Time and Distances Questions in English

(C) The formula for speed is given by $S = \frac{D}{T}$.
Given distance $D = 1.34$ $km = 1340$ $m$.
Given time $T = 4$ $sec$.
Substituting the values into the formula:
$S = \frac{1340 \text{ } m}{4 \text{ } sec} = 335$ $m/sec$.
Therefore,the speed of sound is $335$ $m/sec$.

(C) To convert speed from $km/h$ to $m/s$,we multiply the value by $\frac{5}{18}$.
Given speed $= 45\, km/h$.
Speed in $m/s = 45 \times \frac{5}{18}$.
Speed $= 2.5 \times 5 = 12.5\, m/s$.

(A) Total distance between place $X$ and $Y$ is calculated as: $\text{Distance} = \text{Speed} \times \text{Time} = 55\, km/hr \times 4\, hours = 220\, km$.
New speed after increasing by $5\, km/hr$ is: $55 + 5 = 60\, km/hr$.
Time taken with the new speed is: $\text{Time} = \frac{\text{Distance}}{\text{New Speed}} = \frac{220}{60}\, hours = \frac{11}{3}\, hours = 3\, hours + \frac{2}{3} \times 60\, minutes = 3\, hours + 40\, minutes$.
The initial time was $4\, hours$ $(240\, minutes)$.
The reduction in time is: $240\, minutes - 220\, minutes = 20\, minutes$.

(D) Let the distance between station $A$ and $B$ be $28 \, km$ ($LCM$ of $4$ and $3.5$).
Speed of train from $A = \frac{28}{4} = 7 \, km/h$.
Speed of train from $B = \frac{28}{3.5} = 8 \, km/h$.
The train from $A$ starts at $7 \, am$. By $8 \, am$,it has traveled for $1$ hour.
Distance covered by train from $A$ in $1$ hour $= 7 \, km/h \times 1 \, h = 7 \, km$.
Remaining distance at $8 \, am = 28 - 7 = 21 \, km$.
Since the trains are moving towards each other,their relative speed $= 7 + 8 = 15 \, km/h$.
Time taken to cover the remaining distance $= \frac{21}{15} \, hours = \frac{7}{5} \, hours = 1 \, hour \, 24 \, minutes$.
Therefore,the trains cross each other at $8 \, am + 1 \, hour \, 24 \, minutes = 9:24 \, am$.

(D) When a body covers a certain distance at a speed of $x \text{ km/h}$ and returns the same distance at a speed of $y \text{ km/h}$,the average speed for the entire journey is given by the formula: $\text{Average Speed} = \frac{2xy}{x+y}$.
Here,$x = 20 \text{ km/h}$ and $y = 30 \text{ km/h}$.
Substituting these values into the formula:
$\text{Average Speed} = \frac{2 \times 20 \times 30}{20 + 30} \text{ km/h}$.
$\text{Average Speed} = \frac{1200}{50} \text{ km/h}$.
$\text{Average Speed} = 24 \text{ km/h}$.
Thus,the average speed of the train for the whole journey is $24 \text{ km/h}$.

(A) Let the distance covered by the cyclist be $D$ and the time taken be $T$.
Then,the speed of the cyclist is $S_c = D / T$.
For the jogger,the distance covered is $D_j = D / 2$ and the time taken is $T_j = 2T$.
Therefore,the speed of the jogger is $S_j = D_j / T_j = (D / 2) / (2T) = D / (4T)$.
The ratio of the speed of the jogger to that of the cyclist is $S_j / S_c = (D / 4T) / (D / T) = 1 / 4$.
Thus,the ratio is $1:4$.

(A) Step $1$: Calculate the travel time without stops.
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{999}{55.5} = 18 \text{ hours}$.
Step $2$: Add the duration of the stops.
$\text{Total time} = 18 \text{ hours} + 1 \text{ hour } 20 \text{ minutes} = 19 \text{ hours } 20 \text{ minutes}$.
Step $3$: Calculate the arrival time.
The train starts at $6:00 \text{ am}$. Adding $19 \text{ hours } 20 \text{ minutes}$ to $6:00 \text{ am}$:
$6:00 \text{ am} + 12 \text{ hours} = 6:00 \text{ pm}$.
Remaining time to add: $19 \text{ hours } 20 \text{ minutes} - 12 \text{ hours} = 7 \text{ hours } 20 \text{ minutes}$.
$6:00 \text{ pm} + 7 \text{ hours } 20 \text{ minutes} = 1:20 \text{ am}$ (next day).
Note: The question asks for the time in $\text{pm}$,but based on the calculation,the arrival is at $1:20 \text{ am}$. Given the options,$1:20$ is the correct numerical value.

(B) Let the distance of the school from the house be $x \, km$.
Time taken at $4 \, km/h$ is $t_1 = \frac{x}{4} \, hours$.
Time taken at $3 \, km/h$ is $t_2 = \frac{x}{3} \, hours$.
The difference between the two times is $10 \, minutes$ early and $10 \, minutes$ late,which is $10 + 10 = 20 \, minutes$.
Converting $20 \, minutes$ into hours: $\frac{20}{60} = \frac{1}{3} \, hours$.
According to the problem:
$\frac{x}{3} - \frac{x}{4} = \frac{1}{3}$
Taking the least common multiple of $3$ and $4$,which is $12$:
$\frac{4x - 3x}{12} = \frac{1}{3}$
$\frac{x}{12} = \frac{1}{3}$
$x = \frac{12}{3} = 4 \, km$.
Therefore,the distance of the school from his house is $4 \, km$.

(C) The relative speed of the two trains moving in the same direction is the difference of their speeds.
Relative speed $= (40 - 20) \, km/h = 20 \, km/h$.
To convert the speed from $km/h$ to $m/s$,multiply by $\frac{5}{18}$.
Relative speed in $m/s = 20 \times \frac{5}{18} = \frac{100}{18} = \frac{50}{9} \, m/s$.
The fast train passes a man sitting in the slow train,which means the distance covered is equal to the length of the fast train.
Length of the fast train $=$ Relative speed $\times$ Time.
Length $= \frac{50}{9} \times 5 = \frac{250}{9} \, m$.
Converting to a mixed fraction: $\frac{250}{9} = 27 \frac{7}{9} \, m$.

(C) The boy started at $10:00 \, am$ and was caught at $1:30 \, pm$. The total time taken by the boy is $3 \, hours \, 30 \, minutes = 3.5 \, hours = \frac{7}{2} \, hours$.
Distance covered by the boy = $\text{Speed} \times \text{Time} = 12 \times \frac{7}{2} = 42 \, km$.
The elder brother started $1 \, hour \, 15 \, minutes$ later,so his travel time is $3 \, hours \, 30 \, minutes - 1 \, hour \, 15 \, minutes = 2 \, hours \, 15 \, minutes = 2.25 \, hours = \frac{9}{4} \, hours$.
Since the distance covered by both is the same,let the speed of the scooter be $x \, km/h$.
$x \times \frac{9}{4} = 42$.
$x = \frac{42 \times 4}{9} = \frac{14 \times 4}{3} = \frac{56}{3} = 18 \frac{2}{3} \, km/h$.

(C) The speed of the person from $P$ to $Q$ is $v_1 = 40 \, km/h$.
The speed of the person from $Q$ to $P$ is increased by $50\%$,so $v_2 = 40 + (50\% \text{ of } 40) = 40 + 20 = 60 \, km/h$.
The formula for average speed when the distance traveled is the same for both legs of the journey is $\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$.
Substituting the values: $\text{Average Speed} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \, km/h$.

(B) Let the speed of the second train be $x\, km/h$.
Since the trains are moving in opposite directions, their relative speed is $(x + 60)\, km/h$.
To convert the relative speed into $m/s$, we multiply by $\frac{5}{18}$:
Relative speed $= (x + 60) \times \frac{5}{18}\, m/s$.
The total distance to be covered is the sum of the lengths of both trains:
Total distance $= 180\, m + 270\, m = 450\, m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$450 = (x + 60) \times \frac{5}{18} \times 10.8$.
$450 = (x + 60) \times 5 \times 0.6$.
$450 = (x + 60) \times 3$.
$150 = x + 60$.
$x = 150 - 60 = 90\, km/h$.

(B) Let the two trains meet at a distance of $x\, \text{km}$ from Delhi.
The time taken by Paschim Express to cover $x\, \text{km}$ is $t_1 = \frac{x}{60}\, \text{hours}$.
The time taken by August Kranti Express to cover $x\, \text{km}$ is $t_2 = \frac{x}{80}\, \text{hours}$.
Since August Kranti Express starts $2\, \text{hours}$ later than Paschim Express $(16:30 - 14:30 = 2\, \text{hours})$,the difference in time taken is $2\, \text{hours}$.
Therefore,$\frac{x}{60} - \frac{x}{80} = 2$.
Taking the least common multiple of $60$ and $80$,which is $240$:
$\frac{4x - 3x}{240} = 2$.
$\frac{x}{240} = 2$.
$x = 480\, \text{km}$.
Thus,the two trains will meet at a distance of $480\, \text{km}$ from Delhi.

(C) Let the starting point be $A$.
$1$. The man walks $10 \, km$ South to reach point $B$.
$2$. He turns right (West) and walks $6 \, km$ to reach point $C$.
$3$. He turns right (North) and walks $10 \, km$ to reach point $D$.
$4$. Finally,he turns right (East) and walks $5 \, km$ to reach point $E$.
$5$. The distance from the starting point $A$ is $AE = AB - CD$ (if we consider the horizontal displacement) or more simply,looking at the diagram:
Horizontal distance from $A$ to $E$ is $6 \, km - 5 \, km = 1 \, km$.
Thus,the man is $1 \, km$ away from his starting point.

(A) Let the distance be $d \text{ km}$ and the scheduled time be $t \text{ minutes}$.
When the speed is $40 \text{ km/h}$,the time taken is $\frac{d}{40} \text{ hours} = \frac{60d}{40} \text{ minutes} = 1.5d \text{ minutes}$.
Given that the train is $11 \text{ minutes}$ late,we have: $1.5d = t + 11$ ... $(1)$
When the speed is $50 \text{ km/h}$,the time taken is $\frac{d}{50} \text{ hours} = \frac{60d}{50} \text{ minutes} = 1.2d \text{ minutes}$.
Given that the train is $5 \text{ minutes}$ late,we have: $1.2d = t + 5$ ... $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(1.5d - 1.2d) = (t + 11) - (t + 5)$
$0.3d = 6$
$d = \frac{6}{0.3} = 20 \text{ km}$.
Substituting $d = 20$ into equation $(1)$:
$1.5(20) = t + 11$
$30 = t + 11$
$t = 30 - 11 = 19 \text{ minutes}$.
Thus,the correct time is $19 \text{ minutes}$.

(B) Let the trains meet after $t$ hours.
The train starting from $P$ travels a distance $PR = 24t$.
The train starting from $Q$ travels a distance $QR = 18t$.
Since $P, Q,$ and $R$ are in a straight line with $R$ beyond $Q$,we have:
$PR - QR = PQ$
$24t - 18t = 27$
$6t = 27$
$t = \frac{27}{6} = 4.5 \text{ hours}$.
Therefore,the distance $QR = 18 \times 4.5 = 81 \, km$.

(C) Let the speed of train $A$ be $v_A = 45\, \text{km/h}$ and the speed of train $B$ be $v_B$.
Let $t_1$ be the time taken by train $A$ to reach $Y$ after passing $B$,and $t_2$ be the time taken by train $B$ to reach $X$ after passing $A$.
Given: $t_1 = 4\, \text{hours } 48\, \text{minutes} = 4 + \frac{48}{60} = 4 + \frac{4}{5} = \frac{24}{5}\, \text{hours}$.
Given: $t_2 = 3\, \text{hours } 20\, \text{minutes} = 3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}\, \text{hours}$.
The formula for the ratio of speeds when two objects meet and then travel to their respective destinations is $\frac{v_A}{v_B} = \sqrt{\frac{t_2}{t_1}}$.
Substituting the values: $\frac{45}{v_B} = \sqrt{\frac{10/3}{24/5}} = \sqrt{\frac{10}{3} \times \frac{5}{24}} = \sqrt{\frac{50}{72}} = \sqrt{\frac{25}{36}} = \frac{5}{6}$.
Therefore,$5 \times v_B = 45 \times 6$.
$v_B = \frac{45 \times 6}{5} = 9 \times 6 = 54\, \text{km/h}$.

(A) Let the distance between station $A$ and station $B$ be $x\, km.$
The time taken is $45\, minutes = \frac{45}{60} = \frac{3}{4}\, hours.$
The original speed of the train is $v = \frac{\text{Distance}}{\text{Time}} = \frac{x}{3/4} = \frac{4x}{3}\, km/h.$
According to the problem,if the speed is reduced by $5\, km/h,$ the new speed becomes $(\frac{4x}{3} - 5)\, km/h.$
The new time taken is $48\, minutes = \frac{48}{60} = \frac{4}{5}\, hours.$
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$x = (\frac{4x}{3} - 5) \times \frac{4}{5}$
Multiplying both sides by $5$:
$5x = (\frac{4x}{3} - 5) \times 4$
$5x = \frac{16x}{3} - 20$
Multiplying by $3$ to clear the denominator:
$15x = 16x - 60$
$60 = 16x - 15x$
$x = 60\, km.$
Thus,the distance between the stations is $60\, km.$

(C) Let the fare for the first five kilometres be $Rs. X$.
Total distance of the journey $= 187 \text{ km}$.
The fare for the first $5 \text{ km}$ is $X$.
The remaining distance $= 187 - 5 = 182 \text{ km}$.
The fare for the remaining distance at the rate of $Rs. 13$ per kilometer is $182 \times 13 = 2366$.
According to the problem,the total fare is $Rs. 2402$.
So,$X + 2366 = 2402$.
$X = 2402 - 2366$.
$X = 36$.
Therefore,the value of $X$ is $Rs. 36$.

(D) Speed of the train $= 120\, km/h$.
Converting to $m/s$: $120 \times \frac{5}{18} = \frac{100}{3}\, m/s$.
Let the length of the platform be $x\, m$.
When the train crosses the platform,the total distance covered is $(x + 320)\, m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$x + 320 = \frac{100}{3} \times 24$.
$x + 320 = 100 \times 8 = 800$.
$x = 800 - 320 = 480\, m$.
The man crosses the same platform $(480\, m)$ in $4\, minutes$.
Time in seconds $= 4 \times 60 = 240\, s$.
Speed of the man $= \frac{\text{Distance}}{\text{Time}} = \frac{480}{240} = 2\, m/s$.

(C) Average speed of a tractor $= \frac{575 \text{ km}}{23 \text{ hours}} = 25 \text{ km/h}$.
The speed of a bus is twice the speed of the tractor,so speed of bus $= 25 \times 2 = 50 \text{ km/h}$.
The average speed of a car is $1 \frac{4}{5} = \frac{9}{5}$ times the average speed of a bus,so speed of car $= 50 \times \frac{9}{5} = 90 \text{ km/h}$.
Distance covered by the car in $4 \text{ hours} = \text{speed} \times \text{time} = 90 \times 4 = 360 \text{ km}$.

(A) Let the length of Train $B$ be $x \text{ m}$.
Then the length of Train $A$ is $\frac{x}{2} \text{ m}$.
Time taken by Train $A$ = $25 \text{ seconds}$.
Time taken by Train $B$ = $1 \text{ minute } 15 \text{ seconds} = 60 + 15 = 75 \text{ seconds}$.
Speed of Train $A$ = $\frac{\text{Distance}}{\text{Time}} = \frac{x/2}{25} = \frac{x}{50} \text{ m/s}$.
Speed of Train $B$ = $\frac{\text{Distance}}{\text{Time}} = \frac{x}{75} \text{ m/s}$.
Ratio of speed of Train $A$ to Train $B$ = $\frac{x/50}{x/75} = \frac{75}{50} = \frac{3}{2} = 3:2$.

(D) Let the total distance be $D = 1 \, km$.
Time taken for the first part $(t_1)$ $= \frac{1/5}{8} = \frac{1}{40} \, h$.
Time taken for the second part $(t_2)$ $= \frac{1/10}{25} = \frac{1}{250} \, h$.
Remaining distance $= 1 - (\frac{1}{5} + \frac{1}{10}) = 1 - \frac{3}{10} = \frac{7}{10} \, km$.
Time taken for the remaining part $(t_3)$ $= \frac{7/10}{20} = \frac{7}{200} \, h$.
Total time taken $(T)$ $= t_1 + t_2 + t_3 = \frac{1}{40} + \frac{1}{250} + \frac{7}{200}$.
Finding a common denominator $(1000)$: $T = \frac{25}{1000} + \frac{4}{1000} + \frac{35}{1000} = \frac{64}{1000} = \frac{8}{125} \, h$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{1}{8/125} = \frac{125}{8} = 15.625 \, km/h$.

(B) Let the distance of the school from his house be $x \, km$.
The time taken to reach school at $3 \, km/h$ is $t_1 = \frac{x}{3} \, hours$.
The time taken to reach school at $4 \, km/h$ is $t_2 = \frac{x}{4} \, hours$.
According to the problem,the difference between these two times is $5 \, minutes$ late and $5 \, minutes$ early,which totals $10 \, minutes$.
Converting $10 \, minutes$ to hours: $10 \, minutes = \frac{10}{60} \, hours = \frac{1}{6} \, hours$.
Therefore,the equation is:
$\frac{x}{3} - \frac{x}{4} = \frac{1}{6}$
Taking the least common multiple $(LCM)$ of $3$ and $4$,which is $12$:
$\frac{4x - 3x}{12} = \frac{1}{6}$
$\frac{x}{12} = \frac{1}{6}$
$x = \frac{12}{6} = 2 \, km$.
Thus,the distance of the school from his house is $2 \, km$.

(B) Let the usual speed be $v$ and the usual time taken be $t$.
Since the distance $d$ is constant,$d = v \times t$.
When the man drives at $\frac{3}{4}$ of his original speed,the new speed is $v' = \frac{3}{4}v$.
The new time taken is $t' = t + 20$.
Since $d = v' \times t'$,we have $v \times t = \frac{3}{4}v \times (t + 20)$.
Dividing both sides by $v$,we get $t = \frac{3}{4}(t + 20)$.
$4t = 3t + 60$.
$t = 60 \text{ minutes}$.
Thus,the usual time is $60 \text{ minutes}$.

(D) Let the length of the train be $L$ meters and its speed be $v$ $m/s$.
From statement $II$,we know $L = y$.
From statement $III$,the train crosses a signal pole in $m$ seconds. The distance covered is the length of the train $(y)$. Thus,$v = y/m$ $m/s$.
From statement $I$,the train crosses a platform of length $x$ in $n$ seconds. The total distance covered is $(x + y)$. Thus,$v = (x + y)/n$ $m/s$.
While we can express the speed $v$ using statements $II$ and $III$ (or $I$ and $II$),the variables $x, y, n, m$ are not defined as numerical values. Without specific numerical values for these variables,the speed cannot be calculated as a specific numerical value in $km/h$. Therefore,the question cannot be answered.

(D) Step $1$: Calculate the speed of the car.
$\text{Speed of the car} = \frac{\text{Distance}}{\text{Time}} = \frac{588 \text{ km}}{6 \text{ hours}} = 98 \text{ km/h}$.
Step $2$: Calculate the speed of the train.
The speed of the train is $1 \frac{3}{7} = \frac{10}{7}$ times the speed of the car.
$\text{Speed of the train} = \frac{10}{7} \times 98 \text{ km/h} = 10 \times 14 \text{ km/h} = 140 \text{ km/h}$.
Step $3$: Calculate the distance covered by the train in $13 \text{ hours}$.
$\text{Distance} = \text{Speed} \times \text{Time} = 140 \text{ km/h} \times 13 \text{ hours} = 1820 \text{ km}$.

(D) Average speed is defined as the ratio of the total distance covered to the total time taken.
Total distance $= 39 \, km + 25 \, km = 64 \, km$.
Total time $= 45 \, minutes + 35 \, minutes = 80 \, minutes$.
To convert the time into hours,divide by $60$: $80 \, minutes = \frac{80}{60} \, hours = \frac{4}{3} \, hours$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{64 \, km}{(4/3) \, h} = 64 \times \frac{3}{4} \, km/h = 16 \times 3 \, km/h = 48 \, km/h$.

(B) Speed of the car $= \frac{\text{Distance}}{\text{Time}} = \frac{720 \text{ km}}{9 \text{ h}} = 80 \text{ km/h}$.
Speed of the bus $= \frac{3}{4} \times 80 \text{ km/h} = 60 \text{ km/h}$.
Given the ratio of the speed of the bus to the train is $15:27$,we have $\frac{\text{Speed of bus}}{\text{Speed of train}} = \frac{15}{27}$.
Speed of the train $= \frac{27}{15} \times 60 \text{ km/h} = 27 \times 4 = 108 \text{ km/h}$.
Distance covered by the train in $7 \text{ hours} = \text{Speed} \times \text{Time} = 108 \text{ km/h} \times 7 \text{ h} = 756 \text{ km}$.

(D) For Company $A$: The rate is $1 \text{ paisa}$ per $3 \text{ seconds}$.
Total time spent $= 591 \text{ seconds}$.
Cost for Company $A = \frac{591}{3} = 197 \text{ paisa}$.
For Company $B$: The rate is $45 \text{ paisa}$ per $60 \text{ seconds}$ $(1 \text{ minute})$.
Total time spent $= 780 \text{ seconds}$.
Cost for Company $B = \frac{45}{60} \times 780 = 45 \times 13 = 585 \text{ paisa}$.
Total amount spent $= 197 + 585 = 782 \text{ paisa}$.
Since $100 \text{ paisa} = 1 \text{ Rs.}$,the total amount is $7.82 \text{ Rs.}$

(B) First,convert the speed of the train from $km/h$ to $m/s$:
$108 \, km/h = 108 \times \frac{5}{18} = 30 \, m/s$.
Let the length of the platform be $L \, m$.
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:
$\text{Speed} = \frac{\text{Total Distance}}{\text{Time}}$
$30 = \frac{280 + L}{12}$
$360 = 280 + L$
$L = 360 - 280 = 80 \, m$.
The man crosses the same platform of length $80 \, m$ in $10 \, s$. Assuming the man's length is negligible:
$\text{Speed of the man} = \frac{\text{Length of platform}}{\text{Time}} = \frac{80}{10} = 8 \, m/s$.

(C) Step $1$: Calculate the speed of the tractor.
Speed of the tractor $= \frac{\text{Distance}}{\text{Time}} = \frac{575 \, km}{23 \, hours} = 25 \, km/h$.
Step $2$: Calculate the speed of the bus.
Given that the speed of the bus is twice the speed of the tractor.
Speed of the bus $= 2 \times 25 \, km/h = 50 \, km/h$.
Step $3$: Calculate the speed of the car.
The average speed of the car is $1 \frac{4}{5} = \frac{9}{5}$ times the speed of the bus.
Speed of the car $= \frac{9}{5} \times 50 \, km/h = 90 \, km/h$.
Step $4$: Calculate the distance covered by the car in $4 \, hours$.
Distance $= \text{Speed} \times \text{Time} = 90 \, km/h \times 4 \, hours = 360 \, km$.

(D) Average speed is defined as the ratio of total distance covered to the total time taken.
Total distance $= 39 \, km + 25 \, km = 64 \, km$.
Total time $= 45 \, minutes + 35 \, minutes = 80 \, minutes$.
To convert the time into hours,divide by $60$: Total time $= \frac{80}{60} \, hours = \frac{4}{3} \, hours$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{64 \, km}{\frac{4}{3} \, hours} = 64 \times \frac{3}{4} \, km/h = 16 \times 3 \, km/h = 48 \, km/h$.

(D) Let the speed of the train be $x \, km/h$.
Since the man is walking in the same direction,the relative speed is $(x - 3) \, km/h$.
To convert the relative speed into $m/s$,we multiply by $\frac{5}{18}$. So,relative speed $= (x - 3) \times \frac{5}{18} \, m/s$.
The distance covered by the train to pass the man is equal to its own length,which is $300 \, m$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we have:
$(x - 3) \times \frac{5}{18} = \frac{300}{33}$
$(x - 3) = \frac{300}{33} \times \frac{18}{5}$
$(x - 3) = \frac{60}{11} \times 6 = \frac{360}{11} = 32 \frac{8}{11}$
$x = 32 \frac{8}{11} + 3 = 35 \frac{8}{11} \, km/h$.

(C) Let the speed of the buses be $v_b = 20 \text{ km/h}$ and the speed of the man be $v_m = x \text{ km/h}$.
The distance between two consecutive buses is the distance covered by a bus in $10 \text{ minutes}$.
Distance $D = v_b \times \text{time} = 20 \times (10/60) \text{ km} = 20/6 \text{ km} = 10/3 \text{ km}$.
When the man moves in the opposite direction,the relative speed of the man with respect to the bus is $(v_b + v_m) = (20 + x) \text{ km/h}$.
The man meets the buses at intervals of $8 \text{ minutes}$,which means the distance $D$ is covered by the relative speed in $8 \text{ minutes}$.
$D = (v_b + v_m) \times \text{time interval}$.
$10/3 = (20 + x) \times (8/60)$.
$10/3 = (20 + x) \times (2/15)$.
Multiply both sides by $15/2$:
$(10/3) \times (15/2) = 20 + x$.
$5 \times 5 = 20 + x$.
$25 = 20 + x$.
$x = 5 \text{ km/h}$.

(C) Distance travelled by Car $A = \text{Speed} \times \text{Time} = 65 \text{ km/h} \times 8 \text{ h} = 520 \text{ km}$.
Distance travelled by Car $B = \text{Speed} \times \text{Time} = 70 \text{ km/h} \times 4 \text{ h} = 280 \text{ km}$.
The ratio of distances covered by Car $A$ to Car $B$ is $\frac{520}{280}$.
Simplifying the fraction: $\frac{520}{280} = \frac{52}{28} = \frac{13}{7}$.
Thus, the ratio is $13:7$.

(A) The distance between the starting point and the destination is calculated as: $\text{Distance} = \text{Speed} \times \text{Time} = 30 \, km/h \times 6 \, h = 180 \, km$.
Hema's original speed is: $\text{Speed} = \frac{180 \, km}{4 \, h} = 45 \, km/h$.
Deepa's new speed is $30 + 10 = 40 \, km/h$. The new time taken by Deepa is: $\text{Time} = \frac{180 \, km}{40 \, km/h} = 4.5 \, h = 4 \, hours \, 30 \, minutes$.
Hema's new speed is $45 + 5 = 50 \, km/h$. The new time taken by Hema is: $\text{Time} = \frac{180 \, km}{50 \, km/h} = 3.6 \, h = 3 \, hours \, 36 \, minutes$.
The difference in time taken is: $4 \, hours \, 30 \, minutes - 3 \, hours \, 36 \, minutes = 54 \, minutes$.

(A) Speed of bus $= \frac{480 \, km}{8 \, h} = 60 \, km/h$.
Given that the speed of the bus is $\frac{3}{4}$ of the speed of the train,we have: $60 = \frac{3}{4} \times \text{Speed of train}$.
Speed of train $= \frac{60 \times 4}{3} = 80 \, km/h$.
The ratio of the speed of the train to the car is $16: 15$. Let the speed of the train be $16x$ and the speed of the car be $15x$.
Since $16x = 80$,we find $x = 5$.
Speed of car $= 15 \times 5 = 75 \, km/h$.
Distance covered by the car in $6 \, hours = \text{Speed} \times \text{Time} = 75 \, km/h \times 6 \, h = 450 \, km$.

(A) Bus fare for $1$ person $= Rs. 420$.
Bus fare for $2$ persons $= 420 \times 2 = Rs. 840$.
Train fare for $1$ person $= \frac{3}{4} \times (\text{Bus fare for } 2 \text{ persons}) = \frac{3}{4} \times 840 = 3 \times 210 = Rs. 630$.
Total fare for $2$ persons by bus and $4$ persons by train $= (2 \times 420) + (4 \times 630)$.
$= 840 + 2520 = Rs. 3360$.

(A) Let the length of Train $B$ be $x \,m$. Since the speed of Train $A$ is constant,we can equate the speed calculated from both scenarios.
Speed of Train $A$ when crossing the pole: $v = \frac{\text{Length of Train } A}{\text{Time}} = \frac{240 \,m}{20 \,s} = 12 \,m/s$.
When Train $A$ crosses the stationary Train $B$,the total distance covered is the sum of the lengths of both trains: $D = 240 + x$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we have:
$12 = \frac{240 + x}{50}$
Multiplying both sides by $50$:
$600 = 240 + x$
Solving for $x$:
$x = 600 - 240 = 360 \,m$.
Thus,the length of the stationary Train $B$ is $360 \,m$.

(B) The total distance covered by the bike is calculated using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
Given,$\text{Speed} = 64 \, km/h$ and $\text{Time} = 8 \, hours$.
$\text{Distance} = 64 \times 8 = 512 \, km$.
Now,to cover the same distance of $512 \, km$ in $6 \, hours$,the required speed is calculated as:
$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{512}{6} \, km/h$.
$\text{Speed} \approx 85.33 \, km/h$.
Rounding to the nearest whole number,the approximate speed is $85 \, km/h$.

(B) Let the distance between the house and the school be $x \, km$.
Time taken at $3 \, km/h = \frac{x}{3} \, hours$.
Time taken at $4 \, km/h = \frac{x}{4} \, hours$.
The difference in time between the two scenarios is $5 \, minutes$ late and $5 \, minutes$ early,which equals $10 \, minutes$ or $\frac{10}{60} = \frac{1}{6} \, hours$.
According to the problem:
$\frac{x}{3} - \frac{x}{4} = \frac{1}{6}$
Taking the common denominator:
$\frac{4x - 3x}{12} = \frac{1}{6}$
$\frac{x}{12} = \frac{1}{6}$
$x = \frac{12}{6} = 2 \, km$.
Therefore,the distance is $2 \, km$.

(A) Let the speeds of the cars starting from $A$ and $B$ be $x \, km/h$ and $y \, km/h$ respectively,where $x > y$.
Case $I$: When both cars travel in the same direction,the relative speed is $(x - y) \, km/h$.
Given that they meet in $5 \, hours$ over a distance of $100 \, km$:
$5(x - y) = 100 \Rightarrow x - y = 20$ .....$(1)$
Case $II$: When both cars travel towards each other,the relative speed is $(x + y) \, km/h$.
Given that they meet in $1 \, hour$ over a distance of $100 \, km$:
$1(x + y) = 100 \Rightarrow x + y = 100$ .....$(2)$
Adding equations $(1)$ and $(2)$:
$(x - y) + (x + y) = 20 + 100$
$2x = 120 \Rightarrow x = 60 \, km/h$.
Substituting $x = 60$ in equation $(2)$:
$60 + y = 100 \Rightarrow y = 40 \, km/h$.
The speed of the faster car is $60 \, km/h$.

(C) Speed of the first train $(v_1)$ $= 60 \, km/h$.
Converting to $m/s$: $v_1 = 60 \times \frac{5}{18} = \frac{50}{3} \, m/s$.
Let the speed of the second train be $v_2 \, m/s$.
Since both trains are moving in the same direction,the relative speed is $(v_1 - v_2) = (\frac{50}{3} - v_2) \, m/s$.
The first train covers a relative distance of $120 \, m$ in $18 \, seconds$.
Using the formula: $\text{Relative Speed} = \frac{\text{Distance}}{\text{Time}}$.
$\frac{50}{3} - v_2 = \frac{120}{18} = \frac{20}{3}$.
$v_2 = \frac{50}{3} - \frac{20}{3} = \frac{30}{3} = 10 \, m/s$.
Converting back to $km/h$: $10 \times \frac{18}{5} = 36 \, km/h$.

(A) Let the length of the bus be $l$ units.
Speed of the man $= \frac{l}{18} \text{ units/s}$.
Speed of the bus $= \frac{l}{4} \text{ units/s}$.
Ratio of the speed of the bus to the speed of the man $= \frac{l/4}{l/18} = \frac{18}{4} = \frac{9}{2}$.
Therefore,the ratio is $9:2$.

(A) Step $1$: Calculate the speed of Train $A$. Since it crosses a pole in $20 \text{ seconds}$,the speed is $v = \frac{\text{Length of Train } A}{\text{Time}} = \frac{240 \text{ m}}{20 \text{ s}} = 12 \text{ m/s}$.
Step $2$: Calculate the total distance covered when crossing Train $B$. In $50 \text{ seconds}$,the distance covered is $D = \text{speed} \times \text{time} = 12 \text{ m/s} \times 50 \text{ s} = 600 \text{ meters}$.
Step $3$: The total distance covered when crossing a stationary train is the sum of the lengths of both trains $(L_A + L_B)$.
Therefore,$240 + L_B = 600 \text{ meters}$.
$L_B = 600 - 240 = 360 \text{ meters}$.

(A) Speed of the tractor $= \frac{360\,km}{12\,h} = 30\,km/h$.
Given the ratio of speeds of car,jeep,and tractor is $3:5:2$. Let the speeds be $3x$,$5x$,and $2x$ respectively.
Since the speed of the tractor is $30\,km/h$,we have $2x = 30$,which implies $x = 15$.
Speed of the car $= 3x = 3 \times 15 = 45\,km/h$.
Speed of the jeep $= 5x = 5 \times 15 = 75\,km/h$.
Average speed of the car and the jeep $= \frac{45 + 75}{2} = \frac{120}{2} = 60\,km/h$.

(C) Let the speeds of the car,train,and bus be $5x \text{ km/h}$,$9x \text{ km/h}$,and $4x \text{ km/h}$,respectively.
The average speed of the three is given by $\frac{5x + 9x + 4x}{3} = \frac{18x}{3} = 6x \text{ km/h}$.
Given that the average speed is $72 \text{ km/h}$,we have $6x = 72$,which implies $x = 12 \text{ km/h}$.
The average speed of the car and the train together is $\frac{5x + 9x}{2} = \frac{14x}{2} = 7x$.
Substituting the value of $x$,we get $7 \times 12 = 84 \text{ km/h}$.

(B) The total distance covered by the two trains to cross each other is the sum of their lengths: $180 \, m + 270 \, m = 450 \, m$.
Time taken to cross is $10.8 \, s$.
Relative speed of the two trains in $m/s = \frac{\text{Total Distance}}{\text{Time}} = \frac{450}{10.8} \, m/s = \frac{4500}{108} \, m/s$.
To convert the relative speed from $m/s$ to $km/h$,multiply by $\frac{18}{5}$:
Relative speed $= \frac{4500}{108} \times \frac{18}{5} \, km/h = \frac{900}{6} \, km/h = 150 \, km/h$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds: $V_{rel} = V_1 + V_2$.
$150 \, km/h = 60 \, km/h + V_2$.
Therefore,the speed of the second train $V_2 = 150 \, km/h - 60 \, km/h = 90 \, km/h$.