Boats and Streams, Races and Circular Tracks

Boats and Streams: Downstream and Upstream

A boat has its own speed, the speed it would have on a still lake. This is its speed in still water. A river also carries everything on it along at the speed of the stream (the current, or the rate of flow).

• Going with the stream is downstream. The water helps, so the speeds add.
• Going against the stream is upstream. The water pushes back, so the stream's speed is taken away.

If the boat's speed in still water is $b$ and the stream's speed is $s$:

$$\text{Downstream speed} = b + s, \qquad \text{Upstream speed} = b - s$$

These two lines are relative speed at work; if that idea is new, the Relative Speed, Chases and Trains section explains it first.

"A motorboat moves at 13 km/h in still water and the river flows at 3 km/h."

• Downstream: 13 + 3 = 16 km/h, so 48 km takes 48 ÷ 16 = 3 hours.
• Upstream: 13 − 3 = 10 km/h, so the same 48 km takes 4.8 hours = 4 hours 48 minutes.

Same boat, same 48 km, but the trip against the water is slower. The still-water time (48 ÷ 13) is neither of these, and SSC likes to put it in the options, along with the downstream time when the upstream one is asked.

Things that only float. A raft, a log, a ball dropped in the water or a boat with its engine off has no speed of its own. It moves at the stream's speed, 3 km/h here.

Rowers and swimmers. "A man can row at 7 km/h in still water" gives his $b$. Swimmers follow the same two lines. If $s$ is equal to or more than $b$, he cannot make any headway upstream at all.

Keep the units together. "A boat goes downstream at 16 km/h. How far does it go in 15 minutes?" 15 minutes is $\frac{1}{4}$ hour, so it goes 16 × $\frac{1}{4}$ = 4 km. When speeds are in m/s, add or subtract them in m/s and convert at the end. A swimmer who does 1.5 m/s in still water in a stream of 0.5 m/s goes downstream at 2 m/s and upstream at 1 m/s; in km/h (multiply by $\frac{18}{5}$) that is 7.2 km/h and 3.6 km/h.

Finding the Boat's Speed and the Stream's Speed

Turn the two lines round. Downstream has one extra $s$ and upstream has one $s$ less, so the boat's speed is exactly in the middle and the gap between the two speeds is $2s$:

$$b = \frac{\text{Downstream} + \text{Upstream}}{2}, \qquad s = \frac{\text{Downstream} - \text{Upstream}}{2}$$

"A boat's downstream speed is 26 km/h and its upstream speed is 14 km/h."

• Still water: $\frac{26 + 14}{2}$ = 20 km/h.
• Stream: $\frac{26 - 14}{2}$ = 6 km/h.

Check: 20 + 6 = 26 and 20 − 6 = 14. This check takes three seconds and catches most slips.

From distances and times. Get each speed first, then halve. "A boat goes 65 km downstream in 5 hours and 27 km upstream in 3 hours." Downstream speed = 65 ÷ 5 = 13 km/h, upstream speed = 27 ÷ 3 = 9 km/h. So the boat does 11 km/h in still water and the stream flows at 2 km/h.

From times in minutes. "A man rows 1.5 km downstream in 5 minutes and rows back the same 1.5 km in 9 minutes." 5 minutes is $\frac{1}{12}$ hour, so downstream he does 1.5 × 12 = 18 km/h. 9 minutes is $\frac{3}{20}$ hour, so upstream he does 1.5 ÷ $\frac{3}{20}$ = 10 km/h. His speed in still water is 14 km/h and the stream is 4 km/h.

One speed and $b$ or $s$. If the downstream speed is 22 km/h and the stream flows at 5 km/h, the boat does 22 − 5 = 17 km/h, and its upstream speed is 17 − 5 = 12 km/h. A quick way to see it: upstream = downstream − $2s$.

Different distances in the same time. "In the time a boat goes 40 km downstream, it goes only 24 km upstream. The stream flows at 2 km/h. Find the boat's speed in still water."

In equal times, distances are in the ratio of the speeds, so downstream : upstream = 40 : 24 = 5 : 3. Write the speeds as $5k$ and $3k$. Their gap, $2k$, is twice the stream, so $2k$ = 4 and $k$ = 2. Downstream is 10 km/h, upstream is 6 km/h, and the boat does 8 km/h. Check: 40 ÷ 10 = 4 hours and 24 ÷ 6 = 4 hours, the same time.

[SSC Tip] Two traps come up often. The first is giving downstream − upstream as the stream's speed without halving it (that is $2s$). The second is answering with the downstream speed when the question asks for the speed in still water. Read the last line of the question once more before you mark.

Upstream and Downstream Times in a Ratio

For a fixed distance, time and speed go opposite ways (the inverse ratio from the Speed, Distance and Time Basics section). So for the same stretch of river:

$$\frac{\text{Time upstream}}{\text{Time downstream}} = \frac{b + s}{b - s}$$

"Twice as long upstream." "A man takes twice as long to row a distance upstream as to row it downstream." Then the downstream speed is twice the upstream speed: $b + s = 2(b - s)$, so $b = 3s$ and $b : s$ = 3 : 1.

The general rule. If time upstream : time downstream = $m : n$, then

$$b : s = (m + n) : (m - n)$$

• 2 : 1 gives 3 : 1.
• 3 : 1 gives 4 : 2 = 2 : 1.
• 5 : 3 gives 8 : 2 = 4 : 1.

Do not copy the time ratio as $b : s$. "Twice as long" gives 3 : 1, not 2 : 1.

Using it. "A boat takes 4 hours to go from P to Q upstream and 3 hours to come back. Its speed in still water is 28 km/h. Find the stream's speed and the distance PQ."

The time ratio is 4 : 3, so $b : s$ = (4 + 3) : (4 − 3) = 7 : 1 and the stream flows at 28 ÷ 7 = 4 km/h. The upstream speed is 24 km/h, so PQ = 24 × 4 = 96 km. Check with the trip back: 32 km/h × 3 hours = 96 km.

The other way round. If $b : s$ = 7 : 3, the speeds are downstream : upstream = (7 + 3) : (7 − 3) = 10 : 4 = 5 : 2. For the same distance the times swap, so time upstream : time downstream = 5 : 2.

A ratio and one trip. "A boat's speed in still water and the stream's speed are in the ratio 5 : 1. The boat goes 36 km downstream in 3 hours. How long will it take to come back?"

Downstream speed = 36 ÷ 3 = 12 km/h, and that is 5 + 1 = 6 parts, so 1 part = 2 km/h. The boat does 10 km/h and the stream 2 km/h. Upstream is 4 parts = 8 km/h, so the trip back takes 36 ÷ 8 = 4.5 hours = 4 hours 30 minutes.

Round Trips: Going and Coming Back

A boat goes $d$ km downstream and comes back the same $d$ km upstream. Add the two times:

$$T = \frac{d}{b + s} + \frac{d}{b - s} = \frac{2bd}{b^2 - s^2}$$

Total time. A boat does 7 km/h in still water on a river flowing at 3 km/h. A round trip of 20 km each way takes 20 ÷ 10 + 20 ÷ 4 = 2 + 5 = 7 hours. The formula gives $\frac{2 \times 7 \times 20}{49 - 9} = \frac{280}{40}$ = 7.

Finding the distance. Turn the formula round: $d = \frac{T(b^2 - s^2)}{2b}$. "A man rows at 10 km/h in still water, and the river flows at 4 km/h. He takes 1 hour 40 minutes to row to a place and back. How far is the place?" Here T = $\frac{5}{3}$ hours, so $d = \frac{5}{3} \times \frac{100 - 16}{20} = \frac{5}{3} \times \frac{84}{20}$ = 7 km. Check: 7 ÷ 14 + 7 ÷ 6 = $\frac{1}{2} + \frac{7}{6} = \frac{5}{3}$ hours.

Average speed for the round trip. The two legs are equal distances at $b + s$ and $b - s$, so the harmonic form $\frac{2xy}{x + y}$ from the basics section applies. It simplifies to

$$\text{Average speed} = \frac{(b + s)(b - s)}{b} = \frac{b^2 - s^2}{b}$$

For the 7 km/h boat: $\frac{49 - 9}{7} = \frac{40}{7} = 5\frac{5}{7}$ km/h, not 7 km/h. The boat spends longer on the slow leg, so its average is always below its still-water speed.

Finding the stream's speed. "A boat with a speed of 18 km/h in still water goes 40 km downstream and comes back in 5 hours. Find the stream's speed." $\frac{2 \times 18 \times 40}{324 - s^2} = 5$, so $324 - s^2 = 288$, $s^2 = 36$ and $s$ = 6 km/h. Check: 40 ÷ 24 + 40 ÷ 12 = $1\frac{2}{3} + 3\frac{1}{3}$ = 5 hours.

Finding the boat's speed: test the options. "The stream flows at 4 km/h. A boat goes 45 km downstream and comes back in 7 hours. Find its speed in still water." The equation is a quadratic, but trying options is quicker. With 14 km/h: 45 ÷ 18 + 45 ÷ 10 = 2.5 + 4.5 = 7 hours. Done.

Upstream takes longer by a fixed time. "A boat does 17 km/h in still water on a river flowing at 7 km/h. How much longer does 60 km upstream take than 60 km downstream?" 60 ÷ 10 − 60 ÷ 24 = 6 − 2.5 = 3 hours 30 minutes. In general the extra time is $\frac{2sd}{b^2 - s^2}$.

[SSC Tip] Using $b$ for both legs (time = $\frac{2d}{b}$) is the classic wrong answer. A round trip on a river always takes longer than the same trip on still water, so any option that is shorter than $\frac{2d}{b}$ is out.

Two Trips, Meeting Boats and Floating Objects

[Tier 2] Two trips, two equations. "A boat goes 16 km upstream and 42 km downstream in 5 hours. It goes 40 km upstream and 28 km downstream in 7 hours. Find the speeds."

Call the upstream speed $U$ and the downstream speed $D$, and put $x = \frac{1}{U}$, $y = \frac{1}{D}$. Then

• 16x + 42y = 5
• 40x + 28y = 7

Multiply the first by 5 and the second by 2 to match the $x$ terms: 80x + 210y = 25 and 80x + 56y = 14. Subtract: 154y = 11, so $y = \frac{1}{14}$ and $D$ = 14 km/h. Then 16x = 5 − 3 = 2, so $x = \frac{1}{8}$ and $U$ = 8 km/h. The boat does 11 km/h and the stream 3 km/h. Check: 2 + 3 = 5 hours and 5 + 2 = 7 hours.

In the exam, test the options. An option of 11 km/h and 3 km/h means 8 and 14 km/h, and 16 ÷ 8 + 42 ÷ 14 = 5 at once.

Two boats on the same river. "Towns P and Q are 90 km apart, with P upstream. Boat X leaves P going downstream and boat Y leaves Q going upstream at the same time. Their speeds in still water are 10 km/h and 8 km/h, and the river flows at 3 km/h. When do they meet?" X does 13 km/h and Y does 5 km/h, so they close in at 18 km/h and meet after 90 ÷ 18 = 5 hours, 65 km from P. 18 is just 10 + 8: the stream speeds up one boat exactly as much as it slows the other, so it cancels. The same happens when one boat chases another on the same river.

[Tier 2] Something falls in the water. "A boat going upstream drops a ball as it passes under a bridge. The crew notices 20 minutes later, turns back at the same engine speed and picks up the ball 2 km below the bridge. Find the stream's speed."

Think from the water's point of view. The river carries the boat and the ball alike, so compared with the water the ball is still and the boat moves at its own speed $b$, away and then back. 20 minutes away means 20 minutes back. The ball floated for 40 minutes, which is $\frac{2}{3}$ hour, and went 2 km: the stream flows at 2 ÷ $\frac{2}{3}$ = 3 km/h. The boat's speed never comes in.

[Tier 2] Swimming across a river. SSC rarely asks this, and when it does it is one of two cases. A river is 400 m wide, a man swims at 5 km/h in still water and the stream flows at 3 km/h.

• He heads straight across: he crosses at his own 5 km/h, taking 0.4 ÷ 5 = 0.08 hour = 4.8 minutes, and the river carries him 3 × 0.08 = 0.24 km = 240 m downstream.
• He wants to land exactly opposite: he must aim a little upstream, and his speed straight across is $\sqrt{5^2 - 3^2} = \sqrt{16}$ = 4 km/h, so he takes 0.4 ÷ 4 = 0.1 hour = 6 minutes.

Races: What "Beats By" Means

A race is a time-speed-distance question with special words. In a race of length $L$:

• "A beats B by $x$ m": when A crosses the finish line, B is still $x$ m behind. In the same time A runs $L$ and B runs $L - x$, so

$$A : B = L : (L - x)$$

• "A beats B by $t$ seconds": B finishes $t$ seconds after A. B's time = A's time + $t$.
• "A beats B by $x$ m or $t$ s": both are true at once. The last $x$ m take B exactly $t$ seconds, so B's speed = $\frac{x}{t}$.
• Dead heat: both reach the finish at the same moment.
• "A gives B a start of $x$ m": B starts $x$ m ahead and runs only $L - x$. "A start of $t$ seconds": B starts running $t$ seconds before A.

From a margin in metres. "In a 700 m race A beats B by 140 m." A : B = 700 : 560 = 5 : 4. If A runs at 5 m/s, B runs at 4 m/s. A takes 140 s and B takes 175 s, so A also wins by 35 s. And 140 m at B's 4 m/s is exactly 35 s, which is the "$x$ m or $t$ s" rule.

"By $x$ m or $t$ s." "In a 1 km race A beats B by 60 m or by 15 seconds. How long does A take?" B runs 60 m in 15 s, so B's speed is 4 m/s. B needs 1000 ÷ 4 = 250 s for the race, and A finishes 15 s earlier, in 235 s = 3 minutes 55 seconds. Check: in 235 s B covers 4 × 235 = 940 m, exactly 60 m short.

From two finishing times. "A runs 800 m in 2 minutes 24 seconds and B runs it in 2 minutes 40 seconds. By how many metres does A win?" A takes 144 s and B 160 s, so A wins by 16 s. B's speed is 800 ÷ 160 = 5 m/s, and in those 16 s B covers 5 × 16 = 80 m. Check: at 144 s B has run 5 × 144 = 720 m.

Points games. "In a game of 100 points, A can give B 20 points" means that when A scores 100, B scores 80. Treat it like a 100 m race in which A beats B by 20 m: A : B = 5 : 4.

[SSC Tip] The rule $\frac{x}{t}$ gives the speed of the runner who loses. Turning the time margin into metres with the winner's speed is the trap: in the 800 m race above, 16 s at A's speed of $\frac{50}{9}$ m/s gives $88\frac{8}{9}$ m, and that will be one of the options.

Head Starts and Dead Heats

Draw the course every time: the start line, the point where B starts, and the finish. A head start shortens the distance the other runner has to cover.

A start, and A still wins. "In a 2 km race A gives B a start of 160 m and still beats him by 80 m." While A runs 2000 m, B runs from the 160 m mark to the 1920 m mark, which is 2000 − 160 − 80 = 1760 m. So A : B = 2000 : 1760 = 25 : 22. Leaving out either the start or the margin gives 2000 : 1840 or 2000 : 1920, and both turn up as options.

Dead heat: how long is the race? "A runs $\frac{9}{5}$ times as fast as B. A gives B a start of 96 m. How long should the race be for a dead heat?" In the same time A runs 9 parts and B runs 5 parts. For a dead heat A runs the whole race and B runs the race minus 96 m, so the 96 m is the gap of 4 parts. One part is 24 m, and the race is 9 parts = 216 m. Check: B runs 216 − 96 = 120 m, and 216 : 120 = 9 : 5.

$$L = \text{start} \times \frac{a}{a - b} \quad \text{for speeds } a : b$$

Dead heat: how much start? "In a 1200 m race the speeds of A and B are in the ratio 6 : 5. What start should A give B so that the race ends in a dead heat?" While A runs 1200 m, B runs 1200 × $\frac{5}{6}$ = 1000 m. So B needs a start of 200 m.

A start in metres, a win in seconds. "In a 450 m race A runs at 9 m/s. He gives B a start of 30 m and still beats him by 10 seconds. Find B's speed." A finishes in 50 s, so B finishes 60 s after the start, having run 450 − 30 = 420 m. B's speed = 420 ÷ 60 = 7 m/s. Check: when A finishes, B is at 30 + 7 × 50 = 380 m, and the last 70 m take him 10 s.

A start in seconds. "A runs 400 m in 64 s and B in 80 s. A gives B a start of 6 s. Who wins, and by how much?" B's speed is 400 ÷ 80 = 5 m/s. When A finishes, 64 s after he set off, B has been running for 64 + 6 = 70 s and has covered 350 m. A wins by 50 m. In time: B finishes 80 s after his own start, which is 74 s after A's start, so A wins by 10 s. And 50 m at 5 m/s is 10 s, so the two answers agree.

When the slower runner wins. "In a 500 m race the speeds of A and B are in the ratio 5 : 4, and B has a start of 120 m." While A runs 500 m, B runs 400 m and would be at the 520 m mark, past the finish. So B wins. To find the margin, stop the clock when B finishes: B has run 380 m, and in that time A has run 380 × $\frac{5}{4}$ = 475 m. B wins by 25 m. Measuring at the moment A finishes (B "20 m ahead") is wrong, because B has stopped by then.

Three Runners and Races on a Track

Chain the ratios, do not add the margins. "In a 640 m race A beats B by 64 m, and B beats C by 40 m. By how much does A beat C?"

When A finishes, B has run 576 m. B runs 640 m while C runs 600 m, so while B runs 576 m, C runs 576 × $\frac{600}{640}$ = 540 m. A beats C by 640 − 540 = 100 m, not 104 m. B's 40 m lead over C is measured when B finishes; when A finishes, B is still short of the line, so C is less than 40 m behind B.

For margins $x$ and $y$ in the same race of length $L$:

$$\text{A beats C by } x + y - \frac{xy}{L}$$

Here 64 + 40 − $\frac{2560}{640}$ = 100.

Margins from races of different lengths. "A beats B by 12 m in a 120 m race, and B beats C by 45 m in a 450 m race. In a 600 m race, by how much does A beat C?" A : B = 120 : 108 = 10 : 9 and B : C = 450 : 405 = 10 : 9. Make B's share the same: A : B = 100 : 90 and B : C = 90 : 81, so A : C = 100 : 81. When A runs 600 m, C runs 600 × $\frac{81}{100}$ = 486 m, and A wins by 114 m. Scaling each margin to 600 m and adding (60 m + 60 m) gives the trap answer 120 m.

A beats both; find B against C. "In a 900 m race A beats B by 60 m and C by 102 m. By how much does B beat C in a 900 m race?" When A finishes, B has run 840 m and C 798 m, so B : C = 840 : 798 = 20 : 19. When B runs 900 m, C runs 855 m: B wins by 45 m. The difference 102 − 60 = 42 m is the gap only at the moment A finishes.

[Tier 2] Finding the length of the race. "A beats B by 30 m and C by 49 m, and in a race of the same length B beats C by 20 m." When A finishes, B and C are at $L - 30$ and $L - 49$, so $\frac{L - 30}{L - 49} = \frac{L}{L - 20}$. Cross-multiplying, $L^2 - 50L + 600 = L^2 - 49L$, so $L$ = 600 m. Check: B : C = 570 : 551 and 600 : 580 are both 30 : 29.

Margins in seconds do add. "In a 1 km race A beats B by 12 s, and B beats C by 8 s." C finishes 8 s after B, who finishes 12 s after A, so A beats C by 20 s. This works because all three run the same distance and each margin is a gap between finishing times. Metres do not add, seconds do.

Races on a circular track. "A and B run a 5 km race on a 400 m circular track. Their speeds are in the ratio 5 : 4. How many times does A pass B?" When A finishes 5000 m, B has run 4000 m, so A is 1000 m ahead, which is 2.5 laps. A passes B every time he gains one more full lap: when he is 400 m ahead and when he is 800 m ahead. So A passes B 2 times. Starting together does not count as passing, and the half lap at the end is not a pass.

Circular Tracks: The First Meeting

Two runners start together from the same point on a circular track of length $C$. The relative speed rules from the Relative Speed, Chases and Trains section decide when they first meet.

• Same direction: the faster runner has to gain one full lap on the slower one. Time = $\frac{C}{a - b}$.
• Opposite directions: they run towards each other and together cover one lap. Time = $\frac{C}{a + b}$.

"A track is 650 m long. A runs at 9 m/s and B at 4 m/s."

• Same direction: 650 ÷ (9 − 4) = 130 s.
• Opposite directions: 650 ÷ (9 + 4) = 50 s.

A good check for the same direction: in 130 s A runs 1170 m (1.8 laps) and B runs 520 m (0.8 lap). A has done exactly one lap more, so they are side by side.

Speeds in km/h, track in metres. A cycles at 14.4 km/h and B at 32.4 km/h on a 500 m track, in the same direction. In m/s that is 4 and 9, a gap of 5 m/s, so they meet after 500 ÷ 5 = 100 s = 1 minute 40 seconds. Or stay in km: 0.5 km ÷ 18 km/h = $\frac{1}{36}$ hour, the same 100 s.

Lap times instead of speeds. If A takes $t_1$ and B takes $t_2$ for one round ($t_1$ less than $t_2$):

$$\text{Same direction: } \frac{t_1 t_2}{t_2 - t_1}, \qquad \text{Opposite: } \frac{t_1 t_2}{t_1 + t_2}$$

A takes 36 s and B 45 s a lap. Same direction: $\frac{36 \times 45}{9}$ = 180 s. Opposite: $\frac{36 \times 45}{81}$ = 20 s. If you prefer the units method, take the track as 180 m (LCM of 36 and 45); then A runs 5 m/s and B 4 m/s, and the two answers are 180 ÷ 1 and 180 ÷ 9.

Track given by its diameter or radius. The length is $\pi d$ or $2\pi r$, with $\pi = \frac{22}{7}$. A track of diameter 140 m is $\frac{22}{7} \times 140$ = 440 m long. Cyclists at 6 m/s and 5 m/s going opposite ways meet after 440 ÷ 11 = 40 s. Using $2\pi d$ by mistake doubles the answer.

Not starting at the same point. If B starts 120 m ahead of A on the 650 m track and both run the same way at 4 m/s and 9 m/s, A only has to close the 120 m gap: 120 ÷ 5 = 24 s. After that they meet every 130 s.

Speeds in rounds per hour, and the $n$th meeting. "M walks round a park at 4 rounds an hour and N jogs at 7 rounds an hour, the same way, from the same point." N gains 7 − 4 = 3 rounds an hour, so they meet every $\frac{1}{3}$ hour = 20 minutes. Meetings come at equal gaps, so the 4th meeting is after 4 × 20 = 80 minutes, and in 2 hours 30 minutes they meet 7 times (150 ÷ 20 = 7.5, and only whole meetings count). Going opposite ways they would meet every $\frac{60}{11}$ minutes (4 + 7 = 11 rounds an hour).

Meeting at the Start, Three Runners and Meeting Points

First meeting at the starting point. A runner is at the start only after a whole number of laps: A every $t_1$ seconds, B every $t_2$ seconds. Both are there together for the first time at the LCM of the lap times. The direction does not matter.

"On a 1320 m track A runs at 8 m/s and B at 3 m/s." Lap times: 1320 ÷ 8 = 165 s and 1320 ÷ 3 = 440 s. LCM = 1320 s = 22 minutes. In that time A runs 8 laps and B runs 3.

Two traps sit in the options. The LCM of the speeds (24) gives 1320 ÷ 24 = 55 s, which means nothing. And their first meeting anywhere (same direction, 1320 ÷ 5 = 264 s) is not a meeting at the start.

Lap times that are fractions. LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$. If A takes $4\frac{1}{2} = \frac{9}{2}$ minutes a lap and B takes $7\frac{1}{2} = \frac{15}{2}$ minutes, the LCM is $\frac{45}{2}$ = $22\frac{1}{2}$ minutes: A runs 5 laps and B runs 3. Changing to seconds first (270 s and 450 s, LCM 1350 s) avoids the fractions.

Three runners at the start. Take the LCM of all three lap times. On a 1080 m track, runners at 4, 5 and 6 m/s take 270 s, 216 s and 180 s a lap. LCM = 1080 s = 18 minutes.

[Tier 2] Three riders together anywhere. For riders going the same way, all three are together when A and B are together and B and C are together at the same moment. So take the LCM of the two pair times. On a 1260 m track at 7, 11 and 17 m/s: B gains 4 m/s on A, so they meet every 315 s; C gains 6 m/s on B, so they meet every 210 s. All three first meet after the LCM, 630 s. Check: they have ridden 4410 m, 6930 m and 10710 m, and each is 630 m past a whole number of laps. They are all back at the start only after 1260 s, the LCM of the lap times 180, $\frac{1260}{11}$ and $\frac{1260}{17}$.

How many meeting points? Write the speeds in their lowest ratio $a : b$.

• Same direction: they meet at $a - b$ different points.
• Opposite directions: they meet at $a + b$ different points.

The points are evenly spread round the track, and the runners meet that many times up to the moment they are both back at the start. For the 8 : 3 runners on the 1320 m track, going the same way they meet every 264 s, at the 792 m, 264 m, 1056 m and 528 m marks and then at the start: 5 points, one every 264 m. Going opposite ways they meet at 11 points, one every 120 m.

[SSC Tip] "Meet for the first time" and "meet for the first time at the starting point" are different questions. The first uses relative speed; the second uses the LCM of lap times. Read which one is asked before you start.

Solved Examples

Question 1

The speed of a boat in still water is 16 km/h and the speed of the stream is 4 km/h. How far will the boat travel downstream in 24 minutes? (a) 4.8 km (b) 6.4 km (c) 8 km (d) 9.6 km

Answer. Going downstream the stream helps, so the speed is 16 + 4 = 20 km/h. 24 minutes is $\frac{24}{60} = \frac{2}{5}$ hour. Distance = 20 × $\frac{2}{5}$ = (c) 8 km. If you got 4.8 km, you used the upstream speed, 12 km/h.

Question 2

A boat goes 57 km downstream in 3 hours and 44 km upstream in 4 hours. What is the speed of the stream? (a) 4 km/h (b) 15 km/h (c) 8 km/h (d) 2 km/h

Answer. First get the two speeds. Downstream: 57 ÷ 3 = 19 km/h. Upstream: 44 ÷ 4 = 11 km/h. The gap between them is two lots of the stream, so the stream's speed = $\frac{19 - 11}{2}$ = (a) 4 km/h. The boat itself does $\frac{19 + 11}{2}$ = 15 km/h in still water, which is the trap option. Check: 15 + 4 = 19 and 15 − 4 = 11.

Question 3

A motorboat can go at 24 km/h in still water. It takes three times as long to go a certain distance upstream as to go the same distance downstream. What is the speed of the stream? (a) 8 km/h (b) 16 km/h (c) 6 km/h (d) 12 km/h

Answer. Same distance, so the speeds go the other way: the downstream speed is 3 times the upstream speed. So 24 + s = 3(24 − s), which gives 4s = 48 and s = (d) 12 km/h. With the ratio rule: time up : time down = 3 : 1, so $b : s$ = (3 + 1) : (3 − 1) = 2 : 1, and the stream is half of 24. Check: downstream 36 km/h and upstream 12 km/h, and 36 is three times 12.

Question 4

A man can row at 6 km/h in still water, and the river flows at 2 km/h. It takes him 3 hours to row to a place and come back. How far is the place? (a) 9 km (b) 8 km (c) 6 km (d) 12 km

Answer. Downstream he does 8 km/h and upstream 4 km/h. If the place is d km away, $\frac{d}{8} + \frac{d}{4} = 3$. Over 8 this is $\frac{d + 2d}{8} = 3$, so 3d = 24 and d = (b) 8 km. Check: 8 ÷ 8 = 1 hour down and 8 ÷ 4 = 2 hours up, 3 hours in all.

Question 5

The speed of a boat in still water is 12 km/h. It goes 35 km downstream and comes back to the starting point in 6 hours in all. What is the speed of the stream? (a) 2 km/h (b) 1.5 km/h (c) 2.5 km/h (d) 3 km/h

Answer. The quickest way is to test the options. With a stream of 2 km/h, downstream at 14 km/h takes 35 ÷ 14 = 2.5 hours and upstream at 10 km/h takes 35 ÷ 10 = 3.5 hours, 6 hours in all. So the answer is (a) 2 km/h. By algebra: $\frac{2 \times 12 \times 35}{144 - s^2} = 6$, so $144 - s^2 = 140$ and $s^2 = 4$.

Question 6

A boat covers 39 km downstream in the same time as it covers 21 km upstream. If the speed of the stream is 3 km/h, what is the speed of the boat in still water? (a) 13 km/h (b) 7 km/h (c) 10 km/h (d) 12 km/h

Answer. In equal times, distances are in the ratio of the speeds, so downstream : upstream = 39 : 21 = 13 : 7. Write the speeds as 13k and 7k. Their gap, 6k, is twice the stream, 6 km/h, so k = 1. Downstream is 13 km/h and upstream 7 km/h, and the boat's own speed is the one in the middle: $\frac{13 + 7}{2}$ = (c) 10 km/h. Check: 39 ÷ 13 = 3 hours and 21 ÷ 7 = 3 hours.

Question 7

[Tier 2] A boat goes 20 km upstream and 27 km downstream in 7 hours. It goes 15 km upstream and 45 km downstream in 8 hours. What is the speed of the boat in still water? (a) 2 km/h (b) 7 km/h (c) 9 km/h (d) 5 km/h

Answer. Call the upstream speed $U$ and the downstream speed $D$, and put $x = \frac{1}{U}$ and $y = \frac{1}{D}$. Then 20x + 27y = 7 and 15x + 45y = 8. Multiply the first by 3 and the second by 4 to match the x terms: 60x + 81y = 21 and 60x + 180y = 32. Subtract: 99y = 11, so $y = \frac{1}{9}$ and the downstream speed is 9 km/h. Then 20x = 7 − 3 = 4, so $x = \frac{1}{5}$ and the upstream speed is 5 km/h. Speed in still water = $\frac{9 + 5}{2}$ = (b) 7 km/h, and the stream flows at 2 km/h. Check: 4 + 3 = 7 hours and 3 + 5 = 8 hours. In the exam, testing the options is faster: 7 km/h with a 2 km/h stream means 5 and 9 km/h, and 20 ÷ 5 + 27 ÷ 9 = 7 straight away.

Question 8

A boat goes from P to Q downstream in 3 hours and comes back from Q to P in 5 hours. The speed of the boat in still water is 20 km/h. What is the average speed of the boat for the whole round trip? (a) 20 km/h (b) 15 km/h (c) 17.5 km/h (d) 18.75 km/h

Answer. For the same distance, times 3 : 5 mean speeds 5 : 3, so $b : s$ = (5 + 3) : (5 − 3) = 4 : 1. With $b$ = 20 the stream is 5 km/h, the downstream speed 25 km/h and the upstream speed 15 km/h. PQ = 25 × 3 = 75 km. Average speed = total distance ÷ total time = 150 ÷ 8 = (d) 18.75 km/h. The formula $\frac{b^2 - s^2}{b} = \frac{400 - 25}{20}$ gives the same. 20 km/h is the trap: the boat spends longer on the slow leg, so its average is below its speed in still water.

Question 9

A man rows to a place 48 km away and back in 10 hours. He finds that he can row 3 km downstream in the same time as 2 km upstream. What is the speed of the stream? (a) 10 km/h (b) 4 km/h (c) 2 km/h (d) 1 km/h

Answer. In the same time he covers 3 km down and 2 km up, so downstream speed : upstream speed = 3 : 2. For the same 48 km the times go the other way, 2 : 3, so the 10 hours split as 4 hours down and 6 hours up. Downstream speed = 48 ÷ 4 = 12 km/h and upstream speed = 48 ÷ 6 = 8 km/h. The stream's speed = $\frac{12 - 8}{2}$ = (c) 2 km/h, and he rows at 10 km/h in still water. Check: 4 + 6 = 10 hours.

Question 10

[Tier 2] A motorboat is going upstream. As it passes under a bridge, a ball falls off the boat and floats away with the stream. 15 minutes later the crew notices, turns the boat round and, at the same engine speed, picks up the ball 2 km downstream of the bridge. What is the speed of the stream? (a) 4 km/h (b) 8 km/h (c) 2 km/h (d) 6 km/h

Answer. Look at it from the water. The river carries the boat and the ball together, so compared with the water the ball stays still, and the boat moves away at its own speed for 15 minutes and then comes back at the same speed. Coming back also takes 15 minutes. So the ball floated for 30 minutes and went 2 km: the stream's speed = 2 ÷ $\frac{1}{2}$ = (a) 4 km/h. The boat's speed is never needed. 8 km/h uses only the first 15 minutes.

Question 11

In a 1600 m race, A beats B by 64 m or by 16 seconds. How long does A take to run the race? (a) 6 min 40 s (b) 6 min 56 s (c) 6 min 8 s (d) 6 min 24 s

Answer. The 64 m that B is short take B exactly 16 seconds, so B runs at 64 ÷ 16 = 4 m/s. B needs 1600 ÷ 4 = 400 s for the race, and A finishes 16 s earlier, in 384 s = (d) 6 min 24 s. Check: in 384 s B runs 4 × 384 = 1536 m, which is 64 m short of 1600 m. 6 min 40 s is B's time; you land on it if you take 4 m/s as A's speed.

Question 12

A runs $1\frac{2}{5}$ times as fast as B. If A gives B a start of 84 m, how long should the race be so that both reach the winning post at the same time? (a) 210 m (b) 294 m (c) 117.6 m (d) 588 m

Answer. $1\frac{2}{5} = \frac{7}{5}$, so the speeds are A : B = 7 : 5. In the same time A runs 7 parts and B runs 5 parts. For a dead heat A runs the whole race and B runs the race minus 84 m, so the 84 m is the gap of 2 parts. One part is 42 m and the race is 7 parts = (b) 294 m. Check: B runs 294 − 84 = 210 m, and 294 : 210 = 7 : 5.

Question 13

In a 1 km race, A gives B a start of 140 m and still beats him by 20 m. If A runs at 8 m/s, what is B's speed? (a) 6.88 m/s (b) 6.72 m/s (c) 7.04 m/s (d) 7.84 m/s

Answer. A runs 1000 m in 1000 ÷ 8 = 125 s. In that time B starts 140 m ahead and ends 20 m short of the finish, so B runs 1000 − 140 − 20 = 840 m. B's speed = 840 ÷ 125 = (b) 6.72 m/s.

Question 14

In a 960 m race, A beats B by 96 m and B beats C by 60 m. By how many metres does A beat C? (a) 156 m (b) 144 m (c) 150 m (d) 160 m

Answer. When A finishes 960 m, B has run 864 m. B runs 960 m while C runs 900 m, so while B runs 864 m, C runs 864 × $\frac{900}{960}$ = 810 m. A beats C by 960 − 810 = (c) 150 m. Adding the two margins (156 m) is the trap: B's 60 m lead over C is measured when B finishes, and when A finishes B is still 96 m short of the line, so C is a little less than 60 m behind B.

Question 15

In a 600 m race, A gives B a start of 10 seconds and still beats him by 60 m. B runs at 6 m/s. What is A's speed? (a) 27 km/h (b) 24 km/h (c) 21.6 km/h (d) 7.5 km/h

Answer. When A crosses the line, B is 60 m short, so B has run 540 m. At 6 m/s that took B 90 s. B started 10 s before A, so A ran for 90 − 10 = 80 s. A's speed = 600 ÷ 80 = 7.5 m/s = 7.5 × $\frac{18}{5}$ = (a) 27 km/h.

Question 16

In a 900 m race, the speeds of A and B are in the ratio 4 : 3. A gives B a start of 240 m. Who wins, and by how much? (a) B wins by 15 m (b) A wins by 15 m (c) A wins by 20 m (d) B wins by 20 m

Answer. B has only 900 − 240 = 660 m to run. While A runs 900 m, B runs 900 × $\frac{3}{4}$ = 675 m, more than he needs, so B finishes first. To get the margin, stop the clock when B finishes: B has run 660 m, and A has run 660 × $\frac{4}{3}$ = 880 m. So the answer is (d) B wins by 20 m. "B wins by 15 m" measures the gap at the moment A finishes, but B has already stopped by then.

Question 17

A beats B by 26 m in a 260 m race, and B beats C by 18 m in a 360 m race. By how many metres will A beat C in a 500 m race? (a) 72.5 m (b) 75 m (c) 44 m (d) 70 m

Answer. Turn each margin into a speed ratio. A : B = 260 : 234 = 10 : 9, and B : C = 360 : 342 = 20 : 19. Make B's share the same: A : B = 10 : 9 = 200 : 180 and B : C = 20 : 19 = 180 : 171, so A : C = 200 : 171. When A runs 500 m, C runs 500 × $\frac{171}{200}$ = 427.5 m, so A beats C by (a) 72.5 m.

Question 18

A and B start together from the same point on a circular track 2.7 km long and cycle in the same direction at 15 km/h and 24 km/h. After how many minutes will they meet for the first time? (a) 10.8 minutes (b) $4\frac{2}{13}$ minutes (c) 6.75 minutes (d) 18 minutes

Answer. Same direction, so B gains on A at 24 − 15 = 9 km/h. They meet when B has gained one full lap of 2.7 km: time = 2.7 ÷ 9 = 0.3 hour = (d) 18 minutes. Check: in 18 minutes A rides 4.5 km and B rides 7.2 km, a gap of exactly 2.7 km.

Question 19

Two runners start together from the same point on a circular track and run in opposite directions. One completes a round in 48 seconds and the other in 72 seconds. After how many seconds will they meet for the first time? (a) 144 seconds (b) 60 seconds (c) 28.8 seconds (d) 24 seconds

Answer. Take the track as 144 m (the LCM of 48 and 72). The first runner then does 3 m a second and the second 2 m a second. Running towards each other they close in at 3 + 2 = 5 m a second, so they meet after 144 ÷ 5 = (c) 28.8 seconds. The formula $\frac{t_1 t_2}{t_1 + t_2} = \frac{48 \times 72}{120}$ gives the same.

Question 20

On a circular track of 720 m, A and B start together from the same point and run in the same direction at 4 m/s and 9 m/s. After how much time will they first be together again at the starting point? (a) 2 minutes 24 seconds (b) 12 minutes (c) 36 seconds (d) 20 seconds

Answer. A completes a lap every 720 ÷ 4 = 180 s and B every 720 ÷ 9 = 80 s. Both are at the start together at the LCM of 180 and 80, which is 720 s = (b) 12 minutes. By then A has run 4 laps and B 9 laps.

Question 21

A, B and C start together from the same point on a circular track 2.4 km long and cycle at 8 km/h, 10 km/h and 16 km/h. After how many minutes will all three be together at the starting point for the first time? (a) 18 minutes (b) 126 minutes (c) 1 minute 48 seconds (d) 72 minutes

Answer. Lap times: A takes 2.4 ÷ 8 = 0.3 hour = 18 minutes, B takes 2.4 ÷ 10 = 0.24 hour = 14.4 minutes, and C takes 2.4 ÷ 16 = 0.15 hour = 9 minutes. In seconds these are 1080, 864 and 540, and their LCM is 4320 s = (d) 72 minutes. Check: in 72 minutes A rides 4 laps, B 5 laps and C 8 laps. Rounding 14.4 minutes to 14 gives LCM(18, 14, 9) = 126, which is wrong.

Question 22

Two cyclists start together from the same point on a circular track and ride in opposite directions. Their speeds are in the ratio 8 : 5. At how many different points on the track will they meet? (a) 13 (b) 3 (c) 8 (d) 40

Answer. Take the track as 13 units and the speeds as 8 and 5 units a minute. Riding towards each other they close in at 13 units a minute, so they meet once every minute. After 1 minute the first rider is at mark 8, after 2 minutes at mark 16, which is mark 3, and so on; both are back at the start only after 13 minutes. So the meetings fall at (a) 13 different points, evenly spread round the track. The rule: with the speed ratio $a : b$ in lowest terms, opposite directions give $a + b$ points and the same direction gives $a - b$ points (3 here).

Question 23

A and B run a 2.4 km race on a circular track of 300 m. Their speeds are in the ratio 7 : 4. How many times will A pass B during the race (not counting the start)? (a) 4 (b) 2 (c) 3 (d) 8

Answer. When A finishes 2400 m, B has run 2400 × $\frac{4}{7}$ = $1371\frac{3}{7}$ m. So A ends $1028\frac{4}{7}$ m ahead, which is $3\frac{3}{7}$ laps. A passes B once for every full lap he gains: when he is 300 m, 600 m and 900 m ahead. So the answer is (c) 3.

Question 24

[Tier 2] A, B and C start together from the same point on a 600 m circular track and run in the same direction at 5 m/s, 9 m/s and 15 m/s. After how many seconds will all three be together for the first time anywhere on the track? (a) 600 seconds (b) 300 seconds (c) 150 seconds (d) 60 seconds

Answer. B gains on A at 4 m/s, so A and B are together every 600 ÷ 4 = 150 s. C gains on B at 6 m/s, so B and C are together every 600 ÷ 6 = 100 s. All three are together when both happen at once, at the LCM of 150 and 100 = (b) 300 seconds. Check: at 300 s A has run 1500 m, B 2700 m and C 4500 m; take away whole laps of 600 m and each is at the 300 m mark.