30 106 correct to three
significant figures.
A 30 000B 30 100C 30 110D 30 200
Round off 30 106 correct to three
significant figures.
A 30 000B 30 100C 30 110D 30 200
MODESci2 2 Sci 0 ٨ 9 ?
1
3
3.01 x 1043.01 x 104
3
=
5x
0 0 6
30 10030 100
Express 1.205 x 104 as a single number
A 1 205B 12 050C 1 205 000D 12 050 000
1 . 2 EXP 4
=0
12 050
5
1 02 50
MODECOMP 1 1
Express 4.23 x 10-4 as a single number
A 0. 423B 0. 0423C 0. 00423D 0. 000423
4 . 2 EXP (-) 4
=3
0.000423
Express 52 700 in standard form.
A 5.27 102
B 5.27 104
C 5.27 102
D 5.27 x 10-4
MODE
Sci2 2 Sci 0 ٨ 9 ?
7
3
5.27 x 1045.27 x 104
5
=
5x
2 0 0
5 72 0 0
6900102.3 4x
ABCD
41089.3 x81089.3 x41001.1 x81001.1 x
4
4
44
4
1089.3
10)69.02.3(
1069.0102.3
6900102.3
x
x
xx
x
MODESci2 2 Sci 0 ٨ 9 ?
2
3
3
5x
. EXP 4 +
6 9 0 0 = 3.89 x 1043.89 x 104
6900102.3 4x
76 108.11015.8 xx
ABCD
71035.6 x61035.6 x71097.7 x61097.7 x
6
6
66
1097.7
10)18.015.8(
1018.01015.8
x
x
xx
MODE Sci2 2 Sci 0 ٨ 9 ?
. 1 EXP
3
7. 97 x 10-67. 97 x 10-6
8
1
(-) 6
5x
EXP8. (-)
=
76 108.11015.8 xx
5
- 7
24
3
)104(
1096.2
xx
ABCD
41024.7 x51024.7 x41085.1 x51085.1 x
4
5
8
3
1085.1
10185.0
16
96.2101
1096.2
x
x
x10
6x
x
(-8)-3-
MODE Sci2 2 Sci 0 ٨ 9 ?
. 9 EXP
3
1. 85 x 1041. 85 x 104
2 (-) 3
5x
EXP4
=
6
÷ ( (-)
x2
)
24
3
)104(
1096.2
xx
4