Time and Distances Questions in English

(C) Let the distance between $P$ and $Q$ be $d$.
Speed of the first man $(v_1)$ $= \frac{d}{4} \, \text{km/h}$.
Speed of the second man $(v_2)$ $= \frac{d}{4} \, \text{km/h}$.
The second man starts $2 \, \text{hours}$ earlier. Let the first man meet the second man after $x \, \text{hours}$ from the time he starts from $P$.
In $x \, \text{hours}$,the first man covers a distance of $v_1 \times x = \frac{d}{4}x$.
The second man has been traveling for $(x + 2) \, \text{hours}$ when they meet. The distance covered by the second man is $v_2 \times (x + 2) = \frac{d}{4}(x + 2)$.
The sum of the distances covered by both men must equal the total distance $d$:
$\frac{d}{4}x + \frac{d}{4}(x + 2) = d$
Dividing both sides by $d$ (assuming $d \neq 0$):
$\frac{x}{4} + \frac{x + 2}{4} = 1$
$x + x + 2 = 4$
$2x = 2$
$x = 1 \, \text{hour}$.

(B) Let the time spent in the air be $x \, hrs$ and the time spent in the train be $y \, hrs$.
Given,$x + y = 4 \, hrs$ (Equation $1$).
If he travelled the entire distance by air,he would have saved $\frac{4}{5}$ of the time spent in the train $(y)$.
Given that this time saved is $2 \, hrs$,we have $\frac{4}{5}y = 2$.
Solving for $y$: $y = 2 \times \frac{5}{4} = 2.5 \, hrs$.
Substituting $y = 2.5$ into Equation $1$: $x + 2.5 = 4 \Rightarrow x = 1.5 \, hrs$.
Let $v_a$ be the speed of the aeroplane and $v_t$ be the speed of the train.
The total distance is $v_a x + v_t y = 360$.
If he travelled the whole way by air,the time taken would be $\frac{360}{v_a}$.
The time saved is $4 - \frac{360}{v_a} = 2 \Rightarrow \frac{360}{v_a} = 2 \Rightarrow v_a = 180 \, km/h$.
Distance covered by air $= v_a \times x = 180 \times 1.5 = 270 \, km$.
Distance covered by train $= 360 - 270 = 90 \, km$.

(A) Let the scheduled time be $T$ hours and the original speed be $v \, \text{km/hr}$.
Distance $D = 1500 \, \text{km}$.
So,$v = \frac{1500}{T}$,which implies $T = \frac{1500}{v}$.
Since the plane started $30 \, \text{minutes}$ $(0.5 \, \text{hours})$ late,the time taken in the second case is $(T - 0.5)$ hours at a speed of $(v + 250) \, \text{km/hr}$.
$1500 = (v + 250)(T - 0.5)$.
Substituting $T = \frac{1500}{v}$:
$1500 = (v + 250)(\frac{1500}{v} - 0.5)$.
$1500 = 1500 - 0.5v + \frac{375000}{v} - 125$.
$0.5v + 125 - \frac{375000}{v} = 0$.
Multiplying by $2v$: $v^2 + 250v - 750000 = 0$.
$(v + 1000)(v - 750) = 0$.
Since speed cannot be negative,$v = 750 \, \text{km/hr}$.

(A) Let the original speed of the train be $v \, km/hr$ and the scheduled time be $t \, hr$.
Distance $= 120 \, km$.
Original case: $v = 120 / t \Rightarrow t = 120 / v$.
New case: The train leaves $1 \, hr$ early,so it has $t + 1$ hours to cover the distance at a reduced speed of $(v - 4) \, km/hr$ to arrive on time.
Thus,$v - 4 = 120 / (t + 1)$.
Substitute $t = 120 / v$ into the equation:
$v - 4 = 120 / (120/v + 1) = 120 / ((120 + v) / v) = 120v / (120 + v)$.
$(v - 4)(120 + v) = 120v$.
$120v + v^2 - 480 - 4v = 120v$.
$v^2 - 4v - 480 = 0$.
$(v - 24)(v + 20) = 0$.
Since speed cannot be negative,$v = 24 \, km/hr$.

(D) First,convert the speeds from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$.
Speed of hare $= 12 \times \frac{5}{18} = \frac{10}{3} \, m/s$.
Speed of dog $= 16 \times \frac{5}{18} = \frac{40}{9} \, m/s$.
In $1$ minute ($60$ seconds),the hare runs a distance of $d_1 = \frac{10}{3} \times 60 = 200 \, m$.
At the moment the dog starts chasing,the total distance between them is $D = 100 + 200 = 300 \, m$.
The relative speed of the dog with respect to the hare is $v_{rel} = (16 - 12) \, km/h = 4 \times \frac{5}{18} = \frac{10}{9} \, m/s$.
The time taken by the dog to catch the hare is $t = \frac{D}{v_{rel}} = \frac{300}{10/9} = 300 \times \frac{9}{10} = 270 \, s$.
The distance covered by the hare after being spotted (during the chase) is $d_2 = \text{speed of hare} \times t = \frac{10}{3} \times 270 = 900 \, m$.

(D) First,convert the time taken by the motorcycle into hours: $6 \, hr \, 12 \, min = 6 + \frac{12}{60} = 6 + 0.2 = 6.2 \, hr$.
Calculate the distance to the destination using the motorcycle's speed: $\text{Distance} = \text{Speed} \times \text{Time} = 30 \, km/h \times 6.2 \, hr = 186 \, km$.
Now,calculate the time taken by the train to cover the same distance: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{186 \, km}{24 \, km/h} = 7.75 \, hr$.
Thus,the second person takes $7.75 \, hr$ to reach the destination.

(A) The time taken to travel the distance without stops is calculated as: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{999}{55.5} = 18 \, \text{hours}$.
The train also stops for $1 \, \text{hour} \, 20 \, \text{minutes}$.
Total time taken = $18 \, \text{hours} + 1 \, \text{hour} \, 20 \, \text{minutes} = 19 \, \text{hours} \, 20 \, \text{minutes}$.
The train leaves at $6:00 \, am$. Adding $19 \, \text{hours} \, 20 \, \text{minutes}$ to $6:00 \, am$:
$6:00 \, am + 12 \, \text{hours} = 6:00 \, pm$.
$6:00 \, pm + 7 \, \text{hours} \, 20 \, \text{minutes} = 1:20 \, am$ (the next day).
Therefore,the train reaches $B$ at $1:20 \, am$.

(A) Let the distance between station $A$ and station $B$ be $x \, km$.
Let the initial speed of the train be $v \, km/h$.
Time taken in the first case is $45 \, min = \frac{45}{60} \, h = \frac{3}{4} \, h$.
So,$x = v \times \frac{3}{4} \Rightarrow v = \frac{4x}{3}$.
Time taken in the second case is $48 \, min = \frac{48}{60} \, h = \frac{4}{5} \, h$.
The new speed is $(v - 5) \, km/h$.
So,$x = (v - 5) \times \frac{4}{5} \Rightarrow v - 5 = \frac{5x}{4} \Rightarrow v = \frac{5x}{4} + 5$.
Equating the two expressions for $v$:
$\frac{4x}{3} = \frac{5x}{4} + 5$.
$\frac{4x}{3} - \frac{5x}{4} = 5$.
$\frac{16x - 15x}{12} = 5$.
$\frac{x}{12} = 5 \Rightarrow x = 60 \, km$.

(A) Let the speed of the truck be $x \, km/h.$
Since both vehicles are moving in the same direction,their relative speed is $(45 - x) \, km/h.$
To convert the relative speed into $m/s,$ we multiply by $\frac{5}{18}:$
Relative speed $= (45 - x) \times \frac{5}{18} \, m/s.$
We know that $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}.$
Given $\text{Distance} = 150 \, m$ and $\text{Time} = 30 \, s,$
$30 = \frac{150}{(45 - x) \times \frac{5}{18}}.$
$30 = \frac{150 \times 18}{(45 - x) \times 5}.$
$30 = \frac{30 \times 18}{45 - x}.$
$45 - x = 18.$
$x = 45 - 18 = 27 \, km/h.$
Therefore,the speed of the truck is $27 \, km/h.$

(D) The speed of the car is given as $10\, m/s$.
To convert speed from $m/s$ to $km/h$,we multiply the value by $\frac{18}{5}$.
$\text{Speed} = 10 \times \frac{18}{5} = 2 \times 18 = 36\, km/h$.
Therefore,the speed of the car is $36\, km/h$.

(A) The average speed for a journey where the distance covered in both directions is the same is given by the formula: $\text{Average Speed} = \frac{2xy}{x+y}$,where $x$ and $y$ are the speeds for the two legs of the journey.
Here,$x = 100 \text{ km/h}$ and $y = 150 \text{ km/h}$.
Substituting these values into the formula:
$\text{Average Speed} = \frac{2 \times 100 \times 150}{100 + 150}$
$\text{Average Speed} = \frac{30000}{250}$
$\text{Average Speed} = 120 \text{ km/h}$.

(A) The formula for average speed when the distance traveled is the same for both legs of the journey is given by: $\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$.
Here,$v_1 = 12 \, km/h$ and $v_2 = 18 \, km/h$.
Substituting the values into the formula:
$\text{Average Speed} = \frac{2 \times 12 \times 18}{12 + 18} \, km/h$.
$\text{Average Speed} = \frac{432}{30} \, km/h$.
$\text{Average Speed} = 14.4 \, km/h$.

(B) The total distance traveled is $24 + 24 + 24 = 72 \, km$.
The time taken for each segment is calculated as follows:
Time $t_1 = \frac{24}{6} = 4 \, h$
Time $t_2 = \frac{24}{8} = 3 \, h$
Time $t_3 = \frac{24}{12} = 2 \, h$
Total time taken $= 4 + 3 + 2 = 9 \, h$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{72}{9} = 8 \, km/h$.

(C) Let the distance between $A$ and $B$ be $d \, km$.
The time taken to go from $A$ to $B$ is $t_1 = \frac{d}{12} \, h$.
The time taken to return from $B$ to $A$ is $t_2 = \frac{d}{4} \, h$.
Total distance covered $= d + d = 2d \, km$.
Total time taken $= t_1 + t_2 = \frac{d}{12} + \frac{d}{4} = \frac{d + 3d}{12} = \frac{4d}{12} = \frac{d}{3} \, h$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/3} = 2d \times \frac{3}{d} = 6 \, km/h$.
Alternatively,for two equal distances covered at speeds $v_1$ and $v_2$,the average speed is $\frac{2v_1v_2}{v_1 + v_2} = \frac{2 \times 12 \times 4}{12 + 4} = \frac{96}{16} = 6 \, km/h$.

(A) Let the speed of $A$ be $3x$ and the speed of $B$ be $4x$.
Since the distance $D$ is the same for both, we have $D = \text{speed} \times \text{time}$.
Let $t$ be the time taken by $B$ in hours. Then the time taken by $A$ is $(t + \frac{20}{60}) \, h = (t + \frac{1}{3}) \, h$.
Equating the distances: $3x(t + \frac{1}{3}) = 4x(t)$.
Dividing both sides by $x$: $3(t + \frac{1}{3}) = 4t$.
$3t + 1 = 4t \Rightarrow t = 1 \, h$.
Therefore, the time taken by $A$ is $t + \frac{1}{3} = 1 + \frac{1}{3} = 1\frac{1}{3} \, h$.

(C) Speed of the truck $= \frac{\text{Distance}}{\text{Time}} = \frac{550 \, m}{1 \, min} = \frac{0.55 \, km}{(1/60) \, h} = 0.55 \times 60 \, km/h = 33 \, km/h$.
Speed of the bus $= \frac{\text{Distance}}{\text{Time}} = \frac{33 \, km}{45 \, min} = \frac{33 \, km}{(45/60) \, h} = \frac{33 \times 60}{45} \, km/h = 44 \, km/h$.
The ratio of the speed of the truck to the speed of the bus $= \frac{33}{44} = \frac{3}{4} = 3:4$.

(B) Let the distance between the house and the school be $x \text{ km}$.
Time taken at $2.5 \text{ km/h}$ is $T_1 = \frac{x}{2.5} = \frac{2x}{5} \text{ hours}$.
Time taken at $3 \text{ km/h}$ is $T_2 = \frac{x}{3} \text{ hours}$.
The difference in time is $6 \text{ min late} + 10 \text{ min early} = 16 \text{ min} = \frac{16}{60} \text{ hours} = \frac{4}{15} \text{ hours}$.
According to the problem,$T_1 - T_2 = \frac{4}{15}$.
$\frac{2x}{5} - \frac{x}{3} = \frac{4}{15}$.
$\frac{6x - 5x}{15} = \frac{4}{15}$.
$\frac{x}{15} = \frac{4}{15} \Rightarrow x = 4 \text{ km}$.

(B) Let the distance between the house and the school be $x \text{ km}$.
Initial speed $v_1 = 2.5 \text{ km/h} = \frac{5}{2} \text{ km/h}$.
New speed $v_2 = 2.5 + 1 = 3.5 \text{ km/h} = \frac{7}{2} \text{ km/h}$.
Time taken at speed $v_1$ is $t_1 = \frac{x}{v_1} = \frac{x}{5/2} = \frac{2x}{5} \text{ hours}$.
Time taken at speed $v_2$ is $t_2 = \frac{x}{v_2} = \frac{x}{7/2} = \frac{2x}{7} \text{ hours}$.
The difference in time is $6 \text{ min late} - (-6 \text{ min early}) = 12 \text{ min} = \frac{12}{60} \text{ hours} = \frac{1}{5} \text{ hours}$.
So,$t_1 - t_2 = \frac{1}{5} \Rightarrow \frac{2x}{5} - \frac{2x}{7} = \frac{1}{5}$.
Multiplying by $35$,we get $14x - 10x = 7$.
$4x = 7 \Rightarrow x = \frac{7}{4} \text{ km}$.

(D) Let the original speed be $v \, km/h$ and the time taken be $t = 30 \, h$.
Distance covered $d = v \times t = 30v \, km$.
If the speed is reduced by $\frac{1}{15}^{th}$,the new speed $v' = v - \frac{v}{15} = \frac{14v}{15} \, km/h$.
In the same time $t = 30 \, h$,the new distance covered $d' = v' \times t = \frac{14v}{15} \times 30 = 28v \, km$.
According to the problem,the man goes $10 \, km$ less in the same time:
$d - d' = 10$
$30v - 28v = 10$
$2v = 10$
$v = 5 \, km/h$.
Therefore,the original speed is $5 \, km/h$.

(A) Let the distance covered on foot be $x \, km$.
Then,the distance covered by bicycle is $(80 - x) \, km$.
Time taken to travel on foot = $\frac{\text{distance}}{\text{speed}} = \frac{x}{8} \, h$.
Time taken to travel by bicycle = $\frac{80 - x}{16} \, h$.
The total time taken is $7 \, h$.
So,$\frac{x}{8} + \frac{80 - x}{16} = 7$.
Multiply the entire equation by $16$ to clear the denominators:
$2x + (80 - x) = 7 \times 16$.
$x + 80 = 112$.
$x = 112 - 80 = 32 \, km$.
Therefore,the distance travelled on foot is $32 \, km$.

(B) Let the distance travelled on foot be $x \, km$.
Then the distance travelled by bicycle is $(61 - x) \, km$.
Time taken to travel on foot = $\frac{x}{4} \, h$.
Time taken to travel by bicycle = $\frac{61 - x}{9} \, h$.
According to the problem,the total time taken is $9 \, h$:
$\frac{x}{4} + \frac{61 - x}{9} = 9$
Multiply the entire equation by $36$ (the $LCM$ of $4$ and $9$):
$9x + 4(61 - x) = 9 \times 36$
$9x + 244 - 4x = 324$
$5x = 324 - 244$
$5x = 80$
$x = 16 \, km$.
Thus,the distance travelled on foot is $16 \, km$.

(C) Let the time taken for riding be $R$ and walking be $W$.
Given that walking one way and riding back takes $6\, h \, 15\, min$,which is $6.25\, h$ or $\frac{25}{4}\, h$.
So,$W + R = \frac{25}{4} \dots (1)$
Given that walking both ways takes $7\, h \, 45\, min$,which is $7.75\, h$ or $\frac{31}{4}\, h$.
So,$2W = \frac{31}{4} \Rightarrow W = \frac{31}{8}\, h$.
Substitute $W$ in equation $(1)$:
$R = \frac{25}{4} - \frac{31}{8} = \frac{50 - 31}{8} = \frac{19}{8}\, h$.
The time taken to ride both ways is $2R$:
$2R = 2 \times \frac{19}{8} = \frac{19}{4} = 4.75\, h$.
$4.75\, h = 4\, h \, 45\, min$.

(D) The time taken by each person to complete one round of the circular path is calculated as $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
For $A$: $T_A = \frac{12 \, km}{4 \, km/h} = 3 \, h$.
For $B$: $T_B = \frac{12 \, km}{3 \, km/h} = 4 \, h$.
For $C$: $T_C = \frac{12 \, km}{1.5 \, km/h} = \frac{12 \times 2}{3} = 8 \, h$.
To find when they will meet at the starting point,we need to calculate the Least Common Multiple $(LCM)$ of the times taken by each to complete one round.
$LCM(3, 4, 8) = 24$.
Therefore,they will meet together at the starting place at the end of $24 \, h$.

(C) Let the speed of Ravi be $x \, km/h$. Then,the speed of Ajay is $(x+4) \, km/h$.
Since they start simultaneously,the time taken by both to meet is the same.
Ravi travels a distance of $(60 - 12) = 48 \, km$ to reach the meeting point.
Ajay travels from $A$ to $B$ $(60 \, km)$ and then back $12 \, km$ to the meeting point,covering a total distance of $(60 + 12) = 72 \, km$.
Since time = distance / speed,we have:
$\frac{48}{x} = \frac{72}{x+4}$
$48(x + 4) = 72x$
$48x + 192 = 72x$
$72x - 48x = 192$
$24x = 192$
$x = \frac{192}{24} = 8 \, km/h$.
Thus,Ravi's speed is $8 \, km/h$.

(D) Let the total distance of the journey be $D \, km$.
The time interval between $11:00 \, am$ and $4:30 \, pm$ is $5 \, hours \, 30 \, minutes$,which is $5.5 \, hours$ or $\frac{11}{2} \, hours$.
The distance covered in this interval is $\frac{5}{6}D - \frac{3}{8}D = \frac{20D - 9D}{24} = \frac{11D}{24}$.
Speed of the person = $\frac{\text{Distance}}{\text{Time}} = \frac{11D/24}{11/2} = \frac{11D}{24} \times \frac{2}{11} = \frac{D}{12} \, km/h$.
Now,the time taken to cover the first $\frac{3}{8}D$ of the journey is $\frac{3D/8}{D/12} = \frac{3}{8} \times 12 = 4.5 \, hours$ or $4 \, hours \, 30 \, minutes$.
Since he reached the $\frac{3}{8}$ mark at $11:00 \, am$,the starting time is $11:00 \, am - 4 \, hours \, 30 \, minutes = 6:30 \, am$.

(A) The ratio of speeds of $A$ and $B$ is $A:B = 800 : (800 - 40) = 800 : 760 = 20 : 19$.
The ratio of speeds of $B$ and $C$ is $B:C = 500 : (500 - 5) = 500 : 495 = 100 : 99$.
To find the ratio $A:C$,we multiply the ratios: $\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{20}{19} \times \frac{100}{99} = \frac{2000}{1881}$.
This means when $A$ covers $2000\, m$,$C$ covers $1881\, m$.
When $A$ covers $200\, m$,the distance covered by $C$ is $\frac{1881}{2000} \times 200 = 188.1\, m$.
Therefore,$A$ beats $C$ by $200 - 188.1 = 11.9\, m$.

(B) In a $1000 \, m$ race:
$A$ covers $1000 \, m$,while $B$ covers $1000 - 50 = 950 \, m$.
So,the ratio of speeds of $A$ and $B$ is $A:B = 1000:950 = 20:19$.
$A$ covers $1000 \, m$,while $C$ covers $1000 - 69 = 931 \, m$.
So,the ratio of speeds of $A$ and $C$ is $A:C = 1000:931$.
To find the ratio of speeds of $B$ and $C$,we calculate $\frac{B}{C} = \frac{B}{A} \times \frac{A}{C} = \frac{19}{20} \times \frac{1000}{931}$.
$\frac{B}{C} = \frac{19 \times 50}{931} = \frac{950}{931}$.
This means when $B$ covers $950 \, m$,$C$ covers $931 \, m$.
When $B$ covers $1000 \, m$,$C$ covers $\frac{931}{950} \times 1000 = 980 \, m$.
Therefore,$B$ can give $C$ a start of $1000 - 980 = 20 \, m$.

(A) Let $T_A, T_B, T_C$ be the time taken by $A, B,$ and $C$ to run $1000 \, m$ respectively.
From the first condition,$T_B = T_A + 25$.
From the third condition,$B$ beats $C$ by $30 \, s$,so $T_C = T_B + 30 = (T_A + 25) + 30 = T_A + 55$.
In the race between $A$ and $C$,$A$ covers $1000 \, m$ while $C$ covers $1000 - 275 = 725 \, m$.
Since $A$ finishes the race,$T_A$ is the time taken by $A$. In this time,$C$ covers $725 \, m$.
We know $T_C = T_A + 55$,which means $C$ takes $55 \, s$ more than $A$ to cover the remaining $275 \, m$.
Speed of $C = \frac{275 \, m}{55 \, s} = 5 \, m/s$.
Time taken by $C$ to run $1000 \, m$ is $T_C = \frac{1000 \, m}{5 \, m/s} = 200 \, s$.
Since $T_C = T_A + 55$,we have $200 = T_A + 55$,so $T_A = 145 \, s$.
$145 \, s = 2 \, min \, 25 \, s$.

(C) Let the distance be $x \text{ km}$.
The time taken for the journey by train is $\frac{x}{25} \text{ h}$.
The time taken for the journey on foot is $\frac{x}{4} \text{ h}$.
The total time taken is $5 \text{ h } 48 \text{ min}$.
Convert $48 \text{ min}$ to hours: $\frac{48}{60} \text{ h} = 0.8 \text{ h}$.
So,the total time is $5.8 \text{ h}$.
According to the problem: $\frac{x}{25} + \frac{x}{4} = 5.8$.
Taking the least common multiple $(LCM)$ of $25$ and $4$,which is $100$:
$\frac{4x + 25x}{100} = 5.8$
$\frac{29x}{100} = 5.8$
$29x = 580$
$x = \frac{580}{29} = 20 \text{ km}$.
Thus,the distance is $20 \text{ km}$.

(D) Let the time taken to walk one way be $W$ and the time taken to ride one way be $R$.
According to the problem,walking one way and riding back takes $4 \, h \, 30 \, min$,which is $4.5 \, h$:
$W + R = 4.5 \, h$ --- (Equation $1$)
It is given that riding both ways takes $3 \, h$:
$2R = 3 \, h \Rightarrow R = 1.5 \, h$ --- (Equation $2$)
Substitute the value of $R$ from Equation $2$ into Equation $1$:
$W + 1.5 = 4.5$
$W = 4.5 - 1.5 = 3 \, h$
The time required to walk both ways is $2W$:
$2W = 2 \times 3 = 6 \, h$.

(C) Since the train and the man are moving in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 84 \, km/h + 6 \, km/h = 90 \, km/h$.
To convert the speed from $km/h$ to $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 90 \times \frac{5}{18} = 5 \times 5 = 25 \, m/s$.
The length of the train is the distance covered by the train to pass the man.
Length of the train $= \text{Relative speed} \times \text{Time} = 25 \, m/s \times 4 \, s = 100 \, m$.

(D) The length of the train is $L = 150 \ m$.
The speed of the train is $v_t = 36 \ km/h$.
The speed of the man is $v_m = 9 \ km/h$.
Since both are moving in the same direction,the relative speed is $v_{rel} = v_t - v_m = 36 - 9 = 27 \ km/h$.
To convert the relative speed into $m/s$,we multiply by $\frac{5}{18}$:
$v_{rel} = 27 \times \frac{5}{18} = 3 \times \frac{5}{2} = 7.5 \ m/s$.
The time taken to pass the man is $t = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{150}{7.5} = \frac{1500}{75} = 20 \ s$.

(C) Let the speed of the train be $x \, km/h$.
Since the persons are walking in the same direction as the train,their relative speeds with respect to the train are $(x - 3) \, km/h$ and $(x - 5) \, km/h$ respectively.
To convert these into $m/s$,we multiply by $\frac{5}{18}$.
The length of the train $(L)$ remains constant in both cases.
$L = (x - 3) \times \frac{5}{18} \times 10$
$L = (x - 5) \times \frac{5}{18} \times 11$
Equating the two expressions for $L$:
$(x - 3) \times 10 = (x - 5) \times 11$
$10x - 30 = 11x - 55$
$11x - 10x = 55 - 30$
$x = 25 \, km/h$.
Thus,the speed of the train is $25 \, km/h$.

(C) First, convert the speed of the train from $km/h$ to $m/s$:
Speed $= 60 \times \frac{5}{18} \, m/s = \frac{300}{18} \, m/s = \frac{50}{3} \, m/s$.
Total distance covered by the train while crossing the platform is given by:
Distance $= \text{Speed} \times \text{Time} = \frac{50}{3} \times 30 = 500 \, m$.
This total distance is the sum of the length of the train and the length of the platform:
$\text{Total Distance} = \text{Length of Train} + \text{Length of Platform}$.
$500 \, m = 200 \, m + \text{Length of Platform}$.
Length of the platform $= 500 \, m - 200 \, m = 300 \, m$.

(B) The relative speed of the train with respect to the man moving in the same direction is calculated by subtracting their speeds: $v_{rel} = (63 - 3) \, km/h = 60 \, km/h$.
To convert this speed into $m/s$, multiply by $\frac{5}{18}$: $v_{rel} = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \, m/s$.
The time taken to cross the man is equal to the time taken to cover the length of the train at the relative speed: $t = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{500}{50/3} = 500 \times \frac{3}{50} = 10 \times 3 = 30 \, \text{seconds}$.

(A) Given: Distance $= 600 \, m$,Time $= 5 \, minutes = 5 \times 60 \, seconds = 300 \, seconds$.
Speed in $m/s = \frac{\text{Distance}}{\text{Time}} = \frac{600}{300} = 2 \, m/s$.
To convert speed from $m/s$ to $km/h$,multiply by $\frac{18}{5}$.
Speed in $km/h = 2 \times \frac{18}{5} = \frac{36}{5} = 7.2 \, km/h$.

(B) Let $W$ be the time taken to walk one way and $R$ be the time taken to ride one way.
Given: $W + R = 6$ hours $30$ minutes $= 390$ minutes $... (i)$
If he rides both ways,he gains $2$ hours $10$ minutes compared to walking and riding. This means the time taken to ride both ways is $2R = (W + R) - 2$ hours $10$ minutes.
$2R = 6$ hours $30$ minutes $- 2$ hours $10$ minutes $= 4$ hours $20$ minutes $= 260$ minutes $... (ii)$
From $(ii)$,$R = 130$ minutes.
Substitute $R$ in $(i)$: $W + 130 = 390 \Rightarrow W = 260$ minutes.
The time taken to walk both ways is $2W = 2 \times 260 = 520$ minutes.

(B) The total number of lines to be written is $8190$.
Since the two persons are working towards each other (one from the beginning and one from the end),their relative speed is the sum of their individual rates.
Relative speed $= 200 + 150 = 350$ lines per hour.
To find the time at which they meet,we divide the total number of lines by the relative speed:
Time $= \frac{8190}{350}$ hours.
Time $= \frac{819}{35} = 23.4$ hours.
Therefore,they will meet after $23.4$ hours.

(B) Let the total distance of the journey be $x \, km$.
The first half of the distance is $\frac{x}{2} \, km$,covered at a speed of $21 \, km/h$. The time taken for this part is $t_1 = \frac{x/2}{21} = \frac{x}{42} \, hours$.
The second half of the distance is $\frac{x}{2} \, km$,covered at a speed of $24 \, km/h$. The time taken for this part is $t_2 = \frac{x/2}{24} = \frac{x}{48} \, hours$.
The total time taken is $10 \, hours$,so $t_1 + t_2 = 10$.
$\frac{x}{42} + \frac{x}{48} = 10$
Taking the least common multiple of $42$ and $48$,which is $336$:
$\frac{8x + 7x}{336} = 10$
$\frac{15x}{336} = 10$
$15x = 3360$
$x = \frac{3360}{15} = 224 \, km$.
Therefore,the total distance is $224 \, km$.

(C) The total distance covered by the train in $4 \, hours$ is calculated as: $\text{Distance} = \text{Speed} \times \text{Time} = 45 \, km/h \times 4 \, h = 180 \, km$.
Converting the distance into meters: $180 \, km = 180,000 \, m$.
The telegraph poles are placed at an interval of $50 \, m$.
The number of intervals in the total distance is $\frac{180,000}{50} = 3600$.
Since the train passes the first pole at the starting point $(0 \, m)$,the total number of poles passed is $\text{Number of intervals} + 1 = 3600 + 1 = 3601$.

(C) Let the distance to be covered be $d \text{ km}$ and the scheduled time be $t \text{ hours}$.
Case $1$: Speed $= 4 \text{ km/h}$,Time taken $= t + \frac{15}{60} \text{ hours}$.
Since $\text{Distance} = \text{Speed} \times \text{Time}$,we have $d = 4(t + \frac{1}{4}) = 4t + 1$.
Case $2$: Speed $= 6 \text{ km/h}$,Time taken $= t - \frac{10}{60} \text{ hours}$.
Thus,$d = 6(t - \frac{1}{6}) = 6t - 1$.
Equating the two expressions for $d$: $4t + 1 = 6t - 1 \implies 2t = 2 \implies t = 1 \text{ hour}$.
Substituting $t = 1$ in the first equation: $d = 4(1) + 1 = 5 \text{ km}$.
Alternatively,using the formula: $\text{Distance} = \frac{S_1 \times S_2}{|S_1 - S_2|} \times \Delta t = \frac{4 \times 6}{6 - 4} \times \frac{15 + 10}{60} = \frac{24}{2} \times \frac{25}{60} = 12 \times \frac{5}{12} = 5 \text{ km}$.

(D) Let the time taken for the buses to meet be $t$ hours.
Distance travelled by the first bus is $d_1 = 20t$.
Distance travelled by the second bus is $d_2 = 25t$.
According to the problem,the second bus has travelled $80 \, km$ more than the first bus:
$d_2 - d_1 = 80$
$25t - 20t = 80$
$5t = 80$
$t = 16 \, \text{hours}$.
The total distance between the two bus stations is the sum of the distances travelled by both buses:
$\text{Total Distance} = d_1 + d_2 = 20t + 25t = 45t$.
Substituting $t = 16$:
$\text{Total Distance} = 45 \times 16 = 720 \, km$.

(C) Let the distance travelled be $d\, km$.
The time taken to travel by train is $t_1 = \frac{d}{25}\, hours$.
The time taken to walk back is $t_2 = \frac{d}{4}\, hours$.
The total time taken is $5\, hours\, 48\, minutes = 5 + \frac{48}{60} = 5 + \frac{4}{5} = \frac{29}{5}\, hours$.
According to the problem,$t_1 + t_2 = \frac{29}{5}$.
$\frac{d}{25} + \frac{d}{4} = \frac{29}{5}$.
Taking the least common multiple of $25$ and $4$,which is $100$:
$\frac{4d + 25d}{100} = \frac{29}{5}$.
$\frac{29d}{100} = \frac{29}{5}$.
$d = \frac{29}{5} \times \frac{100}{29} = 20\, km$.
Thus,the distance travelled is $20\, km$.

(A) The initial distance between the policeman and the thief is $200 \text{ meters} = 0.2 \text{ km}$.
The relative speed of the policeman with respect to the thief is $12 \text{ km/h} - 10 \text{ km/h} = 2 \text{ km/h}$.
The time taken by the policeman to catch the thief is given by the formula: $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$.
$\text{Time} = \frac{0.2 \text{ km}}{2 \text{ km/h}} = 0.1 \text{ hours} = \frac{1}{10} \text{ hours}$.
The distance the thief has run before being caught is calculated as: $\text{Distance} = \text{Speed of thief} \times \text{Time}$.
$\text{Distance} = 10 \text{ km/h} \times 0.1 \text{ h} = 1 \text{ km}$.

(B) Speed of the first train $(S_1)$ = $10\, m/s = 10 \times \frac{18}{5} = 36\, km/h$.
Speed of the second train $(S_2)$ = $36 \times (1 + \frac{1}{3}) = 36 \times \frac{4}{3} = 48\, km/h$.
Total distance between Trivandrum and Nagercoil = $68\, km$.
The first train starts at $8:00\, AM$ and the second at $8:20\, AM$. The first train travels alone for $20\, \text{minutes}$ $(1/3\, \text{hour})$.
Distance covered by the first train in $1/3\, \text{hour} = 36 \times \frac{1}{3} = 12\, km$.
Remaining distance to be covered by both trains = $68 - 12 = 56\, km$.
Relative speed of the two trains = $S_1 + S_2 = 36 + 48 = 84\, km/h$.
Time taken to meet after $8:20\, AM$ = $\frac{\text{Remaining Distance}}{\text{Relative Speed}} = \frac{56}{84} = \frac{2}{3}\, \text{hours}$.
The second train travels from Nagercoil for $2/3\, \text{hours}$ at $48\, km/h$.
Distance from Nagercoil = $48 \times \frac{2}{3} = 32\, km$.

(D) Let the time taken for the trains to meet be $t$ hours.
Distance travelled by the first train $D_1 = 50t$.
Distance travelled by the second train $D_2 = 40t$.
According to the problem,the first train travelled $100\, km$ more than the second:
$D_1 - D_2 = 100$
$50t - 40t = 100$
$10t = 100$
$t = 10\, \text{hours}$.
The total distance between $P$ and $Q$ is the sum of the distances covered by both trains:
$\text{Total Distance} = D_1 + D_2 = 50t + 40t = 90t$.
Substituting $t = 10$:
$\text{Total Distance} = 90 \times 10 = 900\, km$.

(C) The thief starts at $2.30$ $p.m.$ and the owner starts at $3.00$ $p.m.$.
In these $30$ minutes ($0.5$ hours),the distance covered by the thief is $60 \times 0.5 = 30\, km$.
Now,the relative speed of the owner with respect to the thief is $75 - 60 = 15\, km/h$.
The time taken by the owner to cover the $30\, km$ gap is $\frac{30\, km}{15\, km/h} = 2$ hours.
Since the owner started at $3.00$ $p.m.$,he will overtake the thief at $3.00 + 2 = 5.00$ $p.m.$

(B) Let the distance covered be $S \, km$ and the initial speed be $v \, km/hr$.
From the first condition,$S = v \times 8$,so $v = S/8$.
From the second condition,the speed is $(v + 4) \, km/hr$ and the time taken is $7.5 \, hours$ (which is $15/2 \, hours$).
Thus,$S = (v + 4) \times 7.5$.
Substituting $v = S/8$ into the second equation:
$S = (S/8 + 4) \times (15/2)$
$S = (S/8 + 32/8) \times (15/2)$
$S = ((S + 32) / 8) \times (15/2)$
$S = (15S + 480) / 16$
$16S = 15S + 480$
$S = 480 \, km$.
Therefore,the distance covered is $480 \, km$.

(D) Let the distance be $S \text{ km}$ and the usual speed be $u \text{ km/hr}$.
According to the first condition:
$\frac{S}{u} - \frac{S}{u+3} = \frac{40}{60} = \frac{2}{3}$
$\frac{3S}{u(u+3)} = \frac{2}{3}$
$9S = 2u(u+3) \quad \dots(1)$
According to the second condition:
$\frac{S}{u-2} - \frac{S}{u} = \frac{40}{60} = \frac{2}{3}$
$\frac{2S}{u(u-2)} = \frac{2}{3}$
$3S = u(u-2) \quad \dots(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{9S}{3S} = \frac{2u(u+3)}{u(u-2)}$
$3 = \frac{2(u+3)}{u-2}$
$3(u-2) = 2(u+3)$
$3u - 6 = 2u + 6$
$u = 12 \text{ km/hr}$
Substituting $u = 12$ in equation $(2)$:
$3S = 12(12-2)$
$3S = 12 \times 10$
$3S = 120$
$S = 40 \text{ km}$.

(A) Let the total distance be $S$ and the average speed be $v_{avg}$.
Time taken for the first one-third distance $(S/3)$ at $10\, km/hr$ is $t_1 = \frac{S/3}{10} = \frac{S}{30}$.
Time taken for the next one-third distance $(S/3)$ at $20\, km/hr$ is $t_2 = \frac{S/3}{20} = \frac{S}{60}$.
Time taken for the last one-third distance $(S/3)$ at $60\, km/hr$ is $t_3 = \frac{S/3}{60} = \frac{S}{180}$.
Total time $T = t_1 + t_2 + t_3 = \frac{S}{30} + \frac{S}{60} + \frac{S}{180} = S \left( \frac{6+3+1}{180} \right) = \frac{10S}{180} = \frac{S}{18}$.
Average speed $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{S}{S/18} = 18\, km/hr$.