Absolute Value Word Problems Homework
Name:_________________________
1
Period:________
Absolute
Value
Word
Problems
1) A
machine
is
used
to
fill
each
of
several
bags
with
16
ounces
of
sugar.
After
the
bags
are
filled,
another
machine
weighs
them.
If
the
bag
weighs
.3
ounces
more
or
less
than
the
desired
weight,
the
bag
is
rejected.
Write
this
equation
to
find
the
heaviest
and
lightest
bag
the
machine
will
approve.
2) The
average
number
of
seeds
in
a
package
of
cucumber
seed
is
25.
The
number
of
seeds
in
the
package
can
vary
by
three.
What
are
the
maximum
and
minimum
number
of
seeds
that
could
be
in
a
package?
3) The
mean
distance
of
the
earth
from
the
sun
is
93
million
miles.
The
distance
varies
by
1.6
million
miles.
What
are
the
maximum
and
minimum
distances
of
the
earth
from
the
sun?
4) Leona
was
in
a
golf
tournament
last
week.
All
of
her
four
rounds
of
gold
were
within
2
strokes
of
par.
If
par
was
72,
what
are
the
maximum
and
minimum
scores
that
Leona
could
have
made
in
the
golf
tournament?
5) Victor
has
a
goal
of
making
$75
per
week
at
his
after--school
job.
Last
month
he
was
within
$6.50
of
his
goal.
What
are
the
maximum
and
minimum
amounts
that
Victor
might
have
made
last
month?
6) Members
of
the
track
team
can
run
400
m
in
an
average
time
of
58.2
seconds.
The
fastest
and
slowest
times
varied
from
the
average
by
6.4
seconds.
What
were
the
maximum
and
minimum
times
for
the
track
team?
7) Amtrak
s
annual
passenger
revenue
for
the
years
1980
2000
is
modeled
approximately
by
the
formula
R
=
--40|x
--
11|+990
where
R
is
the
annual
revenue
in
millions
of
dollars
and
x
is
the
number
of
years
since
January
1,
1980.
In
what
years
was
the
passenger
revenue
$790
million?
ANSWERS:
Name:_________________________
2
Period:________
Absolute
Value
Word
Problems
1) |x
--
16|
=
0.3
x
is
the
bag's
weight.
X=15.7
oz,
16.3
oz
2) - 25 = 3;
x=
22
seeds,
23
seeds
3) - 93 = 1.6;
x=
91.4
million,
94.6
million
4) - 72 = 2;
x=
70
strokes,
74
strokes
5) - 75 = 6.5;
x=
$68.50,
$81.50
6) - 58.2 = 6.4;
x=
51.8
seconds,
64.6
seconds
7) 790
=
--40|x
--
11|
+
990
Solve
for
|x
--
11|
--200
=
--40|x
--
11|
5
=
|x
--
11|
So
x
--
11
is
either
equal
to
5
or
to
--5,
the
two
numbers
with
an
absolute
value
of
5.
X
=
6,
16,
so
in
1986
and
1996.
................
................
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