Time, Rate, and Distance Word Problems
- A freight train starts from Los Angeles and heads for Chicago at 40 mph. Two hours later a passenger train leaves the same station for Chicago traveling at 60 mph. How long will it be before the passenger train overtakes the freight train?
Solution
First draw a picture of the movement
Notice that the distances are equal.
Second make a diagram to put in information:
D = R x T | Time | Rate | Distance |
Freight train | 40 | ||
Passenger train | 60 |
Add information
D = R x T | Time | Rate | Distance |
Freight train | x + 2 | 40 | |
Passenger train | x | 60 |
Add information
D = R x T | Time | Rate | Distance |
Freight train | x + 2 | 40 | 40(x + 2) |
Passenger train | x | 60 | 60x |
Every time rate and distance has some kind of relationship between the distances. This one had the distances equal. That is, the trains traveled the same distance because they started at the same place and traveled until one caught up with the other. This fact was not stated. you have to watch for the relationship. We have two distances in the table above. 40(x + 2) represents the distance for the freight train. 60x represents the distance for the passenger train. Set these two distances equal for your equation:
40(x + 2) = 60x
40x + 80 = 60x
-20x = -80
x = 4 hours
So,
It takes 4 hours for the passenger train to overtake the freight train?
- A car leaves San Francisco for Los Angeles traveling an average of 70 mph. At the same time, another car leaves Los Angeles for San Francisco traveling 60 mph. If it is 520 miles between San Francisco and Los Angeles, how long before the two cars meet, assuming that each maintains its average speed?
Steps
- Find out if you know both speeds or both times. In this problem we find that both speeds are given.
- What is the unknown? The question is, “How long before the two cars meet?” So their time is unknown. But which shall we let x equal? They leave at the same time and meet and meet at the same time, so the traveling times must be equal if neither one stops. Watch for the times being equal when two objects start at the same time and meet at the same time.
- What fact is known about distance? The 520 miles is the total distance, so here is equation information. Remember, don’t put distance information in the distance box unless you have to (for example, when the exact distance each object travels is given)
Solution
First, draw a sketch of movement:
Second, make a diagram and fill in all the information you have determined. And remember, time multiplied by rate equals distance in the table.
Let x = time in hours for the two cars to meet
D = R x T | Time | Rate | Distance |
Car from Los Angeles | x | 60 | 60x |
Car from San Francisco | x | 70 | 70x |
You know that the total distance is 520 miles, so add the two distances in the diagram and set them equal to 520 miles for the equation.
60x + 70x = 520
130x = 52O
Answer
x = 4 hours
Check
60(4) + 70(4) = 520
520=520
- Two planes leave New York at 10 A.M., one heading for Europe at 600 mph and one heading in the opposite direction at 150 mph. At what time will they be 900 miles apart? How far has each traveled?
Steps
- You know both speeds.
- The times must be unknowns and they are equal (or the same) because the planes leave at the same time and travel until a certain time (when they are 900 miles apart).
- The total distance is 900 miles.
Warning! Don’t let x = clock time, that is, the time of day. It stands for traveling time in hours! The rate is in miles per hour, so time must be in hours in all time, rate, and distance problems.
Solution
Solution Sketch
Let x = time in hours for planes to fly 900 miles apart. Remember, x equals the time for the planes, since their times are the same in this problem.
Diagram
D = R x T | Time | Rate | Distance |
Fast Plane | x | 600 | 600x |
Slow Plane | x | 150 | 150x |
Equation
600x + 150x = 900
750x = 900
x = 900/750
Answer
x = 1 1/5 hours
is that what the problem asked for? Always check the question, especially on time, rate, and distance problems. The problem asked what time they would be 900 miles apart. Since the problem says they left at 10 a.m., add 1 1/5 hours to that time. So the answer is 11:12 a.m.
The problem also asked far each would travel. Since we know that rate multiplied by time equals distance,
600x = distance for fast plane and substituting the value of x,
600(1 1/5) = 600(6/5) = 720 miles
150x = distance for slow plane
150(1 1/5) = 150(6/5) = 180 miles
Check
720+180=900 miles
- Moe Tell makes a business trip from his house to Logan in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Moe is gone a total of 6 hours, how fast did he travel on each leg of the trip?
Steps
- This problem doesn’t give us the rates, but it does give us the time both ways. It gives the time to Loganville as 2 hours. The total time is 6 hours, but he didn’t travel during one of those hours. Deducting this hour leaves only 5 hours of total traveling time. That means that if it took 2 hours to travel to Loganville, it must have taken 3 hours to travel back.
- The rates are both unknown. We can let x equal either one. Just be careful that the faster rate goes w/the shorter time. If x equals the rate going, then x – 20 equals the rate returning. If x equals the rate returning, then x+20 equals the rate going. Either way is correct.
Solution
Solution Sketch
Let x = rate in mph going to Loganville
X – 20 = rate in mph returning home.
Diagram
D = R x T | Time | Rate | Distance |
Toward Loganville | 2 | x | 2x |
Toward Home | 3 | x – 20 | 3(x – 20) |
Equation
The distance to Loganville equals the distance back home.
2x = 3(x – 20)
-x = -60
x = 60 mph (rate going)
x – 20 = 40 mph (rate returning)
Check
2(60) = 3(60 – 20)
120 = 3(40)
120 = 120
- Tom and Jerry went on a camping trip with their motorcycles. One day Jerry left camp on his motorcycle to go to the village. Ten minutes later Tom decided to go too. If Jerry was traveling 30 mph and Tom traveled 35 mph, how long was it before Tom caught up with Jerry?
Steps
- We know both speeds.
- We know Tom traveled for 10 minutes less than Jerry. But minutes cannot be used in time, rate, and distance problems because rate is in miles per hour. So we must change 10 minutes into 10/60 or one-sixth of an hour.
- They travel the same distances so the distances are equal.
Solution
Sketch
Let x = time in hours for Tom to catch Jerry.
Diagram
D = R x T | Time | Rate | Distance |
Jerry | x + 1/6 | 30 | 30(x + 1/6) |
Tom | x | 35 | 35x |
Equation
35x = 30(x + 1/6)
35x = 30x + 5
5x = 5
Answer
x =1 hour (Tom’s time)
Alternate Solution
D = R x T | Time | Rate | Distance |
Jerry | x | 30 | 30x |
Tom | x – 1/6 | 35 | 35(x + 1/6) |
If you use the alternate solution, you should be good at fractions!
Also, be sure to answer the question and solve for x – 1/6, Tom’s time.
Note: Can you guess why this problem is here? You must be very careful to have time always in hours or fractions of an hour because rate is in miles per hour.
- Two cars headed for Las Vegas. One is 50 miles ahead of the other on the same road. The one in front is traveling 60 mph while the second car is traveling 70 mph. How long will it be before the second car overtakes the first car?
Steps
1.The speeds are given.
- The times are the unknowns and are equal.
- The distances are not equal; this inequality is the tricky variation we mentioned above. But if we add 50 miles to the distance of one, it will equal the distance of the other.
Solution
Sketch
Let x = time in hours for the second car to overtake the first.
Diagram
Time | Rate | Distance | |
First car | x | 60 | 60x |
Second car | x | 70 | 70x |
Equation
60x + 50 = 70x
-10x = -50
Answer
x = 5 hours
If one car is 50 miles ahead of the other at the same time, it will take the cars the same length of time to reach the spot where they are together (or where one overtakes the other). So each takes x hours. The difficulty is that the two cars do not start from the same point, so their distances are not equal. The sketch shows that you have to add 50 miles to the distance the car first travels to make it equal to the distance the second car travels. Of course, the equation could be:
70x – 50 = 60x or 70 x – 60x = 50
Time Rate and distance problem involving moving air (wind) or moving water (current)
Some more difficult problem have planes flying in a wind or boats traveling in moving water. The only problems of this type which we can solve are those where the objects move directly with or against the wind or water. The plane must have a direct headwind or tailwind and the boat must be going upstream or downstream. In this type of problem the plane’s speed in still air would be increased by a tailwind or decreased by a headwind to determine how fast it actually covers the ground. For example, a plane flies 200 mph in still air. This is called airspeed. If there is a 20-mph headwind blowing, it would decrease the speed over the ground by 20 mph. so the ground speed of the plane would be 200-20 or 180 mph. The ground speed is the rate in time, rate, and distance problems. A headwind reduces the speed of the pane by the velocity of the wind. A tail wind increases the speed of the plane over the ground by the velocity of the wind. A plane with an airspeed of 400 mph with a 30-mph tailwind actually travels over the ground (ground speed) at 430 mph. A current affects a boat the same way.
- The Red Baron takes 5 hours to fly from Los Angeles to Honolulu and 4 1/11 hours to return from Honolulu to Los Angeles. If the wind velocity is 50 mph from the west on both trips, what is the airspeed of the plane? (Airspeed is the speed of the plane in still air)
Steps
- The 2 times are given
- You are asked to find the speed of the plane in still air
- Going to Honolulu you have headwind so subtract the velocity of the wind. Returning to Los Angeles, you have tailwind, so add the velocity to airspeed
- The distances are equal.
Solution
Sketch
Let x = speed of plane in airspeed
x – 50 = speed of plane over the ground to trip from Los Angeles to Honolulu
x + 50 = speed of plane over ground on the trip from Honolulu to Los Angeles
Ground speed determines how long it takes the plane to travel from one place to another.
Diagram
Time | Rate | Distance | |
To Honolulu against the wind | 5 | x – 50 | 5(x – 50) |
To Los Angeles with the wind | 4 1/11 | x + 50 | 4 1/11(x + 50) |
Equation
5(x – 50) = 4 1/11(x + 50)
5(x – 50) = 45/11(x + 50)
5x – 250 = (45x + 2250)/11
Multiply both sides by 11 to clear fractions.
55x – 2750 = 45x + 2250
10x = 500
x = 500mph (airspeed)
Check
5(500 – 50) = 45/11(500 + 50)
2250 = 24750/11
2250 = 2250
The plane will fly at the speed at the same airspeed regardless of the wind velocity. The speed at which it actually covers the ground is the airspeed plus or minus the wind velocity assuming a direct tailwind or headwind.
- In his motorboat, Don Stream can go downstream in 1 hour less time than he can go upstream the same distance. If the current is 5 mph, how fast can Don travel in still water if it takes him 2 hours to travel upstream the given distance?
Steps
- The times are 2 hours upstream and 1 hour downstream.
- The rates are unknown. If you let x equal his rate of traveling in still water, his rate upstream will be x – 5 and his rate downstream will be x + 5. You subtract or add the rate of the current.
- The distance upstream equals the distance downstream.
Solution
Sketch
Let x = rate of boat in mph in still water
x – 5 = rate of boat in mph upstream
x + 5 = rate of boat in mph downstream
Diagram
Time | Rate | Distance | |
Upstream | 2 | x – 5 | 2(x – 5) |
Downstream | 1 | x + 5 | 1(x + 5) |
Equation
2(x – 5) = 1(x + 5)
2x – 10 = x + 5
Answer
x = 15 mph (rate in still water)
- Kay Oss leaves Seattle for New York in her car, averaging 80 mph across open country. One hour later a plane leaves Seattle for New York following the same route and flying 400 mph. How long will it be before the plane overtakes the car?
Solution
Sketch
Let x = time in hours for plane to travel same distance as car.
Diagram
Time | Rate | Distance | |
Car | x + 1 | 80 | 80(x + 1) |
Plane | x | 400 | 400x |
The plane leaves 1 hour later, so the car travels 1 hour longer.
Equation
80 (x + 1) = 400x
80x + 80 = 400
-320x = -80
x = 80/320
Answer
x = 1/4 hour
Check
80(1 + 1/4) = 400(1/4)
80 (5/4) = 400 (1/4)
100 = 100
Alternate Solution
Let x = time in hours car travels before plane overtakes it.
Diagram
Time | Rate | Distance | |
Car | x | 80 | 80x |
Plane | x – 1 | 400 | 400(x – 1) |
The plane takes 1 hour less time than the car.
Equation
80x = 400(x + 1)
80x = 400x – 400
-320x = -400
Answer
x = 5/4 hours
x – 1 = 1/4 hour
- A train averaging 50 mph leaves San Francisco at I P.M. for Los Angeles 440 miles away. At the same time a second train leaves Los Angeles headed for San Francisco on the same track and traveling at an average rate of 60 mph. At what times does the accident occur?
Solution
Sketch
Let x = time in hours for trains to meet.
Diagram
Time | Rate | Distance | |
First train | x | 50 | 50x |
Second train | x | 60 | 60x |
The train leaves at the same time and arrive at the meeting point at the same time, so the times are the same.
Distance traveled for the first train + distance traveled for the second train = total distance.
Equation
50x + 60x = 440
x = 4 hours
Answer
Since the trains left at 1 p.m., they meet 4 hours later at 5 p.m.
Check
50(4) + 60(4) = 440
200 + 240 = 440
- Doug Upp rides his bike at 6 mph to the bus station. He then rides the bus to work, averaging 30 mph. If he spends 20 minutes less time on the bus than on the bike, and the distance from his house to work is 26 miles, what is the distance from his house to the bus station?
Solution
Sketch
Let x = time in hours from home to bus station
Diagram
D = R x T | Time | Rate | Distance |
Bike | x | 6 | 6x |
Bus | x – 1/3 | 30 | 30(x – 1/3) |
Twenty minutes is 1/3 hour.
Equation
6x + 30 (x –1/3) = 26
6x + 30x – 10 = 26
x = 1 hour
Since the problem asked the distance from the house to bus station, we multiply 6 times x,
Answer
6x = 6 miles
Check
6(1) + 30(1 – 1/3) = 26
6 + 20 = 26
- Two planes leave Kansas City at I P.M. Plane A heads east at 450 mph and plane B heads due west at 600 mph. How long will it be before the planes are 2100 miles apart?
Solution
Sketch
Let x = time in hours each plane travels
Diagram
Time | Rate | Distance | |
Plane A | x | 450 | 450x |
Plane B | x | 600 | 600x |
Equation
450 + 600x = 2100
1050x = 2100
Answer
x = 2 hours
- Doomtown is 200 miles due west of Sagebrush, and Joshua is due west of Doomtown. At 9 A.M. Mr. White leaves Sagebrush for Joshua. At I P.M. Mr. Earp leaves Doomtown for Joshua. If Mr. Earp travels at an average speed 20 mph faster than Mr. White and they each reach Joshua at 4 P.M., how fast is each traveling?
Solution
Sketch
Let x = Mr. White’s speed in mph
x + 20 = Mr. Earp’s speed in mph
Diagram
Time | Rate | Distance | |
White | 7 | x | 7x |
Earp | 3 | x + 20 | 3(x + 20) |
Mr. White travels 200 miles farther than Mr. Earp.
Equation
7x = 3(x + 20) + 200
7x = 3x + 60 + 200
4x = 260
x = 65 mph (Mr. White’s speed)
x + 20 = 85 mph (Mr. Earp’s speed)
Check
7(65) = 3(35) + 200
455 = 255 + 200
455 = 455
- Meg A. Bucks left Rome at 8 A.M. and drove her Ferrari at 80 mph from Rome to Sorrento. She then took the boat to Capri for the day, returning to Sorrento 5 hours later. On the return trip from Sorrento to Rome she averaged 60 mph and arrived in Rome at 8 P.M. How far is it from Rome to Sorrento?
Solution
Sketch
Let x = time in hours from Rome to Sorrento.
Diagram
Time | Rate | Distance | |
Go | x | 80 | 80x |
Return | 7 – x | 60 | 60(7 – x) |
The traveling time is the total time (12 hours) less 5 hours out for Capri, or 7 hours. The time on the return trip is the total traveling time minus the time on the trip to Sorrento.
Equation
80x = 60( 7 – x)
80x = 420 – 60x
140x = 420
x = 3 hours
Since the problem asked for the distance from Rome to Sorrento, we multiply 80 times x,
80x = 80(3) = 240 miles
Answer
240 miles
Check
80(3) = 60(7 – 3)
240 = 60(4)
240 = 240
- Superman flies to San Francisco from Santa Barbara in 3 hours. He flies back in 2 hours. If the wind was blowing from the north at a velocity of 40 mph going, but changed to 20 mph from the north returning, what was the airspeed of superman (his speed in still air)?
Solution
Let x = speed of plane still air (airspeed)
40 = velocity of wind from north (headwind) going
x – 40 = ground speed of plane going to San Francisco (SF) against wind
20 = velocity of wind from north (tailwind) returning
x + 20 = ground speed returning to Santa Barbara (SB)
Diagram
Time | Rate | Distance | |
SB to SF | 3 | x – 40 | 3(x – 40) |
SF to SB | 2 | x + 20 | 2(x + 20) |
Equation
Distance going equals distance returning.
3(x – 40) = 2(x = 20)
3x – 120 = 2x + 40
x = 160 mph (airspeed) answer
- Phil T. Rich leaves home for Fresno 400 miles away. After 2 hours, he has to reduce his speed by 20 mph due to rain. If he takes I hour for lunch and gas and reaches Fresno 9 hours after he left home, what was his initial speed?
Solution
Let x = Phil T. Rich’s original speed in mph.
Diagram
Time | Rate | Distance | |
First | 2 | x | 2x |
Second | 6 | x – 20 | 6(x – 20) |
Equation
First distance + second distance = total distance.
2x + 6(x – 20) = 400
8x – 120 = 400
8x = 520
Answer
x = 65 mph
- A highway patrolman spots a speeding car. He clocks it at 70 mph and takes off after it 0.5 mile behind. If the patrolman travels at an average rate of 90 mph, how long will it be before he overtakes the car?
Solution
Let x = time in hours for patrol car to overtake car.
Diagram
Time | Rate | Distance | |
Car | x | 70 | 70x |
Patrol Car | x | 90 | 90x |
Equation
The patrol car travels 0.5 mile farther than the other car.
90x = 70x + 0.5
20x = 0.5
Multiply by 10 to clear the decimal.
200x = 5
Answer
x = 1/40 of an hour
If you wish, change the 1/40 hour to minutes, 1/40 x 60 = 1 1/2 minutes.
Check
90(1/40) = 70(1/40) + 1/2
90/40 = 70/40 + 20/40
90/40 = 90/40
- Boy Scouts hiking in the mountains divide into two groups to hike around Lake Sahara. They leave the same place at 10 A.M., and one group of younger boys hikes east around the lake and the older group hikes west. If the younger boys hike at a rate of 3 mph and the older boys hike at a rate of 5 mph, how long before they meet on the other side of the lake if the trail around the lake is 8 miles long? At what time do they meet?
Solution
Sketch
Let x = time boys take to meet.
Diagram
Time | Rate | Distance | |
Young scouts | x | 3 | 3x |
Older scouts | x | 5 | 5x |
The distance the younger Scouts traveled plus the distance the older Scouts traveled equals the total distance around the lake.
Equation
3x + 5x = 8
8x = 8
Answer
x = 1 hour
Since the problem also asks what time they meet,
10 a.m. + 1 hour = 11 a.m.
Check
3x + 5x = 8
3(1) + 5(1) = 8
3 + 5 = 8
8 = 8
- A boat has a speed of 15 mph in still water. It travels downstream from Greentown to Bluetown in two-fifths of an hour. It then goes back upstream from Bluetown to Redtown, which is 2 miles downstream from Greentown, in three-fifths of an hour. Find the rate of the current.
Solution
Let x = rate of current in mph
15 + x = rate of boat downstream in mph
15 – x = rate of boat upstream in mph
Diagram
Time | Rate | Distance | |
Downstream | 2/5 | 15 + x | 2/5(15 + x) |
Upsteam | 3/5 | 15 – x | 3/5(15 – x) |
Equation
The distance from Greentown to Bluetown is equal to the distance from Bluetown to Redtown plus 2 miles.
2/5(15 + x) = 3/5(15 – x) + 2
Distribute
(30 + 2x)/5 = (45 – 3x)/5 + 2
Multiply each term by the LCD, 5, to clear fractions.
30 + 2x = 45 – 3x + 5(2)
30 + 2x = 45 – 3x + 10
5x = 25
Answer
x = 5 mph
- Calvin can run a mile in 6 minutes. Hobbes can run a mile in 8 minutes. If Hobbes goes out to practice and starts I minute before Calvin starts on his practice run, will Calvin catch up with Hobbes before he reaches the mile marker? (Assume a straight track and average speed.)
Solution
Convert minutes into fractions of an hour first.
6 minutes =6/60 or 1/10 hour to run 1 mile.
8 minutes =8/60 or 2/15 hour to run 1 mile.
Convert the time to rum 1 mile into miles per hour by multiplying it by a quantity which turns the time into 1 hour. This quantity is then equivalent to the miles run in 1 hour, or miles per hour.
1/10 hour x 10 = 1 hour 10 x 1 mile = 10 mph
2/15 hour x 15/2 = 1 hour 15/2 x 1 mile = 7 1/2 mph
Thus Hobbes’s rate was 7 ½ mph and Calvin’s rate was 10 mph. Let x = time in hours for Calvin to catch Hobbes.
Time | Rate | Distance | |
Calvin | x | 10 | 10x |
Hobbes | x + 1/60 | 15/2 | 15/2(x + 1/60) |
Hobbes takes 1 minute more than Calvin. One minute is 1/60 of an hour. Distances are equal and unknown in this problem, and we can use this equality for our equation.
10x = 15/2 (x + 1/60)
10x = 15x/2 + 1/8
80x = 60x + 1
x = 1/20 hour
Answer
Therefore it takes Calvin 1/20 hour to catch up. Thus, Calvin will catch up with Hobbes before he reaches the mile marker.
Check
10x = 15/2 (x + 1/60)
10(1/20) = 15/2 (1/20+ 1/60)
1/2 = 15/2 (3/60 + 1/60)
1/2 = 15/2 (4/60)
1/2 = 1/2
21. Rhoda Davidson rides her bike to the bus station where she barely makes it in time to catch the bus to work. She spends half an hour on her bike and two-thirds of an hour on the bus. If the bus travels 39 mph faster than she travels on her bike, and the total distance from home to work is 40 miles, find the rate of the bike and the rate of the bus.
Solution
Let x = rate of bicycle in mph.
x + 39 = rate of bus in mph.
The two distances are equal.
Time | Rate | Distance | |
Bike | 1/2 | x | 1/2 x |
Bus | x + 39 | x + 39 | 2/3 (x + 39) |
Write 1/2 x as x/2 in your equation.
Equation
Distance on the bike plus the distance on the bus equals the total distance.
x/2 + 2(x + 39)/3 = 40
Remove parentheses.
x/2 + (2x + 78)/3 = 40
Multiply by the LCD of 6, to clear fractions.
6 (x/2) + 6[(2x + 78)/3]= 6(40)
3x + 4x + 156 = 240
7x = 84
Answer
x = 12 mph
x + 39 = 51mph
Check
12/2 + [2(12 + 39)]/3 = 40
6 + 102/3 = 40
6 + 34 = 40
40 = 40
- Harry and Kerry started from the same point at the same time. They traveled in opposite directions on their bicycles. Harry traveled at the rate of 9 km/h, and Kerry traveled at 11 km/h. After how many hours were they 60 km apart?
Solution
Let T = the time
Time | Rate | Distance | |
Harry | T | 9 | 9T |
Kerry | T | 11 | 11T |
Equation
The two distances equal 60
9T + 11T = 60
20T = 60
T = 3
Answer
After 3 hours they are 60 miles apart
- Two trains leave Trackville at the same time. One travels north at 90 km/h. The other travels south at 110 km/h. After how many hours will the trains be 900 km apart?
Solution
Let T = time
Time | Rate | Distance | |
North bound | T | 90 | 90T |
South bound | T | 110 | 110T |
Equation
The two distances equal 900
90T + 110T = 900
200T = 900
T = 4.5
Answer
After 4.5 hours they are 900 miles apart
- Two steamships sailing in opposite directions pass each other. One ship is sailing at 32 knots (nautical miles per hour). The other ship is sailing at 28 knots After how many hours will the ships be 150 nautical miles apart?
Solution
Let T = time
Time | Rate | Distance | |
Ship 1 | T | 32 | 32T |
Ship 2 | T | 28 | 28T |
Equation
The two distances equal 150
32T + 28T = 150
60T = 150
T = 2.5
Answer
After 2.5 hours they are 150 miles apart
- Two jets are traveling toward each other and are 3400 km apart. One jet is flying at 875 km/h. The other jet is flying at 825 km/h. In how many hours will the jets pass each other?
Solution
Let T = time
Time | Rate | Distance | |
Jet 1 | T | 875 | 875T |
Jet 2 | T | 825 | 825T |
Equation
The two distances equal 3400
875T + 825T = 3400
1700T = 3400
T = 2
Answer
After 2 hours they are 3400 miles apart
- A train left Podunk and traveled west at 70 km/h. Two hours later, another train left Podunk and traveled east at 90 km/h. How many hours had the first train traveled when they were 1420 km apart?
Solution
Let T = time
Time | Rate | Distance | |
West bound | T | 70 | 70T |
East bound | T – 2 | 90 | 90(T – 2) |
Equation
The two distances equal 1420
70T + 90(T- 2) = 1420
70T + 90T – 180 = 1420
160T – 180 = 1420
160T = 1600
T = 10
Answer
After 10 hours they are 1420 miles apart
- A train left Podunk and traveled north at 75 km/h. Two hours later, another train left Podunk and traveled in the same direction at 100 km/h. How many hours had the first train traveled when the second train overtook it?
Solution
Let T = time
Time | Rate | Distance | |
North bound | T | 75 | 75T |
South bound | T – 2 | 100 | 100(T – 2) |
Equation
The two distances are equal
75T = 100(T – 2)
75T = 100T – 200
-25T = -200
T = 8
Answer
After 8 hours they have traveled the same distance.
- Joe Spout left a campsite on a trip down the river in a canoe, traveling at 6 km/h. Four hours later, Joe’s lather set out after him in a motorboat. The motorboat traveled at 30 km/h. How long after Joe’s father started did he overtake the canoe?
In the question above, how far had Joe traveled down the river when his father overtook him?
Solution
Let T = time
Time | Rate | Distance | |
Canoe | T | 6 | 6T |
Motor boat | T – 4 | 30 | 30(T – 4) |
Equation
The two distances are the same.
6T = 30T – 120
6T = 30T – 120
-24T = -120
T = 5
Answer
After 5 hours the motor boat overtakes the canoe.
- Two trucks left Bucks Trucks traveling in opposite directions. One truck traveled at a rate of 70 km/h. the other at 80 km/h. After how many hours were the trucks 900 km apart?
Solution
Let T = time
Time | Rate | Distance | |
Truck 1 | T | 70 | 70T |
Truck 2 | T | 80 | 80T |
Equation
The two distances equal 900
70T + 80T = 900
150T = 900
T = 6
Answer
After 6 hours they are 900 miles apart
- A truck left Hucks Trucks and traveled north at 80 km ft One hour later, another truck left Hucks Trucks and traveled south at 60 km/h. How many hours had the first truck traveled when they were 150 km apart?
Solution
Let T = time
Time | Rate | Distance | |
North bound | T | 80 | 80T |
South bound | T – 1 | 60 | 60(T – 1) |
Equation
The two distances equal 150
80T + 60(T – 1) = 150
80T + 60T – 10 = 150
140T -10 = 150
140T = 140
T = 1
Answer
After 1 hour they are 150 miles apart
- Steve Smith left home on his bicycle at 8:00 A.M., traveling at 18 km h. At 10:00 A.M., Steves brother set out after him on a motorcycle, following the same route. The motorcycle traveled at 54 km/h. How long had Steve traveled when his brother overtook him?
Solution
Let T = time
Time | Rate | Distance | |
Bicycle | T | 18 | 18T |
Motorcycle | T – 2 | 54 | 54T |
Equation
The two distances are equal.
18T = 54(T – 2)
18T = 54T -108
-36 = -108
T = 3
Answer
After 3 hours the motorcycle overtakes the bicycle.
- Dr. Pepper left Oakville at 9:00 A.M. and drove to Central City at 60 km/h. H. Salt left Oakville at 11:00 A.M. and traveled the same route to Central City. If both men arrived in Central City at 4:00 P.M., at what rate did H. Salt travel?
Solution
Let R = rate
Time | Rate | Distance | |
Dr. Pepper | 7 hours (9 to 4) | 60 | 7(60) |
H. Salt | 5 hours (11 to 4) | R | 5R |
Equation
The two distances are the same
7(60) = 5R
420 = 5R
R = 84
Answer
- Salt traveled at 84 KM/h
- Two jets are traveling toward each other and are 4000 km apart. The rate of one jet is 100 km h faster than the rate of the other. If the jets pass each other after 2.5 hours, what is the rate of the faster jet?
Solution
Let T = time
Time | Rate | Distance | |
Jet 1 | 2.5 | R | 2.5R |
Jet 2 | 2.5 | R + 100 | 2.5(R + 100) |
Equation
The two distances equal 4000
2.5R + 2.5(R + 100) = 4000
2.5R + 2.5R + 250) = 4000
5R + 250 = 4000
5R = 3750
R = 750
Answer
The rate of the slower jet is 750.
- Ms. Driva Reck drove from her home to a service station at 48 km h. She returned home by bicycle at 16 km h. The entire trip took 4 hours. How far was the service station from Ms. Recks home?
Solution
Let T = time
Time | Rate | Distance | |
Drove | T – 4 | 48 | 48(T – 4) |
Bicycle | T | 16 | 16T |
Equation
The two distances are equal.
16T = 48(T – 4)
16T = 48T – 192
-32T = -192
T = 6
Answer
The service station is 6(16) or 96 km/h
- A plane on a search mission flew east from an airport, turned, and flew west back to the airport. The plane cruised at 300 km/h when flying east, and 400 km/h when flying west. The plane was in the air for 7 hours. How far from the airport did the plane travel?
Solution
Let T = time
Time | Rate | Distance | |
East | T | 300 | 300T |
West | 7 – T | 400 | 400(7 – T) |
Equation
The two distances are equal.
300T = 400(7 – T)
300T = 2800 – 400T
700T = 2800
T = 4
Answer
They traveled 4(300) km or 1200 km from the airport.
- A motorboat can travel upstream on a river at 18 km/h and downstream at 30 km/h. How far upstream can the boat travel if it leaves at 8:00 A.M. and must return by noon?
Solution
Let R = rate
Time | Rate | Distance | |
Upstream | T | 18 | 18T |
Downstream | 4 – T | 30 | 30(4 – T) |
Equation
The two distances are equal.
18T = 30(4 – T)
18T = 120 – 30T
48T = 120
T = 2.5
Answer
The boat travels 2.5(18) or 45 km.
- A boat travels 60 km upstream (against the current) in 5 hours. The boat travels the same distance downstream in 3 hours. What is the rate of the boat in still water? What is the rate of the current? (System of equations Pizzazz)
Solution
Time | Rate | Distance | |
Upstream | 5 | S – R | 60 |
Downstream | 3 | S + R | 60 |
Equation
This is a system of equations problem.
Set up 2 equations.
S = speed of stream
R = rate of the current
First equation 5(S – R) = 60
5S – 5R = 60
Second equation 3(R + S) = 60
3S + 3R = 60
**********************
15S – 15R = 180
15S + 15R = 300
Add the 2 equations
30S = 480
S = 16 This is the speed (rate) of the stream.
Put 16 in for S in one of the Original equations (5(S – R) = 60)
5(16) – R = 60
80 – 5R = 60
-5R = -20
R = 4
Answer
The rate of the stream is 16 and the rate of the current is 4.
- When a plane flies into the wind, it can travel 3000 km in 6 hours. When it flies with the wind, it can travel the same distance in 5 hours. Find the rate of the plane in still air and the rate of the wind. (System of equations Pizzazz)
Time | Rate | Distance | |
With the wind | 5 | P + W | 3000 |
Against the wind | 6 | P – W | 3000 |
Equation
This is a system of equations problem.
Set up 2 equations.
P = speed of plane
W = rate of the wind
First equation 5(P + W) = 3000
5P + 5W = 3000
Second equation 6(P – W) = 3000
6P – 6W = 3000
**********************
30P + 30W = 18000
30P – 30W = 15000
Add the 2 equations
60P = 33000
P = 550 This is the speed (rate) of the plane
Put 550 in for P in one of the Original equations 5(P + W) = 3000
5(550) + 5W = 3000
2750 + 5W = 3000
5W = 250
W=50
Answer
The speed of the plane is 550 and the speed of the wind is 50.
- When Lucy swims with the current, she swims 18km in 2 hours. Against the current, she can swim only 14 km in the same time. How fast can Lucy swim in still water? What is the rate of the current? (System of equations Pizzazz)
Time | Rate | Distance | |
With the current | 2 | L + C | 18 |
Against the current | 2 | L – C | 14 |
Equation
This is a system of equations problem.
Set up 2 equations.
L = speed of Lucy in still water
C = speed of the current
First equation 2(L + C) = 18
2L + 2C = 18
Second equation 2(L – C) = 14
2L – 2C = 14
**********************
2L + 2C = 18
2L – 2C = 14
Add the 2 equations
4L = 32
L = 8 This is Lucy’s speed in still water
Put 8 in for L in one of the Original equations 2(L + C) = 18
2(L + C) = 18
2(8 + C) = 18
16 + 2C = 18
2C = 2
C = 1 This is the speed of the current
Answer
Lucy’s speed in still water is 8 and the sped of the current is 1.
- With the wind, a jet can fly 2500 km in 2 h 30 mm. Against the wind, it can fly only 2000 km in the same time. Find the rate of the jet in still air and the rate of the wind. (System of equations Pizzazz)
Time | Rate | Distance | |
With the wind | 2.5 | J + W | 2500 |
Against the wind | 2.5 | J – W | 2000 |
Equation
This is a system of equations problem.
Set up 2 equations.
J = speed of the Jet in still air.
W = speed of the wind.
First equation 2.5(J + W) = 2500
2.5J + 2.5W = 2500
Second equation 2.5(J – W) = 2000
2.5J – 2.5W = 2000
**********************
2.5J + 2.5W = 2500
2.5J – 2.5W = 2000
Add the 2 equations
5J = 4500
J = 900 This is speed of the Jet.
Put 900 in for J in one of the Original equations 2.5(J + W) = 2500
2.5(J + W) = 2500
2.5(900 + W) = 2500
2250 + 2.5W = 2500
2.5W = 250
W = 100 This is the speed of the wind.
Answer
The speed of the Jet is 900 km/h and the wind speed is 100 km/h
- On an upstream trip, a canoe travels 40 km in 5 hours. Downstream, it travels the same distance in half the time. What is the rate of the canoe in still water and the rate of the current? (System of equations Pizzazz)
Time | Rate | Distance | |
With the wind | 2.5 | J + W | 2500 |
Against the wind | 2.5 | J – W | 2000 |
Equation
This is a system of equations problem.
Set up 2 equations.
J = speed of the Jet in still air.
W = speed of the wind.
First equation 2.5(J + W) = 2500
2.5J + 2.5W = 2500
Second equation 2.5(J – W) = 2000
2.5J – 2.5W = 2000
**********************
2.5J + 2.5W = 2500
2.5J – 2.5W = 2000
Add the 2 equations
5J = 4500
J = 900 This is speed of the Jet.
Put 900 in for J in one of the Original equations 2.5(J + W) = 2500
2.5(J + W) = 2500
2.5(900 + W) = 2500
2250 + 2.5W = 2500
2.5W = 250
W = 100 This is the speed of the wind.
Answer
The speed of the Jet is 900 km/h and the wind speed is 100 km/h
- A duck can fly 2400m in 10 minutes with the wind. Against the wind, it can fly only two thirds of this distance in 10 mm. How fast could the duck fly in still air? What is the rate of the wind? (System of equations Pizzazz)
Time | Rate | Distance | |
With the wind | 10 minutes 1/6 | D + W | 2400 |
Against the wind | 10 minutes 1/6 | D – W | 1600 |
Equation
This is a system of equations problem.
Set up 2 equations.
D = speed of the Duck in still air.
W = speed of the wind.
First equation 10(D + W) = 2400
10D + 10W = 2500
Second equation 10(D – W) = 1600
10D – 10W = 1600
**********************
10D + 10W = 2400
10D – 10W = 1600
Add the 2 equations
20D = 4000
D = 200 This is speed of the Duck.
Put 200 in for D in one of the Original equations 10(D + W) = 2400
10(D + W) = 2400
10(200 + W) = 2400
2000 + 10W = 2400
10W = 400
W = 40 This is the speed of the wind.
Answer
The speed of the Jet is 900 km/h and the wind speed is 100 km/h
- With the wind, a plane flew 1400 km in 4 hours. On the return trip, the pilot was forced to land after 1 h 30 mm, having traveled only 450 km. Find the rate of the plane in still air and the rate of the wind. (System of equations Pizzazz)
Time | Rate | Distance | |
With the wind | 4 | P + W | 1400 |
Against the wind | 1.5 | P – W | 450 |
Equation
This is a system of equations problem.
Set up 2 equations.
P = speed of the Plane in still air.
W = speed of the wind.
First equation 4(P + W) = 1400
4P + 4W = 1400
Second equation 1.5(P – W) = 450
1.5P – 1.5W = 450
**********************
4P + 4W = 1400
3(4P + 4W = 1400)
12P + 12W = 4200
******************
1.5P – 1.5W = 450
8(1.5P – 1.5W = 450)
12P – 12W = 3600
******************
Add the 2 equations
12P + 12W = 4200
12P – 12W = 3600
24P = 7800
P= 325 This is the speed of the plane.
Put 325 in for P in one of the Original equations 4(P + W) = 1400
4(P + W) = 1400
4(325 + W) = 1400
1300 + 4W = 1400
4W = 100
W = 25 This is the speed of the wind.
Answer
The speed of the Jet is 900 km/h and the wind speed is 100 km/h
- A salmon swims 100 m in 8 min downstream. Upstream, it would take the fish 20 min to swim the same distance. What is the rate of the salmon in still water? What is the rate of the current? (System of equations Pizzazz)
Time | Rate | Distance | |
Downstream | 8 minutes | S + C | 100 |
Upstream | 20 minutes | S – C | 100 |
Equation
This is a system of equations problem.
Set up 2 equations.
S = speed of the salmon.
C = speed of the current
First equation 8(S + C) = 100
8S + 8C = 100
Second equation 20(S – C) = 100
20S – 20C = 100
**********************
8S + 8C = 100
5(8S + 8C = 100)
40S + 40C = 500
******************
20S – 20C = 100
2(20S – 20C = 100)
40S – 40C = 200
******************
Add the 2 equations
40S + 40C = 500
40S – 40C = 200
80S = 700
S = 8.75 This is the speed of the salmon.
Put 8.75 in for S in one of the Original equations 8(S + C) = 100
8(8.75 + C) = 100
70 + 8C = 100
8C = 30
C = 3.75
Answer
The speed of the Salmon is 8.75 and the speed of the current is 3.75.