Transcript KFM03Pr00_Decimals and Percentages

OBJECTIVES
1.
Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE
decimal fractions.
2.
With an approved calculator; ADD, SUBTRACT, MULTIPLY, and
DIVIDE decimal fractions.
3.
Without a calculator, CALCULATE the average of a series of
numbers.
4.
With an approved calculator, CALCULATE the average of a series
of numbers.
5.
Without a calculator, EXPRESS the solution of addition,
subtraction, multiplication, and division operations using the
appropriate number of significant digits.
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 1 / Rev 1
DECIMAL PLACE RELATIONSHIPS
7 3 6 1 . 2 9
8
Place title
Place Value
Thousandths
1/1,000
Hundredths
1/100
Tenths
1/10
Units
1
Tens
10
Hundreds
100
Thousands
1,000
Fig 3-1
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 2 / Rev 1
EXAMPLE
The magnitude of 7361.298 is:
Digit Place Value
7  1,000
3 
100
6 
10
1 
1
2 
1/10
9 
1/100
8 
1/1,000
=
=
=
=
=
=
=
7,000
300
60
1
2/10 = 200/1,000
9/100 = 90/1,000
8/1,000 = 8/1,000
Ex 3-1
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 3 / Rev 1
EXAMPLE
Sum =
298
7,361 +
1,000
=
298
7,361
1,000
=
7,361.298
Ex 3-1
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 4 / Rev 1
EXAMPLE
3
Find the decimal equivalent of
4
3
34 
4
0.75
4 3.00
28
20
20
0
3
 0.75
4
Ex 3-2
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 5 / Rev 1
EXAMPLE
2 is:
The decimal equivalent of
3
2
= 2  3 = 0.6666…..
3
where the ….. indicates successive 6’s.
Ex 3-3
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 6 / Rev 1
EXAMPLE
Write the common fraction that is equivalent to the
decimal fraction 0.375.
375
375  125 3
0.375 


1,000 1,000  125 8
Recall from the previous chapter to reduce the
fraction to lowest terms, factor each term into its
smallest component.
375
3555

1,000 2  2  2  5  5  5
Ex 3-4
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 7 / Rev 1
EXAMPLE
Then cancel out each factor that occurs in each
term.
3x5x5x5
2x2x2x5x5x5
Then multiply the factors in each term together.
3
3

2x2x2 8
3
Thus simplified, 0.375 in a fractional form is .
8
Ex 3-4
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 8 / Rev 1
EXAMPLE
Find the sum of 39.62, 41.093, and 0.0327.
39.62
41.093
+ 0.0327
80.7457
Ex 3-5
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 9 / Rev 1
EXAMPLE
Subtraction is done in the same manner as with
whole numbers.
32.100
– 16.379
15.721
Ex 3-6
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 10 / Rev 1
EXAMPLE
Multiply 16.2 and 1.15. Multiply without concern for
the decimal.
162

115
810
162
162
18630
Ex 3-7
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 11 / Rev 1
EXAMPLE
Multiply 16.2 and 1.15.
2 162
16.2  16

10 10
15 115
1.15  1

100 100
162 115 18,630


 18.630
10 100 1,000
Ex 3-8
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 12 / Rev 1
EXAMPLE
Divide 41.05 by 2.5. Divide without concern for the
decimal.
1642
25 4105
25
160
150
105
100
50
50
0
Ex 3-9
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 13 / Rev 1
EXAMPLE
Divide 41.05 by 2.5.
5
4,105
41.05  41

100 100
5 25
2 .5  2

10 10
4,105
100  4,105  10  41,050
25
100 25 2,500
10
41,050
2  5  5  821

2,500 2  2  5  5  5  5
1,642

 16.42
100
© 2003 General Physics Corporation
Ex 3-10
ABC/ Mathematics / Chapter 3 / TP 3 - 14 / Rev 1
EXAMPLE
Example 3-12
Convert 0.25 to a percentage.
0.25  100% = 25%
Example 3-13
Convert 2 to a percentage.
2  100% = 200%
Example 3-14
Convert 1.25 to a percentage.
1.25  100% = 125%
Ex 3-12
Ex 3-13
Ex 3-14
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 15 / Rev 1
EXAMPLE
Find the average of the following recorded
temperatures: 600° F, 596° F, 597° F, 603° F.
Step 1. After making sure that the individual quantities
to be averaged have the same units, add the individual
numbers of quantities to be averaged.
600 + 596 + 597 + 603 = 2,396
Step 2. Count the number of numbers or quantities to
be averaged.
The number of items is 4.
Step 3. Divide the sum found in Step 1 by the number
counted in Step 2.
2,396 ÷ 4 = 599° F
Ex 3-36
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 16 / Rev 1
EXAMPLE
The level in a tank is recorded once a day. The
recorded level in the tank since the last addition over
the last several days has been 500 gals, 490 gals,
487 gals, 485 gals, and 480 gals. Calculate the
average tank level.
Step 1. All levels have the same units (gals) so the
individual numbers can be averaged.
ADD the individual numbers.
500 + 490 + 487 + 485 + 480 = 2,442
Step 2. Count the quantities to be averaged.
The number of items is 5.
Ex 3-37
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 17 / Rev 1
EXAMPLE
Step 3. Divide the sum found in Step 1 by the
quantities in Step 2.
2,442 ÷ 5 = 488.4
The average tank level for the recorded days is
488.4 gals.
Ex 3-37
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 18 / Rev 1
EXAMPLE
Given the following price list of replacement pumps,
find the average cost.
$10,200; $11,300; $9,900; $12,000; $18,000; $7,600
Step 1. After making sure that the individual quantities
to be averaged have the same units, add the individual
numbers or quantities to be averaged.
10,200 + 11,300 + 9,900 + 12,000 + 18,000 + 7,600 =
69,000
Step 2. Count the numbers or quantities to be
averaged.
Total number of prices is 6.
Ex 3-38
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 19 / Rev 1
EXAMPLE
Step 3:. Divide the sum found in Step 1 by the number
counted in Step 2.
69,000 ÷ 6 = 11,500
Thus, the average price of the replacement pump is
$11,500.
Ex 3-38
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 20 / Rev 1
EXAMPLE
The Purchasing department ordered replacement
valves for the upcoming outage
• 4 valves cost $145.25 each
• 2 valves cost $137.85 each
• 3 valves cost $150.00 each
$
145.25
$
145.25
$
145.25
Step 1. The valves have the same $
145.25
units ($). Add the quantities to be $
137.85
averaged.
$
137.85
$
150.00
$
150.00
$
150.00
$
1,306.70 Ex 3-39
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 21 / Rev 1
EXAMPLE
Step 2. Count the quantities to be averaged.
4+2+3=9
Step 3. Divide the sum found in Step 1 ($1,306.70)
by the number counted in Step 2 (9).
$1,306.70 ÷ 9 = $145.19
The average price for the valves ordered is $145.19.
Alternately, Step 1 could have been performed as
(4)($145.25) + (2)($137.85) + (3)($150.00)
= ($581.00) +($275.70) + ($450.00)
= $1,306.70
Ex 3-39
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 22 / Rev 1
EXAMPLE
Calculate the average of the following lengths.
2 ft, 30 in, 1.5 ft, 18 in
Step 1. The items to be averaged have different
units (ft and in). Determine which unit would be
preferred. (In this case it doesn’t matter.)
2 ft = 24 in
30 in = 2.5 ft
1.5 ft =18 in
18 in = 1.5 ft
Ex 3-40
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 23 / Rev 1
EXAMPLE
You could convert all to feet, and sum and average.
Step 1.
2 ft + 2.5 ft + 1.5 ft + 1.5 ft = 7.5 ft
Step 2
4 items
Step 3
7.5 ft ÷ 4 = 1.875 ft
Ex 3-40
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 24 / Rev 1
EXAMPLE
Alternately, you could convert to inches, sum and
average.
Step 1.
24 in + 30 in + 18 in + 18 in = 90 in.
Step 2
4 items
Step 3
90 in ÷ 4 = 22.5 in
Ex 3-40
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 25 / Rev 1
THREE TEMPERATURE SCALES
102.0
110
600
101.0
100
500
100.0
90
400
99.0
80
300
98.0
70
200
97.0
60
100
Outdoor
Thermometer
Oven
Thermometer
Fever
Thermometer
Fig 3-2
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 26 / Rev 1
EXAMPLE
Least
Significant
Digits
12,345
Most
Significant
Digits
(left)
1
5
Number
of
Significant
Digits
5
123.45
1
5
5
1,986
1
6
4
37.806
3
6
5
201
2
1
3
300.7
3
7
4
500.08
5
8
5
0.00875
8
5
3
Measured
Number
Ex 3-41
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 27 / Rev 1
EXAMPLE
Most
Significant
Digits
(left)
Least
Significant
Digits
Number
of
Significant
Digits
900.030
9
right-most 0
6
0.090
9
right-most 0
2
200
2
2
1
200.
2
right-most 0
3
200.0
2
right-most 0
4
200.00
2
right-most 0
5
Measured
Number
Ex 3-41
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 28 / Rev 1
EXAMPLE
Yard Stick
Length
8.0”
Width
5.0”
Calculated Area
40 sq. in.
Significant Digits
2
Correct Answer
40. sq. in.
Ex 3-42
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 29 / Rev 1
EXAMPLE
Ruler
Length
8.2”
Width
4.8”
Calculated Area
39.36”.
Significant Digits
2
Correct Answer
39 sq. in.
Ex 3-43
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 30 / Rev 1
EXAMPLE
Micrometer
Length
8.164”
Width
4.795”
Calculated Area
39.14638 sq. in.
Significant Digits
4
Correct Answer
39.15 sq. in.
Ex 3-44
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 31 / Rev 1
EXAMPLE
Round to two figures
0.193
0.19
Round to five figures
157,632
157,630
Round to three figures
7,591
7,590
Round to four figures
0.98764
0.9876
Ex 3-45
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 32 / Rev 1
EXAMPLE
Round to one figure
0.193
0.2
Round to three figures
157,632
158,000
Round to two figures
7,591
7,600
Round to four figures
0.98764
0.9877
Ex 3-46
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 33 / Rev 1
EXAMPLE
Round to two figures
0.1853
0.19
Round to four figures
195,753
195,800
Round to one figure
7,591
8,000
Round to three figures
0.98751
0.988
Round to two figures
18,501
19,000
Round to four figures
19,555,005
19,560,000
Ex 3-47
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 34 / Rev 1
EXAMPLE
Round to two figures
0.18500
0.18
Round to four figures
195,750
195,800
Round to one figure
7,500
8,000
Round to one figure
6,500
6,000
Round to three figures
0.98450
0.984
Ex 3-48
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 35 / Rev 1
EXAMPLE
Add 0.0146, 0.0950, and 0.43
Step 1
Set up to do the math as normal.
0.0146
0.0950
+ 0.043
Step 2
Identify the least significant digits in each term to be
added or subtracted.
0.0146
the six is the least significant digit
0.0950
the zero is the least significant digit
+ 0.043
the three is the least significant digit
Ex 3-49
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 36 / Rev 1
EXAMPLE
Step 3
Draw a vertical line to the right of the term with the
least accuracy (the least significant digit that is
farthest to the left).
0.0146
0.0950
+ 0.043
Step 4
Do the math as normal.
0.0146
0.0950
+ 0.043
0.1526
Ex 3-49
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 37 / Rev 1
EXAMPLE
Step 5
Round the digit to the left of the line following the
rounding rules.
0.1526 rounds to 0.153 (Rounding Rule 1)
Ex 3-49
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 38 / Rev 1
EXAMPLE
Example 3-50
Add 850 and 5.90
850
+
5.90
855.90
855.90 rounds to 860 (Rounding Rule 3)
Example 3-51
Subtract 5.6 from 875
875
–
5.6
869.4
869.4 rounds to 869 (Rounding Rule 1)
© 2003 General Physics Corporation
Ex 3-50
EX 3-51
ABC/ Mathematics / Chapter 3 / TP 3 - 39 / Rev 1
EXAMPLE
Example 3-52
Subtract 85 from 1,000,000
1,000,000
–
85
999,915
999,915 rounds to 1,000,000 (Rounding Rule 2)
Example 3-53
Subtract 0.375 from 0.5
0.5
– 0.375
0.125
0.125 rounds to 0.1 (Rounding Rule 1)
© 2003 General Physics Corporation
Ex 3-52
EX 3-53
ABC/ Mathematics / Chapter 3 / TP 3 - 40 / Rev 1
EXAMPLE
The altimeter in an airplane reads to the nearest 100 ft. A
cargo plane is cruising at 35,500 ft. Inside the cargo
plane are crates. The crates are each 4 ft tall. These
crates are stacked five high. On top of the highest crate
is a ball bearing which measures 0.350 inches in
diameter. How far is the top of the ball bearing from the
ground?
Add 35,500 ft + (5 × 4 ft) + 0.350 in.
If we convert 0.350 in. to ft
the problem becomes:
35,500
ft
20
ft
+
0.029166 ft
35,520.029166 ft
Ex 3-54
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 41 / Rev 1
EXAMPLE
The altimeter is the least accurate measurement,
and it controls the accuracy of the answer. Since
the plane’s altitude is only accurate to within 100 ft,
this controls the accuracy of the addition problem.
By rule 2 above, the sum has the same accuracy as
the least accurate measurement. The altimeter
cannot distinguish between 35,500 ft and
35,520.029 ft.
Therefore the answer is 35,500 ft, not 35,520.029 ft.
Ex 3-54
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 42 / Rev 1
EXAMPLE
A scale used to weigh trucks is marked in tons (T).
In an hour, the following weights are recorded:
18T, 22T, 17T, 19T, 25T, 30T, 11T, 8T.
a. Calculate the total weight of the trucks
weighed.
The total is 150 T.
b. Calculate the average weight of all the trucks
weighed.
The average is 18.75 T = 19 T
Since the scale can only measure to the nearest ton,
the average cannot be more accurate than the
scale.
Ex 3-55
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 43 / Rev 1
EXAMPLE
Multiply 3.3 and 0.025.
Step 1
3.3 has two significant digits (rule 3).
0.025 has three significant digits (rule 3).
3.3  0.025 = 0.0825
Step 3
The answer contains two significant digits.
Step 4
The most significant digit is 8.
Step 5
0.0825 rounds to 0.082 (rounding rule 3b)
Ex 3-56
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 44 / Rev 1
EXAMPLE
Multiply 1,780.0 and 0.050.
Step 1
1,780.0 has five significant digits (rule 3).
0.050 has three significant digits (rule 3).
Step 2
(1,780)(0.050) = 89
Step 3
Answer has three significant digits.
Step 4
The most significant digit is the 8 followed
by two other significant digits. 89.0
Step 5
89 rounds to 89.0 (rounding rule 1)
(Note decimal and zero)
© 2003 General Physics Corporation
Ex 3-57
ABC/ Mathematics / Chapter 3 / TP 3 - 45 / Rev 1
EXAMPLE
Divide 23.5 into 180,000
Step 1
23.5 has three significant digits (rule 3)
180,000 has two significant digits (rule 2)
Step 2
180,000 ÷ 23.5 = 7,659.5745
Step 3
Answer will contain two significant digits.
Step 4
The most significant digit is the first 7 followed by
one other significant digit.
Step 5
7,659.5745 rounds to 7,700 (rounding rule 3a).
(Note no decimal.)
Ex 3-58
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 46 / Rev 1
EXAMPLE
Divide 888 by 464.
Step 1
888 has three significant digits (rule 2).
464 has three significant digits (rule 2).
Step 2
888 ÷ 464 = 1.9137931
Step 3
Answer will have three significant digits.
Step 4
The most significant digit is one and is
followed by two more significant digits.
Step 5
1.9137931 rounds to 1.91 (rounding rule 2).
Ex 3-59
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 47 / Rev 1
EXAMPLE
Calculate the area of a rectangle measuring 4.473 in
by 6.238 in.
4.473 in 6.238 in
= 27.902574 in2
= 27.90 in2
Both measurements have the
same accuracy, four significant
digits.
.
Ex 3-60
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 48 / Rev 1
EXAMPLE
Calculate the area of a rectangle measuring 9.825 in
by 3.0 in.
9.825 in 3.0 in
= 29.475 in2
= 29 in2
The measurement with the least accuracy
has two significant digits. Therefore the
answer must have two significant digits.
Ex 3-61
© 2003 General Physics Corporation
ABC/ Mathematics / Chapter 3 / TP 3 - 49 / Rev 1