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Math 63 Section 1.1: Operations with Real Numbers
Section 1.1
1. 16
4.8 x 105
4.3 x 10-4
7.5 x 103
4.2 x 10-8
6.2 x 107
16.
7"
16
2. A = 12.672 in
C = 7.134 in
B = 3.934 in
D = 2.67 in
7"
8
3"
H=
16
5"
R = 20 8
3. S = 1
4.
maximum
17.351 in
18.27 cm
21
26 32 in
320.06 cm
17.
Width between centers = 58
1"
8
1"
31 2
3"
4
5. Width = 76
Height =
6. a)
7"
16
18.
3"
16
15"
B = 5 16
3"
C = 11
8
7. A = 3
1"
16
5"
104 8
8. Total Rise = 68
Total Run =
9. A = 109
3"
16
7"
,
8
B = 122
this design will produce 84
parts.
10. A = 102.506 in
B = 75.39 in, this design will produce 96
parts.
23
11. Area = 89 32 square inches
Perimeter = 39
12.
13.
14.
23"
Width = 2 32
1"
Height = 1 8
11"
A = 2 16
1"
B=3 8,
3"
32 4
1"
4
15. a) 6,320,000,000 V
b) .0000000098 A
c) .00000047 F
d) 14,200,000 W
480 kHz
430 µV
7.5 kW
42 nF
62MH
minimum
17.283 in
17.97 cm
19
26 32 in
319.94 cm
Difference
9
- 32 in
- .05 mm
+ .041 in
7"
b) Yes there is 8 to spare, although this
is not much when divided among 9
spaces.
480 x 103
430 x 10-6
7.5x 103
42 x 10-9
62 x 106
19. 27 Rows, 41 Columns, this design will
produce 1107 parts.
20. 32 Rows, 35 Columns, this design will
produce 1120 parts.
21.
Design
Length
Tolerance
3.7 in
±
36.62 in
±
80.5 in
±
100.29 in
±
2
100
2
100
2
100
2
100
Fraction
70
in
3
in
36
in
80
in
100
100
Max
72
in
3
in
36
in
80
in
100
62
100
50
100
29
100
100
Min
3
in
36
in
80
in
100
64
100
52
100
31
100
3
4
in
1
1 in
4
5
8 in
8
8
5
16
in
1
15 in
8
1
20 in
4
Tolerance
1
7
8
8
± in
1
± in
8
1
± in
8
±
Maximum
Dimension
1
16
in
1
± in
8
1
± in
8
in
3
1 in
8
3
8 in
4
3
8 in
8
1
15 in
4
3
20 in
8
Minimum
Dimension
5
8
100
in
60
100
48
100
in
in
27
100
22.
Design
Dimension
68
in
in
1
1 in
8
1
8 in
2
1
8 in
4
15 in
1
20 in
8
in
Math 63 Section 1.1: Operations with Real Numbers
5
8
1
4
1. A carpenter needs a board that is 79 " long that she will cut from a board is 96 " long. If the
saw blade has a
3
" kerf blade (kerf is the thickness of wood that the blade removes with a
16
cut), calculate the length of the left over piece.
Approach:
Step 1. Add the length of the board needed (79-5/8”) to the kerf (3/16”).
Step 2. Subtract the sum in step one from the original length of the board.
Step 1 –
Total of kerf plus length
of board needed (T)
Step 2 –
Length of Board
Leftover (L)
13 3

16 16
13
T  79
in
16
1
13
L  96  79
4
16
7
L  16
in
16
T  79
2. Calculate the missing dimensions A, B, C, and D. Measurements are in inches.
Find A
Approach: Add the two dimensions (3.247 in
and 9.425 in) which make up the length of A.
A  3.247  9.425
A  12.672 in
Find B
Approach:
Step 1. Add the two dimensions on
the same side as B (4.868 in and 3.870 in).
Step 2. Subtract the sum in step 1
from the length found for A (12.672 in).
B  12.672   4.868  3.870 
B  12.672  8.738
B  3.934 in
Find C
Approach: C is the same length as its opposite side; total the three dimensions that make up the length
of the opposite side (1.844 in, 2.721 in, 2.569 in).
C  1.844  2.721  2.569
C  7.134 in
Find D
Approach: Subtract 4.464 in from the length found for C (7.134 in).
D  7.134  4.464
D  2.67 in
Math 63 Section 1.1: Operations with Real Numbers
3. Calculate the length of the nonthreaded portion of the bolt (S).
Calculate the size of the
overhang (H). Measurements
are in inches.
Find S
Approach: Subtract the length of the threaded portion of the bolt (3-1/4 in) from the
length of the length of the bolt (5-1/8 in).
1
1
S  5 3
8
4
7
S  1 in
8
Find H
Approach:
7 1
Step 1. Subtract the diameter of the threaded portion of the bolt (1/2 in) from H      2
8 2
the diameter of the head of the bolt (7/8 in).
Step 2. To get the size of the overhang, divide the answer from step 1 by two.
3
H  2
8
3
H
in
16
4. If the drawing of the part below
is scaled up by a factor of five,
find the new dimensions.
Measurements are in inches.
Find the new Radius (R)
1
8
Approach: Multiply the 4 in radius by 5.
1
R  4 5
8
5
R  20 in
8
Find the new distance between the centers of the circles (D)
3
4
Approach: Multiply the distance 11 in by 5.
3
D  11  5
4
3
D  58 in
4
Math 63 Section 1.1: Operations with Real Numbers
5. If the drawing of the link below is scaled up by a factor of seven, find the new dimensions for the
overall width and height. Measurements are in inches. Note: An R in drafting denotes the
radius of a circle.
Find the new overall width (W)
Approach:
Find the new overall height (H)
Approach:
1. Multiply the radius
1. Multiply the radius
 1 
 2 in  by 2 and add the
 4 
distance between the centers
of the circles).
2. Multiply the answer from step
1 by 7.
3
 1
W  2 26 7
8
 4
3
 1
W  4 6 7
8
 2
7
W  10  7
8
1
W  76 in
8
 1 
 2 in  by 2.
 4 
2. Multiply the answer from step
1 by 7.
 1 
H   2  27
 4 
1
H  4 7
2
1
H  31 in
2
Math 63 Section 1.1: Operations with Real Numbers
6.
a) A fence is to have seven boards
between two posts so that the space
between each board is the same.
Calculate the distance between each
board. Answer as a fraction.
b) If the carpenter decided the
distance between boards was too
large, would it be possible to add an
extra board, decreasing the space
between each board? Explain. Note:
dashes are commonly placed
between whole numbers and
7
8
fractions so 21 " is not mistaken
for
217
" .
8
a) Find the distance between each board (D)
Approach:
5
 7
D   21  7  2   8
8
 8
3
 7
D   21  18   8
8
 8
Step 3. Divide the answer from step 2 by 8 (because there are eight spaces D  3 1  8
between the boards and posts).
2
7
D
in
16
 5 
Step 1. Multiply 7 boards times the width of each board  2 in  .
 8 
 7 
Step 2. Subtract the answer from step 1 from the total length  21 in  .
 8 
b) Can you add an extra board?
Approach:
 5
 8


Step 1. Multiply 8 boards times the width of a single board  2 in  and
 7 
compare the answer to the overall length of  21 in  .
 8 
5
8  2  21 in
8
Step 2. If you can add the extra board, see how much space there would be leftover to allow for
spacing between the boards and posts.
There is enough room for 8 boards, but this only leaves
That’s less than
7
in to be spread across 9 spaces.
8
1
in between each board. Although possible, this really isn’t very much room.
8
Math 63 Section 1.1: Operations with Real Numbers
7. Calculate the missing dimensions
A, B, and C. Measurements are in
inches.
Find A
Approach:
1. Total the dimensions of the side
5
1 
 3
opposite A  2 in, 2 in, 2 in  .
8
2 
 4
2. Subtract the sum of the two known
dimensions, on the same side as A,
from the answer in step 1.
5
1  7
1
 3
A  2  2  2 2  2 
8
2   16
4
 4
7
11
A 7 4
8
16
3
A3
in
16
Find B
Approach: Multiply the radius
7
 3 
1 in  by 2, then add 2 in .
16
 4 
3
7
B  2 1  2
4
16
1
7
B3 2
2
16
15
B5
in
16
Find C
Approach: Add 2 times the radius
 3 
1 in  plus all three dimensions on the
 4 
5
1 
 3
same side as C  2 in, 2 in, 2 in  .
8
2 
 4
3  3
5
1
C  2 1   2  2  2 
4  4
8
2
1
7
C 3 7
2
8
3
C  11 in
8
Math 63 Section 1.1: Operations with Real Numbers
9
16
5
8
8. A stair stringer is cut with a 7 " rise and a 11 " run. Code stipulates that all rises must be the
same height and all runs must be the same length to avoid being a trip hazard. Calculate the
total rise and total run for the stringer with nine steps.
Find Total Rise
Find Total Run
9
Approach: Multiply 9 times a rise of 7 in .
16
Approach: Multiply 9 times a run of 11 in .
Total Rise = 9  7
= 68
9
16
1
in
16
5
8
Total Run  9  11
=104
5
8
5
in
8
Math 63 Section 1.1: Operations with Real Numbers
9. A manufacturer wants to
layout the parts on a
sheet of steel in 14 rows
and 6 columns with a ½”
space between each row
and column. Calculate
the width (A) and length
(B) of the sheet that will
be needed to produce
the parts. How many
parts will this design
produce?
Measurements are in
inches.
Find the width (A)
Approach:
1. Find the width of 6 parts by multiplying the
5 
 15
in + 2  4 in 
32 
 32
width of one part  9
times 6.
2. Find the length of all the spaces by
multiplying 5 times
1
in .
2
3. Find the total width (A) by adding the
answers from steps 1 and 2.
4.
5 
1
 15
A  9  2 4 6  5
32 
2
 32
10 
1
 15
A  9  8 6  2
32 
2
 32
1
 25 
A  17   6  2
2
 32 
11
1
A  106  2
16
2
3
A  109
in
16
How many parts will this design produce (N)?
This design will produce a maximum of 84 parts.
Find the length (B)
Approach:
5. Find the length of 14 parts by multiplying


the length of one part  2  4
5 
in 
32 
times 14.
6. Find the length of all the spaces by
1
2
multiplying 13 times in .
7. Find the total length (B) by adding the
answers from steps 1 and 2.
5 
1

B   2  4   14  13 
32 
2

5
1
B  8  14  6
16
2
3
1
B  116  6
8
2
7
B  122 in
8
N  6  14
N  84 parts
Math 63 Section 1.1: Operations with Real Numbers
10. A manufacturer wants to layout the parts on a sheet of plastic in 12 rows and 8 columns with a
.15-inch space between each row and column. Calculate the width (A) and length (B) of the
sheet of that will be needed to produce the parts. How many parts will this design produce?
Measurements are inches.
Find the width (A)
Approach:
4. Fid the width of 8 parts by
multiplying the width of one
part (12.682 in) times 8.
5. Find the distance of all the
spaces by multiplying 7 times
0.15 in.
6. Get the total width (A) by
adding the answers from
steps 1 and 2.
Find the length (B)
Approach:
1. Fid the length of 12 parts by
multiplying the length of one
part (6.145 in) times 12.
2. Find the distance of all the
spaces by multiplying 11
times 0.15 in.
3. Get the total length (B) by
adding the answers from
steps 1 and 2.
A  12.682  8  0.15  7
A  101.456  1.05
B  6.145 12  0.15 11
B  73.74  1.65
A  102.506 in
B  75.39 in
How many parts will this design produce (N)?
This design will produce a maximum of 96 parts.
N  12  8
N  96 parts
Math 63 Section 1.1: Operations with Real Numbers
11. Calculate the area
and perimeter of a
rectangular sheet of
glass that is 12-3/8” x
7-1/4”.
Find the Perimeter (P)
Approach: Use the formula P  2l  2w , where l = 12-3/8 in and w = 7-1/4 in
l = length
w = width
P  2l  2 w
3
1
P  2  12  2  7
8
4
3
1
P  24  14
4
2
1
P  39 in
4
Find the Area (A)
Approach: Use the formula A  l w , where l = 12-3/8 in and w = 7-1/4 in
l = length
w = width
Alw
3 1
A  12  7
8 4
23
A  89
sq in
32
Math 63 Section 1.1: Operations with Real Numbers
12. If the drawing of the link below is scaled down by a factor of four (divide by four), find the new
dimensions for the overall width and height. Measurements are in inches.
Find the new height (H)
Approach:
3. Determine the current height.
4. Divide the answer from step 1 by 4.
Find the new width (W)
Approach:
1. Determine the current width.
2. Divide the answer from step 1 by 4.
1

H  22  4
4

1
H  4 4
2
1
H  1 in
8
1
3

W  22  6 4
4
8

3
 1
W  4 6 4
8
 2
7
W  10  4
8
23
W2
in
32
Math 63 Section 1.1: Operations with Real Numbers
13. A steel plate is to have five 7/8” diameter holes drilled so that the space between the edges of
each hole and between the edge of each hole and the edge of the plate is the same. Calculate
the distance from the edge of the plate to the center of the first hole (A), then the distance
between the center of each hole (B). Answer as a fraction.
Find the distance from the edge of the plate to center of the
first hole (A)
Approach:
1. Determine the total of the 5 diameters (each 7/8 in).
2. Subtract the value from step 1 from the total length
(17-7/8 in).
3. Divide the answer from step 2 by 6 (6 is the number
of spaces between the five holes (4) plus the two
spaces from the outside holes to the edge). This will
give the distance from the edge of the plate to the
closest edge of the circle (also the distance between
each hole).
4. Add half the diameter to the answer from step 3 in
order to get the distance from the edge of the plate
to the center of the first hole (A).
 7  7 
1 7
A  17   5     6  
2 8
 8  8 
3
7
 7
A  17  4   6 
8
16
 8
1
7
A  13  6 
2
16
1 7
A2 
4 16
11
A2
in
16
Find the distance between the centers of
each hole (B)
Approach: Use the answer from Step 3 in
finding A and add two times the radius
(7/8 in).
1 7
B2 
4 8
1
B  3 in
8
Math 63 Section 1.1: Operations with Real Numbers
14. The face frame of a cabinet is made of vertical stiles and horizontal rails. Calculate the width of
the rails in the design below so that all nine rails are the same length.
Find the width of a single rail (W)
Approach:
1. Find the total width of all of the stiles.
2. Subtract the result from step 1 from the total width (107 in).
3. Divide the result from step 2 by 3 to get the width of a single rail.

3 
 3 5 5
W  107   2  1  1  2    3
4 
 4 8 8

3

W  107  8   3
4

1
W  98  3
4
3
W  32 in
4
Math 63 Section 1.1: Operations with Real Numbers
15. In electronics large or small quantities are written using
metric prefixes. For example:
3.2 μA = 3.2 x 10-6 A = 3.2 x .000001 A = .0000032 A
8.3 kΩ = 8.3 x 103 Ω = 8.3 x 1000 Ω = 8300 Ω
otice that the power of 10 moves the decimal left or right
according to the sign of the number.
Common Electronics Units
Unit
Symbol
W
Watt
Ω
Ohm
V
Volt
A
Amp
F
Farad
H
Henry
Hz
Hertz
Expand the following without prefixes:
a) 6.32 GV
b) 9.8 nA
c) .47 μF
d) 14.2 MW
Metric Prefix
Tera
Giga
Mega
kilo
milli
micro
nano
pico
Table of Common Metric Prefixes
Symbol
Power of 10
Factor
T
1012
1,000,000,000,000
G
109
1,000,000,000
M
106
1,000,000
3
k
10
1,000
m
10-3
.001
-6
μ
10
.000001
n
10-9
.000000001
-12
p
10
.000000000001
Approach: Change the metric prefix to its factor equivalent, then multiply by the value given.
Note: You can go directly from the exponential notation to the final value by moving the decimal in
accordance with the value of the exponent. If the exponent is positive, move the decimal right. If the
exponent is negative, move the decimal left. This is a much faster approach.
6.32 GV  6.32 109 V
a)
 6.32  1, 000, 000, 000 V
.47 μF  .47 106
c)
= .47  0.000001
= 6,320, 000, 000 V
= 0.00000047 F
14.2 MW  14.2 106
9.8 nA  9.8 109
b)
= 9.8  0.000000001
= 0.0000000098 A
d)
= 14.2 1, 000, 000
= 14, 200, 000 W
Math 63 Section 1.1: Operations with Real Numbers
16. Scientific Notation is a number between 1 and 10 times a power of ten.
Engineering Notation is a modification of Scientific Notation where the number is between 1
and 1000 and the power of 10 is always a multiple of three, making it easy to write the number
using a metric prefix. Consider the two examples in the table and fill in the rest using the table
of common metric prefixes.
Metric Prefix
Tera
Giga
Mega
kilo
milli
micro
nano
pico
Standard
Form
12,300 Ω
.000072 A
480,000 Hz
.00043 V
7,500 W
.000000042 F
62,000,000 H
Table of Common Metric Prefixes
Symbol
Power of 10
Factor
12
T
10
1,000,000,000,000
G
109
1,000,000,000
6
M
10
1,000,000
k
103
1,000
-3
m
10
.001
μ
10-6
.000001
n
10-9
.000000001
p
10-12
.000000000001
Scientific
Notation
1.23 x 104
7.2 x 10-5
4.8 x 105
4.3 x 10-4
7.5 x 103
4.2 x 10-8
6.2 x 107
Engineering
Notation
12.3 x 103
72 x 10-6
480 x 103
430 x 10-6
7.5 x 103
42 x 10-9
62 x 106
Metric
Prefix
12.3 kΩ
72 μA
480 kHz
430 μV
7.5 kW
42 nF
62 MH
Math 63 Section 1.1: Operations with Real Numbers
17. Consider a rod being cut to a length. The length can never be perfect and so must be cut for
accuracy based on a desired tolerance. Tolerance is the amount that a measurement can vary
above or below a design size.
Approach:
For maximum length, add the tolerance to the design length.
For minimum length, subtract the tolerance from the design
length.
Note: for the last problem in the table, first change meters to
centimeters, then apply the tolerance as before. (1 meter = 100
centimeters, therefore, 3.2 meters = 320 centimeters)
Design
Length
Tolerance
Maximum
Length
Minimum
Length
17.317 in
± .034 in
17.351 in
17.283 in
18.12 cm
± .15 cm
18.27 cm
17.97 cm
5
±
26 8 in
3.2 meters
1
32
in
± .06 cm
26
21 in
32
320.06 cm
26
19 in
32
319.94 cm
18. Calculate the difference between the design size and the actual size in the chart. Actual sizes that
are less than the design size should be denoted with negative numbers.
Approach: Subtract the actual size from the design size.
14
3
29
 13
16
32
Design Size
3
14 16 in
Actual Size
13
29
32
in
Difference

9 in
32
18.43  18.38
18.43 mm
18. 38 mm
-0.05 mm
17.972 in
18.013 in
0.041 in
17.972  18.013
Math 63 Section 1.1: Operations with Real Numbers
19. Challenge problem: A manufacturer
wants to layout parts on a sheet of
steel that is 120” x 68”. If the part is
2.34” wide and 1.87” tall, calculate the
number of rows and columns that will
fit if there is a .6” space between each
part. How many parts will this design
produce? Note: the drawing is not to
scale.
Find the number of parts this design will produce (N)
Approach:
1. Find the number of parts (A) that can fit horizontally (120 inches) if each part is 2.34 in long.
2. Find the number of parts (B) that can fit vertically (68inches) if each part is 1.87 in wide.
3. Multiply the answers from steps 1 and 2 to find the total number of parts (N).
Note: There is no space between the part and the edge of the sheet. There are only gaps between the
parts. Therefore, there will be one less gap than there are parts (e.g. 15 parts across means 14 gaps).
Step 1. Find the number of parts horizontally
A = number of parts
Step 2. Find the number of parts vertically
B = number of parts
Each part is 2.34 plus a gap of 0.6,
but we need to subtract the gap
at the end.
120  (2.34  0.6) A  0.6
120  2.94 A  0.6
120.6  2.94 A
Each part is 1.87 plus a gap of 0.6,
but we need to subtract the gap
at the end.
68  (1.87  0.6) B  0.6
68  2.47 B  0.6
68.6  2.47 B
120.6
A
2.94
A  41.02 parts
68.6
B
2.47
B  27.77 parts
Therefore, the maximum number of parts
horizontally is 41.
Therefore, the maximum number of parts
vertically is 27.
Step 3. The maximum number of parts this design will produce is 1107.
N  41  27
N  1107 parts
Math 63 Section 1.1: Operations with Real Numbers
20. Challenge Problem: A manufacturer wants to
layout parts on a sheet of steel that is 108” x 60”.
If the part is 2 5" wide and
16
1
1"
8
tall, calculate the
number of rows and columns that will fit if there
is a
3"
4
space between each part. How many
parts will this design produce? Note: the drawing
is not to scale.
Find the number of parts this design will produce (N) You can use fractions or decimals to calculate.
Approach:
1. First find the number of parts (A) that can fit horizontally (108 in) if each part is 2-5/16 in long.
2. Find the number of parts (B) that can fit vertically (60 in) if each part is 1-1/8 in wide.
3. Multiply the answers from steps 1 and 2 to find the total number of parts (N).
Note: There is no space between the part and the edge of the sheet. There are only gaps between the
parts. Therefore, there will be one less gap than there are parts (e.g. 15 parts across means 14 gaps).
Step 1. Find the number of parts horizontally
A = number of parts, A – 1 = number of gaps
Step 2. Find the number of parts vertically
B = number of parts, B – 1 = number of gaps
3
5
 A  1  2 A
4
16
3
3
5
108  A   2 A
4
4
16
1
3
108  3 A 
16
4
3
1
108  3 A
4
16
3
1
108  3  A
4 16
25
A  35
parts
49
60 
108 
3
1
 B  1  1 B
4
8
60  0.75  B  1  1.125B
60  0.75B  0.75  1.125B
60  1.875B  0.75
60.75  1.875B
B  32.4 parts
Therefore, the maximum number of parts
vertically is 32.
Therefore, the maximum number of parts
horizontally is 35.
Step 3: The maximum number of parts this design will produce is 1120.
N  35  32
N  1120 parts
Math 63 Section 1.1: Operations with Real Numbers
Math 63 Section 1.1: Operations with Real Numbers
21. Consider the diagram. Convert the given
1
dimensions to fractions to the nearest 100 inch (do
2
not simplify). Using a tolerance of ± 100 inch,
determine the maximum and minimum allowable
dimensions.
Approach:
1. To rewrite the design length as a fraction, take the decimal out two places (if necessary),
then rewrite as a mixed number.
2
to the mixed number. Do not simplify.
100
2
3. To get the minimum length, subtract
from the mixed number. Do not simplify.
100
2. To get the maximum length, add
Design
Length
Tolerance
3.7 in

2
in
100
3
16.62 in

2
in
100
80.5 in

100.29 in

Fraction
Maximum
Length
70
100
3
16
62
100
2
in
100
80
2
in
100
100
Minimum
Length
72
100
3
68
100
16
64
100
16
60
100
50
100
80
52
100
80
48
100
29
100
10
31
100
10
27
100
Math 63 Section 1.1: Operations with Real Numbers
22. Consider the diagram. Determine the maximum and minimum allowable dimensions for the given
tolerances in the table below. Answers should be simplified.
Note: the symbol Ø indicates the diameter of the circle.
Side View
End View
Approach:
1. To get the maximum length, add the tolerance to the design dimension.
2. To get the minimum length, subtract the tolerance to the design dimension
Note: All answers should be fully simplified.
Design
Dimension
Tolerance
3
4
1
1
4
5
8
8
5
8
16
1
15
8
1
20
4
1
 in
8
1
 in
8
1
 in
8
1
 in
16
1
 in
8
1
 in
8
Maximum Minimum
Dimension Dimension
7
8
3
1
8
3
8
4
3
8
8
1
15
4
3
20
8
5
8
1
1
8
1
8
2
1
8
4
15
20
1
8