Example Distance Problems
Ex. 1 D1 = D2 (Round Trip Travel)
Richie hopped onto his bike and pedaled to the
park at 15 miles per hour. Then he sped back
at 20 miles per hour. If the total trip took 5¼
hours, how far was it to the park?
Rate
x Time
= Distance
pedaled
15
x
15x
sped
20
5¼ - x
20 (5¼ - x)
Distance 1 = Distance 2
15x = 20 (5¼ - x)
15x = 20 (5.25 – x) Use Distributive Property
15x = 105 – 20x
Get variables on same side
+20x
+ 20x
35x = 105
Get variable alone by ÷
35
35
x =
3
This is the time you pedaled.
Substitute the value of x into Distance 1 and
you will know how far it is to the park.
15 (3) = 45 miles
Example 2 D1 + D2 = N (the total number of
miles is given)
At noon the horse and buggy headed north at 8
miles per hour. Two hours later, the car
headed south from the same location at 35
miles per hour. At what time will the horse and
buggy be 102 miles from the car?
Rate
x Time
= Distance
Horse and
8
x
8x
buggy
Car
35
X - 2
35 ( x – 2)
8 x + 35 ( x – 2) = 102
8x + 35x – 70 = 102 Use Distrib.Property
43x – 70 = 102
Combine Like Terms
+ 70 = +70
Addition Prop.of Equality
43x
=
172
43
43
Division Prop. of Equality
x
= 4
You have solved the time
(in hours for the horse
and buggy)
Look at the problem and find your start time.
12:00 noon + 4 hours would be 4:00 p.m.
(EXACT TIME).
Example 3- D1 + D2 = N (Total number of
miles is given in the problem)
Kurt and Lois walked to the dock at 2 miles per
hour, jumped into the boat and motored to
Dylan’s at 10 miles per hour. If the total
distance was 16 miles and the trip took 3 hours
in all, how far did they go by boat? * This
time you are looking for Distance 2.
walked
motored
Rate x Time
2
x
10
3- x
2x + 10(3-x) = 16
2x + 30 -10x =16
-30
-30
= Distance
2x
10 ( 3 - x)
(Distributive property)
(numbers on one side and
variables on the other)
2x-10x
=-14 Combine like terms
-8x
=-14
-8
-8
x
= 7/4
Walking time is 1.75
Substitute the value of x into Distance 2 since
this is the value you are identifying.
D2 = 10 (3 – 1.75)
D2 = 10 (1.25) = 12.5 miles
Example 4 D1 + D2 = N
The trains were 225 miles apart at 2:00 p.m.
and were headed toward each other. If they
met at 4:15 p.m. and one was traveling 20 miles
per hour faster than the other, what was the
speed of each train?
* This time you are
looking for the rate of train 1 and train 2.
Rate
x Time = Distance
Train 1
x
2.25
2.25x
Train 2
x + 20
2.25
2.25 ( x + 20)
When determining time look at your start time
2:00 and the meeting time 4:15. They traveled
2 hours and 15 minutes. 15 minutes is what
portion of 1 hour (60 minutes)? 15/60 = ¼ =
.25.
So the total time for each train was
2.25. The rate of train 2 was 20 miles faster
than the rate of train 1, which is unknown.
2.25x + 2.25(x + 20) = 225
2.25x + 2.25x + 45 = 225 (distributive prop.)
-45
-45
4.5x
=180
4.5
4.5
x
= 40, so Train 1’s rate =
40mph and Train 2’s rate is 40 + 20 = 60mph
Example 5 D1 ≠D2 (Used when one person is
ahead of another person)
Victor was hauling a wide load out of Boston.
At 7:00 a.m. he headed north from exit 31 at
an average speed of 35 mph. Cameron headed
north from the same exit at 10:00 a.m. in his
Pontiac. By 6:00 p.m. the same day Cameron
was 15 miles ahead of Victor.What was
Cameron’s average speed (rate)?
Victor
Cameron
Rate
35
x
x Time
11
8
= Distance
35(11) + 15
Add the miles
Cameron is ahead to
Victor’s distance
8x
To determine time go from start time to end
time: Victor- 7:00am to 6:00p.m = 11 hours
Cameron-10:00a.m.-6:00 p.m. = 8 hours
Once you add the extra miles to Victor the
distances become equal.
35 (11) + 15 = 8x
385 + 15
= 8x
400
= 8x
8
8
50
= x, so Cameron’s rate of speed is 50
mph