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

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 $s$ be the speed of the boat in still water and $a$ be the speed of the stream.

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

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

If the speed of the boat travelling downstream and upstream are given as $x$ and $y$ respectively, then speed of the boat is $\dfrac{x + y}{2}$ and the speed of the stream is $\dfrac{x - y}{2}$. 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,

$\dfrac{\text{Downstream Time}}{\text{Upstream Time}} = \dfrac{\text{Upstream Speed}}{\text{Downstream Speed}}$

Example 28

A boat can travel at $10$ km/hr in still water. It takes $5$ 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 $2$ 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 $10 + 2 = 12$ km/hr and speed upstream $= 10 - 2 = 8$ km/hr.

$\dfrac{D}{12} + \dfrac{D}{8} = 5$

⇒ $\dfrac{5D}{24} = 5$

⇒ $D = 24$ km

Answer: $24$ km

Example 29

Water in a stream is flowing at $1$ m/s. If it takes Harry $50$% 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 $x$. If the time taken to swim downstream is $t$, then time taken to upstream is 1.5t.

$\dfrac{\text{Downstream Time}}{\text{Upstream Time}} = \dfrac{\text{Upstream Speed}}{\text{Downstream Speed}}$

⇒ $\dfrac{t}{1.5t} = \dfrac{x - 1}{x + 1}$

⇒ $x + 1 = 1.5x - 1.5$

⇒ $x = 5$

$x = 5$ m/s

Answer: $5$ 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.

$1$) A gives B a head start of $\bm{x}$ metres: This means A lets B run for $x$ metres, before starting to run.

$2$) A gives B a head start of $\bm{t}$ seconds: This means A lets B run for $t$ seconds, before starting to run.

$3$) A starts x metres behind B: This means A starts $x$ metres behind B and therefore, has to cover $x$ metres more than B to finish the race.

$4$) A beats B by $\bm{x}$ metres: This means when A completes the race, B is $x$ metres behind A.

$5$) A beats B by $\bm{t}$ seconds: This means B takes $t$ more seconds than A to complete the race.

Example 30

In a $100$ metre race, Anand beats Aravind by $10$ metres and Gukan by $20$ metres. Then, in a $900$ metre race, Aravind will beat Gukan by how many metres?

Solution

When Anand covers $100$m, Aravind and Gukan cover $90$m and $80$m respectively.

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

In a $900$m race, when Aravind completes $900$m, Gukan would have covered 800m.

$\therefore$ Aravind beats Gukan by $900 - 800 = 100$m

Answer: $100$ metres

Example 31

In a $400$ metre race Robert beats Rahim by $80$ metres. When Robert starts $x$ metres behind the start line, they finish at the same time. What is $x$?

Solution

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

Robert starts $x$ 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 $400 + x$ and $400$ respectively.

$\dfrac{400 + x}{400} = \dfrac{5}{4}$

⇒ $x = 100$ metres

Answer: $100$ 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	$360^0$	$30^0$
1 Minute	$6^0$	$0.5^0$

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 $12$'o clock as the reference or $0^0$ point.

At $6:40$ or $40$ minutes past $6$'o clock, the angle formed with $12$'o clock by

Minute hand $= 40 \times 6^0 = 240^0$

Hour hand $= (6 \times 30^0) + (40 \times 0.5^0) = 200^0$

[Explanation: The hour hand would have been at $180^0$ mark at $6$'o clock and would have started moving by $0.5^0$ per minute for the next $40$ minutes.]

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

Answer: $40^0$

Example 33

How many minutes after 4'o clock will the two hands of a clock be at an angle of $60^0$ 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 $60^0$ 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 $= 6x$
Angle formed by hour hand $= 4 \times 30 + 0.5x = 120 + 0.5x$

$120 + 0.5x - 6x = 60$

⇒ $x = \dfrac{60}{5.5} = \dfrac{120}{11} = 10\dfrac{10}{11}$ minutes

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