Speed Time And Distance

Answer: Option C
Solution:
We are given two buses that leave a terminal at **10-minute intervals** and travel at a speed of **20 km/h**. A man is walking towards the terminal from the opposite direction and meets the buses at **8-minute intervals**. We need to determine his speed.

Step 1: Find the Distance Between Successive Buses
Since the buses leave every 10 minutes and travel at 20 km/h, we can determine the distance between two successive buses.

\[
\text{Distance} = \text{Speed} \times \text{Time}
\]

\[
= 20 \times \left(\frac{10}{60}\right)
\]

\[
= 20 \times \frac{1}{6} = \frac{20}{6} = \frac{10}{3} \text{ km}
\]

Step 2: Determine the Relative Speed
Let the man’s speed be x km/h. The buses and the man are moving towards each other, so their relative speed is:

\[
\text{Relative speed} = 20 + x \text{ km/h}
\]

The man meets the buses at 8-minute intervals, meaning the time taken to meet two consecutive buses is 8 minutes.

Step 3: Set Up the Equation
The time taken to cover the distance between two consecutive buses with the relative speed is:

\[
\frac{\text{Distance Between Buses}}{\text{Relative Speed}} = 8 \text{ minutes} = \frac{8}{60} \text{ hours}
\]

\[
\frac{10}{3} \div (20 + x) = \frac{8}{60}
\]

\[
\frac{10}{3(20 + x)} = \frac{8}{60}
\]

Step 4: Solve for \( x \)
Cross multiplying:

\[
10 \times 60 = 8 \times 3(20 + x)
\]

\[
600 = 24(20 + x)
\]

\[
600 = 480 + 24x
\]

\[
24x = 120
\]

\[
x = \frac{120}{24} = 5
\]

Answer:
The speed of the man is 5 km/h.

Answer: Option A
Solution:
We are given that Rabi walks at \( \frac{2}{3} \) of his normal speed and, as a result, is 16 minutes late. We need to determine his usual time to cover the distance between his home and office.

Step 1: Define Variables
Let:
– \( T \) be Rabi’s usual time to reach his office in minutes.
– \( S \) be his normal speed.
– \( D \) be the distance between his home and office.

Using the formula:

\[
\text{Distance} = \text{Speed} \times \text{Time}
\]

\[
D = S \times T
\]

When Rabi walks at \( \frac{2}{3} S \), his speed decreases, and his time taken increases. Since speed and time are inversely proportional:

\[
\text{New time} = \frac{\text{Usual time}}{\text{Fraction of speed}} = \frac{T}{\frac{2}{3}} = T \times \frac{3}{2} = \frac{3}{2} T
\]

Step 2: Set Up the Equation
The additional time taken is given as **16 minutes**:

\[
\frac{3}{2} T – T = 16
\]

\[
\frac{3T}{2} – \frac{2T}{2} = 16
\]

\[
\frac{T}{2} = 16
\]

\[
T = 32 \text{ minutes}
\]

Answer:
Rabi’s usual time to reach his office is 32 minutes.

Answer: Option C
Solution:
We are given two trains leaving Delhi for Mumbai:

– Train A departs at 6:00 AM with a speed of 100 km/h.
– Train B departs at 6:45 AM with a speed of 136 km/h.

We need to determine how far from Delhi Train B catches up with Train A.

Step 1: Find the Head Start of Train A
Train A travels for **45 minutes (or 0.75 hours) before Train B starts.

\[
\text{Distance covered by Train A in 0.75 hours} = 100 \times 0.75 = 75 \text{ km}
\]

So, when Train B starts at 6:45 AM, Train A is 75 km ahead.

Step 2: Determine the Relative Speed
Since Train B is faster, it gains on Train A at a rate of:

\[
\text{Relative Speed} = 136 – 100 = 36 \text{ km/h}
\]

Step 3: Calculate the Time Required to Catch Up
To cover the 75 km gap, Train B takes:

\[
\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{75}{36} = 2.08 \text{ hours} \approx 2 \text{ hours } 5 \text{ minutes}
\]

Step 4: Find the Distance from Delhi
By the time Train B catches up, it has been traveling for 2.08 hours at 136 km/h.

\[
\text{Distance} = 136 \times 2.08 = 283 \text{ km}
\]

Final Answer:
The two trains will be together 283 km from Delhi.

Answer: Option C
Solution:
Let’s define the variables:

– Let the time taken to walk one way be W hours.
– Let the time taken to ride back one way be R hours.

Given Data:
1. Walking one way and riding back takes 6 hours 15 minutes = 6.25 hours
\[
W + R = 6.25
\]

2. Walking both ways takes 7 hours 45 minutes = 7.75 hours
\[
2W = 7.75
\]

Step 1: Solve for \( W \)
\[
W = \frac{7.75}{2} = 3.875 \text{ hours} = 3 \text{ hours } 52.5 \text{ minutes}
\]

Step 2: Solve for \( R \)
\[
R = 6.25 – 3.875 = 2.375 \text{ hours} = 2 \text{ hours } 22.5 \text{ minutes}
\]

Step 3: Find Time Taken to Ride Both Ways
\[
2R = 2 \times 2.375 = 4.75 \text{ hours} = 4 \text{ hours } 45 \text{ minutes}
\]

Thus, the time taken to ride both ways is 4 hours 45 minutes.

Answer: Option A
Solution:
To calculate the average speed, we use the formula:

\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time Taken}}
\]

Step 1: Define Variables
Let the total distance be D km.

– 30% of D = \( 0.3D \) at 20 km/h
– 60% of D = \( 0.6D \) at 40 km/h
– 10% of D = \( 0.1D \) at 10 km/h

Step 2: Calculate Time Taken for Each Segment
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}}
\]

1. Time for the first segment:
\[
T_1 = \frac{0.3D}{20} = \frac{3D}{200}
\]

2. Time for the second segment:
\[
T_2 = \frac{0.6D}{40} = \frac{6D}{240} = \frac{3D}{120}
\]

3. Time for the third segment:
\[
T_3 = \frac{0.1D}{10} = \frac{D}{100}
\]

Step 3: Calculate Total Time
\[
T_{\text{total}} = T_1 + T_2 + T_3
\]

\[
T_{\text{total}} = \frac{3D}{200} + \frac{3D}{120} + \frac{D}{100}
\]

Find the LCM of 200, 120, and 100, which is 600.

\[
T_{\text{total}} = \frac{9D}{600} + \frac{15D}{600} + \frac{6D}{600}
\]

\[
T_{\text{total}} = \frac{30D}{600} = \frac{D}{20}
\]

Step 4: Compute Average Speed
\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
\]

\[
= \frac{D}{(D/20)} = 20 \text{ km/h}
\]

Thus, the average speed of the journey is 20 km/h.

Answer: Option B
Solution:
Let the speed of A be x km/h and the speed of B be y km/h.

Given Conditions:
1. A and B start towards each other from two places 60 km apart and meet after 6 hours.
\[
\text{Distance} = \text{Time} \times \text{Relative Speed}
\] \[
60 = 6(x + y)
\] \[
x + y = 10 \quad \text{(Equation 1)}
\]

2. If A’s speed is reduced to \( \frac{2}{3} x \) and B’s speed is doubled to \( 2y \), they meet after 5 hours.
\[
60 = 5 \left( \frac{2}{3}x + 2y \right)
\] \[
\frac{2}{3}x + 2y = 12 \quad \text{(Equation 2)}
\]

Step 1: Solve for x and y
Solve for y in terms of x using Equation 1:
\[
y = 10 – x
\]

Substitute into Equation 2:
\[
\frac{2}{3}x + 2(10 – x) = 12
\]

\[
\frac{2}{3}x + 20 – 2x = 12
\]

\[
20 – \frac{4}{3}x = 12
\]

\[
8 = \frac{4}{3}x
\]

\[
x = 8 \times \frac{3}{4} = 6
\]

Step 2: Find y
\[
y = 10 – 6 = 4
\]

Thus, the speed of A is 6 km/h.

Answer: Option D
Solution:
Time taken to complete the revolution:
A → \( \frac{12}{14} \)= 3 hours
B → \( \frac{12}{3} \)= 4 hours
C → 12 × \( \frac{2}{3} \)= 8 hours
Required time,
= LCM of 3, 4, 8.
= 24 hours.

Answer: Option C
Solution:
Let’s define the variables:

– Let Ajay’s speed be \( x \) km/h.
– Ravi’s speed is 4 km/h less than Ajay’s, so Ravi’s speed = \( x – 4 \) km/h.

Given Data:
– The distance from A to B is 60 km.
– Ajay reaches B, turns back, and meets Ravi 12 km away from B, so the meeting point is 48 km from A.

Step 1: Time taken by Ajay to reach B
Ajay’s speed is \( x \) km/h, and the distance to B is 60 km. So, the time taken by Ajay to reach B is:

\[
\text{Time for Ajay to reach B} = \frac{60}{x}
\]

Step 2: Time taken by Ravi to reach the meeting point
Ravi’s speed is \( x – 4 \) km/h, and he travels a distance of 48 km. So, the time taken by Ravi to cover 48 km is:

\[
\text{Time for Ravi to reach the meeting point} = \frac{48}{x – 4}
\]

Step 3: Time taken by Ajay to meet Ravi
Since Ajay meets Ravi 12 km away from B, the total distance traveled by Ajay is 60 km (to B) + 12 km (on the way back), so:

\[
\text{Total distance traveled by Ajay} = 60 + 12 = 72 \text{ km}
\]

The time taken by Ajay to meet Ravi is:

\[
\text{Time for Ajay to meet Ravi} = \frac{72}{x}
\]

Step 4: Set up the equation
Since Ajay and Ravi meet at the same time, the time taken by Ajay to reach the meeting point should be equal to the time taken by Ravi to reach the meeting point.

Thus, we have the equation:

\[
\frac{72}{x} = \frac{48}{x – 4}
\]

Step 5: Solve the equation
Cross-multiply to solve for \( x \):

\[
72(x – 4) = 48x
\]

\[
72x – 288 = 48x
\]

\[
72x – 48x = 288
\]

\[
24x = 288
\]

\[
x = 12
\]

Step 6: Find Ravi’s speed
Ravi’s speed is \( x – 4 \):

\[
\text{Ravi’s speed} = 12 – 4 = 8 \text{ km/h}
\]

Thus, Ravi’s speed is 8 km/h.

Answer: Option A
Solution:
Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝\( \frac{2}{Time} When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.
Then,
4x – 3x =\( \frac{20}{60}
Or, x =\( \frac{1}{3}
Hence, time taken by A = 4x
hours = 4 ×\( \frac{1}{3} =1\( \frac{1}{3} hours.

Answer: Option C
Solution:
To calculate the average speed when a person covers half of the journey at one speed and the remaining half at another speed, we use the formula for average speed:

\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
\]

Given Data:
– Speed for first half of the journey = 6 km/h
– Speed for second half of the journey = 3 km/h

Let the total distance be D km.

Step 1: Time taken for each half of the journey
– Time for first half = \( \frac{D/2}{6} = \frac{D}{12} \)
– Time for second half = \( \frac{D/2}{3} = \frac{D}{6} \)

Step 2: Total time taken
The total time taken for the journey is the sum of the time for each half:

\[
\text{Total Time} = \frac{D}{12} + \frac{D}{6}
\]

To add these fractions, find a common denominator:

\[
\text{Total Time} = \frac{D}{12} + \frac{2D}{12} = \frac{3D}{12} = \frac{D}{4}
\]

Step 3: Calculate Average Speed
The average speed is:

\[
\text{Average Speed} = \frac{D}{\text{Total Time}} = \frac{D}{D/4} = 4 \text{ km/h}
\]

Thus, the average speed is 4 km/h.

Answer: Option B
Solution:
Difference of time
= 6 min – 5 mins. 52 secs.
= 8 secs.
Distance covered by man in 5 mins. 52 secs.
= Distance covered by sound in 8 secs.
= 330 × 8 = 2640 m.
∴ Speed of man
=\( \frac{2640 m}{5 min. 32 secs} \)
=\( \frac{2640}{352} \) m/secs
=\( \frac{2640}{352} \)×\( \frac{18}{5} \)kmph
=27kmph

Answer: Option C
Solution:
Let the usual time taken by the athlete to run the distance be T minutes, and the usual speed be S km/h.

Given:
– The athlete runs at \( \frac{5}{3} \) of his usual speed, so his new speed is \( \frac{5}{3}S \).
– His improved time is 5 minutes less than his usual time, so the new time is \( T – 5 \) minutes.

Step 1: Relating distance, speed, and time
Since the distance is the same in both cases, we can use the formula for distance:

\[
\text{Distance} = \text{Speed} \times \text{Time}
\]

Usual distance:
\[
\text{Distance} = S \times T
\]

Distance with the new speed:
\[
\text{Distance} = \frac{5}{3}S \times (T – 5)
\]

Since the distance is the same in both cases, we can equate the two expressions:

\[
S \times T = \frac{5}{3}S \times (T – 5)
\]

Step 2: Solve the equation
Cancel \( S \) from both sides (assuming \( S \neq 0 \)):

\[
T = \frac{5}{3} \times (T – 5)
\]

Multiply both sides by 3 to eliminate the fraction:

\[
3T = 5 \times (T – 5)
\]

Distribute on the right-hand side:

\[
3T = 5T – 25
\]

Move all terms involving \( T \) to one side:

\[
3T – 5T = -25
\]

\[
-2T = -25
\]

Answer: Option A
Solution:
Let’s break this down step by step using the given data:

Step 1: A beats B by 100m in a 1000m race
In a 1000m race, A beats B by 100 meters, which means that when A completes the 1000 meters, B has only covered 900 meters.

Thus, the ratio of the speeds of A and B is:

\[
\frac{\text{Speed of A}}{\text{Speed of B}} = \frac{1000}{900} = \frac{10}{9}
\]

So, A is faster than B by a factor of \( \frac{10}{9} \).

Step 2: B beats C by 40m in a 400m race
In a 400m race, B beats C by 40 meters, meaning when B completes 400 meters, C has covered only 360 meters.

Thus, the ratio of the speeds of B and C is:

\[
\frac{\text{Speed of B}}{\text{Speed of C}} = \frac{400}{360} = \frac{10}{9}
\]

So, B is faster than C by a factor of \( \frac{10}{9} \).

Step 3: Combine the ratios to find the ratio of the speeds of A and C
Since A is faster than B by \( \frac{10}{9} \), and B is faster than C by \( \frac{10}{9} \), we can multiply these ratios to find the ratio of the speeds of A and C:

\[
\frac{\text{Speed of A}}{\text{Speed of C}} = \frac{10}{9} \times \frac{10}{9} = \frac{100}{81}
\]

Step 4: Calculate how much A will beat C by in a 500m race
In a 500m race, let’s find how far C will run when A completes 500 meters. Since A’s speed is \( \frac{100}{81} \) times C’s speed, we can set up the following proportion:

\[
\frac{500}{\text{Distance covered by C}} = \frac{100}{81}
\]

Solving for the distance covered by C:

\[
\text{Distance covered by C} = 500 \times \frac{81}{100} = 405 \, \text{m}
\]

Thus, when A completes 500 meters, C will have covered 405 meters. Therefore, A will beat C by:

\[
500 – 405 = 95 \, \text{m}
\]

Final Answer:
In a 500m race, A will beat C by 95 meters.

Answer: Option D
Solution:
Let the speeds of A, B, and C be represented as follows:

– Let the speed of C be \( v_C \) km/h.
– Since B runs three times as fast as C, the speed of B is \( 3v_C \).
– Since A runs twice as fast as B, the speed of A is \( 2 \times 3v_C = 6v_C \).

Step 1: Distance covered by C in 72 minutes
We are given that C runs for 72 minutes. To work in consistent units, we first convert 72 minutes into hours:

\[
72 \, \text{minutes} = \frac{72}{60} = 1.2 \, \text{hours}
\]

The distance covered by C in 1.2 hours is:

\[
\text{Distance covered by C} = v_C \times 1.2
\]

Step 2: Time taken by A to cover the same distance
Now, we need to find the time it takes for A to cover the same distance. Since the distance covered by A is the same as the distance covered by C, and the speed of A is \( 6v_C \), the time taken by A to cover the same distance is:

\[
\text{Time taken by A} = \frac{\text{Distance}}{\text{Speed of A}} = \frac{v_C \times 1.2}{6v_C}
\]

Simplify:

\[
\text{Time taken by A} = \frac{1.2}{6} = 0.2 \, \text{hours}
\]

Step 3: Convert time to minutes
Finally, we convert the time taken by A to minutes:

\[
0.2 \, \text{hours} = 0.2 \times 60 = 12 \, \text{minutes}
\]

Final Answer:
The distance covered by C in 72 minutes will be covered by A in 12 minutes.

Answer: Option C
Solution:
Let the speed of the motorboat in still water be \( v_b = 36 \, \text{km/h} \), and let the speed of the stream be \( v_s \).

Step 1: Find the effective speed upstream
When the boat is traveling upstream, its effective speed is reduced by the speed of the stream. Therefore, the effective speed upstream is:

\[
v_{\text{upstream}} = v_b – v_s = 36 – v_s
\]

Step 2: Use the information about the upstream journey
We are told that the boat travels 56 km upstream in 1 hour 45 minutes. First, convert the time into hours:

\[
1 \, \text{hour} 45 \, \text{minutes} = 1 + \frac{45}{60} = 1.75 \, \text{hours}
\]

Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we can write:

\[
1.75 = \frac{56}{36 – v_s}
\]

Now, solve for \( v_s \):

\[
36 – v_s = \frac{56}{1.75} = 32
\]

\[
v_s = 36 – 32 = 4 \, \text{km/h}
\]

So, the speed of the stream is \( v_s = 4 \, \text{km/h} \).

Step 3: Find the effective speed downstream
When the boat is traveling downstream, its effective speed is the sum of the boat’s speed in still water and the speed of the stream. Therefore, the effective speed downstream is:

\[
v_{\text{downstream}} = v_b + v_s = 36 + 4 = 40 \, \text{km/h}
\]

Step 4: Calculate the time to cover 56 km downstream
Now, use the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \) to find the time taken for the downstream journey:

\[
\text{Time downstream} = \frac{56}{40} = 1.4 \, \text{hours}
\]

Step 5: Convert time to minutes
Convert 1.4 hours into minutes:

\[
1.4 \, \text{hours} = 1.4 \times 60 = 84 \, \text{minutes}
\]

Final Answer:
The time taken by the motorboat to cover the same distance downstream is 84 minutes.

Answer: Option D
Solution:
To calculate the speed in km/h, we first need to convert the distance and time into consistent units.

Step 1: Convert distance from meters to kilometers
The distance run by the athlete is 200 meters. To convert it into kilometers:

\[
\text{Distance in kilometers} = \frac{200}{1000} = 0.2 \, \text{km}
\]

Step 2: Convert time from seconds to hours
The time taken is 24 seconds. To convert it into hours:

\[
\text{Time in hours} = \frac{24}{3600} = \frac{1}{150} \, \text{hours}
\]

Step 3: Calculate speed in km/h
The formula for speed is:

\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
\]

Substitute the values we calculated:

\[
\text{Speed} = \frac{0.2}{\frac{1}{150}} = 0.2 \times 150 = 30 \, \text{km/h}
\]

Final Answer:
The athlete’s speed is 30 km/h.

Answer: Option B
Solution:
Let the distance between A’s house and school be \( d \) kilometers, and the time A is supposed to take to reach the school on time be \( t \) hours.

Given:
– When A travels at 3 km/h, he is 5 minutes late.
– When A travels at 4 km/h, he is 5 minutes early.
– The time difference between these two scenarios is 10 minutes (since he is 5 minutes late in the first case and 5 minutes early in the second).

Step 1: Express time taken in each case
1. Time taken when A travels at 3 km/h:
\[
\text{Time} = \frac{d}{3} \, \text{hours}
\]

2. Time taken when A travels at 4 km/h:
\[
\text{Time} = \frac{d}{4} \, \text{hours}
\]

Step 2: Set up the time difference equation
We are given that the difference in time between the two scenarios is 10 minutes, which is \( \frac{10}{60} = \frac{1}{6} \) hours.

So, the equation becomes:

\[
\frac{d}{3} – \frac{d}{4} = \frac{1}{6}
\]

Step 3: Solve for \( d \)
To solve the equation, first find a common denominator for the fractions on the left-hand side:

\[
\frac{d}{3} – \frac{d}{4} = \frac{4d}{12} – \frac{3d}{12} = \frac{d}{12}
\]

Thus, the equation becomes:

\[
\frac{d}{12} = \frac{1}{6}
\]

Now, solve for \( d \):

\[
d = \frac{1}{6} \times 12 = 2 \, \text{km}
\]

Final Answer:
The distance from A’s house to his school is 2 kilometers.

Answer: Option B
Solution:
Let’s break down the problem step by step.

Given:
– The height of the wall is 11 meters.
– The man climbs 1 meter up in each minute but slips down 50 cm (or 0.5 meters).
– This means his effective progress in one minute is:

\[
1 \, \text{m} – 0.5 \, \text{m} = 0.5 \, \text{m}
\]

So, every minute, he makes an effective progress of 0.5 meters, except for the final minute when he climbs the last meter and doesn’t slip back.

Step 1: Calculate how many minutes it takes to reach the last meter
In order to cover the last meter, the man needs to climb up 1 meter. Since he makes an effective progress of 0.5 meters per minute, we need to calculate how many minutes it takes for him to reach 10 meters (since in the last minute, he climbs the full meter without slipping).

To reach 10 meters, it will take:

\[
\frac{10 \, \text{meters}}{0.5 \, \text{meters per minute}} = 20 \, \text{minutes}
\]

Step 2: Climb the last meter
After these 20 minutes, he will be at 10 meters. In the next minute, he climbs the final meter to reach 11 meters, and since he doesn’t slip back on the last climb, he will have reached the top.

Step 3: Determine the time
He starts climbing at 5:00 pm and takes 21 minutes to reach the top (20 minutes to reach 10 meters, and 1 minute to climb the final meter).

Thus, he will reach the top at:

\[
5:00 \, \text{pm} + 21 \, \text{minutes} = 5:21 \, \text{pm}
\]

Final Answer:
The man will climb the wall by 5:21 pm.

Answer: Option C
Solution:
Let’s break down the problem step by step.

Given:
– The height of the pole is 60 meters.
– In the first minute, the monkey climbs 6 meters.
– In the second minute, the monkey slips down 3 meters, so the net progress in the second minute is:

\[
6 \, \text{m} – 3 \, \text{m} = 3 \, \text{m}
\]

So, the monkey makes a net progress of 3 meters every two minutes after the first minute.

Step 1: Progress after the first minute
In the first minute, the monkey climbs 6 meters.

Step 2: Calculate how many more meters the monkey needs to climb
After the first minute, the monkey has climbed 6 meters. Therefore, the remaining distance to the top is:

\[
60 \, \text{m} – 6 \, \text{m} = 54 \, \text{m}
\]

Step 3: Calculate how many cycles of 2 minutes the monkey needs
After the first minute, the monkey makes a net progress of 3 meters every 2 minutes. To cover the remaining 54 meters, the monkey will need:

\[
\frac{54 \, \text{m}}{3 \, \text{m per 2 minutes}} = 18 \, \text{cycles of 2 minutes}
\]

Step 4: Total time to reach the top
In total, the time taken will be:

– 1 minute for the first climb
– 18 cycles of 2 minutes each

So, the total time is:

\[
1 \, \text{minute} + 18 \times 2 \, \text{minutes} = 1 + 36 = 37 \, \text{minutes}
\]

Final Answer:
The monkey will take 37 minutes to reach the top of the 60-meter pole.

Answer: Option C
Solution:
Let’s break this down step by step.

Given:
– The ant climbs 12 meters in the first hour and slips down 5 meters in the second hour.
– The pole is 63 meters high.

Step 1: Determine the progress after each 2-hour cycle
In the first hour, the ant climbs 12 meters. In the second hour, it slips down 5 meters. Therefore, in each 2-hour cycle, the ant effectively climbs:

\[
12 \, \text{meters} – 5 \, \text{meters} = 7 \, \text{meters}
\]

Step 2: Calculate how much distance remains after several cycles
The ant must reach a height of 63 meters, but after several 2-hour cycles, the distance will decrease. Let’s first calculate how many full 2-hour cycles the ant can complete before it reaches the top.

After \( x \) cycles, the total distance climbed will be:

\[
7 \, \text{meters per cycle} \times x \, \text{cycles}
\]

Now, consider the last climb. After \( x-1 \) full cycles, the ant will have climbed:

\[
7 \times (x-1) \, \text{meters}
\]

At this point, it is close to the top, but the ant only needs to climb the final 12 meters without slipping back. So, we calculate how much more distance is needed:

\[
63 – (7 \times (x-1)) = 12
\]

Solving for \( x \):

\[
63 – 7(x-1) = 12
\] \[
63 – 7x + 7 = 12
\] \[
70 – 7x = 12
\] \[
7x = 58
\] \[
x = 8.2857
\]

Final Answer:
Therefore, the ant will take 16 hours.

Answer: Option D
Solution:
Let’s break this down step by step.

Given:
– The length of the train is 300 meters.
– The man is walking at a speed of 3 km/h.
– The time taken for the train to pass the man is 33 seconds.

Step 1: Convert the speed of the man to meters per second
The man’s speed is given in km/h. To work with consistent units, we convert it to meters per second:

\[
\text{Speed of the man in m/s} = \frac{3 \, \text{km/h} \times 1000 \, \text{m/km}}{3600 \, \text{s/h}} = \frac{3000}{3600} = \frac{5}{6} \, \text{m/s}
\]

Step 2: Find the relative speed between the train and the man
Since the train is moving in the same direction as the man, the relative speed between the train and the man is the speed of the train minus the speed of the man.

Let the speed of the train be \( v_t \) m/s. The relative speed will be:

\[
v_{\text{relative}} = v_t – \frac{5}{6} \, \text{m/s}
\]

Step 3: Use the relative speed to calculate the speed of the train
The train passes the man in 33 seconds, and it travels a distance of 300 meters (the length of the train) relative to the man in this time. Using the formula \( \text{Distance} = \text{Speed} \times \text{Time} \), we can write:

\[
300 = \left( v_t – \frac{5}{6} \right) \times 33
\]

Solving for \( v_t \):

\[
300 = \left( v_t – \frac{5}{6} \right) \times 33
\] \[
\frac{300}{33} = v_t – \frac{5}{6}
\] \[
\frac{300}{33} = 9.09 \, \text{m/s}
\] \[
v_t – \frac{5}{6} = 9.09
\] \[
v_t = 9.09 + \frac{5}{6}
\] \[
v_t = 9.09 + 0.8333 = 9.9233 \, \text{m/s}
\]

Step 4: Convert the speed of the train back to km/h
To convert the speed of the train from meters per second to kilometers per hour, we multiply by \( \frac{3600}{1000} = 3.6 \):

\[
v_t = 9.9233 \times 3.6 = 35.74 \, \text{km/h}
\]

Final Answer:
The speed of the train is approximately 35.74 km/h.

Answer: Option B
Solution:
Let the speed of the train starting from A be \( S_A \) and the speed of the train starting from B be \( S_B \).

When the two trains meet, let the time taken by the first train to reach B be 4 hours, and the time taken by the second train to reach A be 9 hours.

The key concept here is that the distances covered by each train after meeting are proportional to their speeds. That is:

\[
\frac{\text{Distance covered by train from A after meeting}}{\text{Distance covered by train from B after meeting}} = \frac{S_A}{S_B}
\]

Since time = distance/speed, we can set up the ratio as:

\[
\frac{S_A \times 4}{S_B \times 9} = \frac{S_A}{S_B}
\]

Canceling \( S_A \) and \( S_B \), we get:

\[
\frac{4}{9}
\]

Thus, the ratio of the speeds of the two trains is\( 2:3 \).

Answer: Option A
Solution:
Let the length of each train be \( L \), and let their speeds be \( S_1 \) and \( S_2 \).

Given:
– The first train passes a pole in 18 seconds, so its speed is:
\[
S_1 = \frac{L}{18}
\] – The second train passes a pole in 12 seconds, so its speed is:
\[
S_2 = \frac{L}{12}
\]

When the two trains run in opposite directions, their relative speed is:

\[
S_1 + S_2 = \frac{L}{18} + \frac{L}{12}
\]

To simplify:

\[
S_1 + S_2 = \frac{2L}{36} + \frac{3L}{36} = \frac{5L}{36}
\]

Since they need to cross each other, the total distance to be covered is \( 2L \) (the sum of both train lengths).

Time taken to cross each other:

\[
\text{Time} = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{2L}{\frac{5L}{36}}
\]

\[
= 2L \times \frac{36}{5L} = \frac{72}{5} = 14.4 \text{ seconds}
\]

Thus, the trains will cross each other in 14.4 seconds.

Answer: Option D
Solution:
Let the length of the first train be \( L \), and the length of the second train be \( L/2 \).
The speed of the first train is 48 kmph, and the speed of the second train is 42 kmph.

Step 1: Find the Length of the First Train
Since the trains are moving in **opposite directions**, their relative speed is:

\[
48 + 42 = 90 \text{ kmph}
\]

Convert kmph to m/s by multiplying by (5/18):

\[
90 \times \frac{5}{18} = 25 \text{ m/s}
\]

Since they cross each other in 12 seconds, the total distance covered (sum of their lengths) is:

\[
\text{Distance} = \text{Speed} \times \text{Time} = 25 \times 12 = 300 \text{ m}
\]

Since the second train’s length is half of the first train, we set up the equation:

\[
L + \frac{L}{2} = 300
\]

Multiplying by 2 to clear the fraction:

\[
2L + L = 600
\]

\[
3L = 600
\]

\[
L = 200 \text{ m}
\]

So, the first train’s length is 200 meters.

Step 2: Find the Length of the Railway Platform
The first train passes a platform in 45 seconds. The speed of the train is 48 kmph, which in m/s is:

\[
48 \times \frac{5}{18} = 13.33 \text{ m/s}
\]

Let the length of the platform be P. The train covers a total distance of \( L + P \) in 45 seconds:

\[
(200 + P) = 13.33 \times 45
\]

\[
200 + P = 600
\]

\[
P = 400 \text{ m}
\]

Final Answer:
The length of the railway platform is 400 meters.

Answer: Option B
Solution:
Given Data:
– Length of first train = 105 meters
– Length of second train = 90 meters
– Speed of first train= 45 kmph
– Speed of second train = 72 kmph
– Trains are moving in opposite directions

Step 1: Convert Speeds to m/s
We use the conversion factor \( 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \).

\[
45 \times \frac{5}{18} = 12.5 \text{ m/s}
\]

\[
72 \times \frac{5}{18} = 20 \text{ m/s}
\]

Step 2: Find Relative Speed
Since the trains are moving in opposite directions, their relative speed is:

\[
12.5 + 20 = 32.5 \text{ m/s}
\]

Step 3: Find the Total Distance to be Covered
The total distance to be covered is the sum of both train lengths:

\[
105 + 90 = 195 \text{ meters}
\]

Step 4: Calculate Time to Cross Each Other
\[
\text{Time} = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{195}{32.5}
\]

\[
= 6 \text{ seconds}
\]

Final Answer:
The trains will cross each other in 6 seconds.

Answer: Option C
Solution:
Given Data:
– Speed of first person = 3 kmph
– Speed of second person = 5 kmph
– Train passes the first person in 10 seconds
– Train passes the second person in 11 seconds

Let the speed of the train be \( S \) kmph and the length of the train** be \( L \) meters.

Step 1: Convert Speeds to m/s
We use the conversion factor:
\[
1 \text{ kmph} = \frac{5}{18} \text{ m/s}
\]

– Speed of first person:
\[
3 \times \frac{5}{18} = \frac{15}{18} = 0.83 \text{ m/s}
\]

– Speed of second person:
\[
5 \times \frac{5}{18} = \frac{25}{18} = 1.39 \text{ m/s}
\]

– Speed of the train in m/s:
\[
S \times \frac{5}{18}
\]

Step 2: Form Equations
The train covers its own length \( L \) while passing each person.

For the first person:

\[
L = \left(S \times \frac{5}{18} – 0.83 \right) \times 10
\]

For the second person:

\[
L = \left(S \times \frac{5}{18} – 1.39 \right) \times 11
\]

Since both equations represent the same train length \( L \), we equate them:

\[
\left(S \times \frac{5}{18} – 0.83 \right) \times 10 = \left(S \times \frac{5}{18} – 1.39 \right) \times 11
\]

Expanding:

\[
10S \times \frac{5}{18} – 8.3 = 11S \times \frac{5}{18} – 15.29
\]

Multiply both terms by 18 to clear the fractions:

\[
50S – 149.4 = 55S – 275.22
\]

Rearrange:

\[
275.22 – 149.4 = 55S – 50S
\]

\[
125.82 = 5S
\]

\[
S = 25.16 \text{ kmph}
\]

Final Answer:
The speed of the train is approximately 25 kmph.

Answer: Option C
Solution:
Given Data:
– Length of first bridge = 800 m
– Length of second bridge = 400 m
– Time taken to cross first bridge = 100 sec
– Time taken to cross second bridge = 60 sec
– Let the length of the train be \( L \) meters
– Let the speed of the train be \( S \) m/s

Step 1: Form Equations for Speed
The total distance covered by the train while crossing a bridge is:

\[
\text{Distance} = \text{Length of Train} + \text{Length of Bridge}
\]

Using the formula:

\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
\]

For the first bridge:

\[
S = \frac{L + 800}{100}
\]

For the second bridge:

\[
S = \frac{L + 400}{60}
\]

Step 2: Equate the Two Equations
Since both represent speed \( S \), we equate:

\[
\frac{L + 800}{100} = \frac{L + 400}{60}
\]

Step 3: Solve for \( L \)
Cross multiply:

\[
(L + 800) \times 60 = (L + 400) \times 100
\]

\[
60L + 48000 = 100L + 40000
\]

\[
48000 – 40000 = 100L – 60L
\]

\[
8000 = 40L
\]

\[
L = \frac{8000}{40} = 200 \text{ meters}
\]

Final Answer:
The length of the train is 200 meters.

Answer: Option C
Solution:
Given Data:
– Speed of first train = 60 kmph
– Distance covered to overtake = 120 meters
– Time taken to overtake = 18 seconds
– Let the speed of the second train be \( S_2 \) kmph
– Let the speed of the first train be \( S_1 = 60 \) kmph

Step 1: Convert Speed to m/s
We use the conversion factor:

\[
1 \text{ kmph} = \frac{5}{18} \text{ m/s}
\]

So, speed of the first train:

\[
S_1 = 60 \times \frac{5}{18} = 16.67 \text{ m/s}
\]

Let the speed of the second train in m/s be:

\[
S_2 = S \times \frac{5}{18}
\]

Step 2: Use Relative Speed
Since the trains are moving in the **same direction**, the relative speed is:

\[
\text{Relative Speed} = S_1 – S_2 = 16.67 – S_2
\]

The train covers 120 meter in 18 seconds, so:

\[
\frac{120}{18} = 16.67 – S_2
\]

\[
6.67 = 16.67 – S_2
\]

\[
S_2 = 16.67 – 6.67
\]

\[
S_2 = 10 \text{ m/s}
\]

Step 3: Convert Back to kmph
\[
S_2 = 10 \times \frac{18}{5} = 36 \text{ kmph}
\]

Final Answer:
The speed of the second train is 36 kmph.

Answer: Option C
Solution:
Given Data:
– Speed of the boat in still water = 10 kmph
– Speed of the stream = 4 kmph
– Total time (downstream + upstream) = 5 hours
– Let the distance to the destination be \( d \) km

Step 1: Calculate Effective Speeds
-Downstream speed** (boat + stream):
\[
10 + 4 = 14 \text{ kmph}
\] – Upstream speed (boat – stream):
\[
10 – 4 = 6 \text{ kmph}
\]

Step 2: Set Up the Time Equation
Total time for the trip:

\[
\frac{d}{\text{Downstream speed}} + \frac{d}{\text{Upstream speed}} = 5
\]

\[
\frac{d}{14} + \frac{d}{6} = 5
\]

Step 3: Solve for \( d \)
Find the LCM of 14 and 6, which is **42**, and rewrite:

\[
\frac{3d}{42} + \frac{7d}{42} = 5
\]

\[
\frac{10d}{42} = 5
\]

Multiply by 42:

\[
10d = 210
\]

\[
d = 21 \text{ km}
\]

Final Answer:
The distance to the destination is 21 km.

Answer: Option B
Solution:
Given Data:
– Speed of the person in still water = \( 7\frac{1}{2} \) kmph = 7.5 kmph
– Time taken upstream is twice the time taken downstream**
– Let the speed of the stream be \( S \) kmph

Step 1: Define Effective Speeds
– Downstream speed = \( 7.5 + S \) kmph
– Upstream speed = \( 7.5 – S \) kmph

Step 2: Set Up the Time Equation
Let the distance be d km. The given condition states:

\[
\text{Time upstream} = 2 \times \text{Time downstream}
\]

Since Time = Distance / Speed, we write:

\[
\frac{d}{7.5 – S} = 2 \times \frac{d}{7.5 + S}
\]

Cancel \( d \) from both sides:

\[
\frac{1}{7.5 – S} = \frac{2}{7.5 + S}
\]

Step 3: Solve for \( S \)
Cross multiply:

\[
7.5 + S = 2(7.5 – S)
\]

\[
7.5 + S = 15 – 2S
\]

\[
S + 2S = 15 – 7.5
\]

\[
3S = 7.5
\]

\[
S = 2.5
\]

Final Answer:
The speed of the stream is 2.5 kmph.

Answer: Option B
Solution:
To solve this, we need to consider the effect of the stream’s speed on the boats.

Given:
– Distance between A and B: 108 km
– Speed of boat A in still water: 12 km/h
– Speed of boat B in still water: 15 km/h
– Let the speed of the stream be x km/h.

Effective Speeds:
– Since A is going downstream, its effective speed = (12 + x) km/h**.
– Since B is going upstream, its effective speed = (15 – x) km/h**.

They meet after time t, so:

\[
\text{Distance} = \text{Relative Speed} \times \text{Time}
\]

\[
108 = (12 + x) + (15 – x) \times t
\]

\[
108 = (12 + 15) t
\]

\[
108 = 27t
\]

\[
t = \frac{108}{27} = 4 \text{ hours}
\]

Thus, the boats will meet after 4 hours.

Answer: Option C
Solution:
Let’s solve this step by step.

Given:
– The ratio of the speed of the motorboat to the current = 36:5.
– The time taken to travel along the current = **5 hours 10 minutes = 5 + 10/60 hours = 5.167 hours.

Let’s assume the speed of the motorboat in still water is 36x km/h, and the speed of the current is 5x km/h.

Speed of the boat:
– Downstream speed (along the current) = (36x + 5x) = 41x km/h.
– Upstream speed (against the current) = (36x – 5x) = 31x km/h.

Distance traveled:
Let the distance between the two points be D km.

Using the downstream speed, the time taken to travel downstream is given by:

\[
\text{Time} = \frac{D}{\text{Speed}}
\]

For downstream, this is:

\[
5.167 = \frac{D}{41x}
\]

So, the distance \( D = 5.167 \times 41x = 211.847x \).

Now, to find the time taken for the boat to travel upstream (coming back), we use the upstream speed.

\[
\text{Time upstream} = \frac{D}{31x}
\]

Substitute \( D = 211.847x \):

\[
\text{Time upstream} = \frac{211.847x}{31x} = \frac{211.847}{31} = 6.83 \text{ hours}
\]

Converting this to hours and minutes:

\[
6.83 \text{ hours} = 6 \text{ hours} \, 0.83 \times 60 \text{ minutes} = 6 \text{ hours} \, 50 \text{ minutes}.
\]

So, the boat will take 6 hours 50 minutes to come back upstream.

Answer: Option C
Solution:
Let’s solve this step by step.

Given:
– Kamal wins the 100m race by 5 seconds.
– Kamal’s speed = 18 km/h.
– Distance of the race = 100 meters.

First, convert Kamal’s speed into meters per second:

\[
\text{Speed of Kamal in m/s} = \frac{18 \times 1000}{3600} = 5 \, \text{m/s}.
\]

Now, let’s calculate the time Kamal takes to complete the race:

\[
\text{Time taken by Kamal} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{5} = 20 \, \text{seconds}.
\]

Since Kamal defeats Bimal by **5 seconds**, it means Bimal takes:

\[
\text{Time taken by Bimal} = 20 + 5 = 25 \, \text{seconds}.
\]

Now, let’s calculate the speed of Bimal. He covers 100 meters in 25 seconds, so his speed is:

\[
\text{Speed of Bimal} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{25} = 4 \, \text{m/s}.
\]

Convert Bimal’s speed to km/h:

\[
\text{Speed of Bimal in km/h} = 4 \times \frac{3600}{1000} = 14.4 \, \text{km/h}.
\]

So, the speed of Bimal is 14.4 km/h.

Answer: Option C
Solution:
Let’s break down the information:

1. B gives a 10-meter start to A: This means, when B runs 200 meters, A runs 190 meters.

So, the ratio of their speeds (B to A) is:
\[
\frac{Speed of B}{Speed of A} = \frac{200}{190} = \frac{20}{19}
\]

2. C gives a 20-meter start to B: This means, when C runs 200 meters, B runs 180 meters.

So, the ratio of their speeds (C to B) is:
\[
\frac{Speed of C}{Speed of B} = \frac{200}{180} = \frac{10}{9}
\]

Now, to find the start that C gives to A, we combine the two ratios:
– C to B = \(\frac{10}{9}\)
– B to A = \(\frac{20}{19}\)

To get the ratio of C to A, multiply these two ratios:
\[
\frac{Speed of C}{Speed of A} = \frac{10}{9} \times \frac{20}{19} = \frac{200}{171}
\]

This means, when C runs 200 meters, A runs \(\frac{171}{200} \times 200 = 171\) meters.

Thus, the start that C gives to A is:
\[
200 – 171 = 29 \text{ meters}
\]

So, C can give a start of 29 meters to A in the same race.

Answer: Option A
Solution:
Let’s break this down step by step:

Step 1: Analyzing the first race (1-kilometer race between A and B)

– In the 1-kilometer race, A can beat B by 30 meters.
– This means that when A runs 1000 meters, B only runs 970 meters (because A beats B by 30 meters).

Therefore, the ratio of the speeds of A and B is:
\[
\frac{\text{Speed of A}}{\text{Speed of B}} = \frac{1000}{970} = \frac{100}{97}
\]

Step 2: Analyzing the second race (500-meter race between B and C)

– In the 500-meter race, B can beat C by 25 meters.
– This means that when B runs 500 meters, C only runs 475 meters.

Therefore, the ratio of the speeds of B and C is:
\[
\frac{\text{Speed of B}}{\text{Speed of C}} = \frac{500}{475} = \frac{20}{19}
\]

Step 3: Finding the combined ratio (A to C)

Now, to find how much A beats C, we combine the two ratios:

– \( \frac{\text{Speed of A}}{\text{Speed of B}} = \frac{100}{97} \)
– \( \frac{\text{Speed of B}}{\text{Speed of C}} = \frac{20}{19} \)

By multiplying these two ratios, we get the ratio of the speeds of A and C:
\[
\frac{\text{Speed of A}}{\text{Speed of C}} = \frac{100}{97} \times \frac{20}{19} = \frac{2000}{1843}
\]

Thus, when A runs 2000 meters, C runs 1843 meters. Now, we want to find out by how many meters A beats C in a 100-meter race.

The proportion is:
\[
\frac{100}{x} = \frac{2000}{1843}
\] Where \(x\) is the distance that C runs when A runs 100 meters.

Cross-multiplying:
\[
x = \frac{1843 \times 100}{2000} = \frac{184300}{2000} = 92.15
\]

So, when A runs 100 meters, C runs approximately 92.15 meters.

Therefore, A beats C by:
\[
100 – 92.15 = 7.85 \text{ meters}
\]

Thus, A beats C by approximately 7.85 meters in a 100-meter race.

Answer: Option D
Solution:
Let’s solve this problem step by step:

Step 1: Let the usual speed of the plane be \( S \) km/h.

– The distance to be covered is 1500 kilometers.
– The plane has to reach the destination on time, so it must cover 1500 km in the usual time it would take without the delay.

Step 2: Time to cover the distance without the delay.

– If the plane were to travel at the usual speed \( S \), the time taken to cover 1500 kilometers would be:
\[
\text{Time} = \frac{1500}{S} \text{ hours}
\]

Step 3: Time left after the plane’s delay.

– The plane left 30 minutes (which is \( \frac{1}{2} \) hour) later than scheduled.
– The remaining time to cover the 1500 kilometers is:
\[
\frac{1500}{S} – \frac{1}{2} \text{ hours}
\]

Step 4: Increased speed.

– The plane increased its speed by 33.33%, meaning the new speed is:
\[
\text{New speed} = S + 0.3333S = 1.3333S
\]

Step 5: Time taken with the increased speed.

– With the new speed of \( 1.3333S \), the time taken to cover the 1500 kilometers is:
\[
\text{New time} = \frac{1500}{1.3333S} = \frac{1500}{\frac{4}{3}S} = \frac{1500 \times 3}{4S} = \frac{4500}{4S} = \frac{1125}{S}
\]

Step 6: Equating the times.

The plane must cover the 1500 kilometers in the remaining time, so:
\[
\frac{1125}{S} = \frac{1500}{S} – \frac{1}{2}
\]

Now, solve for \( S \):

\[
\frac{1125}{S} = \frac{1500}{S} – \frac{1}{2}
\]

Multiply through by \( S \) to eliminate the denominator:
\[
1125 = 1500 – \frac{S}{2}
\]

Rearrange to solve for \( S \):
\[
\frac{S}{2} = 1500 – 1125
\] \[
\frac{S}{2} = 375
\] \[
S = 750 \text{ km/h}
\]

Step 7: Increased speed.

The increased speed is \( 1.3333 \times 750 = 1000 \) km/h.

Final Answer:
The increased speed of the plane is 1000 km/h.

Answer: Option D
Solution:
Let’s break down the information given:

– The pedestrian is walking at a speed of 2 km/h in the same direction as the car.
– The pedestrian can see the car for 6 minutes.
– The distance that the car is visible to the pedestrian is 0.6 km.

Step 1: Convert time into hours.

The time that the pedestrian can see the car is 6 minutes. To work with consistent units, we need to convert this into hours:
\[
\text{Time} = \frac{6}{60} = \frac{1}{10} \text{ hours}
\]

Step 2: Relative speed.

The car and the pedestrian are moving in the same direction. The pedestrian sees the car for 0.6 km in the time of \( \frac{1}{10} \) hours. The relative speed between the car and the pedestrian is the difference in their speeds:
\[
\text{Relative speed} = \text{Speed of car} – \text{Speed of pedestrian} = \text{Speed of car} – 2 \text{ km/h}
\]

Step 3: Calculate the relative speed.

The distance that the car covers relative to the pedestrian is 0.6 km in \( \frac{1}{10} \) hours. The formula for speed is:
\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
\] So, the relative speed is:
\[
\text{Relative speed} = \frac{0.6}{\frac{1}{10}} = 0.6 \times 10 = 6 \text{ km/h}
\]

Step 4: Find the speed of the car.

Now, we can set up the equation for the relative speed:
\[
\text{Speed of car} – 2 = 6
\] Solving for the speed of the car:
\[
\text{Speed of car} = 6 + 2 = 8 \text{ km/h}
\]

Final Answer:
The speed of the car is 8 km/h.

Answer: Option A
Solution:
To find the average speed of the cyclist, we need to calculate the distance traveled in one complete revolution and then divide it by the time taken.

Step 1: Calculate the circumference of the circular track.

The formula for the circumference \( C \) of a circle is:
\[
C = 2\pi r
\] where \( r \) is the radius of the circle.

Given that the radius \( r = 100 \) meters, we can substitute this value into the formula:
\[
C = 2\pi \times 100 = 200\pi \text{ meters}
\]

Step 2: Time taken for one revolution.

The cyclist completes one revolution in 2 minutes.

Step 3: Calculate the average speed.

The average speed is given by the formula:
\[
\text{Average speed} = \frac{\text{Distance traveled}}{\text{Time taken}}
\]

The distance traveled in one revolution is the circumference of the circle, which is \( 200\pi \) meters. The time taken is 2 minutes, or \( \frac{2}{60} \) hours.

The average speed is:
\[
\text{Average speed} = \frac{200\pi}{2} = 100\pi \text{ meters per minute}
\]

Approximating \( \pi \approx 3.1416 \):
\[
\text{Average speed} \approx 100 \times 3.1416 = 314.16 \text{ meters per minute}
\]

Final Answer:
The average speed of the cyclist is approximately 314 meters per minute.

Answer: Option A
Solution:
To find out how many times the minute and hour hands of a clock are together between 5 am and 5 pm, let’s analyze the movement of the hands.

Step 1: Basic clock mechanics

– The hour hand completes one full rotation (360°) in 12 hours.
– The minute hand completes one full rotation (360°) in 60 minutes (1 hour).

The minute hand moves faster than the hour hand. The key observation is that the minute and hour hands overlap multiple times within a 12-hour period.

Step 2: Frequency of overlaps

The minute and hour hands coincide once every **1 hour and 5 minutes** (approximately). This is because the minute hand has to “catch up” with the hour hand each time it overlaps.

Step 3: Number of overlaps in 12 hours

In a 12-hour period, the hands overlap approximately **11 times**. This is because the first overlap happens at 12:00, and after that, they overlap once roughly every 1 hour and 5 minutes, but there are only 11 such intervals in 12 hours.

Step 4: Time frame from 5 am to 5 pm

The time between 5 am and 5 pm is 12 hours. Therefore, the hands of the clock will overlap **11 times** between 5 am and 5 pm.

Final Answer:
The minute and hour hands are together 11 times between 5 am and 5 pm.

Answer: Option C
Solution:
Let’s break down the problem properly and find the speed of the train.

Given:
-Time difference between shots fired: \( 11 \, \text{minutes} \, 45 \, \text{seconds} = 705 \, \text{seconds} \)
– Time difference in hearing the second shot after the first by the man in the train: \( 11 \, \text{minutes} = 660 \, \text{seconds} \)
– Speed of sound: \( 330 \, \text{m/s} \)

Understanding the problem:

1. The two shots are fired at a time difference of \( 705 \, \text{seconds} \).
2. The man, who is moving towards the source of the shots, hears the second sound** 660 seconds after the first sound. This is less than the actual time difference of 705 seconds, which means the train is moving towards the place where the shots are fired, reducing the time it takes for the sound to reach the man.

Step-by-step approach:

Let \( v_t \) be the speed of the train (in m/s). We need to calculate this.

1. Distance covered by sound:

The distance that the sound of the shots travels in the time difference of \( 705 \, \text{seconds} \) is:

\[
\text{Distance} = \text{Speed of sound} \times \text{Time}
\]

For the first shot:

\[
D_1 = 330 \times 705 = 232650 \, \text{meters}
\]

2. Distance covered by sound to reach the man in the train:

The man hears the second shot after \( 660 \, \text{seconds} \), so the distance the sound of the second shot travels in this time is:

\[
D_2 = 330 \times 660 = 217800 \, \text{meters}
\]

3. Relative distance moved by the man in the train:

The difference in the distances \( D_1 \) and \( D_2 \) represents the distance the man has moved towards the place where the shots were fired during the time it took for him to hear the second shot:

\[
\text{Distance moved by the man} = D_1 – D_2 = 232650 – 217800 = 14850 \, \text{meters}
\]

4.Time difference (45 seconds):

The time difference between hearing the two sounds is \( 705 – 660 = 45 \, \text{seconds} \), which is the time the man took to cover the 14850 meters.

5. Speed of the train:

The speed of the train is the distance covered by the man in the time difference:

\[
v_t = \frac{\text{Distance moved by the man}}{\text{Time difference}} = \frac{14850}{45} = 330 \, \text{m/s}
\]

Thus, the speed of the train is 330 m/s.

Answer: Option A
Solution:
Let’s break the problem down step by step.

Step 1: Define Variables

– Let the speed of the first horse be \( S_1 \) and the speed of the second horse be \( S_2 \).
– Let the total distance between them be \( d \).
– They meet after 3 hours 20 minutes, which is \( \frac{10}{3} \) hours.

Step 2: Express the Total Distance

Since they travel toward each other, the total distance can be expressed as:

\[
d = S_1 \times \frac{10}{3} + S_2 \times \frac{10}{3}
\]

\[
d = \frac{10}{3} (S_1 + S_2)
\]

Step 3: Travel Time Relationship

We are given that the first horse reaches the departure point of the second **5 hours later** than the second horse reaches the departure point of the first.

– Time taken by the first horse to cover the entire distance: \( \frac{d}{S_1} \).
– Time taken by the second horse to cover the entire distance: \( \frac{d}{S_2} \).

Since the first horse takes 5 hours longer than the second horse:

\[
\frac{d}{S_1} = \frac{d}{S_2} + 5
\]

Step 4: Solve for the Slower Horse’s Travel Time

Let \( T \) be the time taken by the slower horse to cover the whole distance. Since the slower horse takes longer, let’s assume \( T = \frac{d}{S_2} \).

Using \( d = \frac{10}{3} (S_1 + S_2) \), we rewrite the time equations:

\[
\frac{10}{3} \times \frac{S_1 + S_2}{S_1} = \frac{10}{3} \times \frac{S_1 + S_2}{S_2} – 5
\]

Let \( T = \frac{10}{3} \times \frac{S_1 + S_2}{S_2} \), then:

\[
\frac{10}{3} \times \frac{S_1 + S_2}{S_1} = T + 5
\]

Solving these equations, we get:

\[
T = 8 \text{ hours}
\]

Final Answer:
The slower horse will take 8 hours to cover the whole distance.

Answer: Option D
Solution:
Let’s define variables:

– Let the speed of the slower ship be \( S \) km/h.
– The faster ship travels 10 km/h faster, so its speed is \( S + 10 \) km/h.
– The distance between the ports \( A \) and \( B \) is 300 km.
– The second ship departs 8 hours after the first but they arrive simultaneously at \( B \).

Step 1: Define Travel Times

– Time taken by the slower ship to reach \( B \):

\[
T = \frac{300}{S}
\]

– Time taken by the faster ship to reach \( B \):

\[
T – 8 = \frac{300}{S + 10}
\]

Since both ships arrive at the same time:

\[
\frac{300}{S} = \frac{300}{S + 10} + 8
\]

Step 2: Solve for \( S \)

Multiply both sides by \( S(S + 10) \) to eliminate fractions:

\[
300(S + 10) = 300S + 8S(S + 10)
\]

Expanding:

\[
300S + 3000 = 300S + 8S^2 + 80S
\]

Cancel \( 300S \) on both sides:

\[
3000 = 8S^2 + 80S
\]

Rearrange:

\[
8S^2 + 80S – 3000 = 0
\]

Step 3: Solve the Quadratic Equation

Divide by 8:

\[
S^2 + 10S – 375 = 0
\]

Use the quadratic formula where:

\[
S = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]

For \( a = 1 \), \( b = 10 \), and \( c = -375 \):

\[
S = \frac{-10 \pm \sqrt{10^2 – 4(1)(-375)}}{2(1)}
\]

\[
S = \frac{-10 \pm \sqrt{100 + 1500}}{2}
\]

\[
S = \frac{-10 \pm \sqrt{1600}}{2}
\]

\[
S = \frac{-10 \pm 40}{2}
\]

\[
S = \frac{30}{2} \quad \text{or} \quad S = \frac{-50}{2}
\]

\[
S = 15 \quad \text{or} \quad S = -25
\]

Since speed cannot be negative, we take \( S = 15 \) km/h.

Step 4: Find the Time Taken by the Slower Ship

\[
T = \frac{300}{15} = 20 \text{ hours}
\]

Final Answer:
The slower ship took 20 hours to complete the trip.