TIME SPEED AND DISTANCE

Q1. Ram and Ravi are 100 km apart and started to walk towards each other at 10 am. Ram walked at the rate of 5 km/hr and Ravi at 10 km/hr. at what time will they meet?

   Ans.  Relative speed = 5 + 10 = 15 km/hr

            Time to walk 100 km = 100/15 = 6.66 hours, which is 6 hours and 40 min(approx)

            As they started at 10 am, therefore they meet after
10 + 6 hours and 40 min = 4:40 pm

    Q2.  Hari runs after Sam who is 200 m ahead of him. If speed of Hari is 15 km/hr and that of Sam is 10 km/hr, what will be the distance covered by Hari when he catches Sam?

   Ans.  Relative speed = 15 × 10 =  5 km/hr 

            Distance to be covered = 200 m = .2 km

            Time taken = .2/5= 0.04 hrs

            Distance covered = 15 × 0.04 = 0.6 km = 600 meters

    Q3.  Two trains, 200 and 160 meters long take a minute to cross each other while traveling in the same direction and take only 10 seconds when they cross in opposite directions. What are the speeds at which the trains are traveling?

   Ans.  Distance covered (sum of lengths of the train)

                            = 200 + 160 = 360 meters.

            Let Train 1 be traveling at X m/sec and Train 2 be traveling at Y m/sec. (considering X > Y)

            Now while traveling in same direction Time =  (L₁ + L₂)/(X – Y)

            Therefore 60 =  360/(X – Y), 60X – 60Y = 360

            Now while traveling in opposite direction Time

                                    = (L₁ + L₂)/(X + Y)

            Therefore 10 = =  360/(X + Y), 10X +10Y = 360

            Solving the two equations       X = 21 m/sec

                                                            Y = 15 m/sec

    Q4.  A train traveling at 72 km/hr crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

   Ans.  Students may recall from the theory of the chapter, When the train crosses a man standing on a platform, the distance covered by the train is equal to the length of the train and when the same train crosses a platform, the distance covered by the train is equal to the length of the train plus the length of the platform.

            Train takes 18 seconds to cross a man, which is basically to cover length of the train

            Train takes 30 seconds to cross the platform, which is to cover length of the train and length of the platform. Therefore the extra 12 seconds are to cover the length of the platform. Therefore Length of platform

            = (72 x 12)/3600 = 0.24 km = 240 meter.

    Q5.  Two trains starting from the same station and traveling in opposite directions are 228 km apart in 3 hours. Had they been traveling in same direction they would have been 33 km apart in the same time. What are their speeds?

   Ans.  They are 228 km apart in 3 hrs traveling in opposite direction

            In one hour they will be 228/3 = 76 km = Sum of speeds of two trains        

            Assume speeds to be X km/hr and Y km/hr

            Therefore X + Y = 76 km/hr

            In the other case X – Y = 33/3 = 11 km/hr

            Solving X = 43.5 km/hr, Y = 32.5 Km/hr

    Q6.  Two trains traveling in the opposite directions pass each other in 8 seconds. But when they travel in same direction at the same rates, the man in the faster train passes the other in 30 seconds. Find the lengths of the trains when their speeds are 45 km/hr and 35 km/hr.

   Ans.  When trains travel in opposite direction,

                        Time    = (L₁ + L₂)/(X + Y)

            Therefore,       8/3600 = (L₁ + L₂)/(45 + 35)

            Lengths of train, L1 + L2         = 0.17778 km

                                                 = 177.78 meter

            When trains travel in same direction, the man in the faster train passes the other in 30 seconds, the man only passes the length of the smaller train.

            Therefore distance traveled by trains in 30 seconds = length of the smaller train

            Relative speed = 45 – 35 = 10 km/hr

            Distance traveled        =  30/3600× 10

                                     = 0.08334 km = 83.34 meter

            Therefore length of longer train

                                     = 177.78 – 83.34

                                     = 94.44 meters

    Q7.  Two trains 150 miles distant travel towards each other along the same track, the first train at 60 km/hr, the second at 90 km/hr. A fly buzzes back and forth between the two trains until they collide. If the fly’s speed is 110km/hr, how far will it travel and how many rounds will it take?

   Ans.  Since the trains start 150 km apart and have a relative speed of 90 + 60 = 150 km/hr, they will meet in exactly 1 hour. The bird is flying at a speed of 110 km/hr, so in 1 hour it flies 110 km. The interesting part remains the number of rounds, which actually is an interesting physics problem with a infinite series being formed and the answer is infinity.

    Q8.  Two trains A and B each of length 100m travel in opposite directions in parallel tracks.  The speeds are 20m/s and 30m/s respectively.  A boy sitting in the front end of train A throws a ball to a boy sitting in the front end of train B when they are at the closest distance.  The speed of the ball is 2m/s.  The ball, instead of reaching the boy, hits the rear end of the train.  Find the distance between the parallel tracks.

   Ans.  The boy throws the ball when trains are at closest distance, which is when front ends of trains meet; therefore boy throws perpendicular to the train, in line with the distance between the tracks.

            Suppose from the time ball leaves boy’s hands to hitting the train’s rear is T

            In this time T, Train A has traveled 20T         

            Train B has traveled 30T

            Since ball hits the rear of train B, with length of trains being 30T + 20 T = 100, therefore T = 2 seconds

            Since the ball travels at 2 m/s, and ball hitting the other train means that ball has traveled the distance between two tracks, assuming distance between tracks is D

            Therefore 2T = D, D = 4 meters

    Q9.  Ram and Mohan start at the same time from A to B to go to B and A, a distance of 42 km at the rates of 4 km/hr and 3 km/hr. They meet at X, then go to B and A and return immediately and meet again at Y. find the distance XY

   Ans.  First Ram and Mohan meet at X, Distance = 42 km

            First Ram and Mohan meet at Y,

              Distance = 42 × 3 = 126 km

              Speed of Ram = 4 km/hr,

              Speed of Mohan = 3 km/hr

            Ram travels AX distance and Mohan BX

            Relative speed = 4 + 3 = 7 km/hr

            Time when they meet first = 42/7 = 6 hrs

            In 6 hrs Ram travels 6 × 4 = 24 km = AX

            Therefore Mohan travels = 42 – 24 = 18 = BX

            Time when they meet second = 126/7 = 18 hrs

            In 18 hrs Ram travels 18 × 4  = 72 km

                                                      = AX + XB + BY

            Since AX + XB = 42, BY = 30 and AY = 12

            Since AX = 24, and AY = 12. XY = 12 km

  Q10.  There are two boats that start out on opposite sides of a river at the same time.  Each one is heading across the river to the other side.  They each go a constant speed throughout the entire problem (so ignore having to slow down to turn around, and ignore current, etc.), but they are not necessarily the same speed as each other.  When each boat reaches its opposite bank, it immediately turns around and heads back to where it started. The boats thus pass each other twice. The first time they pass, they are 700 yards from one of the banks of the river.  The second time they pass, they have each turned around after reaching their respective opposite shores and have started back toward where they each began.  When they pass the second time, they are 300 yards from the other bank of the river.  How wide is the river? 

   Ans.  Assume total distance = X, and speeds of boats being S1 and S2 when the two boats meet for the first time,

            Distance travel by boat I, D1 = 700 yards

            Distance travel by boat II, D2 = X – 700 yards

            Here Time is the same, therefore

                               T = D₁/S₁ and T = D₂/S₂

            Therefore D₁/S₁           =  D₂/S₂

            Also               D₁/D₂     =  S₁/S₂ = 700/(700 – x)

            Now, Boat I then continues on to the bank, which is (X – 700) yards away, and then it turns around and goes back 300 yards. After the first meeting, boat I travels (X – 700) + 300, which is (X – 400) yards. After the first meeting Boat II travels 700 yards to the bank and then turns around and travels back (X – 300) yards, which is (X + 400) yards, where it then meets Boat I again.

            When the two boats meet for the second time after first meeting,

            Distance travel by boat I, D1 = X – 400 yards

            Distance travel by boat II, D2 = X + 400 yards

            Here Time is the same, therefore

                               T = D₁/S₁  and T = D₂/S₂

            Therefore D₁/S₁          =  D₂/S₂

            Also                 D₁/D₂   =  S₁/S₂ = (x – 400)/(x + 400)

            The first and second time ratios are the same, therefore

                                    700/(x – 700) = (x – 400)/(x + 400)

            Solving X = 1800 meters

  Q11.  Ravi can swim with the stream at the rate of 10 km/hr and 5 km/hr against the stream; find his speed in still water.

   Ans.  Upstream speed = 10 km/hr, Downstream speed is 5 km/hr

            Using direct formula 1/2 (10 + 5) = 7.5 km/hr

  Q12.  Hari swims 20 km downstream a river in 5 hours and returns in 10 hours. What is his speed and speed of the stream?

   Ans.  Speed downstream

                            = 20/5  = 4 km/hr

                            = Speed of Hari + Speed of stream

            Speed downstream

                            = 20/10 = 2 km/hr

                            = Speed of Hari – Speed of stream

            From the two equations speed of Hari = 3 km/hr

            Therefore speed of stream = 4 – 3 = 1 km/hr

  Q13.  A boat travels from point A to point B upstream and returns from point B to point A downstream. If the round trip takes the boat 5 hours and the distance between point A and point B is 120 km and the speed of the stream is 10 km/hr, what is the speed of the boat?

   Ans.  Total distance = 120 × 2 = 240, Total time = 5hrs

            Average speed = 240/5 = 48 km/hr

            Assuming speed of boat = X

            Speed downstream

                            = Speed of Boat + Speed of stream

                            = X + 10

            Speed downstream

                            = Speed of Boat – Speed of stream

                            = X – 10

            Since average speed = 48, and using formula of average speed

            48 = 2(X + 10 )(X – 10)/(X + 10) + (X – 10)

            Solving for X, X = 50 km/hr

  Q14.  Mohan can beat Ravi by 50 m in a 1700 m race; Ravi can beat Shyam by 20 m in a 1700 m race. If Mohan and Shyam run 1700 m, by how much will Mohan win?

   Ans.  The students should be able to do this faster using simple logic:

            Mohan can beat Ravi by 50 m, Mohan travels 1700, when Ravi travels 1650

            Ravi can beat Shyam by 20m, Ravi travels 1700, when Shyam travels 1680

            Therefore when Ravi travels 1 Shyam travels

            Therefore when Ravi travels 1650 Shyam travels 1680/1700

            1680/1700  × 1650 = 1630.5

            Mohan beats Shyam by 1700 – 1630.5 = 69.5 meters

  Q15.  In a kilometer race, A can give B a start of 100 m or 15 seconds. How long does A take to complete the race?

   Ans.  In a 1000 meter race A gives B a start of 100 m or 15 seconds. This means that B takes 15 seconds to run 100 m. Therefore, B will take 150 (1000/100 x 15) seconds to run the stretch of 1000 meters. As A takes 15 seconds less than B, he will take 135 seconds to run the 1000 m.

  Q16.  Ravi can give anand a start of 20 seconds in a kilometer race. Ravi can give ranjan a start of 200 meters in the same kilometer race. And anand can give ranjan a start of 20 seconds in the same kilometer race. How long does Ravi take to run the kilometer?

   Ans.  Ravi can give anand a start of 20 seconds in a km race.

            Let anand takes X seconds to run a km, then Ravi will take X – 20 seconds

            Now, Anand can give ranjan a start of 20 seconds in a km race.

            Let ranjan takes Y seconds to run a kilometer, then anand will takeY – 20 seconds

            From the two X = Y – 20

            Therefore Ravi can give ranjan a start of 40 seconds

            Now, Ravi can give ranjan a start 200 meters or 40 seconds in a km race.

            This essentially means that ranjan runs 200 meters in 40 seconds.

            Therefore, ranjan will take 200 seconds to run a km.

            Then ravi will take 200 – 40 = 160 seconds

  Q17.  Rohit and Ravi walk around a circular path of circumference 1000 meters. Rohit walks at speed of 100 m/min and Ravi at a speed of 50 m/min. if they start at the same point and walk in the same direction, when will they be together again? When will they meet at the starting point?

   Ans.  Rohit gains 50 meters in a minute over Ravi, so it will gain 1000 meters in 1000/50 = 20 minutes, so they will meet after 20 minutes

            Time taken by Rohit to complete one round

                        = 1000/100 = 10 min

            Time taken by Ravi to complete one round

                        = 1000/50 = 20 min

            They meet again at starting point, LCM of 10 and 20 = 20 minutes

  Q18.  Three cyclists Raman, Mohan and Nitin ride around a circular course 85 km around at the rate of 8, 12 and 20 km an hour. Raman and Mohan ride in the same direction and Nitin in the opposite direction. In how many hours will they meet again?

   Ans.  Mohan will meet Raman in every

85/(12 – 8) = 85/ 4 hrs

            Mohan and Nitin will meet in every

            85/(12 + 20 )   = 85/32 hrs

            Raman and Nitin will meet in every    85/(8 + 20)

                            = 85/28 hrs

            They all will meet at the LCM of the three 85/4, 85/32 and 85/28

            LCM is 85/4 hrs

  Q19.  At what time between 3 and 4 will the hands of a watch coincide? 

   Ans.  At 3, the hour hand and the minute hand are at right angle and are 15 minute away, the minute hand gains 1 in 12/11 minute, it will gain 15 in 12/11 × 15 = 16.3 minutes.

            Therefore hands will coincide 16.3 min past 3.

  Q20.  A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time that the clock will show exactly 14 days from the time it was set right?

   Ans.  The clock loses 1% time during the first week. In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 × 24 = 168 hours in a week. Therefore time lost from noon to noon = 1/100 × 168 = 1.68 hrs (behind)

            The clock gains 2% time during the first week. Therefore time gained from noon-first week to noon-second week = 2/100 × 168 = 3.36 hrs (Ahead)

            Net change      = 3.36 – 1.68

                             = 1.68 (ahead of 12 noon)

            1.68 hours = 1 hour + 40 minutes + 48 seconds. Therefore time is 1:40:48 P.M.

Q21. A truck travels 15 mph for the first half of the distance of a trip. The driver wants to average 30 mph for the whole trip. How fast must he travel for the second half of the trip?

Ans21. This is more of a puzzle, assume any distance for the trip. If you said 30 miles, then the first half took 1 hour. But to average 30 mph for the total trip would require an hour. Therefore it has no solution and it is impossible!

Q22.   A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship. Find the distance when they meet from starting point.

Ans22. Let’s say the ship is travelling x miles/hr and plane took t hours to meet the ship

                        Then, xt is the distance ship traveled after plane started, Let 10xt is the distance plane traveled. So we have, xt + 180 = 10xt

                        Solving xt = 20, therefore distance traveled by the ship = 180 + 20 = 200 miles

Q23.   A can complete a piece of work in 4 days, B takes double the time taken by A, C takes double that of B, and D takes double that of C, to complete the same task. They are paired in-groups of two each, one pair takes 2/3rds the time needed by the second pair to complete the work. Which is the first pair? (CAT 2001)

Ans23. A takes 4 days, B takes 8, C takes 16, D takes 32, In one day A does 1/4th, B does 1/8th, C does 1/16th, D does  1/32th of work.

                        Now A and D work in a day = ¼ + 1/32 = 9/32 , so finish work in 32/9 days

                        A and C work in a day  ¼ + 1/16 = 5/16 , so finish work in  16/5 days

                        A and B work in a day  ¼ + 1/8 = 3/8, so finish work in 8/3 days

                        B and C work in a day 1/8 + 1/16 = 3/16, so finish work in 16/3 days

                        We can already see a relationship between pair of A and D, and B and C,

                        16/3 X 2/3 = 32/9, so these are the pairs and B and C finish work faster.

Q24.   Three small pumps and a large pump are filling a tank. Each small pump works at 2/3rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of time, it would have taken the large pump alone.(CAT 2001)

Ans24. Let the rate of large pump is 1 liter/hr, therefore the rate of small pump is 2/3 liter/hr. Assume there is a 3 liter tank, all four will fill it in 1 hour. Alone large pump will take 3 hours, ratio = 1/3

Q25.   Six technicians working at the same rate complete work of one server in 10 hours. If they start at 11 am and one additional technician per hour been added, beginning at 5 PM, at what time the server will be complete? (CAT 2002)

Ans25. The work requirement is 10 × 6 = 60 Man hours

                        Man hours till 5 pm = 6 × 6 = 36, Man hours required more = 24. Next hour will contribute 7, next will 8, and next 9, making for the entire 24(9 + 8 + 7), therefore work will finish by 8:00 PM.

Q26. Two boats travelling at 5 and 10 km/hr respectively, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in km) 1 minute before they collide?   (CAT 2004)

Ans26. Relative speed = 15(10 + 5) km/hr, time to travel 20 km = 20/15 = 4/3 hours = 80 minutes

                        So in 80 minutes they will travel 20 km

                        In 1 minute they will travel 20/80 = 4 km

                        In 79 minutes they will travel 79/4 = 19.75 km

                        Since total is 20, they were apart by 20 – 19.75 = 0.25 km

Q27. A sprinter starts running on a circular path of radius r meters. Her average speed (in meter/min) is Pr, during the first 30 seconds, πr/2, during next one minute, πr/4, during next two minutes, πr/8 , during next four minutes and so on. What is the ratio of the time taken for the nth round trip that for the previous round?     (CAT 2004)

Ans27. The circumference of circle is 2Πr, so distance traveled in first 30 seconds in , in next one minute is again Πr/2(Πr/2 for one minute), in the next two minutes is also Πr/2(Πr/4 for two minutes), and so on as per question. Now for the 1st round the time is 71/2 minutes, and basically four intervals, for 2nd round the intervals will be 16 times the intervals as it will be Πr/2(Πr/16, in next 8 minutes), so the ratio will be 16.

Q28.   A company has a job to prepare certain number of cans and there are three machines A, B and C for this job. A can complete the job in three days. B can complete the job in four day and C can complete the job in six days. How many days the company will take to complete the job if all the machines are used simultaneously? (CAT 1998)

Ans28. A can do the job in 3 days, which means 1/3rd job in a day

B can do the job in 4 days, which means 1/4^th  job in a day.

C can do the job in 6 days, which means 1/6th job in a day.

In one day they all can do 1/3 + ¼ + 1/6 = 3/4th of job, so they will do the job in 4/3 days.

Q29.   Two full tanks, one shaped like a cylinder and other like a cone contains jet fuel. The cylindrical tank holds 500 liters more than the conical tank. After 200 liters of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many liters of fuel did the cylindrical tank had when it was full? (CAT 2000)

Ans29. Let conical tank hold X liter of fuel, then cylindrical holds X + 500, also after 200 liters from each has been used (X + 300) = 2(X – 200)

                         X = 700 L, and cylindrical tank has 1200 liters

Q30.   A and B walk up an escalator. The escalator moves at a constant speed. A takes 3 steps for every 2 of B’s steps. A gets to the top of escalator in 25 steps, while B takes 20 steps to reach the top. If escalator was turned off, how many steps would they have to take to walk up? (CAT 2001)

Ans30. A is obviously faster than B, let A reaches the top in 1 hour, in 25 steps as given, and let escalator contributed X steps (in one hour), so total steps, T = X + 25. If A takes 25 steps in an hour, B takes 20 steps in 6/5 hrs(A takes 3 steps B takes 2), and escalator contributes  5/6X steps, so total steps, T = 6/5X  + 20,

                        Equating X + 25 =  6/5X + 20

                        Solving X = 25, therefore, total steps = 50

Q31.   At his usual rowing rate Kapil can travel 12 miles downstream in a certain river in 6 hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour. (CAT 2001)

Ans31. Let Kapils speed in still water = X miles/hr

            And the speed of the stream = Y miles/hr

            Time difference = 6 = 12/(X – Y) – 12(X+Y)

Time difference for second case = 1 = 12/(2X – Y) – 12(2X + Y).

Solving Y = 8/3 miles per hour

Q32.   A car is moving at a rate of 50 miles per hour and the radius of its wheels is 2.5 feet. Find the number of revolutions per minute the wheels are rotating. (1 mile = 5280 feet)

Ans32. We have a car traveling 50 miles an hour with a tire radius of 2.5 ft. We wish to find the rotational velocity of the tires in revolutions per minute. First let’s convert the car’s speed to ft/min, 50 miles/hour * (5280 ft/mile) * (1 hr/60 min) = 4400 ft/min

We know that this is some multiple of the number of circumferences of the tire (since the tire rolls smoothly along). Let’s find out how many circumferences (or revolutions) this is:

C = 2 × pi × r = 2 × pi × 2.5 ft = 5pi ft

Let’s convert this speed to revolutions per minute = 4400/5 X 3.14 =  280.2 revolutions per min.

Q33.   A man is walking across a railroad bridge that goes from point A to point B. He starts at point A, and when he is  of the way across the bridge, he hears a train approaching. The train’s speed is 60 mph (miles per hour).  The man can run fast enough so that if he turns and runs back toward point A, he will meet the train at A, and if he runs forward toward point B, the train will overtake him at B. How fast can the man run?

Ans33. Let M stand for the man’s speed in mph.  When the man runs toward point A, the relative speed of the train with respect to the man is the train’s speed plus the man’s speed (60 + M). When he runs toward point B, the relative speed of the train is the train’s speed minus the man’s speed (60 - M). When he runs toward the train the distance he covers is 3 units. When he runs in the direction of the train the distance he covers is 5 units. Therefore (60 + M)/(60 – M) = 5/3, solving M = 15 mph

Special Examples

Q1: A truck travels 15 mph for the first half of the distance of a trip. The driver wants to average 30 mph for the whole trip. How fast must he travel for the second half of the trip?

Ans1: This is more of a puzzle, assume any distance for the trip. If you said 30 miles, then the first half took 1 hour. But to average 30 mph for the total trip would require an hour. Therefore it has no solution and it is impossible!

Q2: A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship. Find the distance when they meet from starting point.

Ans2: Let's say the ship is travelling x miles/hr and plane took t hours to meet the ship

Then, xt is the distance ship traveled after plane started, Let 10xt is the distance plane traveled. So we have, xt + 180 = 10xt

Solving xt = 20, therefore distance traveled by the ship = 180 + 20 = 200 miles

Q3: A can complete a piece of work in 4 days, B takes double the time taken by A, C takes double that of B, and D takes double that of C, to complete the same task. They are paired in-groups of two each, one pair takes 2/3rds the time needed by the second pair to complete the work. Which is the first pair? (CAT 2001)

Ans3:  A takes 4 days, B takes 8, C takes 16, D takes 32, In one day A does 1/4^th, B does 1/8^th, C does 1/16^th, D does 1/32th of work.

Now A and D work in a day = ¼+1/32 = 9/32, so finish work in 32/9 days

 A and C work in a day = ¼+1/16 = 5/16, so finish work in 16/5 days

            A and B work in a day = ¼+1/8 = 3/8, so finish work in 8/3 days

            B and C work in a day = 1/8+1/16 = 3/16, so finish work in 16/3 days

            We can already see a relationship between pair of A and D, and B and C,

            16/3 x 2/3 = 32/9, so these are the pairs and B and C finish work faster.

Q4: Three small pumps and a large pump are filling a tank. Each small pump works at 2/3^rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of time, it would have taken the large pump alone.(CAT 2001)

Ans4: Let the rate of large pump is 1 liter/hr, therefore the rate of small pump is 2/3 liter/hr. Assume there is a 3 liter tank, all four will fill it in 1 hour. Alone large pump will take 3 hours, ratio = 1/3

Q5: Six technicians working at the same rate complete work of one server in 10 hours. If they start at 11 am and one additional technician per hour been added, beginning at 5 PM, at what time the server will be complete? (CAT 2002)

Ans5: The work requirement is 10 x 6 = 60 Manhours

            Manhours till 5 pm = 6 x 6 = 36, Manhours required more = 24

Next hour will contribute 7, next will 8, and next 9, making for he entire 24(9+8+7), therefore work will finish by 8:00 PM

Q6: Two boats travelling at 5 and 10 km/hr respectively, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in km) 1 minute before they collide? (CAT 2004)

Ans6: Relative speed = 15(10+5) km/hr, time to travel 20 km = 20/15 = 4/3 hours = 80 minutes

            So in 80 minutes they will travel 20 km

            In 1 minute they will travel 20/80 = 4 km

            In 79 minutes they will travel 79/4 = 19.75 km

            Since total is 20, they were apart by 20 – 19.75 = 0.25 km

Q7. A sprinter starts running on a circular path of radius r meters. Her average speed (in meter/min) is Pr, during the first 30 seconds, Pr/2, during next one minute, Pr/4, during next two minutes, Pr/8, during next four minutes and so on. What is the ratio of the time taken for the nth round trip that for the previous round? (CAT 2004)

Ans7.  The circumference of circle is 2Pr, so distance traveled in first 30 seconds in Pr/2, in next one minute is again Pr/2(Pr/2 for one minute), in the next two minutes is also Pr/2(Pr/4 for two minutes), and so on as per question. Now for the 1^st round the time is 71/2 minutes, and basically four intervals, for 2^nd round the intervals will be 16 times the intervals as it will be Pr/2(Pr/16, in next 8 minutes), so the ratio will be 16.

 Q8: A company has a job to prepare certain number of cans and there are three machines A, B and C for this job. A can complete the job in three days. B can complete the job in four day and C can complete the job in six days. How many days the company will take to complete the job if all the machines are used simultaneously? (CAT 1998)

Ans8: A can do the job in 3 days, which means 1/3^rd job in a day

            B can do the job in 4 days, which means 1/4^th job in a day

            C can do the job in 6 days, which means 1/6^th job in a day

In one day they all can do 1/3 + ¼ + 1/6 = 3/4^th of job, so they will do the job in 4/3 days

Q9: Two full tanks, one shaped like a cylinder and other like a cone contains jet fuel. The cylindrical tank holds 500 liters more than the conical tank. After 200 liters of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many liters of fuel did the cylindrical tank had when it was full? (CAT 2000)

Ans9: Let conical tank hold X liter of fuel, then cylindrical holds X+500, also after 200 liters from each has been used (X+300) = 2(X-200)

                                                 X = 700 L, and cylindrical tank has 1200 liters

Q10: A and B walk up an escalator. The escalator moves at a constant speed. A takes 3 steps for every 2 of B’s steps. A gets to the top of escalator in 25 steps, while B takes 20 steps to reach the top. If escalator was turned off, how many steps would they have to take to walk up? (CAT 2001)

Ans10: A is obviously faster than B, let A reaches the top in 1 hour, in 25 steps as given, and let escalator contributed X steps (in one hour), so total steps, T = X + 25. If A takes 25 steps in an hour, B takes 20 steps in 6/5 hrs(A takes 3 steps B takes 2), and escalator contributes 5/6X steps, so total steps, T = 6/5X + 20, Equating

                                    X + 25 = 6/5X + 20

                                    Solving X = 25, therefore, total steps = 50

Q11: At his usual rowing rate Kapil can travel 12 miles downstream in a certain river in 6 hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour. (CAT 2001)

Ans11: Let Kapils speed in still water = X miles/hr

            And the speed of the stream = Y miles/hr

            Time difference = 6 = 12/(X-Y) - 12(X+Y)

            Time difference for second case = 1 = 12/(2X-Y) - 12(2X+Y)

            Solving Y = 8/3 miles per hour

Q12: A car is moving at a rate of 50 miles per hour and the radius of its wheels is 2.5 feet. Find the number of revolutions per minute the wheels are rotating. (1 mile = 5280 feet)

Ans12: We have a car traveling 50 miles an hour with a tire radius of 2.5 ft. We wish to find the rotational velocity of the tires in revolutions per minute. First let's convert the car's speed to ft/min, 50 miles/hour * (5280 ft/mile) * (1 hr/60 min)

              =  4400 ft/min

We know that this is some multiple of the number of circumferences of the tire (since the tire rolls smoothly along). Let's find out how many circumferences (or revolutions) this is:

            C = 2 x pi x r = 2 x pi x 2.5 ft = 5pi ft

Let's convert this speed to revolutions per minute = 4400 / 5 x 3.14 =  280.2 revolutions per min.

Q13: A man is walking across a railroad bridge that goes from point A to point B. He starts at point A, and when he is 3/8 of the way across the bridge, he hears a train approaching. The train's speed is 60 mph (miles per hour).  The man can run fast enough so that if he turns and runs back toward point A, he will meet the train at A, and if he runs

forward toward point B, the train will overtake him at B. How fast can the man run?

Ans13: Let M stand for the man's speed in mph.  When the man runs toward point A, the relative speed of the train with respect to the man is the train's speed plus the man's speed (60 + M).  When he runs toward point B, the relative speed of the train is the train's speed minus the man's speed (60 - M). When he runs toward the train the distance he covers is 3 units.  When he runs in the direction of the train the distance he covers is 5 units. Therefore (60 + M)/(60 - M) = 5/3, solving M = 15 mph