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Arithmetic II

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Time & Speed
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CAT 2025 Lesson : Time & Speed - Boats, Races & Clock

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9. Streams & Boats

Questions of this kind will involve a stream flowing at a constant speed and a boat travelling at a constant speed in still water. The speed at which the boat travels downstream and upstream differ.

Let ss be the speed of the boat in still water and aa be the speed of the stream.

When travelling downstream, which is with the stream, the speed of the boat increases to s+as + a.

When travelling upstream, which is against the stream, the speed of the boat reduces to s−as - a.

If the speed of the boat travelling downstream and upstream are given as xx and yy respectively, then speed of the boat is x+y2 \dfrac{x + y}{2} and the speed of the stream is x−y2\dfrac{x - y}{2}. This is a simple deduction from the concept detailed above.

Also when the distance covered by a boat while going upstream is the same as the distance covered by the boat while going downstream then,

Downstream TimeUpstream Time=Upstream SpeedDownstream Speed\dfrac{\text{Downstream Time}}{\text{Upstream Time}} = \dfrac{\text{Upstream Speed}}{\text{Downstream Speed}}

Example 28

A boat can travel at 1010 km/hr in still water. It takes 55 hours for the boat to travel from point A to point B and return back to point A. If there is a stream flowing from point A to point B at a constant speed of 22 km/hr, then what is the distance (in km) between points A and B?

Solution

Let D be the distance between points A and B. Speed downstream is 10+2=1210 + 2 = 12 km/hr and speed upstream =10−2=8= 10 - 2 = 8 km/hr.

D12+D8=5 \dfrac{D}{12} + \dfrac{D}{8} = 5

⇒ 5D24=5 \dfrac{5D}{24} = 5

⇒ D=24 D = 24 km

Answer: 2424 km

Example 29

Water in a stream is flowing at 11 m/s. If it takes Harry 5050% more time to swim upstream as compared to downstream, then what is Harry's speed (in m/s) while swimming in still water?

Solution

Let the speed of Harry in still water be xx. If the time taken to swim downstream is tt, then time taken to upstream is 1.5t.

Downstream TimeUpstream Time=Upstream SpeedDownstream Speed\dfrac{\text{Downstream Time}}{\text{Upstream Time}} = \dfrac{\text{Upstream Speed}}{\text{Downstream Speed}}

⇒ t1.5t=x−1x+1 \dfrac{t}{1.5t} = \dfrac{x - 1}{x + 1}

⇒ x+1=1.5x−1.5 x + 1 = 1.5x - 1.5

⇒ x=5 x = 5

x=5x = 5 m/s

Answer: 55 m/s

10. Common Types of Questions

10.1 Races

In races we will have runners at different speeds. Very often we have the following mentioned in the questions.

11) A gives B a head start of x\bm{x} metres: This means A lets B run for xx metres, before starting to run.

22) A gives B a head start of t\bm{t} seconds: This means A lets B run for tt seconds, before starting to run.

33) A starts x metres behind B: This means A starts xx metres behind B and therefore, has to cover xx metres more than B to finish the race.

44) A beats B by x\bm{x} metres: This means when A completes the race, B is xx metres behind A.

55) A beats B by t\bm{t} seconds: This means B takes tt more seconds than A to complete the race.

Example 30

In a 100100 metre race, Anand beats Aravind by 1010 metres and Gukan by 2020 metres. Then, in a 900900 metre race, Aravind will beat Gukan by how many metres?

Solution

When Anand covers 100100m, Aravind and Gukan cover 9090m and 8080m respectively.

For Aravind and Gukan,
Ratio of distance covered == Ratio of Speed =90:80=9:8= 90 : 80 = 9 : 8

In a 900900m race, when Aravind completes 900900m, Gukan would have covered 800m.

∴\therefore Aravind beats Gukan by 900−800=100900 - 800 = 100m

Answer: 100100 metres

Example 31

In a 400400 metre race Robert beats Rahim by 8080 metres. When Robert starts xx metres behind the start line, they finish at the same time. What is xx?

Solution

For Robert and Rahim,
Ratio of speeds == Ratio of distance covered =400:320=5:4= 400 : 320 = 5 : 4

Robert starts xx metres behind and completes at the same time as Rahim. Therefore, ratio of distances should equal the ratio of speeds. Distances travelled by Robert and Rahim are 400+x400 + x and 400400 respectively.

400+x400=54 \dfrac{400 + x}{400} = \dfrac{5}{4}

⇒ x=100 x = 100 metres

Answer: 100100 metres

10.2 Clocks

The hour hand and minute hand of an analog clock go around in circles. This movement is similar to that of a circular track. These questions almost always pertain to angles formed by the hour and minute hands.

The angles covered by the minute and hour hand for 1 hour and 1 minute are provided below.

           \space \space \space \space \space \space \space \space \space \space \space Minute Hand Hour Hand
1 Hour 3600360^0 30030^0
1 Minute 606^0 0.500.5^0

Example 32

What is the angle formed by the hour and minute hand when the time is 6:40?

Solution

We will solve this by using 1212'o clock as the reference or 000^0 point.

At 6:406:40 or 4040 minutes past 66'o clock, the angle formed with 1212'o clock by

Minute hand =40×60=2400= 40 \times 6^0 = 240^0

Hour hand =(6×300)+(40×0.50)=2000= (6 \times 30^0) + (40 \times 0.5^0) = 200^0

[Explanation: The hour hand would have been at 1800180^0 mark at 66'o clock and would have started moving by 0.500.5^0 per minute for the next 4040 minutes.]

Angle between the minute and hour hand = 240−200=400240 - 200 = 40^0

Answer: 40040^0

Example 33

How many minutes after 4'o clock will the two hands of a clock be at an angle of 60060^0 with each other for the first time?

Solution

Let x be the number of minutes past 4'o clock when for the first time the minute hand forms an angle of 60060^0 with the hour hand.

As this is for the first time, the angle formed by hour hand should be greater than that of the minute hand.

Angle formed by minute hand =6x= 6x
Angle formed by hour hand =4×30+0.5x=120+0.5x= 4 \times 30 + 0.5x = 120 + 0.5x

120+0.5x−6x=60120 + 0.5x - 6x = 60

⇒ x=605.5=12011=101011 x = \dfrac{60}{5.5} = \dfrac{120}{11} = 10\dfrac{10}{11} minutes

Answer: 10101110\dfrac{10}{11} minutes

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