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=============================================================^15
: :^15
: :^15
: Example problem #1 :^15
: :^15
: :^15
=============================================================^15
1. Two cyclists head in opposite directions on a straight^14
road. One cycles at 10 mph and the other at 15 mph. In^14
how many hours will they be 75 miles apart?^14
(a) 3^14
(b) 4^14
(c) 2^14
Remember that the formula for distance is
DISTANCE = RATE x TIME^15
There are two distances to look at: the distance of the first
cyclist and the distance of the second cyclist. Since the two
cyclists are going in opposite directions from the same starting
point, you have to add the two separate distances.
STARTING
POINT
|
v
| |^15
|<====================O============================>|^15
| |^15
\ / \ /
\ DISTANCE #1 / \ DISTANCE #2 /
(first cyclist) (second cyclist)
When added together, the two distances equal the total distance.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
For example, if you travel north for 15 miles, and your friend
travels south for 18 miles, then you are 33 miles apart.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
( 15 miles ) + ( 18 miles ) = TOTAL DISTANCE
33 miles = TOTAL DISTANCE
If you travel north at 5 miles per hour for 3 hours, and your
friend travels at 9 miles per hour for 2 hours, the two of you
will end up 33 miles apart:
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE
(5 mph x 3 hrs ) + (9 mph x 2 hrs ) = TOTAL DISTANCE
( 5 x 3 ) + ( 9 x 2 ) = TOTAL DISTANCE
15 miles + 18 miles = TOTAL DISTANCE
33 miles = TOTAL DISTANCE
Now, let's solve the original problem:
Two cyclists head in opposite directions on a straight^14
road. One cycles at 10 mph and the other at 15 mph. In^14
how many hours will they be 75 miles apart?^14
The first sentence tells us we are going in opposite directions,
so we will need to ADD two distances to come up with a total.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
Look at the second sentence of the problem. The rates of the two
cyclists are given: 10 mph and 15 mph. Substitute these rates
into our formula.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(10 mph x TIME#1) + (15 mph x TIME#2) = TOTAL DISTANCE^3
In the third sentence, the total distance is given. Substitute
this into our formula.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(10 mph x TIME#1) + (15 mph x TIME#2) = TOTAL DISTANCE
(10 mph x TIME#1) + (15 mph x TIME#2) = 75 miles^3
The question asked in the third sentence
In how many hours^14
means that both cyclists have been traveling the same amount of
time. Let the time they have been traveling = X. Substitute
this into our formula.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(10 mph x TIME#1) + (15 mph x TIME#2) = TOTAL DISTANCE
(10 mph x TIME#1) + (15 mph x TIME#2) = 75 miles
(10 mph x X hrs ) + (15 mph x X hrs ) = 75 miles^3
The variables have been taken care of in the formula so all that
is left to do is to solve.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(10 mph x TIME#1) + (15 mph x TIME#2) = TOTAL DISTANCE
(10 mph x TIME#1) + (15 mph x TIME#2) = 75 miles
(10 mph x X hrs ) + (15 mph x X hrs ) = 75 miles
( 10 x (X) ) + ( 15 x (X) ) = 75^3
10X + 15X = 75^3
25X = 75^3
25X 75^3
--- = --^3
25 25^3
X = 3^3
The answer is 3 hours. In 3 hours, the first cyclist will have
traveled 30 miles (3 x 10), and the second cyclist will have
traveled 45 miles (3 x 15). Together they will have traveled
30 + 45 = 75 miles.
=============================================================^15
: :^15
: :^15
: Example problem #2 :^15
: :^15
: :^15
=============================================================^15
2. Two cars traveling in opposite directions were 330 miles^14
apart after 3 hours. If one car travels 10 mph faster^14
than the other, find the rate of each car.^14
(a) 50 and 60 mph^14
(b) 55 and 60 mph^14
(c) 40 and 55 mph^14
Remember that the formula for distance is
DISTANCE = RATE x TIME^15
There are two distances to look at: the distance of the first car
and the distance of the second car. Since the two cars are going
in opposite directions from the same starting point, you have to
add the two separate distances.
STARTING
POINT
|
v
| |^15
|<====================O============================>|^15
| |^15
\ / \ /
\ DISTANCE #1 / \ DISTANCE #2 /
(first car) (second car)
When added together, the two distances equal the total distance.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
Look at the first sentence of the problem:
Two cars traveling in opposite directions were 330 miles^14
apart after 3 hours.^14
It tells us we are going in opposite directions, so we will need
to ADD two distances to come up with a total.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
It also tells us what the TOTAL DISTANCE is and the time that
each of the cars was traveling. Substitute this into our
formula.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(RATE#1 x 3 hrs ) + (RATE#2 x 3 hrs ) = 330 miles^3
Look at the second sentence of the problem:
If one car travels 10 mph faster than the other^14
The rates of the two cyclists are given: (X) mph and (X + 10)
mph. Substitute these rates into our formula.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(RATE#1 x 3 hrs ) + (RATE#2 x 3 hrs ) = 330 miles
(X mph x 3 hrs ) + ((X+10) mph x 3 hrs) = 330 miles^3
The variables have been taken care of in the formula so all that
is left to do is to solve.
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE^15
(RATE#1 x 3 hrs ) + (RATE#2 x 3 hrs ) = 330 miles
(X mph x 3 hrs ) + ((X+10) mph x 3 hrs) = 330 miles
( (X) x 3 ) + ((X + 10) x 3 ) = 330^3
3X + 3(X + 10) = 330^3
3X + 3X + 30 = 330^3
6X + 30 = 330^3
- 30 - 30^3
---------------^3
6X = 300^3
6X 300^3
-- = ---^3
6 6^3
X = 50^3
Since X = 50, the rate of the first car is 50 mph, and the rate
of the second car is 50 + 10 or 60 mph.
=============================================================^15
: :^15
: :^15
: Example problem #3 :^15
: :^15
: :^15
=============================================================^15
3. How long will it take a runner traveling 10 mph to lap^14
a runner traveling 8 mph on a 2-mile oval track?^14
(a) 2 hours^14
(b) 1 hour^14
(c) 1.5 hours^14
Remember that the formula for distance is
DISTANCE = RATE x TIME^15
Usually, if the two distances are going in opposite
directions, you will need to add them together.
FOR OPPOSITE DIRECTIONS:^3
Starting
Point
|
v
| |^15
|<====================O========================>|^15
| |^15
\ / \ /
\- Distance #1 -/ \--- Distance #2 ---/
DISTANCE #1 + DISTANCE #2 = TOTAL DISTANCE^15
Conversely, when there are two distances going in the same
direction, you will need to subtract one distance from the other.
FOR SAME DIRECTION:^3
|^15
O===============distance #1===================>|^15
|^15
| |^15
O======distance #2======>| |^15
| |^15
\ /
\ /
(How far apart)
Longer distance - shorter distance = How far apart^15
DISTANCE #1 - DISTANCE #2 = How far apart^15
To "lap" the second runner, the first runner has to run around
the track one extra time and catch up to the second runner. In
other words, the first runner has to run 2 extra miles (the
length of the track). There are two distances to look at: the
distance of the first runner and the distance of the second
runner. Since the two runners are going in the same direction
from the same starting point, you have to subtract the two
separate distances. After one distance is subtracted from the
other, the result has to equal the extra lap, or 2 miles (the
distance needed to "lap" the other runner).
DISTANCE #1 - DISTANCE #2 = ONE EXTRA LAP^15
(RATE#1 x TIME#1) - (RATE#2 x TIME#2) = ONE EXTRA LAP^15
From the problem, we know what the rates of the two runners are,
and we also know what the distance of one lap is. Substitute
these numbers into the formula
DISTANCE #1 - DISTANCE #2 = ONE EXTRA LAP^15
(RATE#1 x TIME#1) - (RATE#2 x TIME#2) = ONE EXTRA LAP^15
(10 mph x TIME#1) - (8 mph x TIME#2) = 2 miles^3
Both runners will be running for the same amount of time, so
their times will be the same. Let each of their times = X. The
times must be in hours since the rates are measured in miles per
HOUR.
DISTANCE #1 - DISTANCE #2 = ONE EXTRA LAP^15
(RATE#1 x TIME#1) - (RATE#2 x TIME#2) = ONE EXTRA LAP^15
(10 mph x TIME#1) - (8 mph x TIME#2) = 2 miles
(10 mph x X hrs ) - (8 mph x X hrs ) - 2 miles^3
The hard work is done. Solve the equation for X
DISTANCE #1 - DISTANCE #2 = ONE EXTRA LAP^15
(RATE#1 x TIME#1) - (RATE#2 x TIME#2) = ONE EXTRA LAP^15
(10 mph x TIME#1) - (8 mph x TIME#2) = 2 miles
(10 mph x X hrs ) - (8 mph x X hrs ) = 2 miles
10X - 8X = 2^3
2X = 2^3
2X 2^3
---- = ---^3
2 2^3
X = 1^3
The first runner will lap the second runner in 1 hour.
=============================================================^15
: :^15
: :^15
: Example problem #4 :^15
: :^15
: :^15
=============================================================^15
4. A jogger leaves the starting point traveling 5 mph. One^14
hour later, a runner leaves the same point and travels at^14
9 mph. How long does it take the runner to overtake the^14
jogger?^14
(a) 1.75 hours^14
(b) 2.25 hours^14
(c) 1.25 hours^14
(d) none of the above^14
Remember the formula for distance:
DISTANCE = RATE x TIME^15
When the problem states
How long does it take the runner to overtake the jogger?^14
you need to know when their two distances are the same.
"Overtake" means catching up, or when the distance ran is the
same as the distance jogged. The two distances are equal
RUNNER'S DISTANCE = JOGGER'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
Looking at the first two sentences of the problem gives you the
rate of the jogger and the rate of the runner
RUNNER'S DISTANCE = JOGGER'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(9 mph x TIME#1) = (5 mph x TIME#2)^3
The second sentence also states that the runner took off running
one hour later than the jogger started jogging. That means that
the runner has been at it for one hour less than the jogger. If
you let the jogger's time = (X), then the runner's time = (X-1).
Substitute this into the formula
RUNNER'S DISTANCE = JOGGER'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(9 mph x TIME#1) = (5 mph x TIME#2)
(9 mph x (X-1) hrs) = (5 mph x (X) hrs)^3
Solve the equation
RUNNER'S DISTANCE = JOGGER'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(9 mph x TIME#1) = (5 mph x TIME#2)
(9 mph x (X-1) hrs) = (5 mph x (X) hrs)
9(X - 1) = 5X^3
9X - 9 = 5X^3
+ 9 + 9^3
--------------------^3
9X = 5X + 9^3
- 5X - 5X^3
--------------------^3
4X = 9^3
4X 9^3
---- = ---^3
4 4^3
X = 2.25^3
The jogger had been jogging 2.25 hours when the runner caught up.
Since the runner had been running for one hour less than the
jogger, the runner had been running (2.25 - 1) hours or 1.25
hours. The answer is 1.25 hours.
=============================================================^15
: :^15
: :^15
: Example problem #5 :^15
: :^15
: :^15
=============================================================^15
5. A salesman leaves home on a business trip driving 35 mph.^14
One hour later, his wife starts out after him at 55 mph.^14
How long does it take her to catch him?^14
(a) 2 hours and 45 min.^14
(b) 1 hour and 15 min.^14
(c) 2 hours and 15 min.^14
(d) none of the above^14
Remember the formula for distance:
DISTANCE = RATE x TIME^15
When the problem states
How long does it take her to catch him?^14
you need to know when their two distances are the same. "Catch
him" means catching up, or when the salesman's distance is the
same as the wife's distance. The two distances are equal
SALESMAN'S DISTANCE = WIFE'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
Looking at the first two sentences of the problem gives you the
rate of the jogger and the rate of the runner
SALESMAN'S DISTANCE = WIFE'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(35 mph x TIME#1) = (55 mph x TIME#2)^3
The second sentence also states that the wife took off one hour
later than the salesman. That means that the wife has been at it
for one hour less than the salesman. If you let the salesman's
time = (X), then the wife's time = (X-1). Substitute this into
the formula
SALESMAN'S DISTANCE = WIFE'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(35 mph x TIME#1) = (55 mph x TIME#2)
(35 mph x (X) hrs) = (55 mph x (X-1) hrs)^3
Solve the equation
SALESMAN'S DISTANCE = WIFE'S DISTANCE^15
(RATE#1 x TIME#1) = (RATE#2 x TIME#2)^15
(35 mph x TIME#1) = (55 mph x TIME#2)
(35 mph x (X) hrs) = (55 mph x (X-1) hrs)
35X = 55(X - 1)^3
35X = 55X - 55^3
+ 55 + 55^3
-----------------------^3
35X + 55 = 55X^3
- 35X - 35X^3
-----------------------^3
55 = 20X^3
55 20X^3
---- = ---^3
20 20^3
2.75 = X^3
The salesman had been driving 2.75 hours when his wife caught up.
Since the wife had been driving for one hour less than the
salesman, the wife had been driving (2.75 - 1) hours or 1.75
hours. The answer is 1.75 hours.