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

1off 430 618 to the
700
Round
nearest thousand

431 000
Round off 30 106 correct to
the
nearest hundred.

0
3 0106
0<5
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
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
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)
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)
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
Before Getting started……
 MODES
Before starting a calculation, you must
enter the correct mode as indicated in
the table below

MODE

1

MODE

2
MODE

3

MODE

1

MODE

2

MODE

2
Arithmetic Calculations
Use the MODE key to enter the COMP
when you want to perform basic
calculations.
MODE

COMP
1

1
FIX, SCI, RND
(Fix)

: Number of Decimal Places

1

(Sci)
2

: Number of Significant
Digits

(Norm) : Exponential of significant
Digits
3
Round off 5.316
to 1 decimal place
5x
MODE
Fix
1

1

Fix 0 ٨ 9 ? 1

5 . 3
= 5.3

1

6
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
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
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
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
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
ositive numbers greater than 1 can be
ounded off to a given number of significa
gures
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
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
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 are
not 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 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
Example 1
Express 3.15 x 105 as a single
number

3 . 1 5
= 315000

EXP

5
3

.

1

5

EXP

5

5x

= 3.15 x 105
Norm 1^2 ?

MODE Norm

2 315000

3

3
Example 2
Express 4.23 x 10-4 as a single
number

4 . 2 3
= 0.000423

EXP

(-)

4
4

.

2

3

EXP

(-)

4

5x

= 4.23 x 10-4
Norm 1^2 ?

MODE Norm

2 0.000423

3

3
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
Round off 30 106 correct to
three
significant figures.

0
3 0106
0<5
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
Round off 0.05098 correct to
three
significant figures.

1 0 +1

0. 0 5098
8>5

00
0 . 0 51 9 8
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
 To clear the Sci specification……
5X

Press

MODE

Norm
3

Norm 1 ⱱ 2 ?

3
1

 To continue the Sci specification……
ON

Press
Round off 0.0724789 correct to
four significant figures.

+1

8

0. 0 724789
8>5

8
0. 0 724789
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
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
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
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
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
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
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
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
=
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
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
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
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 10

4

=3 . 4 0 5 0
Move the decimal point 4 places to the right

=34050
Express 3.405 x 104 as a single number

MODE

COMP
1

1

3 . 4 0
= 34 050

5

EXP

4
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
Express 7.53 x 10-4 as a single number

7 . 5 3
= 0.000753

EXP

(-)

4
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
3.25 X 105 = 325000
-5
7.14 X 10 = 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

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
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 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
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
=
0.0000398_
3.98x10

−5

2.9 x10

− 0.29 x10

= (3.98 − 0.29) x10
= 3.69 x10

−5

−5

−5

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

÷
−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
−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
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

6.4 x10 6
=
1.6x10 3
6.4
=
x103
1.6
3
= 4 x10 m
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

Chapter 1 standard form

  • 1.
    Created By: MohdSaid B Tegoh
  • 2.
    Required Basic MathematicalSkills 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 30106 correct to the nearest hundred. 0 3 0106 0<5
  • 4.
    Round off 14.78 tothe 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 MathematicalSkills 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 to1 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 to2 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.
  • 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.
  • 11.
    Arithmetic Calculations Use theMODE 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 to1 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 1 5 . 3 = 5.3 1 6
  • 14.
    Round off 5.316 to2 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 to2 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 to1 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 4 . 3 = 4.4 4.4 8 7 1
  • 17.
    Required Basic MathematicalSkills 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 andvery 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 greaterthan 1 can be ounded off to a given number of significa gures
  • 20.
    The rules fordetermining 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 fordetermining 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 fordetermining 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 fordetermining 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 fordetermining 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 fordetermining 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 fordetermining 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 numberof 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 3.15x 105 as a single number 3 . 1 5 = 315000 EXP 5
  • 29.
    3 . 1 5 EXP 5 5x = 3.15 x105 Norm 1^2 ? MODE Norm 2 315000 3 3
  • 30.
    Example 2 Express 4.23x 10-4 as a single number 4 . 2 3 = 0.000423 EXP (-) 4
  • 31.
    4 . 2 3 EXP (-) 4 5x = 4.23 x10-4 Norm 1^2 ? MODE Norm 2 0.000423 3 3
  • 32.
    hod of roundingoff 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 30106 correct to three significant figures. 0 3 0106 0<5
  • 34.
    Round off 30106 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 0.05098correct to three significant figures. 1 0 +1 0. 0 5098 8>5 00 0 . 0 51 9 8
  • 36.
    Round off 0.05098correct 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 clearthe Sci specification…… 5X Press MODE Norm 3 Norm 1 ⱱ 2 ? 3 1  To continue the Sci specification…… ON Press
  • 38.
    Round off 0.0724789correct to four significant figures. +1 8 0. 0 724789 8>5 8 0. 0 724789
  • 39.
    Round off 0.0724789correct 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 followingtable (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 usestandard 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 greaterthan 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 lessthan 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 000in 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 000in 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 instandard 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 instandard 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 instandard 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 followingnumbers 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 thestandard 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 x104 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 x104 as a single number MODE COMP 1 1 3 . 4 0 = 34 050 5 EXP 4
  • 53.
    Express 7.53x 10-4as 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 x10-4 as a single number 7 . 5 3 = 0.000753 EXP (-) 4
  • 55.
    Express the followingin 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 32.5X 106 = 3.25 X 107 = 0.325 X 108 =
  • 58.
    431 X 10-8 43.1X 10-7 = 4.31 X 10-6 = 0.431 X 10-5 =
  • 59.
    Two numbers instandard form can be added or subtracted if both numbers have the same index
  • 60.
    s MA R T ax 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 instandard 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 x10 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 x104 – 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 − 0.29x10 = (3.98 − 0.29) x10 = 3.69 x10 −5 −5 −5 −6
  • 68.
  • 69.
    When two numbersin 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 ax 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 103x 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 x10-2 . 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 x10-2 . 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 ofa 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 0.05098correct 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 0.05098correct 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 0.08305correct 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 0.08305correct 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 30106 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 30106 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 x104 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 x10-4 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 700in 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 7.97x10 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 0٨ 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 (4x10 ) −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 (4x10 ) −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