Relative Speed: Same Direction and Opposite Directions

When two things move at the same time, what matters is how fast the gap between them changes. That rate is the relative speed.

  • Opposite directions (towards each other, or away from each other): the gap changes at the sum of the speeds.
  • Same direction: the gap changes at the difference of the speeds.

Time=gap to be closed or openedrelative speed\text{Time} = \frac{\text{gap to be closed or opened}}{\text{relative speed}}

Towards each other. Two cars are 210 km apart and drive towards each other at 40 km/h and 30 km/h. Each hour the gap shrinks by 70 km, so they meet after 210 ÷ 70 = 3 hours. Check with positions: the first car has gone 120 km and the second 90 km, and 120 + 90 = 210.

Away from each other. The same cars start from one point and drive in opposite directions. After 2122\frac{1}{2} hours they are 70 × 2122\frac{1}{2} = 175 km apart.

Same direction. If both drive the same way from one point, the gap grows by 40 − 30 = 10 km every hour, so after 2122\frac{1}{2} hours they are only 25 km apart.

Units. Relative speed is still a speed, so the units rules from Speed, Distance and Time Basics apply. Two joggers 1.2 km apart run towards each other at 7 km/h and 8 km/h. They close the gap at 15 km/h, so they meet after 1.2 ÷ 15 = 225\frac{2}{25} hour = 4.8 minutes, which is 4 minutes 48 seconds.

One hour before they meet. "How far apart are the cars 1 hour before they meet?" The starting distance does not matter: in the last hour they close one hour's worth of relative speed, which is 70 km. Fifteen minutes before meeting they are 70 × 14\frac{1}{4} = 17.5 km apart.

After the gap closes. In the same direction, once the faster one draws level, the gap opens again at the same rate. If a 30 km gap closes at 10 km/h, they are level after 3 hours, and after 5 hours the faster car is 20 km ahead.

Where relative speed is used. Meeting from two ends, chases, and trains passing people or other trains are all relative speed questions, and they fill the rest of this section.

Meeting from Two Ends: When and Where

"P and Q are 300 km apart. A leaves P for Q at 45 km/h, and at the same time B leaves Q for P at 55 km/h."

When. The gap closes at 45 + 55 = 100 km/h, so they meet after 300 ÷ 100 = 3 hours.

Where. In 3 hours A covers 135 km, so they meet 135 km from P (and 165 km from Q). Read which town the distance is asked from, because 165 km is usually one of the choices too. Ratio view: in equal times the distances are in the ratio of the speeds, 45 : 55 = 9 : 11, so the meeting point splits PQ as 9 : 11:

distance from P=300×99+11=135 km\text{distance from P} = 300 \times \frac{9}{9 + 11} = 135 \text{ km}

One starts later. Change the question: P and Q are 345 km apart, A leaves P at 7 am at 45 km/h, and B leaves Q at 8 am at 55 km/h.

  • By 8 am A has gone 45 km, so the gap is 300 km.
  • From 8 am the gap closes at 100 km/h, which takes 3 hours, so they meet at 11 am.
  • A has driven 4 hours, so the meeting point is 180 km from P. B has driven 3 hours, 165 km from Q. 180 + 165 = 345, so the answer checks.

The slip is to divide 345 by 100 and forget that A drove alone for the first hour.

"One has covered x km more". "Two trains start towards each other at the same time at 55 km/h and 45 km/h. When they meet, the faster one has covered 25 km more than the other. How far apart were the stations?" The extra distance grows at 55 − 45 = 10 km/h, so 25 km more takes 2122\frac{1}{2} hours. In that time they cover 100 × 2.5 = 250 km together. As a formula,

D=x×a+ba−b=25×10010=250 kmD = x \times \frac{a + b}{a - b} = 25 \times \frac{100}{10} = 250 \text{ km}

Ratio view: the distances are 11 : 9, the difference is 2 parts = 25 km, and the whole is 20 parts = 250 km.

Clock times. When a question gives clock times, write down where each person is at one common moment (8 am above), then use relative speed from that moment. This one habit handles nearly every "starts later" twist, including two people who start together but one stops for a while: find the gap at the moment both are moving again.

After They Meet: The Square-Root Rule

"A and B start at the same time from P and Q towards each other. After meeting, A takes t1t_1 hours more to reach Q, and B takes t2t_2 hours more to reach P." Then

speed of Aspeed of B=t2t1time before meeting=t1t2\frac{\text{speed of A}}{\text{speed of B}} = \sqrt{\frac{t_2}{t_1}} \qquad \text{time before meeting} = \sqrt{t_1 t_2}

Why it works. Say they meet after T hours, with speeds a and b. The part A still has to cover after meeting is exactly what B covered before meeting, so at1=bTa t_1 = b T. In the same way bt2=aTb t_2 = a T. Dividing one by the other gives at1bt2=ba\frac{a t_1}{b t_2} = \frac{b}{a}, so a2b2=t2t1\frac{a^2}{b^2} = \frac{t_2}{t_1}. Multiplying them instead gives T2=t1t2T^2 = t_1 t_2.

Using it. After meeting, A takes 3 hours and B takes 1 hour 20 minutes, which is 43\frac{4}{3} hours.

  • ab=43÷3=49=23\frac{a}{b} = \sqrt{\frac{4}{3} \div 3} = \sqrt{\frac{4}{9}} = \frac{2}{3}, so A : B = 2 : 3.
  • They met after 3×43=4\sqrt{3 \times \frac{4}{3}} = \sqrt{4} = 2 hours.
  • If A drives at 36 km/h, then B drives at 54 km/h, and PQ = 36 × (2 + 3) = 180 km. Check from B's side: 54×(2+43)=54×10354 \times \left(2 + \frac{4}{3}\right) = 54 \times \frac{10}{3} = 180.

A, who takes longer after the meeting, is the slower one. So the time ratio is flipped and then square-rooted.

The ratio method behind it. Before the meeting both drive for the same time, so the distances before meeting are in the ratio of the speeds, 2 : 3. After the meeting each covers the other's part: A covers 3 parts at speed 2 and B covers 2 parts at speed 3, so the times after meeting are 32:23\frac{3}{2} : \frac{2}{3} = 9 : 4, which is 3 hours : 1131\frac{1}{3} hours. That is where the square comes from.

[Tier 2] The second meeting. If both turn straight back on reaching the far end, then by their second meeting they have together covered 3 times PQ (once to meet, then twice more between them). Take PQ = 150 km and speeds 40 and 50 km/h. The first meeting is after 150 ÷ 90 = 1231\frac{2}{3} hours, and the second after 450 ÷ 90 = 5 hours. A has covered 200 km: all 150 to Q and 50 back, so the second meeting is 100 km from P. B has covered 250 km: 150 to P and 100 back, also 100 km from P.

[SSC Tip] The two traps are leaving out the square root (4 : 9 instead of 2 : 3 in the case above) and flipping the ratio.

Chases with a Head Start in Distance: Police and Thief

In a chase both run the same way, so the gap closes at the difference of the speeds.

time to catch=head startfaster speed−slower speed\text{time to catch} = \frac{\text{head start}}{\text{faster speed} - \text{slower speed}}

"A thief is 150 m ahead of a policeman. The thief runs at 8 km/h and the policeman at 11 km/h. When is the thief caught, and how far does he run?"

Convert first. The gap is in metres, so turn the relative speed into m/s: 11 − 8 = 3 km/h = 3×518=563 \times \frac{5}{18} = \frac{5}{6} m/s.

  • Time = 150 ÷ 56\frac{5}{6} = 180 s = 3 minutes.
  • Thief's distance: 8 km/h = 209\frac{20}{9} m/s, and 209×180\frac{20}{9} \times 180 = 400 m.
  • Policeman's distance = 400 + 150 = 550 m.

The ratio short cut. While the chase lasts, both run for the same time, so their distances are in the ratio of their speeds, 8 : 11. The policeman runs 3 parts more, and those 3 parts are the 150 m head start, so 1 part = 50 m. The thief runs 8 × 50 = 400 m and the policeman 11 × 50 = 550 m. In one line:

thief’s distance=head start×thief’s speeddifference of speeds=150×83=400 m\text{thief's distance} = \text{head start} \times \frac{\text{thief's speed}}{\text{difference of speeds}} = 150 \times \frac{8}{3} = 400 \text{ m}

No unit change is needed here, because only a ratio of speeds is used.

How far apart after some time. "How far apart are they after 2 minutes?" In 120 s the gap closes by 56×120\frac{5}{6} \times 120 = 100 m, so they are 50 m apart. This works only before the catch, so first check that the time asked is less than 3 minutes.

The thief notices late. "A policeman sees a thief 130 m away and runs after him at 12 km/h. The thief notices only 15 seconds later and then runs at 9 km/h." Find the gap at the moment both are running. In 15 s the policeman covers 103×15\frac{10}{3} \times 15 = 50 m, so the gap is 80 m. It closes at 3 km/h = 56\frac{5}{6} m/s, which takes 96 s more. Check: after 111 s in all, the policeman has run 103×111\frac{10}{3} \times 111 = 370 m, and the thief is at 130 + 2.5 × 96 = 130 + 240 = 370 m.

[SSC Tip] Two traps: using speeds in km/h with a gap in metres (an answer 3.6 times off), and adding the speeds instead of subtracting. In a chase the speeds are always subtracted. Also read whose distance is asked: the thief's or the policeman's.

Chases with a Head Start in Time: Two from the Same Place

Often the one being chased simply leaves earlier. Turn the head start in time into a head start in distance, then chase as before. Count each person's driving hours separately: the wrong options give both the same number of hours.

Thief and owner. "A thief drives off in a car at 10:15 am at 48 km/h. The owner finds out at 11 am and follows at 64 km/h. When and where is the thief caught?"

  • Head start: 45 minutes at 48 km/h = 36 km.
  • The gap closes at 64 − 48 = 16 km/h, which takes 2142\frac{1}{4} hours, so the catch is at 1:15 pm.
  • Place: the owner drives 2142\frac{1}{4} hours at 64 km/h, so 144 km from the start. Check: the thief has driven 3 hours at 48 km/h, also 144 km.

Two trains from one station. A goods train leaves at 7 am at 42 km/h, and an express leaves the same station at 9 am at 70 km/h on a parallel track.

  • By 9 am the goods train is 84 km ahead.
  • The gap closes at 28 km/h, which takes 3 hours, so the express passes it at 12 noon, 210 km from the station.
  • At 2 pm the express is 2 × 28 = 56 km ahead (positions 350 km and 294 km).

Finding the chaser's speed. "A car leaves a town at 54 km/h. A second car leaves the town 1 hour later on the same road and catches the first one 3 hours after it started. Find its speed." At the catch both are at the same place. The first car has driven 4 hours, so 216 km, and the second covered those 216 km in 3 hours: 72 km/h.

Ratio view: same distance, times 4 : 3, so speeds 3 : 4, and the second car runs at 54 × 43\frac{4}{3} = 72 km/h.

Opposite directions from one point. When they start at different times and go opposite ways, find each one's own distance and add. A car leaves at 9 am going east at 40 km/h, and another leaves the same place at 10 am going west at 50 km/h. At 1 pm the first has driven 4 hours and the second 3 hours, so they are 160 + 150 = 310 km apart.

Trains: Crossing a Pole, a Man, a Platform or a Bridge

A train has length, and that changes what "crossing" means. The train has crossed something only when its last coach has passed it. So the distance covered is:

The train crosses Distance covered
a pole, a tree, a signal post or a standing man length of the train
a platform, a bridge or a tunnel length of the train + length of the platform

Anything with no length given is treated as a point.

Pole. A 240 m train runs at 72 km/h = 20 m/s. It passes a pole in 240 ÷ 20 = 12 s.

Platform. The same train crosses a 360 m platform in (240 + 360) ÷ 20 = 30 s.

Train length from two crossings. "A train passes a pole in 15 s and crosses a 200 m platform in 25 s." The extra 10 s are spent on the platform's 200 m, so the speed is 20 m/s, and the train is 20 × 15 = 300 m long. Ratio view: the times 15 : 25 = 3 : 5 are also train : (train + platform), so the platform is 2 parts = 200 m and the train is 3 parts = 300 m.

Two platforms. "A train crosses a 150 m platform in 14 s and a 275 m platform in 19 s." The extra 125 m take 5 s, so the speed is 25 m/s = 90 km/h. Train length = 25 × 14 − 150 = 200 m.

Length of a bridge. A 230 m train at 63 km/h = 17.5 m/s takes 32 s to cross a bridge. It covers 17.5 × 32 = 560 m, so the bridge is 560 − 230 = 330 m.

[Tier 2] Fully inside a tunnel. A 200 m train at 25 m/s enters a 700 m tunnel. From the engine entering to the last coach coming out takes 900 ÷ 25 = 36 s. The whole train is inside the tunnel only from the moment the last coach goes in to the moment the engine comes out, which is (700 − 200) ÷ 25 = 20 s.

[SSC Tip] SSC always has three wrong options ready: the time without the train's own length, the time without the platform, and the time with km/h left unconverted. Write "distance = train + platform, speed in m/s" before you calculate.

Trains and Moving People, Cars or Passengers

When a train passes something that is itself moving, use the relative speed. The distance is still the train's length (plus the other thing's length, if it has one).

A man walking. A 180 m train runs at 57 km/h.

  • Man walking at 3 km/h in the same direction: relative speed 57 − 3 = 54 km/h = 15 m/s, so the train passes him in 180 ÷ 15 = 12 s.
  • Man walking at 3 km/h towards the train: 57 + 3 = 60 km/h = 503\frac{50}{3} m/s, so 180 ÷ 503\frac{50}{3} = 10.8 s.

Finding the train's speed. "A 150 m train passes a man walking at 6 km/h in the same direction in 10 s. Find its speed." Relative speed = 150 ÷ 10 = 15 m/s = 54 km/h, so the train runs at 54 + 6 = 60 km/h. The trap option is 54 km/h, the relative speed itself.

A car on the road beside the track. A car with no length given is treated like the walking man. A train overtaking a car is the same-direction case; a car coming the other way is the opposite case.

A man sitting in a train. "A passenger in a train at 44 km/h sees a 300 m goods train going the other way pass him in 10 s." Only the goods train's length counts, because he is a point. Relative speed = 30 m/s = 108 km/h, so the goods train runs at 108 − 44 = 64 km/h. If the goods train were overtaking him in the same direction, you would add his speed to the relative speed instead.

[Tier 2] Two people walking. "A train passes a man walking at 3 km/h in its own direction in 9 s, and another man walking at 3 km/h towards it in 8 s. Find its speed and length." The length is the same both times:

(s−3)×9=(s+3)×8⇒9s−27=8s+24⇒s=51 km/h(s - 3) \times 9 = (s + 3) \times 8 \Rightarrow 9s - 27 = 8s + 24 \Rightarrow s = 51 \text{ km/h}

Length = 48 km/h for 9 s = 403×9\frac{40}{3} \times 9 = 120 m. Check: 54 km/h = 15 m/s for 8 s is also 120 m. You can solve for s in km/h because the factor 518\frac{5}{18} appears on both sides and cancels.

[SSC Tip] For anything that moves, decide same or opposite direction first, then take the relative speed, then convert. Using the train's own speed and ignoring the walker is the usual mistake, and with a slow walker it gives an option that looks very close to the right one.

Two Trains Crossing Each Other

When two trains cross, the crossing is over when their back ends pass each other. So the distance is the sum of the two lengths, and the speed is the relative speed:

time=L1+L2relative speed\text{time} = \frac{L_1 + L_2}{\text{relative speed}}

Opposite directions. A 150 m train at 60 km/h and a 100 m train at 30 km/h: relative speed 90 km/h = 25 m/s, so they cross in 250 ÷ 25 = 10 s.

Same direction. The faster one overtakes the slower. Relative speed 30 km/h = 253\frac{25}{3} m/s, so it takes 250 ÷ 253\frac{25}{3} = 30 s.

Finding a length. Work backwards: relative speed × time = L1+L2L_1 + L_2. Two trains running towards each other at 50 km/h and 40 km/h cross in 14 s. The relative speed is 25 m/s, so the two lengths add up to 350 m. If one is 150 m, the other is 200 m.

Speeds from two crossing times. Two trains of equal length take 12 s to cross each other in opposite directions and 60 s in the same direction. The distance is the same both times, so the relative speeds are in the inverse ratio of the times:

a+ba−b=6012=5⇒a+b=5a−5b⇒4a=6b⇒a:b=3:2\frac{a + b}{a - b} = \frac{60}{12} = 5 \Rightarrow a + b = 5a - 5b \Rightarrow 4a = 6b \Rightarrow a : b = 3 : 2

Crossing a man and each other. "Two trains in opposite directions pass a man on the platform in 28 s and 18 s, and cross each other in 22 s. Find the ratio of their speeds." With speeds a and b the lengths are 28a and 18b, and 28a+18b=22(a+b)28a + 18b = 22(a + b), so 6a=4b6a = 4b. In general,

a:b=(t3−t2):(t1−t3)=(22−18):(28−22)=4:6=2:3a : b = (t_3 - t_2) : (t_1 - t_3) = (22 - 18) : (28 - 22) = 4 : 6 = 2 : 3

Ratio of lengths. Length = speed × time to pass a pole. If train A passes a pole in 15 s and train B in 20 s, and their speeds are 3 : 2, the lengths are 3 × 15 : 2 × 20 = 45 : 40 = 9 : 8.

Trains from two stations. Two trains leaving two stations towards each other are just the meeting question with trains in place of people; their lengths do not matter for when and where they meet. The square-root rule for the times after meeting works for them too.

Solved Examples

Question 1

Two cyclists are 77 km apart. They ride towards each other at 13 km/h and 9 km/h. After how long will they meet? (a) 19 hours 15 minutes (b) 3 hours 30 minutes (c) 7 hours (d) 3 hours 50 minutes

Answer. They ride towards each other, so the gap closes at 13 + 9 = 22 km/h. Time = 77 ÷ 22 = 3.5 hours = (b) 3 hours 30 minutes. Check: in 3.5 hours one rides 45.5 km and the other 31.5 km, and 45.5 + 31.5 = 77. 3 hours 50 minutes is a misreading of 3.5 hours.

Question 2

Two buses leave the same depot at the same time and travel in the same direction at 42 km/h and 57 km/h. How far apart are they after 2 hours 20 minutes? (a) 231 km (b) 30 km (c) 45 km (d) 35 km

Answer. Same direction, so the gap grows at 57 − 42 = 15 km/h. In 2132\frac{1}{3} hours that is 15×7315 \times \frac{7}{3} = (d) 35 km. Check: one bus is at 133 km and the other at 98 km.

Question 3

Towns A and B are 480 km apart. A car leaves A for B at 70 km/h, and at the same time a bus leaves B for A at 50 km/h. How far from A do they meet? (a) 280 km (b) 200 km (c) 240 km (d) 300 km

Answer. The gap closes at 70 + 50 = 120 km/h, so they meet after 480 ÷ 120 = 4 hours. The car has gone 70 × 4 = (a) 280 km from A, and the bus 200 km from B; 280 + 200 = 480. Ratio view: the speeds are 7 : 5, so A's share of 480 is 712\frac{7}{12} of it.

Question 4

Stations P and Q are 381 km apart. A train leaves P for Q at 8:30 am at 54 km/h. Another train leaves Q for P at 10 am at 66 km/h. At what time do they meet? (a) 1:10 pm (b) 12:00 noon (c) 12:30 pm (d) 1:30 pm

Answer. By 10 am the first train has run 1121\frac{1}{2} hours on its own, which is 81 km, so the gap is 381 − 81 = 300 km. From 10 am the gap closes at 54 + 66 = 120 km/h, which takes 300 ÷ 120 = 2122\frac{1}{2} hours. They meet at (c) 12:30 pm. Check: the first train runs 4 hours (216 km) and the second 2122\frac{1}{2} hours (165 km), and 216 + 165 = 381. Dividing the full 381 km by 120 leaves out the first train's early start and gives about 1:10 pm.

Question 5

Two trains start at the same time from two stations and run towards each other at 64 km/h and 56 km/h. When they meet, the faster train has run 36 km more than the other. How far apart are the stations? (a) 540 km (b) 480 km (c) 576 km (d) 600 km

Answer. The faster train pulls ahead of the other's distance at 64 − 56 = 8 km/h, so 36 km more takes 36 ÷ 8 = 4.5 hours. In 4.5 hours the two trains together cover 120 × 4.5 = (a) 540 km. Check: 64 × 4.5 = 288 km and 56 × 4.5 = 252 km, and the difference is 36 km. Ratio view: the distances are 8 : 7, 1 part = 36 km, and the whole is 15 parts.

Question 6

Two cars start at the same time from P and Q and drive towards each other. After they meet, the car from P takes 4 hours 30 minutes more to reach Q, and the car from Q takes 8 hours more to reach P. If the car from P runs at 56 km/h, what is the speed of the other car? (a) 742374\frac{2}{3} km/h (b) 31.5 km/h (c) 48 km/h (d) 42 km/h

Answer. Use speed of P-carspeed of Q-car=84.5=169=43\frac{\text{speed of P-car}}{\text{speed of Q-car}} = \sqrt{\frac{8}{4.5}} = \sqrt{\frac{16}{9}} = \frac{4}{3}. So the other car runs at 56 × 34\frac{3}{4} = (d) 42 km/h. Check: they met after 4.5×8\sqrt{4.5 \times 8} = 6 hours, and PQ = 56 × 10.5 = 588 km = 42 × 14.

Question 7

A thief is 180 m ahead of a policeman when the chase starts. The thief runs at 11 km/h and the policeman at 14 km/h. How far will the thief have run when he is caught? (a) 840 m (b) 660 m (c) 720 m (d) 600 m

Answer. Both run for the same time, so their distances are in the ratio 11 : 14. The policeman runs 3 parts more, and those 3 parts are the 180 m head start, so 1 part = 60 m. The thief runs 11 × 60 = (b) 660 m and the policeman 840 m. Check: the gap closes at 3 km/h = 56\frac{5}{6} m/s, so the chase lasts 216 s, and 11 km/h = 5518\frac{55}{18} m/s for 216 s is 660 m.

Question 8

A policeman sees a thief 160 m ahead of him and starts chasing him. The thief runs at 7 km/h and the policeman at 13 km/h. How long will the policeman take to catch the thief? (a) 262326\frac{2}{3} seconds (b) 2 minutes (c) 1 minute 36 seconds (d) 1 minute 20 seconds

Answer. The gap closes at 13 − 7 = 6 km/h = 6×518=536 \times \frac{5}{18} = \frac{5}{3} m/s. Time = 160 ÷ 53\frac{5}{3} = 96 s = (c) 1 minute 36 seconds. Check: the policeman runs 6518\frac{65}{18} m/s × 96 = 34623346\frac{2}{3} m, and the thief 160 + 3518×96\frac{35}{18} \times 96 = 34623346\frac{2}{3} m from the policeman's start.

Question 9

A thief drives off in a stolen car at 2 pm at 54 km/h. The owner finds out at 2:40 pm and chases him at 72 km/h on the same road. At what time does the owner catch the thief? (a) 4:40 pm (b) 4:00 pm (c) 3:20 pm (d) 5:20 pm

Answer. By 2:40 pm the thief has driven 40 minutes, which is 54 × 23\frac{2}{3} = 36 km. The owner closes this gap at 72 − 54 = 18 km/h, so he needs 2 hours, and the catch is at (a) 4:40 pm. Check: the thief has driven 2 hours 40 minutes (144 km) and the owner 2 hours (144 km). The 2 hours start at 2:40 pm, not 2 pm, so 4:00 pm is wrong.

Question 10

Car A leaves a town at 8 am at 48 km/h. Car B leaves the same town at 10:30 am on the same road and overtakes car A at 3:30 pm. What is the speed of car B? (a) 60 km/h (b) 90 km/h (c) 64 km/h (d) 72 km/h

Answer. At 3:30 pm both cars are at the same place. Car A has driven 7127\frac{1}{2} hours, so 48 × 7.5 = 360 km. Car B covered the same 360 km in 5 hours, so its speed is (d) 72 km/h. Ratio view: same distance, times 7.5 : 5 = 3 : 2, so speeds 2 : 3, and 48 × 32\frac{3}{2} = 72.

Question 11

A train 330 m long runs at 66 km/h. How long does it take to pass a signal post? (a) 5 seconds (b) 18 seconds (c) 20 seconds (d) 15 seconds

Answer. A signal post has no length, so the train covers only its own 330 m. Speed = 66×518=55366 \times \frac{5}{18} = \frac{55}{3} m/s. Time = 330 × 355\frac{3}{55} = (b) 18 seconds.

Question 12

A train 250 m long runs at 90 km/h. How long will it take to cross a platform 350 m long? (a) 14 seconds (b) 10 seconds (c) 24 seconds (d) 20 seconds

Answer. To cross the platform the train covers its own length plus the platform: 250 + 350 = 600 m. Speed = 90 × 518\frac{5}{18} = 25 m/s. Time = 600 ÷ 25 = (c) 24 seconds.

Question 13

A train passes a pole in 12 seconds and crosses a platform 240 m long in 28 seconds. What is the length of the train? (a) 240 m (b) 120 m (c) 200 m (d) 180 m

Answer. The platform's 240 m take the extra 28 − 12 = 16 seconds, so the speed is 240 ÷ 16 = 15 m/s. The train passes the pole in 12 s, so its length is 15 × 12 = (d) 180 m. Ratio view: the times 12 : 28 = 3 : 7 are train : (train + platform), so the platform is 4 parts = 240 m, 1 part = 60 m, and the train is 3 parts.

Question 14

A train 400 m long runs at 76 km/h. In how many seconds will it pass a man walking at 4 km/h in the same direction as the train? (a) 20 seconds (b) 18 seconds (c) 19 seconds (d) 24 seconds

Answer. Same direction, so the relative speed is 76 − 4 = 72 km/h = 20 m/s. The train covers its own 400 m: 400 ÷ 20 = (a) 20 seconds. 18 seconds uses 76 + 4 = 80 km/h, which would be right if the man walked towards the train.

Question 15

A train 160 m long overtakes a car moving at 36 km/h on a road beside the track. It passes the car completely in 16 seconds. What is the speed of the train? (a) 36 km/h (b) 46 km/h (c) 72 km/h (d) 54 km/h

Answer. The car has no length given, so the train covers 160 m relative to the car in 16 s: relative speed = 10 m/s = 36 km/h. The train is faster than the car by 36 km/h, so it runs at 36 + 36 = (c) 72 km/h.

Question 16

A passenger sitting in a train running at 36 km/h sees a goods train 350 m long, coming from the opposite direction, pass him completely in 14 seconds. What is the speed of the goods train? (a) 90 km/h (b) 54 km/h (c) 126 km/h (d) 60 km/h

Answer. The passenger is a point, so only the goods train's 350 m count. Relative speed = 350 ÷ 14 = 25 m/s = 25 × 3.6 = 90 km/h. The trains move towards each other, so their speeds add up to 90 km/h, and the goods train runs at 90 − 36 = (b) 54 km/h. 90 km/h is only the relative speed of the two trains.

Question 17

Two trains, 140 m and 160 m long, run on parallel tracks in opposite directions at 63 km/h and 45 km/h. How long do they take to cross each other completely? (a) 60 seconds (b) 5135\frac{1}{3} seconds (c) 12 seconds (d) 10 seconds

Answer. Opposite directions, so the relative speed is 63 + 45 = 108 km/h = 30 m/s. To cross completely they cover both lengths, 140 + 160 = 300 m. Time = 300 ÷ 30 = (d) 10 seconds.

Question 18

Two trains, 200 m and 175 m long, run in the same direction at 75 km/h and 48 km/h. How long will the faster train take to pass the slower one completely? (a) 50 seconds (b) 262326\frac{2}{3} seconds (c) 231323\frac{1}{3} seconds (d) 45 seconds

Answer. Same direction, so the relative speed is 75 − 48 = 27 km/h = 7.5 m/s. The faster train has to gain both lengths, 200 + 175 = 375 m. Time = 375 ÷ 7.5 = (a) 50 seconds.

Question 19

Two trains run in opposite directions at 64 km/h and 44 km/h and cross each other completely in 12 seconds. One train is 170 m long. How long is the other? (a) 360 m (b) 170 m (c) 190 m (d) 210 m

Answer. Relative speed = 64 + 44 = 108 km/h = 30 m/s. In 12 seconds they cover 30 × 12 = 360 m, which is the two lengths together. So the other train is 360 − 170 = (c) 190 m.

Question 20

Two trains of equal length take 10 seconds to cross each other when running in opposite directions, and 30 seconds when running in the same direction. What is the ratio of their speeds? (a) 3 : 1 (b) 2 : 1 (c) 3 : 2 (d) 1 : 3

Answer. The distance (both lengths) is the same both times, so the relative speeds are in the inverse ratio of the times: (a + b) : (a − b) = 30 : 10 = 3 : 1. So a + b = 3a − 3b, which gives 2a = 4b and a : b = (b) 2 : 1. Check with speeds 2 and 1: the relative speeds are 3 and 1, so the times are 1 : 3. 3 : 1 is the ratio of the times, not of the speeds.

Question 21

[Tier 2] A train passes a man walking at 2 km/h in the same direction in 10 seconds, and another man walking at 4 km/h towards it in 9 seconds. What is the speed of the train? (a) 54 km/h (b) 60 km/h (c) 56 km/h (d) 58 km/h

Answer. The train's length is the same both times. With the train's speed s km/h: (s − 2) × 10 = (s + 4) × 9, so 10s − 20 = 9s + 36 and s = (c) 56 km/h. (The 518\frac{5}{18} cancels from both sides, so km/h is fine here.) Length check: 54 km/h = 15 m/s for 10 s is 150 m, and 60 km/h = 503\frac{50}{3} m/s for 9 s is also 150 m.