Rate Problems

Relative Motion Rate

An object travels relatively faster when it is moving along with an outside force than when it is traveling under its own power. onversely, an object will move relatively slower when it is moving against an outside force than when it is moving under its own power.

Mistake 7

Did Sylvia walk faster than 4 mph while walking the 2 miles, at a constant rate, from her home to the park? The questions is asking if Sylvia's rate/speed is greater than 4mph 1) It took Sylvia longer than 45 minutes to walk from her home to the park. Because rate is inversely proportional to time, and it took longer than 45 minutes, Sylvia's rate will be smaller than 2miles/45 minutes Converting 2 miles/45 minutes to mph, you obtain that Slyvina's rate is less than 8/3mph This is sufficient because it is less than 4mph 2)It took Sylvia longer than 120 minutes to walk from her home to the park. Because it took Sylvia longer than 2h to complete 2 miles, or 1h to complete 1m, her rate is smaller than 1mph which is less than 4mph and thus it is sufficient

Time

Distance / Rate

Rate

Distance / Time

To remember

From point A, Tom begins running at a constant rate of 8 miles per hour.Shandra starts from the same location a half-hour later and runs at a constant rate of 12 miles per hour. If Shandra runs on the same path as Tom, how long will it take for Shandra to catch up to Tom? Because Shandra starts 30 minutes after, Tom has 30 minutes of advantage. In order to show this in the equation, Tom's Time will be T + 0.5 and Shandra's time will be T

Diverging Rate Problems

In a diverging rate question, if two objects start at the same place and move in opposite directions, then the sum of the distances that two objects travel must equal the total distance traveled between the two objects. Also, when two objects start at the same point and move away from each other, the distance they cover is a function of the sum of their speeds.

Round trip problem

In a round-trip problem, the distance an object travels from the starting point to the destination equals the distance the object travels back from the destination to the starting point In a round-trip problem when only the total time traveled is provided, consider letting the time to a destination equal some variable t and the time back equal (total trip time - t)

Converging Rate Questions

It is important to note that when two objects converge (meet), the total distance will be equal to the sum of the individual distances each object travels.

Mistake 1

Melissa walked from home to school at an average rate of 4 mph. She realized when she got to school that she had forgotten her math book, so she ran home at an average rate of 10 mph to get it. What was her average speed for this round trip? My mistake was that I did a simple average to get her average speed, however, since we are dealing with rations, that approach is wrong Average Rate = Total Distance / Total Time Total Distance = Distance 1 + Distance 2 Total Time = (Distance 1 / Rate 1) + (Distance 2 * Rate 2) Because we don't know the distance, but we know that D1 is equal to D2 we can assume both to be x Total Distance = x Total time = x / 4 + x /10 If you work on the equation, the final result is 5.6mph

Mistake 8

Sorhab and Ramin both jog the same 3-mile loop.If x is the time it takes Ramin to jog the loop, and y is the time it takes Sorhab to jog the loop, is x + y > 60 minutes? 1) Sorhab jogs at a constant rate that is faster than 6 mph 2) Ramin runs twice as fast as Sorhab Here I need to remember that the fact that Sorhab runs faster than 6mph means that the final rate of Sorhab will be less than the actual value. Why? If Sorhab runs faster than 6 mph, his jogging time, which is inversely proportional to rate, will be less than 1/2 hour.

Difference of speeds

When one object is traveling faster than another object but we do not know the speed of either, we must account for the difference between their speeds

Objects leaving at the same time

When two objects leave at the same time of day and converge at a constant rate, they will have traveled for the same time at the instant that they meet.

Mistake 12

Each day John drives from home to school and then back home, taking the same route each way. If it takes him a total of 3 hours to drive to school and then back home, what is John's average rate for the round trip? 1) John travels from home to school at a rate of 30 mph 2) John travels from school to home at a rate of 45 mph Average Rate = Total Distance / Total time Average Rate = 2D / (T1 + T2) T1 + T2 = 3 hours R1: 30mph T1: D / R1 = D/30 Mistake#1 I didn't realize that I could get the the time (t) by using every the time rate distance equation

There are eight types of Rate-Time-Distance problems that are important for you to understand.

1. Elementary Rate Questions 2. Average Rate Questions 3. Converging Rate Questions 4. Diverging Rate Questions 5. Round Trip Rate Questions 6. Catch up Rate Questions 7. Relative Motion Questions 8. If / Then Questions

Mistake 11

A spacecraft can travel M mph when there is no wind. If the spacecraft can travel W miles when it travels with the wind in the same amount of time that it takes to travel X miles against the wind, what is the speed of the wind? We have information about the rate of the spacecraft with favourable and with negative wind and we also have information about the distance of the spacecraft with favourable and negative wind. The only unknown variable is T for both s= speed of wind Rate wind favourable: m + s Rate wind negative: m - s Distance wind favourable: W Distance wind negative: X We can get the time of both by T = D / R Since we are told that the time the spacecraft travels in both cases is equal, we can set the two expressions equal to each other and determine the speed of the wind

Mistake 6

An airplane can fly at a constant rate of 200 miles per hour without any wind. At this rate, it can travel 800 miles flying with a certain wind of constant speed in the same time that it can travel 600 miles against the same constant wind speed. What is the speed of the wind. Mistake #1 The fact that the mind is increasing the speed by 600 and decreasing the speed from 800 to 600 doesn't explain the speed of the wind. In order to calculate the speed, we will need to do the following: Speed of the plane with wind: 200 + w Speed of the plane with wind: 200 - w We know the distance for in both cases Distance with wind in favour: 800 Distance with wind against: 600 If we have R and D we can calculate T T = D / R T with wind in favour = 800/(200 + w) T with wind against = 600 / (200 - w) Because the prompt mention that the time for the case with wind in favour, and the time for the case with the wind against is the same, we can do equal each other 800/(200 + w) = 600 / (200 - w) W = 200/7mph

Mistake 26

Bob drove to a baseball game. For the first 150 miles of the trip, he drove at an average speed of 60 miles per hour, but for the last half of his trip, he drove at an average speed of X miles per hour. If Bob’s average speed for the entire drive to the game was 70 miles per hour and the distance to the game was 300 miles, what is the value of X? This is a simple problem in terms of data organization, however requires multiple calculations with large numbers that can lead to mistakes. This is only a reminder that you should have short-term focus

Mistake 25

Carl and Margaret left from the same location and traveled in opposite directions in a straight line. If Carl left at 4:00 pm and traveled at a constant rate of 50 mph and Margaret left at 5:30 pm and traveled at a constant rate of 60 mph, at what time were they 625 miles apart? Because they travel in opposite directions, we know the sum of their individual distances will be 625 at a specific time of the day Carl Rate: 50mph Margarete Rate: 60mph Carl Time: t + 3/2 (because drives for 1.5h more) Margarete Time: T Carl Distance: 50(t + 3/2)= 50t+150 Margarete Distance: 60t Because they travel in opposite directions, we know the sum of their individual distances will be 625 at a specific time of the day 50t+150 + 60t= 625 t = 5 Therefore, they will find each other at 17:30 + 5h = 22.30h

Mistake 9

Drivers A and B both drove the same 360-mile route. If a is the time, in hours, it took driver A to complete the route and b is the time, in hours, it took driver B to complete the route, is a + b > 4.5? 1) Driver A drove at a constant speed that was less than 72 mph 360miles / 72mph = 5 hours If driver A's speed was less than 72 mph, as statement one tells us, then it will take A longer than 5 hours, since A is driving more slowly. So a > 5, and since both a and b must be positive numbers, it cannot be true that a + b < 4.5. 2) Driver B drove at a constant speed that was less than half of driver A's constant speed. Not sufficient

Mistake 20

How long did it take for Mary to run a race? 1) Had she run at an average speed of 2 miles per hour greater than she actually did, she would have run the race in 5 hours and 30 minutes. 2) Had she run at an average rate that was 20 percent greater, she would have finished the race in 5 hours. We are interested in finding T, which is D/R 1) We can set up the standard equation 5.5 = D / (2 + r) 11 + 5.5r = D We don't have enough information 2) We can set up the standard equation 5 = D / 1.2r 6r = D 6 = D / r Because D/r is equivalent to time, 6 is the number of hours it took Mary to ruin the race. Statement 2 is sufficient

Mistake 15

Howard and Alyssa are f feet apart, on a straight city street. It takes Howard x minutes to travel f feet and Alyssa to travel f feet where x<y. If Howard and Alyssa begin walking towards each other at the same time, how many more feet will Howard have traveled than Alyssa when they pass each other? The first thing to note is that this is a complex problem and requires multiple steps 1) Recognize what information we have and what piece of information we need to solve for We have: Howard's rate: f/x Alyssa's rate: f/y We know that they start walking towards each other at the same time, thus the time variable is the same for both (t) We also see that we have a converging rate problem in which the total distance is f feet - "Howard and Alyssa are f feet apart" Thus Howard's distance + Alyssa's distance = f The answer that we need to find is the difference between Howard's distance and Alyssa's distance 2) Let's organize the data Howard's distance + Alyssa's distance = f (f / x) + (f / y) = f If we add time to the equation (remember time is the same for both) (ft / x) + (ft / y) = f We can eliminate the f's - this is a mistake I made, didn't realize I could get rid of the f's (t / x) + (t / y) = f We can solve for t and obtain (xy) / (x+y) Now in order to get the distance of both we can replace t by (xy) / (x+y) Howard's distance = (fy) / (x + y) Alyssa's distance = (fx) / (x + y) The difference of the distances, and the answer to the question is ((fy) / (x + y)) - ((fx) / (x + y)) = f ((y-x)/(x+y))

Mistake 29

If a certain car takes 2 minutes to travel through a tunnel, is the tunnel no more than 800 meters long? (1 kilometer = 1,000 meters) 1) The car travels at a constant speed that is less than 30 kilometers per hour. 2) The car travels at a constant speed that is greater than 25 kilometers per hour In this question, I must do a lot of work before immersing into the statements In order to conclude if a tunnel is no more than 0.8km, we should look at the rate of the car 2 minutes = 1 / 30 hours 800m = 0.8km Rate = Distance / Time Rate = 0.8 / (1 /30) = 24km If the rate of the car is equal greater than 24km, the distance of the tunnel will be greater than 800 meters if the rate of the car is lower than 24km, the distance of the tunnel will be less than 800m 1) The car travels at a constant speed that is less than 30 kilometers per hour. Not sufficient because the car could drive at a speed lower or greater than 24km 2) The car travels at a constant speed that is greater than 25 kilometers per hour Sufficient, because we know that the speed is greater than 24km, we know that the distance of the tunnel is greater than 800m

Mistake 13

If it takes a minute and a half for an elevator to go from the bottom floor to the top floor of a skyscraper and every floor has a height of 20 feet, does the skyscraper have more than 16 floors? 1) The elevator travels at a constant rate that is less than 5 feet per second 2) The elevator travels at a constant rate that is greater than 4 feet per second. Here I made many mistakes, this is why I will only include the right solution: The first thing we need to notice is the units of the prompt are different from the units on the statements. Thus, we need to change the units of the prompt from seconds to minutes In prompt 1, less than 5 fps is equivalent to less than 300 feets per minute In prompt 2, greater than 4 feet per second is equivalent to greater than 240 feets per minute Additionally, we know that it takes a minute and a half, or 3/2 minutes, to travel from the bottom to the top of the skyscraper. We want to determine whether the skyscraper has more than 16 floors, or is taller than 16 floors or 20 feet per floor = 320 feet. To make the calculations easier, suppose the skyscraper has exactly 16 floor. In order for the elevator to travel from the 1st to the 16th floor, it has to travel (16 - 1) = 15 floors up or 15 floors x 20 feet = 300 feet With this information, we can calculate the rate of the elevator and we can conclude that if it higher than 200, the building will have 16 floors There are more parts of the problem but they are hard to put into quizlet

Mistake 19

In the morning, Thomas drove from Manchester to Mayfield. In the evening, he drove the same road from Mayfield back to Manchester. Was Thomas' average speed for the entire trip less than 100 miles per hour? 1) In the morning, Thomas drove at an average speed of at least 10 miles per hour while traveling from Manchester to Mayfield. 2) In the evening, Thomas drove at an average speed of no more than 50 miles per hour while traveling from Mayfield to Manchester. 1) Thomas drove 10 miles per hour or faster. If Thomas drove at 10 miles per hour, from Manchester to Mayfield, his average speed would always be less than 100 miles per hour for the entire trip. Even if his speed on his way back was 300mph. However, if he drove at 100 miles per hour from Manchester to Mayfield and back, his average speed would be 100 miles per hour for the entire trip. Because of this contradiction, statement one alone is not sufficient. 2) Statement two alone is sufficient to answer the question. f Thomas averaged 50 miles per hour while driving from Mayfield to Manchester (one-half of the trip distance), his average speed for the entire trip can never be 100 miles per hour. Consider the most extreme (and impossible) situation in which he travels from Manchester to Mayfield in zero seconds. His total time would D/50 + 0 and his average speed would be Average rate = Total Distance / Total Time Average rate = (2D) / (D/50) + (0) = 100mph Notice that at zero seconds (an impossibility), his average speed will be 100 miles per hour. We know that it will take Thomas some positive amount of time to drive from Mayfield to Manchester. This means the denominator of the fraction will increase. Even if this increase is extremely small, the average speed would be less than 100 miles per hou

Mistake 16

It takes Chuck m minutes to drive n miles from town A to town B at a constant rate, while it takes Diane h hours to cycle the same distance at a constant rate. If Chuck leaves Town A for Town B at the same time Diane leaves Town B for Town A along the same road, how far will Chuck have driven when he passes Diane? My mistake was that I didn't regognize that this was a converging rates problem. In converging rate problems, the total distance will be equal to the sum of the individual distances each object travels. Thus the Chuck's distance + Diane's distance = total distance Chuck's distance + Diane's distance = n Chuck's distance: 60nt/m Diane's distance: nt/h (60nt/m) + (nt/h) = n If we eliminate the n's (60t/m) + (t/h) = 1 If we solve for t (hm)/(60h + m) If we replace T from Charle's distance formula 60nt/m with (hm)/(60h + m), we obtain the distance that Charle's final distance

Mistake 10

Joe and Mike sign up for a motorbike race. When the race begins, Joe rides with the rest of the riders at a constant rate of 24 miles per hour, but Mike has a hard time getting his motorbike started. He finally leaves 20 minutes after the start of the race. If Mike travels at a constant rate of 36 miles per hour, how long will it take Mike to catch up to Joe? Mistake #1 20 minutes is 1/3 of an hour because 1h has 60 minutes Comment #1 If Mike leaves 20 minutes after Joe, this means that Joe can ride the bike for 20 more minutes. Thus Joe's time => t + 20

Mistake 21

John walked from his house to the store at a constant rate of 4 miles per hour. He then walked home along the same path at a constant rate of 3 miles per hour. How many miles is the store from John's house 1) It took John one hour to walk from his house to the store 2) If the store were 60 miles from John's house, it would have taken John 15 hours to walk to the store We need to find out D From the stem we know the rate at which John walked to the store, but we don't have any information about T, which is necessary to calculate D 1) If it took 1h to walk from his house to the store, we have T, and thus we can calculate D = R x T = D = 4 x 1 = 4 2) In statement 2 we get a hypothetical distance (D) and a hypothetical time (T). With this information we can calculate R (D/R) = 60 / 15 = 4mph. This is the same information from the stem, and it doesn't help to calculate the actual distance. Not sufficient.

Mistake 17

Jung-Su drove r miles per hour to a car show 264 miles away. Had he driven 2 mph faster, he would have arrived at the show 24 minutes sooner. What is the value of r squared * 2r? The mistake that I made here is assume that I had information about time. We have no information about time We have two independent pieces of information 1) Case if actual rate 2) Case if faster rate Since we also have the distance for both cases (264), we can get the time for both cases (T=D/R) Time Actual Rate = 264/r Time Faster Rate = 264/r+2 Because Jung-Su would have needed 24 fewer minutes to travel to the car show if he had traveled at the faster rate, we can take Jung-Su's actual time, set it equal to the faster time plus 24 minutes, and then determine rate r Actual time = Faster time + 24 minutes (264/r) = 264/r+2 + (2/5) We use 2/5 because it is the equivalent of 24min in hours Doing this operation, you get that r squared * 2r = 1320

Mistake 23

Max's car has a different fuel efficiency (in mpg) when it is driven in the city than when it is driven on the highway. He remembers the fuel efficiency is 30 mpg when the car is driven on the highway, but he doesn't remember the fuel efficiency when it is driven in the city. If one day he drives 40 miles in the city and 100 miles on the highway and the car uses 6 gallons of gas, what is his car's fuel efficiency when it is driven in the city? Here I made several mistakes 1) Didn't notice the difference between a traditional miles per hour problem and this problem, miles per gallon. When we work on miles per gallon, we no longer have time as a function, we have gallons. See that mpg vs mph, the g replaces the h 2) I didn't understand what " the car uses 6 gallons of gas" really meant. This means that 6 = gallons used in highway + gallons used in city Taking both points into account, the solution is simple Highway R = 30mpg D = 100 G = Unknown G = (D / R) = 10/3 City R = R D = 40 G = Unknown G = ( D / R) = 40 / R Because we know that the sum of both gallons equals 6 (40 / R) + (10 / 3) = 6 R = 15

MIstake 27

On a cross-country trip, did the plane fly more than 2,400 miles? 1) The plane flew for more than 2 hours 2) During the first half of the plane's flying time, it averaged 1,000 mph; during the second half, it averaged 3,000 mph 1) Doesn't give us any information about the rate/speed of. the planes so we can't calculate the distance 2) Although we now have the rates, we don't know anything about the travel time. We only know that there are 2 halfs, meaning the plane travelled for 2 periods worth the same time. Combining both: We can set a case where the plane travels for 2h, which is an extreme case as we know that the plane travels for more than 2h. Half 1: Time= 1h R= 1000 D = 1000 Half 2: Time= 1h R= 3000 D = 3000 Total distance if the plane flies for 2h = 4000 Because we know that the plane travels for more than 2h, the total distance covered by the plane will be greater than 4000, which is in turn greater than 2400

Mistake 30

Plane A is flying in a clockwise circular path above Town X. Plane B is also flying in a clockwise circular path above Town X at a different altitude and rate from plane A. If the circumference of the path of Plane A is equal to the circumference of the path of Plane B, and at 5 p.m. both planes are directly above the same location over Town X, how many circular paths must the planes make so that both planes are again directly above the same location at the same time above Town X. We can understand the completion of the circumference as one unit of distance. At the moment, they find themselves in the same spot, and the question is asking us how many completion of circumferences will both planes need to do to be in the same spot. 1) The circumference of the circular path traveled by both planes is 4pi 2) Plane A travels at a constant rate of 1/8 of the length of the circular path per hour, and Plane B travels at a constant rate of 1/6 of the length of the circular path per hour 1) Not much information here. Since both planes go through the same circumference, we can't get any information about the rates 2) Since plane A travels 1/8 the length of the circumference per hour and plane B travels 1/6 the length of the circumference per hour, note that plane B is "gaining" on plane A at the rate of 1 / 6 - 1 / 8 = 1 / 24 of the length of the circumference per hour. In order for the planes to be aligned again, the number of these 1/24 increments must add to 1. At 1 Plane B will have covered 2x the distance of Plane A, but because they are flying on circumferences, and they are at different altitudes, they can be above each other. Since there are twenty-four of these increments in the whole number 1, the planes will be aligned again after 24 hours. This means plane A will have made 3 circular rotations (24/8) and plane B will have made 4 (24/6) circular rotations when they are aligned again.

Mistake 3

Question: The distance between Newark and Boston by train is 245 miles. Train A departs Newark at 12:00 pm, traveling to Boston at a constant speed of 50 miles per hour. Train B departs Boston at 12:30 pm and heads toward Newark at a constant speed of 60 miles per hour. If the trains are traveling on parallel tracks, at what time will they pass each other? Solution Mistake #1 Distance Train A + Distance Train B = Total Distance 50(t+0.5) + 60t = 240 T = 2 Mistake #2 If T= 2, this means that because train A leaves at T + 0.5, Train A will find Train B 2.5h after departure, 12 + 2.5 = 14.5h The same approach can be applied to Train B If If T= 2, and Train B leaves at 12:30, Train B will find Train A 2h after departure, 12:30 + 2 = 14.5

Mistake 18

Sally leaves her house at 8:00 a.m. and rides her bike to school at a constant rate. At 8:15 a.m., Sally's father realizes she has forgotten a book on the kitchen table and rides after her at a constant rate that is 5 mph greater than Sally's. If her father caught Sally at 8:45 a.m., how fast was he traveling? The main mistake that I made is that time was an incognita (T). Instead, they already give us the time Sally's time = 3/4 hours (8:45-8) Dad's time = 1/2 hours (8:45-8:15) We also have the rate: Sally's rate = r Dad's rate = r + 5 Given that we have the time, and the rate, and we know that the distance is the same (it's a catch up problem, so both drive the same distance), we can calculate the rates by equalling the distances equations Sally's distance = 3/4r Dad's distance = (1/2r) + (5/2) Dad's distance = Sally's distance 3/4r = (1/2r) + (5/2) r = 10 r + 5 = 15

Mistake 22

Sally traveled from Pittsburgh to Cleveland, driving X miles in Y hours. Ted drove from Columbus to Cleveland, driving a distance that was 50% greater than Sally's distance but in 1/3 hour less. What was Ted's average speed? The mistake that I made here was that instead of subtracting 1/3 hours from Tom's time, I added it to Sally's The solution is the following: Sally D = x T = Y Tom D = 1.5x T = Y - 1/3 R = D / T R = 1.5x / (Y - 1/3) R = 9x / (6Y - 2)

Average Distance

The average rate is calculated by dividing the total distance that an object travels by the total time that it takes the object to travel that distance. Average Distance = Total Time * Average Rate

Mistake 14

The distance between Town P and Town Q is 100 miles. A train starts from Town P and moves toward town Q at a constant rate. A bird starts at the same time from Town Q and flies straight toward the moving train at a constant rate. On reaching the train, the bird instantaneously turns back and returns straight to Town Q at a constant rate. The bird makes these trips from Q to the train and back to Town Q continuously until the train reaches Town Q, at which point the bird lands on top of the train and takes a well-deserved rest. What is the total distance the bird flies? 1) ) The bird flies at a constant rate of 18 miles per hour 2) The train travels at a constant rate of 10 miles per hour. We must determine how many miles the bird travelled between town P and Q 1) Not sufficient because we only have the rate of the bird, no info about the speed of the train 2) If the train travels at a constant rate of 10 miles per hour, it will take the train 10h to reach town Q ((T= D / R) = (100/10) However, we don't have any info about the rate of the bird 3) Combining both, because we know that the bird will fly for 10h and we know that he travels at a constant rate of 18mph, the bird will fly 180m in total

Catch-up and pass rate questions

The time that it will take a faster object to "catch up and pass" a slower object can be quickly determined by Time = Change in distance / Change in rate The change in distance is simply the extra distance that the faster object must travel over that of the slower object. The change in rate is simply (the rate of the faster object) - (the rate of the slower object).

Distance

Time x Rate

Mistake 24

Train A traveled from city X to city Y, and train B traveled from city Y to city X, each train traveling at its own constant speed. If both trains started their journeys at the same time and traveled toward each other on parallel tracks, how many hours did it take for train B to complete its journey from city Y to city X? 1) Train A's speed was 1/4 of train B's speed 2) Train A and train B passed each other 2 hours after they began their journeys From statement 1 we know that Train B travels at a rate 4x greater than train A. However, we don't have any information about distance or time From statement 2 we know that both trains travel for 2h until they meet. But we don't know anything about the distance between the cities and the rates Combining both - this is what I missed to see If trains meet 2h after the departure, the sum of their distances will be the distance from city Y to city X. If train B travels 4x faster than Train A, then while train A will have covered 1D, Train B will have covered 4D. Again, because the sum of their distances will be the distance from city Y to city X 1D + 4D = 5D 5D is the total distance from city X to City Y Train B has travelled 4D / 5D in 2h thus, Train B has covered 4/5ths of the way in 2h. Looking at the fraction we can see that every fifth is equivalent to 30 min, and because it needs to reach 5D, the total needed will be 2h + 30 min

Mistake 28

Train M leaves General Station traveling towards Mountain Station, and train Q leaves Mountain Station traveling towards General Station. If the two stations are 120 miles apart and the two trains leave at the same time and travel on parallel tracks, in how many hours will train M arrive at Mountain Station 1) Train M is traveling a constant rate that is 10 miles per hour faster than train Q's constant rate. 2) Trains M and Q pass each other in 1.5 hours. Given that they travel in opposite directions, whenever they cross, they will have travelled 120 miles, distance from one station to the other 1) We have information about the rates of the trains but we don't have information about the time 2) We know that both trains will travel for 1.5h before meeting each other at a distance of 120 miles. However, because we don't know the rates of the trains we won't know in how long they got there 3) Combining both Rate Q: R Rate M: R + 10 Time Q: 1.5h Time M: 1.5h Distance Q: 1.5R DistanceM: 1.5R + 15 Because we know that the sum of their distances is 120miles 1.5R + 1.5R + 15 = 120 R = 35 Thus train M's Rate = 35 + 15 = 50 and train M's Distance to cover = 120m If we have the distance and rate, we can get the time T = D / R T = 120 / 5 = 2hours and 2/3 hours = 2h and 40min

Mistake 4

Two boats, X and Y, leave at the same time from ports that are 100 miles apart and converge at a constant rate on parallel paths. If one boat is traveling at a constant rate that is 10 miles per hour faster than the constant rate of the other boat and if they pass each other in 2 hours, what is the sum of the rates of the boats? Mistake #1 The distance of every boat is not 100, the distance is R x T Distance Boat X = 2(r+10) Distance Boat Y = 2r And the combined distances of Boat X and Y is equal to 100 2(r+10) + 2r = 100

Mistake 5

Vladimir is driving to Sara's house at a constant rate, and Sara is driving to Vladimir's house at a constant rate. Vladimir leaves his house at the same time as Sara does, and they travel on the same road, heading toward each other in parallel lanes. In how many minutes will Vladimir arrive at Sara's house 1) Vladimir's rate is twice Sara's rate 2) Vladimir and Sara pass each other in 10 minutes Things to note: 1) If they leave the house at the same time, then the variable T is equivalent for both, thus, the time driven is the same 1) Vladimir's rate is twice Sara's rate We know that Vladimir is twice as fast as Sara but we don't have any info on the total distance driven 2) Vladimir and Sara pass each other in 10 minutes This means that both had driven by 10 minutes when they saw each other - T is equivalent. However, not enough information. 3) Combining both we get that Vladimir's distance is 20r and Sara's distance is 10r. Since Vladimir's travelled is 20r, and the total distance to Sara's house is 30r (10r + 20r), he travelled 10 minutes in 2/3r, we know that in 15 minutes he will be able to cover 3/3r of time, meaning that it will take him 15 minutes to get to Sara's house

Time zone changes

When solving a Rate-Time-Distance problem with time-zone changes, convert the time zone of the destination to the time zone of the origin to accurately compute the time traveled When conducting time zone change problems, make sure to adjust the time of the initial location to the final location, so you just need to conduct a simple operation

Rate problems: One object is relatively faster than the other

When the first object is x times as fast as the second object, let the rate of the second object be r and the rate of the first object be xr. If the rate of the first object is x% of the second object, let the rate of the first object be (x/100)r, and the rate of the second object r

Objects leave at different times

When two objects leave at different times and converge at a constant rate, the travel time of the object that leaves later can be represented by some variable t; the travel time of the object that leaves earlier can be represented by t + the difference between their departure times Remember, we always add time to the object leaving first because the total travel time will be greater for that object

Catch up problems

When two objects start from the same point and catch up to each other, both objects will have traveled the same distance when they meet.