Time and Distances Questions in English

(C) Average speed is defined as the total distance traveled divided by the total time taken.
Total distance $= 50 \, km + 40 \, km + 90 \, km = 180 \, km$.
Time taken for the first part $= \frac{50 \, km}{25 \, kmph} = 2 \, hours$.
Time taken for the second part $= \frac{40 \, km}{20 \, kmph} = 2 \, hours$.
Time taken for the third part $= \frac{90 \, km}{15 \, kmph} = 6 \, hours$.
Total time taken $= 2 + 2 + 6 = 10 \, hours$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{180 \, km}{10 \, hours} = 18 \, kmph$.

(C) The formula for average speed is given by: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.
First,calculate the total distance covered:
Distance $1 = 5 \, km/hr \times 6 \, hours = 30 \, km$.
Distance $2 = 4 \, km/hr \times 12 \, hours = 48 \, km$.
Total Distance $= 30 + 48 = 78 \, km$.
Next,calculate the total time taken:
Total Time $= 6 \, hours + 12 \, hours = 18 \, hours$.
Now,calculate the average speed:
$\text{Average Speed} = \frac{78}{18} \, km/hr$.
Dividing both numerator and denominator by $6$,we get:
$\text{Average Speed} = \frac{13}{3} \, km/hr = 4 \frac{1}{3} \, km/hr$.

(B) First,calculate the speed of the person.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{10 \frac{1}{5} \text{ km}}{3 \text{ hours}} = \frac{51/5}{3} \text{ km/h} = \frac{51}{5 \times 3} \text{ km/h} = \frac{17}{5} \text{ km/h}$.
Now,calculate the distance covered in $5 \text{ hours}$ using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
Distance $= \frac{17}{5} \text{ km/h} \times 5 \text{ hours} = 17 \text{ km}$.

(D) Step $1$: Calculate the speed of the train.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{110 \, m}{3 \, s} = \frac{110}{3} \, m/s$.
Step $2$: Calculate the total distance to be covered to cross the platform.
Total distance $= \text{Length of train} + \text{Length of platform} = 110 \, m + 165 \, m = 275 \, m$.
Step $3$: Calculate the time taken.
Time $= \frac{\text{Total distance}}{\text{Speed}} = \frac{275}{110/3} = \frac{275 \times 3}{110} = 2.5 \times 3 = 7.5 \, s$.

(D) The formula for distance covered when a train crosses a tunnel is: (Length of the train $+$ Length of the tunnel) $=$ Speed $\times$ Time.
First,convert the speed from $km/hr$ to $m/s$:
Speed $= 72 \times \frac{5}{18} = 20 \, m/s$.
Next,convert the time from minutes to seconds:
Time $= 1 \, minute = 60 \, seconds$.
Let the length of the tunnel be $x \, m$.
According to the formula:
$x + 700 = 20 \times 60$
$x + 700 = 1200$
$x = 1200 - 700$
$x = 500 \, m$.
Thus,the length of the tunnel is $500 \, metres$.

(C) The total distance covered by the train to cross the platform is the sum of the length of the train and the length of the platform.
Total distance $= 200 \, m + 200 \, m = 400 \, m$.
Time taken $= 20 \, s$.
Speed of the train $= \frac{\text{Total distance}}{\text{Time taken}} = \frac{400 \, m}{20 \, s} = 20 \, m/s$.
To convert the speed from $m/s$ to $km/hr$,multiply by $\frac{18}{5}$.
Speed $= 20 \times \frac{18}{5} \, km/hr = 4 \times 18 \, km/hr = 72 \, km/hr$.

(B) The total distance covered by the two trains to cross each other is the sum of their lengths: $D = 125 \, m + 125 \, m = 250 \, m$.
The time taken to cross is $t = 6 \, s$.
The relative speed of the trains moving in opposite directions is the sum of their individual speeds: $V_{rel} = V_1 + V_2$.
Given $V_1 = 65 \, km/h$. Let $V_2 = x \, km/h$.
Converting relative speed to $m/s$: $V_{rel} = (65 + x) \times \frac{5}{18} \, m/s$.
Using the formula $Distance = Speed \times Time$:
$250 = (65 + x) \times \frac{5}{18} \times 6$
$250 = (65 + x) \times \frac{5}{3}$
$65 + x = 250 \times \frac{3}{5}$
$65 + x = 50 \times 3 = 150$
$x = 150 - 65 = 85 \, km/h$.
Therefore,the speed of the other train is $85 \, km/h$.

(C) Let the usual speed be $s$ and the usual time be $t$. The distance $d$ is constant.
$d = s \times t$
When the speed becomes $\frac{3}{5}s$,the time taken is $t + 2 \frac{1}{2} = t + \frac{5}{2}$.
Since distance is constant,$s \times t = \frac{3}{5}s \times (t + \frac{5}{2})$.
Dividing both sides by $s$,we get $t = \frac{3}{5}(t + \frac{5}{2})$.
$t = \frac{3}{5}t + \frac{3}{5} \times \frac{5}{2}$.
$t - \frac{3}{5}t = \frac{3}{2}$.
$\frac{2}{5}t = \frac{3}{2}$.
$t = \frac{3}{2} \times \frac{5}{2} = \frac{15}{4} = 3 \frac{3}{4} \text{ hours}$.

(C) Let the normal speed of the train be $s$ and the usual time taken be $t$.
Since distance $d$ is constant, we have $d = s \times t$.
When the train runs at $\frac{7}{11}$ of its normal speed, the time taken is $22\, \text{hours}$.
So, $d = (\frac{7}{11} s) \times 22$.
Equating the two expressions for distance: $s \times t = \frac{7}{11} s \times 22$.
Dividing both sides by $s$: $t = \frac{7}{11} \times 22 = 7 \times 2 = 14\, \text{hours}$.
The time saved is the difference between the actual time taken and the usual time: $22 - 14 = 8\, \text{hours}$.

(C) Let the usual speed be $s$ and the usual time taken be $t$ hours.
Distance $d = s \times t$ .....$(i)$
When the speed is $\frac{3}{4}s$, the time taken is $(t + 2)$ hours.
Distance $d = \frac{3}{4}s \times (t + 2)$ .....$(ii)$
Since the distance $d$ is the same in both cases, equate $(i)$ and $(ii)$:
$s \times t = \frac{3}{4}s \times (t + 2)$
Divide both sides by $s$ (assuming $s \neq 0$):
$t = \frac{3}{4}(t + 2)$
$4t = 3(t + 2)$
$4t = 3t + 6$
$t = 6 \, \text{hours}$.
Therefore, the time taken at his usual speed is $6 \, \text{hours}$.

(A) The formula for average speed when the distance covered in both directions is the same is given by $\text{Average Speed} = \frac{2xy}{x+y}$,where $x$ and $y$ are the speeds for the two parts of the journey.
Here,$x = 12 \, km/hr$ and $y = 18 \, km/hr$.
Substituting these values into the formula:
$\text{Average Speed} = \frac{2 \times 12 \times 18}{12 + 18}$
$\text{Average Speed} = \frac{432}{30}$
$\text{Average Speed} = \frac{72}{5} = 14 \frac{2}{5} \, km/hr$.

(B) Let the speeds of the two trains be $v_1$ and $v_2$. Let $t_1$ and $t_2$ be the times taken by the trains to reach their destinations after passing each other.
Given $t_1 = 4 \ hours$ and $t_2 = 9 \ hours$.
The relationship between speeds and time taken after meeting is given by the formula: $\frac{v_1}{v_2} = \sqrt{\frac{t_2}{t_1}}$.
Substituting the values: $\frac{v_1}{v_2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
Therefore,the ratio of the speeds of the two trains is $3:2$.

(B) Let the speed of $A$ be $v_A$ and the speed of $B$ be $v_B$.
Let $t_1 = 16 \text{ hours}$ be the time taken by $A$ to reach $Q$ after meeting $B$.
Let $t_2 = 25 \text{ hours}$ be the time taken by $B$ to reach $P$ after meeting $A$.
The formula for the ratio of speeds when two objects meet and then take $t_1$ and $t_2$ time to reach their respective destinations is given by $\frac{v_A}{v_B} = \sqrt{\frac{t_2}{t_1}}$.
Substituting the given values: $\frac{v_A}{v_B} = \sqrt{\frac{25}{16}}$.
Therefore,$\frac{v_A}{v_B} = \frac{5}{4}$.
The ratio of their speeds is $5:4$.

(D) Step $1$: Convert the speed of the train from $km/h$ to $m/s$.
Speed $= 120 \times \frac{5}{18} = \frac{600}{18} = \frac{100}{3} \, m/s$.
Step $2$: Calculate the length of the platform $(x)$.
Total distance covered by the train = Length of train + Length of platform = $320 + x$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$320 + x = \frac{100}{3} \times 24$
$320 + x = 100 \times 8 = 800$
$x = 800 - 320 = 480 \, m$.
Step $3$: Calculate the speed of the man.
The man crosses the platform $(480 \, m)$ in $4 \, min$.
Time in seconds $= 4 \times 60 = 240 \, s$.
Speed of man $= \frac{\text{Distance}}{\text{Time}} = \frac{480}{240} = 2.0 \, m/s$.

(C) The formula for average speed is $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.
Total distance $= 39 \, km + 25 \, km = 64 \, km$.
Total time $= 45 \, minutes + 35 \, minutes = 80 \, minutes$.
To convert the time into hours,divide by $60$: $\text{Total time} = \frac{80}{60} \, h = \frac{4}{3} \, h$.
$\text{Average speed} = \frac{64 \, km}{(80/60) \, h} = \frac{64 \times 60}{80} \, km/h = 48 \, km/h$.

(B) Let the length of the platform be $x \, m$.
First,convert the speed of the train from $km/h$ to $m/s$:
Speed of train $= 108 \times \frac{5}{18} = 30 \, m/s$.
When the train crosses the platform,the total distance covered is the sum of the length of the train and the length of the platform:
Distance $= 280 + x$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$:
$30 = \frac{280 + x}{12}$.
$360 = 280 + x$.
$x = 80 \, m$.
The length of the platform is $80 \, m$.
The boy crosses the same platform in $10 \, seconds$.
Speed of the boy $= \frac{\text{Length of platform}}{\text{Time taken}} = \frac{80}{10} = 8 \, m/s$.

(B) Speed of the truck $= \frac{224 \ km}{4 \ hours} = 56 \ km/h$.
Average speed of the bike $= \frac{1}{4} \times 56 \ km/h = 14 \ km/h$.
Distance covered by the bike in $7 \ hours = \text{Speed} \times \text{Time} = 14 \ km/h \times 7 \ hours = 98 \ km$.

(B) Let the actual distance be $d \, km$ and the time taken be $t \, hours$.
According to the problem,when the speed is $10 \, km/h$,the distance is $d$.
So,$10 = \frac{d}{t} \Rightarrow t = \frac{d}{10} \dots (i)$
When the speed is $14 \, km/h$,the distance covered is $d + 20 \, km$.
So,$14 = \frac{d + 20}{t} \Rightarrow t = \frac{d + 20}{14} \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{d}{10} = \frac{d + 20}{14}$
$14d = 10(d + 20)$
$14d = 10d + 200$
$4d = 200$
$d = 50 \, km$
Thus,the actual distance travelled is $50 \, km$.

(A) Given that the adult fare is $102$.
Since the adult fare is three times the child fare,the child fare is calculated as $\frac{102}{3} = 34$.
To find the total fare for $2$ adults and $3$ children,we calculate: $2 \times (102) + 3 \times (34)$.
This equals $204 + 102 = 306$.
Therefore,the total fare is $306$.

(A) The average speed is defined as the total distance traveled divided by the total time taken.
Total distance = $75 \, km + 25 \, km + 50 \, km = 150 \, km$.
Time taken for the first part = $\frac{75 \, km}{25 \, km/h} = 3 \, h$.
Time taken for the second part = $\frac{25 \, km}{5 \, km/h} = 5 \, h$.
Time taken for the third part = $\frac{50 \, km}{25 \, km/h} = 2 \, h$.
Total time = $3 \, h + 5 \, h + 2 \, h = 10 \, h$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{150 \, km}{10 \, h} = 15 \, km/h$.

(B) The total distance to be covered by the train to cross the platform is the sum of the length of the train and the length of the platform.
Total distance $= 700 \, m + 500 \, m = 1200 \, m$.
The speed of the train is $54 \, km/h$. Converting this into $m/s$:
Speed $= 54 \times \frac{5}{18} = 15 \, m/s$.
Using the formula,$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
$t = \frac{1200 \, m}{15 \, m/s} = 80 \, sec$.
Therefore,the time taken is $80 \, sec$.

(C) The total driving time is calculated as: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1000}{50} = 20 \, hrs$.
He takes a rest after every $3 \, hours$ of driving. The number of rest intervals is calculated by dividing the total driving time by the interval duration: $\frac{20}{3} = 6.66$.
Since he only rests after completing full $3 \, hour$ segments,he will take $6$ rests in total (at the end of $3, 6, 9, 12, 15,$ and $18$ hours).
Total rest time $= 6 \times 20 \, minutes = 120 \, minutes = 2 \, hours$.
Total time taken $= \text{Driving time} + \text{Rest time} = 20 + 2 = 22 \, hrs$.

(C) Step $1$: Calculate the time taken by the car.
Time = Distance / Speed = $330 \, km / 55 \, km/h = 6 \, hours$.
Step $2$: Determine the distance and time for the bike.
Distance covered by bike = $330 \, km - 15 \, km = 315 \, km$.
Time taken by bike = $6 \, hours - 1 \, hour = 5 \, hours$.
Step $3$: Calculate the average speed of the bike.
Average speed = Distance / Time = $315 \, km / 5 \, hours = 63 \, km/h$.

(A) The formula for distance is: $\text{Distance} = \text{Speed} \times \text{Time}$.
Given,$\text{Speed} = 40 \, km/h$ and $\text{Time} = 9 \, hours$.
Therefore,$\text{Distance} = 40 \times 9 = 360 \, km$.
Now,to find the time taken at a speed of $60 \, km/h$ for the same distance:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{360}{60} = 6 \, hours$.

(C) To find the time taken by the train to cross a man, we use the formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
Here, the distance to be covered is the length of the train, which is $75 \, m$.
The speed of the train is $20 \, km/h$. We convert this into $m/s$ by multiplying by $\frac{5}{18}$:
$\text{Speed} = 20 \times \frac{5}{18} = \frac{100}{18} = \frac{50}{9} \, m/s$.
Now, substitute the values into the formula:
$t = \frac{75}{50/9} = \frac{75 \times 9}{50} = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5 \, seconds$.

(A) To find the time taken by the train to cross an electric pole,we use the formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
$1$. Convert the speed from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$:
$\text{Speed} = 144 \times \frac{5}{18} = 8 \times 5 = 40 \, m/s$.
$2$. The distance covered by the train to cross the pole is equal to its own length,which is $100 \, m$.
$3$. Calculate the time:
$t = \frac{100 \, m}{40 \, m/s} = 2.5 \, \text{seconds}$.
Therefore,the train will cross the electric pole in $2.5 \, \text{seconds}$.

(C) Let the length of the train be $x\, m$. Then,the length of the platform is $2x\, m$.
Total distance covered by the train to cross the platform $= \text{Length of train} + \text{Length of platform} = x + 2x = 3x\, m$.
Speed of the train $= 60\, km/h = 60 \times \frac{5}{18}\, m/s = \frac{300}{18}\, m/s = \frac{50}{3}\, m/s$.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$3x = \left(\frac{50}{3}\right) \times 32.4$.
$3x = 50 \times 10.8$.
$3x = 540$.
$x = 180\, m$.
Length of the platform $= 2x = 2 \times 180 = 360\, m$.

(A) The speed of the train is given as $66 \, km/h$.
To convert the speed from $km/h$ to $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 66 \times \frac{5}{18} \, m/s = \frac{330}{18} \, m/s$.
When a train crosses a signal pole,the distance covered is equal to the length of the train $(d)$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$d = (66 \times \frac{5}{18}) \times 18$.
$d = 66 \times 5 = 330 \, m$.
Therefore,the length of the train is $330 \, m$.

(C) Speed of the first train $(S_1) = \frac{120\, m}{10\, s} = 12\, m/s$.
Speed of the second train $(S_2) = \frac{120\, m}{15\, s} = 8\, m/s$.
When two trains travel in opposite directions,their relative speed is the sum of their individual speeds: $S_{rel} = S_1 + S_2 = 12 + 8 = 20\, m/s$.
The total distance to be covered to cross each other is the sum of their lengths: $D = 120\, m + 120\, m = 240\, m$.
Time taken to cross each other $= \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{240\, m}{20\, m/s} = 12\, s$.

(A) The formula for average speed when the distance covered in both directions is the same is given by $\text{Average Speed} = \frac{2xy}{x+y}$,where $x$ and $y$ are the speeds for the two legs of the journey.
Here,$x = 70 \, km/hr$ and $y = 55 \, km/hr$.
Substituting the values into the formula:
$\text{Average Speed} = \frac{2 \times 70 \times 55}{70 + 55}$
$= \frac{2 \times 70 \times 55}{125}$
$= \frac{7700}{125}$
$= 61.6 \, km/hr$.

$(A)$ Let the distance be $d\, km$ and the scheduled time be $t\, hr$.
Case $1$: Speed $= 10\, km/hr$, Time taken $= t + \frac{15}{60} = t + \frac{1}{4}\, hr$.
Using $d = \text{speed} \times \text{time}$, we get $d = 10(t + \frac{1}{4}) \Rightarrow d = 10t + 2.5$ ... $(I)$.
Case $2$: Speed $= 10 + 2 = 12\, km/hr$, Time taken $= t + \frac{5}{60} = t + \frac{1}{12}\, hr$.
Using $d = \text{speed} \times \text{time}$, we get $d = 12(t + \frac{1}{12}) \Rightarrow d = 12t + 1$ ... $(II)$.
Equating $(I)$ and $(II)$:
$10t + 2.5 = 12t + 1$
$2.5 - 1 = 12t - 10t$
$1.5 = 2t \Rightarrow t = 0.75\, hr$.
Substituting $t$ in $(I)$:
$d = 10(0.75) + 2.5 = 7.5 + 2.5 = 10\, km$.

(A) Let the total distance be $2D \, km$.
Time taken for the first half $= \frac{D}{21} \, hrs$.
Time taken for the second half $= \frac{D}{24} \, hrs$.
Total time $= \frac{D}{21} + \frac{D}{24} = 10 \, hrs$.
Taking $D$ as common: $D \left( \frac{24 + 21}{504} \right) = 10$.
$D \left( \frac{45}{504} \right) = 10$.
$D = \frac{10 \times 504}{45} = \frac{5040}{45} = 112 \, km$.
Total distance $= 2D = 2 \times 112 = 224 \, km$.

(D) Since the two men are walking in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= (3 + 3.5) \, km/h = 6.5 \, km/h$.
The time taken is $3 \, h$.
The distance between them after $3 \, h$ is calculated as: $\text{Distance} = \text{Relative speed} \times \text{Time}$.
Distance $= 6.5 \, km/h \times 3 \, h = 19.5 \, km$.

(C) Given:
Speed of the man $= 15 \, km/h$
Time taken $= 5 \, minutes = \frac{5}{60} \, hours = \frac{1}{12} \, hours$
Distance (Length of the bridge) $= \text{Speed} \times \text{Time}$
Distance $= 15 \times \frac{1}{12} \, km$
Distance $= \frac{15}{12} \, km = 1.25 \, km$
To convert the distance into meters,multiply by $1000$:
Distance $= 1.25 \times 1000 \, m = 1250 \, m$
Therefore,the length of the bridge is $1250 \, m$.

(D) To convert speed from $km/hr$ to $m/s$,we multiply the value by $\frac{5}{18}$.
Given speed $= 180 \, km/hr$.
Speed in $m/s = 180 \times \frac{5}{18} \, m/s$.
$= 10 \times 5 \, m/s = 50 \, m/s$.
Therefore,the correct option is $D$.

(D) The speed of the train is given as $60\, km/hr$.
To convert the speed into $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \, m/s$.
Since the train crosses a pole,the distance covered is equal to the length of the train $(d)$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$d = \frac{50}{3} \times 30$.
$d = 50 \times 10 = 500 \, m$.
Therefore,the length of the train is $500 \, m$.

(C) When two objects move in the same direction,their relative speed is the difference of their individual speeds.
Relative speed $= (50 - 30) \text{ km/hr} = 20 \text{ km/hr}$.
To convert the speed from $\text{km/hr}$ to $\text{m/s}$,multiply by $\frac{5}{18}$:
Relative speed $= 20 \times \frac{5}{18} \text{ m/s} = \frac{100}{18} \text{ m/s}$.
The faster train crosses a man in the slower train,which means the distance covered is equal to the length of the faster train.
Distance $= \text{Relative speed} \times \text{Time}$.
Length of the faster train $= \left( 20 \times \frac{5}{18} \right) \text{ m/s} \times 18 \text{ s} = 100 \text{ meters}$.

(A) Let the length of the train be $L_t$ and the length of the platform be $L_p$.
Convert the speed of the train from $km/hr$ to $m/s$: $25 \times \frac{5}{18} = \frac{125}{18} \, m/s$.
When the train passes a man walking in the opposite direction at $5 \, km/hr$ $(5 \times \frac{5}{18} = \frac{25}{18} \, m/s)$,the relative speed is $\frac{125}{18} + \frac{25}{18} = \frac{150}{18} = \frac{25}{3} \, m/s$.
The length of the train $L_t = \text{Relative Speed} \times \text{Time} = \frac{25}{3} \times 12 = 100 \, m$.
When the train passes the platform,the total distance covered is $L_t + L_p$.
$L_t + L_p = \text{Speed of train} \times \text{Time} = \frac{125}{18} \times 18 = 125 \, m$.
Substituting $L_t = 100 \, m$,we get $100 + L_p = 125$,so $L_p = 25 \, m$.
The sum of the length of the train and the platform is $L_t + L_p = 100 + 25 = 125 \, m$.

(B) The formula for time is $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
Given,$\text{Distance} = \frac{4}{5} \, km$ and $\text{Speed} = 45 \, km/hr$.
$\text{Time} = \frac{4/5}{45} \, hr = \frac{4}{5 \times 45} \, hr = \frac{4}{225} \, hr$.
To convert hours into seconds,multiply by $3600$ $(1 \, hr = 60 \times 60 \, sec = 3600 \, sec)$.
$\text{Time} = \frac{4}{225} \times 3600 \, sec$.
$\text{Time} = 4 \times 16 \, sec = 64 \, sec$.

(C) The total distance covered by the train to cross the platform is the sum of the length of the train and the length of the platform.
Total distance $= 570 \, m + 570 \, m = 1140 \, m$.
Time taken $= 15 \, s$.
Speed $= \frac{\text{Total distance}}{\text{Time taken}}$.
Speed $= \frac{1140}{15} \, m/s = 76 \, m/s$.

(B) The initial speed of the boy is $p \, km/h$.
Due to the slippery ground,the speed is reduced by $q \, km/h$.
Therefore,the actual speed at which the boy runs is $(p - q) \, km/h$.
The distance to be covered is $1 \, km$.
The time taken to cover this distance is $r$ hours.
Using the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
Substituting the given values: $(p - q) = \frac{1}{r}$.
Thus,$\frac{1}{r} = p - q$.

(A) Let the distance between the village and the school be $d \, km$.
Time taken to go to school $= \frac{\text{distance}}{\text{speed}} = \frac{d}{3} \, hours$.
Time taken to return from school $= \frac{\text{distance}}{\text{speed}} = \frac{d}{2} \, hours$.
Total time taken $= \frac{d}{3} + \frac{d}{2} = 5 \, hours$.
Taking the least common multiple $(LCM)$ of $3$ and $2$,which is $6$:
$\frac{2d + 3d}{6} = 5$
$\frac{5d}{6} = 5$
$5d = 30$
$d = 6 \, km$.
Therefore,the distance between the village and the school is $6 \, km$.

(B) Let the distance between the house and the school be $d \, km$.
Let the scheduled time to reach the school be $t \, hours$.
Case $1$: Speed $= 5 \, km/h$,Time taken $= t + \frac{10}{60} \, hours$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$,we have $d = 5(t + \frac{1}{6})$.
Case $2$: Speed $= 6 \, km/h$,Time taken $= t - \frac{15}{60} \, hours$.
Using the formula,we have $d = 6(t - \frac{1}{4})$.
Equating both expressions for $d$:
$5(t + \frac{1}{6}) = 6(t - \frac{1}{4})$
$5t + \frac{5}{6} = 6t - \frac{6}{4}$
$t = \frac{5}{6} + \frac{3}{2} = \frac{5+9}{6} = \frac{14}{6} = \frac{7}{3} \, hours$.
Now,substitute $t$ back into the distance equation:
$d = 5(\frac{7}{3} + \frac{1}{6}) = 5(\frac{14+1}{6}) = 5(\frac{15}{6}) = 5(2.5) = 12.5 \, km$.

(C) Let the length of the train be $L \, m$ and its speed be $v \, m/s$.
When the train passes a station,the total distance covered is the sum of the length of the train and the length of the station.
For the first station: $v = \frac{L + 162}{18} \dots (I)$
For the second station: $v = \frac{L + 120}{15} \dots (II)$
Equating the speeds from $(I)$ and $(II)$:
$\frac{L + 162}{18} = \frac{L + 120}{15}$
Dividing the denominators by $3$,we get:
$\frac{L + 162}{6} = \frac{L + 120}{5}$
Cross-multiplying:
$5(L + 162) = 6(L + 120)$
$5L + 810 = 6L + 720$
$L = 810 - 720$
$L = 90 \, m$
Thus,the length of the train is $90 \, m$.

(B) The total distance to be covered to cross each other is the sum of the lengths of the two trains: $D = 120 \,m + 80 \,m = 200 \,m$.
Since the trains are moving in the same direction,their relative speed is the difference of their individual speeds: $v_{rel} = 50 \,km/h - 40 \,km/h = 10 \,km/h$.
Convert the relative speed into $m/s$: $v_{rel} = 10 \times \frac{5}{18} \,m/s = \frac{50}{18} \,m/s = \frac{25}{9} \,m/s$.
The time taken to cross each other is given by $t = \frac{D}{v_{rel}}$.
$t = \frac{200}{25/9} = 200 \times \frac{9}{25} = 8 \times 9 = 72 \,sec$.

(B) Let the length of each train be $x \, m$.
The speed of the first train is $v_1 = \frac{x}{4} \, m/s$.
The speed of the second train is $v_2 = \frac{x}{5} \, m/s$.
When the trains move in opposite directions,their relative speed is $v_{rel} = v_1 + v_2 = \frac{x}{4} + \frac{x}{5} = \frac{5x + 4x}{20} = \frac{9x}{20} \, m/s$.
To cross each other,the total distance to be covered is the sum of their lengths,which is $x + x = 2x \, m$.
The time taken to cross each other is $t = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{2x}{9x/20} = 2x \times \frac{20}{9x} = \frac{40}{9} \, seconds$.

(C) Let the speed of the train be $s \, km/h$.
Since the man is walking in the opposite direction,the relative speed is $(s + 3) \, km/h$.
To convert the relative speed into $m/s$,we multiply by $\frac{5}{18}$.
Relative speed $= (s + 3) \times \frac{5}{18} \, m/s$.
The distance covered by the train to cross the man is equal to its own length,which is $240 \, m$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we have:
$(s + 3) \times \frac{5}{18} = \frac{240}{10}$
$(s + 3) \times \frac{5}{18} = 24$
$s + 3 = 24 \times \frac{18}{5}$
$s + 3 = 4.8 \times 18$
$s + 3 = 86.4$
$s = 86.4 - 3 = 83.4 \, km/h$.

(C) Let the time taken by train $A$ to reach the meeting point be $t$ hours after $8 \, a.m.$
Since train $B$ starts at $9 \, a.m.$,it travels for $(t - 1)$ hours.
The distance covered by train $A$ is $60t$ and the distance covered by train $B$ is $75(t - 1)$.
The sum of the distances covered by both trains equals the total distance between the cities:
$60t + 75(t - 1) = 330$
$60t + 75t - 75 = 330$
$135t = 405$
$t = 3 \, \text{hours}$.
Since train $A$ started at $8 \, a.m.$,the meeting time is $8 \, a.m. + 3 \, \text{hours} = 11 \, a.m.$

(B) The total distance between the two men is $1200 \, m$.
Since they are walking towards each other,their relative speed is the sum of their individual speeds.
Relative speed $= 5 \, m/min + 10 \, m/min = 15 \, m/min$.
Let the time taken to meet be $t$ minutes.
Using the formula: $\text{Distance} = \text{Relative Speed} \times \text{Time}$.
$1200 = 15 \times t$.
$t = \frac{1200}{15} = 80 \, minutes$.
Therefore,they will meet after $80 \, minutes$.

(B) When two objects move in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 77 \, km/h + 67 \, km/h = 144 \, km/h$.
To convert the speed from $km/h$ to $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 144 \times \frac{5}{18} \, m/s = 8 \times 5 = 40 \, m/s$.
The total distance to be covered to cross each other is the sum of the lengths of the two trains:
Total distance $= 160 \, m + 140 \, m = 300 \, m$.
Time taken $= \frac{\text{Total distance}}{\text{Relative speed}} = \frac{300}{40} \, s = 7.5 \, s$.