Problems on Trains Questions in English

(D) Relative speed of the two trains moving in the same direction $= (42 - 30) \; km/h = 12 \; km/h$.
Convert the relative speed into $m/s$ by multiplying by $\frac{5}{18}$:
Relative speed $= 12 \times \frac{5}{18} = \frac{60}{18} = \frac{10}{3} \; m/s$.
Total distance to be covered to pass the other train $= (60 + 84) \; m = 144 \; m$.
Time taken $= \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{144}{10/3} = \frac{144 \times 3}{10} = \frac{432}{10} = 43.2 \; seconds$.

(C) Let the lengths of the two trains be $L_1$ and $L_2$ in meters.
Relative speed in opposite direction $= 45 + 36 = 81 \; km/h$.
Convert to $m/s$: $81 \times \frac{5}{18} = 22.5 \; m/s$.
Since they pass each other in $8 \; seconds$,the sum of their lengths is $L_1 + L_2 = 22.5 \times 8 = 180 \; m$.
Relative speed in the same direction $= 45 - 36 = 9 \; km/h$.
Convert to $m/s$: $9 \times \frac{5}{18} = 2.5 \; m/s$.
$A$ person in the faster train observes the other train passing in $30 \; seconds$. This means the length of the slower train $(L_2)$ is covered at the relative speed: $L_2 = 2.5 \times 30 = 75 \; m$.
Therefore,the length of the faster train is $L_1 = 180 - 75 = 105 \; m$.
The lengths are $105 \; m$ and $75 \; m$.

(B) The train starting from $P$ travels for $1 \; hour$ (from $7 \; am$ to $8 \; am$) at a speed of $20 \; km/h$.
Distance covered by the first train by $8 \; am = 20 \; km/h \times 1 \; h = 20 \; km$.
Remaining distance between $P$ and $Q$ at $8 \; am = 110 \; km - 20 \; km = 90 \; km$.
Since the trains are moving towards each other,their relative speed $= 20 \; km/h + 25 \; km/h = 45 \; km/h$.
Time taken to meet after $8 \; am = \frac{\text{Remaining Distance}}{\text{Relative Speed}} = \frac{90 \; km}{45 \; km/h} = 2 \; hours$.
Therefore,the trains will meet at $8 \; am + 2 \; hours = 10 \; am$.

(B) The length of the train is $75 \; m$.
For the first person:
Relative speed $= \frac{\text{Distance}}{\text{Time}} = \frac{75}{7.5} = 10 \; m/s$.
Converting to $km/h$: $10 \times \frac{18}{5} = 36 \; km/h$.
Since the train is overtaking the person,the relative speed is (Speed of train - Speed of person).
$36 = \text{Speed of train} - 6 \implies \text{Speed of train} = 42 \; km/h$.
For the second person:
Time taken $= 6 \frac{3}{4} = 6.75 \; s = \frac{27}{4} \; s$.
Relative speed $= \frac{75}{27/4} = \frac{300}{27} = \frac{100}{9} \; m/s$.
Converting to $km/h$: $\frac{100}{9} \times \frac{18}{5} = 40 \; km/h$.
Let the speed of the second person be $v_2$.
$40 = 42 - v_2 \implies v_2 = 2 \; km/h$.

(C) Let the distance between two stations $A$ and $B$ be $x \; km$ and the speed of the train be $y \; km/h$.
Since $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$,we have:
$\frac{x}{y} = \frac{45}{60} = \frac{3}{4} \implies x = \frac{3}{4}y \quad (1)$
When the speed is reduced by $5 \; km/h$,the new speed is $(y - 5) \; km/h$ and the time taken is $48$ minutes:
$\frac{x}{y - 5} = \frac{48}{60} = \frac{4}{5} \implies x = \frac{4}{5}(y - 5) \quad (2)$
Equating $(1)$ and $(2)$:
$\frac{3}{4}y = \frac{4}{5}(y - 5)$
$15y = 16(y - 5)$
$15y = 16y - 80$
$y = 80 \; km/h$
Substituting $y = 80$ in $(1)$:
$x = \frac{3}{4} \times 80 = 60 \; km$
Thus,the distance is $60 \; km$ and the speed is $80 \; km/h$.

(D) The relative speed of the faster train with respect to the slower train is calculated as the difference between their speeds because they are moving in the same direction.
Relative speed $= 40 \; km/hr - 20 \; km/hr = 20 \; km/hr$.
To convert the speed from $km/hr$ to $m/s$,we multiply by $\frac{5}{18}$.
Relative speed $= 20 \times \frac{5}{18} = \frac{100}{18} = \frac{50}{9} \; m/s$.
Let the length of the faster train be $x \; meters$.
Since the train passes a man in the slower train,the distance covered is equal to the length of the faster train $(x)$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$x = \frac{50}{9} \; m/s \times 5 \; s = \frac{250}{9} \; m$.
Converting the improper fraction to a mixed fraction: $\frac{250}{9} = 27 \frac{7}{9} \; m$.

(B) The speed of the train is $72 \; km/hr$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$:
$\text{Speed} = 72 \times \frac{5}{18} = 20 \; m/s$.
When a train crosses a bridge,the total distance covered is the sum of the length of the train and the length of the bridge.
$\text{Total distance} = 110 \; m + 132 \; m = 242 \; m$.
Using the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
$t = \frac{242}{20} = 12.1 \; seconds$.
Therefore,the train takes $12.1 \; seconds$ to cross the bridge.

(D) The speed of the train is given as $60 \; km/hr$.
To convert the speed into $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \; m/s$.
The time taken to cross the pole is $9 \; seconds$.
The length of the train is equal to the distance covered by the train in $9 \; seconds$ while crossing the pole.
$\text{Length} = \text{Speed} \times \text{Time} = \left(\frac{50}{3}\right) \times 9 = 50 \times 3 = 150 \; m$.

(C) Step $1$: Calculate the speed of the train in $m/s$.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{12 \;km}{10 \;min}$.
Convert $12 \;km$ to $12000 \;m$ and $10 \;min$ to $600 \;s$.
Speed = $\frac{12000 \;m}{600 \;s} = 20 \;m/s$.
Step $2$: Calculate the length of the train.
When a train passes a telegraph post,it covers a distance equal to its own length.
Length of train = $\text{Speed} \times \text{Time taken to pass the post}$.
Length = $20 \;m/s \times 6 \;s = 120 \;m$.

(B) Let the length of the platform be $x \;m$.
When the train crosses a signal pole,it covers its own length. Therefore,the speed of the train is:
$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{300 \;m}{18 \;s} = \frac{50}{3} \;m/s$.
When the train crosses a platform,it covers the sum of its own length and the length of the platform. Therefore:
$\text{Speed} = \frac{300 + x}{39} \;m/s$.
Equating the two expressions for speed:
$\frac{300 + x}{39} = \frac{50}{3}$.
Multiplying both sides by $39$:
$300 + x = \frac{50}{3} \times 39$.
$300 + x = 50 \times 13$.
$300 + x = 650$.
$x = 650 - 300 = 350 \;m$.
Thus,the length of the platform is $350 \;m$.

(C) Let the time taken by the slower train to cross the faster train be $t$ seconds.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds:
Relative speed $= 60 + 90 = 150 \; km/hr$.
The total distance to be covered to cross each other is the sum of their lengths:
Total length $= 1.1 + 0.9 = 2 \; km$.
Time taken to cross each other is given by $t = \frac{\text{Total length}}{\text{Relative speed}}$.
$t = \frac{2 \; km}{150 \; km/hr} = \frac{1}{75} \; hours$.
To convert the time into seconds,multiply by $3600$:
$t = \frac{1}{75} \times 3600 \; seconds = 48 \; seconds$.

(B) Let the time taken by the train to cross the running man be $t$ seconds.
Since the man is running in the opposite direction to the train,the relative speed is the sum of their individual speeds.
Relative speed $= 60 + 6 = 66 \;km/h$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 66 \times \frac{5}{18} = 11 \times \frac{5}{3} = \frac{55}{3} \;m/s$.
The distance to be covered to pass the man is equal to the length of the train,which is $110 \;m$.
Time taken $t = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{110}{55/3} = 110 \times \frac{3}{55} = 2 \times 3 = 6 \;seconds$.

(B) When two trains move in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 45 \; km/hr + 30 \; km/hr = 75 \; km/hr$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 75 \times \frac{5}{18} = \frac{375}{18} = \frac{125}{6} \; m/s$.
To pass the driver of the faster train,the slower train needs to cover a distance equal to its own length,which is $500 \; m$.
Time taken $(t) = \frac{\text{Distance}}{\text{Relative speed}}$.
$t = \frac{500}{\frac{125}{6}} = 500 \times \frac{6}{125} = 4 \times 6 = 24 \; seconds$.

(B) Speed of the first train $= \frac{120 \; m}{10 \; s} = 12 \; m/s$.
Speed of the second train $= \frac{120 \; m}{15 \; s} = 8 \; m/s$.
When two trains travel in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 12 \; m/s + 8 \; m/s = 20 \; m/s$.
To cross each other,the total distance covered must be the sum of the lengths of both trains.
Total distance $= 120 \; m + 120 \; m = 240 \; m$.
Time taken to cross each other $= \frac{\text{Total distance}}{\text{Relative speed}} = \frac{240 \; m}{20 \; m/s} = 12 \; s$.

(B) Let the time taken by the train to cross the man be $t$ seconds.
Since the man is walking in the same direction as the train,the relative speed is the difference between their speeds.
Relative speed $= (63 - 3) \;km/h = 60 \;km/h$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \;m/s$.
The distance to be covered to cross the man is equal to the length of the train,which is $500 \;m$.
Using the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
$t = \frac{500}{50/3} = \frac{500 \times 3}{50} = 10 \times 3 = 30 \;seconds$.

(D) Relative speed of the trains moving in opposite directions is the sum of their individual speeds: $v_r = 60 + 40 = 100 \; km/hr$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$: $v_r = 100 \times \frac{5}{18} = \frac{500}{18} = \frac{250}{9} \; m/s$.
The total distance to be covered to cross each other is the sum of the lengths of the two trains: $L = 140 + 160 = 300 \; m$.
The time taken to cross each other is given by $t = \frac{L}{v_r}$.
Substituting the values: $t = \frac{300}{250/9} = \frac{300 \times 9}{250} = \frac{6 \times 9}{5} = \frac{54}{5} = 10.8 \; seconds$.

(C) Let the speed of each train be $v \; m/s$.
Since the trains are moving in opposite directions,their relative speed is $v_r = v + v = 2v \; m/s$.
The total distance covered to cross each other is the sum of their lengths: $D = 120 + 120 = 240 \; m$.
The time taken to cross is $t = 12 \; s$.
Using the formula $v_r = \frac{D}{t}$,we get:
$2v = \frac{240}{12} = 20 \; m/s$.
Therefore,$v = 10 \; m/s$.
To convert the speed from $m/s$ to $km/h$,we multiply by $\frac{18}{5}$:
$v = 10 \times \frac{18}{5} = 36 \; km/h$.

(D) Let the speed of the train be $v \; m/s$ and the length of the train be $L \; m$.
When the train passes a telegraph post,it covers its own length in $8 \; seconds$:
$L = v \times 8 \quad ... (1)$
When the train passes a bridge of length $264 \; m$,it covers its own length plus the bridge length in $20 \; seconds$:
$L + 264 = v \times 20 \quad ... (2)$
Subtracting equation $(1)$ from equation $(2)$:
$(L + 264) - L = 20v - 8v$
$264 = 12v$
$v = \frac{264}{12} = 22 \; m/s$
To convert the speed from $m/s$ to $km/hr$,multiply by $\frac{18}{5}$:
$v = 22 \times \frac{18}{5} = \frac{396}{5} = 79.2 \; km/hr$.

(D) First,convert the speed of the train from $km/h$ to $m/s$:
Speed $= 72 \times \frac{5}{18} \; m/s = 20 \; m/s$.
Let the length of the goods train be $x \; m$.
When a train crosses a platform,the total distance covered is the sum of the length of the train and the length of the platform.
Total distance $= (x + 250) \; m$.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$
$x + 250 = 20 \times 26$
$x + 250 = 520$
$x = 520 - 250$
$x = 270 \; m$.
Therefore,the length of the goods train is $270 \; m$.

(A) Speed of the train $= 45 \; km/hr$.
To convert the speed into $m/s$, multiply by $\frac{5}{18}$:
Speed $= 45 \times \frac{5}{18} = 12.5 \; m/s$.
Total distance to be covered to pass the bridge $=$ Length of the train $+$ Length of the bridge.
Total distance $= 360 \; m + 140 \; m = 500 \; m$.
Time taken $(t) = \frac{\text{Total Distance}}{\text{Speed}}$.
$t = \frac{500}{12.5} = 40 \; seconds$.

(A) Let the length of each train be $l$ meters.
Total length to be covered by the faster train to pass the slower train $= l + l = 2l$ meters.
Relative speed of the faster train with respect to the slower train $= 46 - 36 = 10 \; km/h$.
Convert the relative speed into $m/s$: $10 \times \frac{5}{18} = \frac{50}{18} = \frac{25}{9} \; m/s$.
Using the formula: $\text{Relative Speed} = \frac{\text{Total Distance}}{\text{Time}}$.
$\frac{25}{9} = \frac{2l}{36}$.
$2l = \frac{25 \times 36}{9}$.
$2l = 25 \times 4 = 100$.
$l = 50 \; m$.
Thus,the length of each train is $50 \; m$.

(A) Let the length of the second train be $l \;m$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 120 + 80 = 200 \;km/hr$.
To convert the speed into $m/s$,multiply by $\frac{5}{18}$:
Relative speed $= 200 \times \frac{5}{18} = \frac{1000}{18} = \frac{500}{9} \;m/s$.
The total distance covered to cross each other is the sum of the lengths of both trains,which is $(270 + l) \;m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$270 + l = \frac{500}{9} \times 9$.
$270 + l = 500$.
$l = 500 - 270 = 230 \;m$.
Therefore,the length of the other train is $230 \;m$.

(C) Let the speed of the slower train be $v$ and the speed of the faster train be $2v$ in $m/s$.
Since the trains are moving in opposite directions,their relative speed is $v + 2v = 3v$.
The total distance to be covered to cross each other is the sum of their lengths: $100 \; m + 100 \; m = 200 \; m$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we have:
$3v = \frac{200}{8}$
$3v = 25$
$v = \frac{25}{3} \; m/s$.
The speed of the faster train is $2v = 2 \times \frac{25}{3} = \frac{50}{3} \; m/s$.
To convert this speed into $km/h$,multiply by $\frac{18}{5}$:
$\text{Speed} = \frac{50}{3} \times \frac{18}{5} = 10 \times 6 = 60 \; km/h$.

(B) Let the speeds of the two trains be $v_1$ and $v_2$. Let them meet at a point $P$ after time $T$.
After meeting, the first train takes $t_1 = 9 \; \text{hours}$ to reach Patna, and the second train takes $t_2 = 16 \; \text{hours}$ to reach Howrah.
The formula for the ratio of speeds when they meet and then reach their destinations is given by $\frac{v_1}{v_2} = \sqrt{\frac{t_2}{t_1}}$.
Substituting the given values: $\frac{v_1}{v_2} = \sqrt{\frac{16}{9}}$.
Therefore, $\frac{v_1}{v_2} = \frac{4}{3}$.
The ratio of their speeds is $4:3$.

(B) Let the speed of the train be $x \; km/hr$.
Since the man is running in the same direction as the train,the relative speed of the train with respect to the man is $(x - 5) \; km/hr$.
To convert this speed into $m/s$,we multiply by $\frac{5}{18}$:
Relative speed $= (x - 5) \times \frac{5}{18} \; m/s$.
The train covers its own length of $125 \; m$ in $10 \; seconds$ to pass the man.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$:
$(x - 5) \times \frac{5}{18} = \frac{125}{10} = 12.5$.
Now,solve for $x$:
$x - 5 = \frac{12.5 \times 18}{5} = 2.5 \times 18 = 45$.
$x = 45 + 5 = 50 \; km/hr$.
Therefore,the speed of the train is $50 \; km/hr$.

(B) Let the length of the train be $l \; \text{m}$.
The speed of the train is constant. When passing a pole, the train covers its own length $l$ in $15 \; \text{seconds}$.
Speed $v = \frac{l}{15} \; \text{m/s}$.
When passing a platform of $100 \; \text{m}$, the train covers a total distance of $(l + 100) \; \text{m}$ in $25 \; \text{seconds}$.
Speed $v = \frac{l + 100}{25} \; \text{m/s}$.
Equating the speeds:
$\frac{l}{15} = \frac{l + 100}{25}$
Simplify the fractions by dividing by $5$:
$\frac{l}{3} = \frac{l + 100}{5}$
Cross-multiplying gives:
$5l = 3(l + 100)$
$5l = 3l + 300$
$2l = 300$
$l = 150 \; \text{m}$.
Thus, the length of the train is $150 \; \text{m}$.

(C) Let the length of the tunnel be $x \; m$.
The speed of the train is $78 \; km/hr$.
To convert the speed into $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 78 \times \frac{5}{18} = \frac{13 \times 5}{3} = \frac{65}{3} \; m/s$.
The time taken to cross the tunnel is $1 \; minute = 60 \; seconds$.
When a train crosses a tunnel,the total distance covered is the sum of the length of the train and the length of the tunnel.
$\text{Total Distance} = (800 + x) \; m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$800 + x = \frac{65}{3} \times 60$
$800 + x = 65 \times 20$
$800 + x = 1300$
$x = 1300 - 800 = 500 \; m$.
Thus,the length of the tunnel is $500 \; m$.

(B) First,calculate the speed of the train.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{240 \; m}{24 \; s} = 10 \; m/s$.
When passing a platform,the total distance covered by the train is the sum of the length of the train and the length of the platform.
Total distance = $240 \; m + 650 \; m = 890 \; m$.
Time taken = $\frac{\text{Total distance}}{\text{Speed}} = \frac{890 \; m}{10 \; m/s} = 89 \; s$.

(A) To cross an electric pole,the train must cover a distance equal to its own length.
Length of the train $(d)$ = $100 \; m$.
Speed of the train $(v)$ = $144 \; km/hr$.
First,convert the speed from $km/hr$ to $m/s$ by multiplying by $\frac{5}{18}$:
$v = 144 \times \frac{5}{18} \; m/s = 8 \times 5 \; m/s = 40 \; m/s$.
Using the formula,$\text{Time} (t) = \frac{\text{Distance}}{\text{Speed}}$:
$t = \frac{100}{40} \; s = 2.5 \; s$.
Therefore,the train will cross the pole in $2.5 \; seconds$.

(B) To pass a tree, the train must cover a distance equal to its own length.
Length of the train = $280 \; m$.
Speed of the train = $63 \; km/hr$.
First, convert the speed from $km/hr$ to $m/s$ by multiplying by $\frac{5}{18}$:
Speed = $63 \times \frac{5}{18} = \frac{7 \times 5}{2} = 17.5 \; m/s$.
Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{280}{17.5}$.
Time = $\frac{280 \times 18}{63 \times 5} = \frac{5040}{315} = 16 \; seconds$.

(B) Let the speeds of the first and second trains be $x \; m/s$ and $y \; m/s$ respectively.
Length of the first train $= x \times 27 = 27x \; \text{meters}$.
Length of the second train $= y \times 17 = 17y \; \text{meters}$.
When the trains cross each other in opposite directions, the relative speed is $(x + y) \; m/s$ and the total distance covered is the sum of their lengths $(27x + 17y) \; \text{meters}$.
Time taken to cross each other $= \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{27x + 17y}{x + y} = 23$.
Multiplying both sides by $(x + y)$, we get:
$27x + 17y = 23(x + y)$
$27x + 17y = 23x + 23y$
Rearranging the terms to solve for the ratio $x/y$:
$27x - 23x = 23y - 17y$
$4x = 6y$
$\frac{x}{y} = \frac{6}{4} = \frac{3}{2}$.
Thus, the ratio of their speeds is $3:2$.

(C) Let the time taken by the train to pass the jogger be $t$ seconds.
Relative speed of the train with respect to the jogger $= 45 - 9 = 36 \; km/hr$.
Converting the relative speed to $m/s$: $36 \times \frac{5}{18} = 10 \; m/s$.
The total distance to be covered by the train to completely pass the jogger is the sum of the initial distance between them and the length of the train: $240 \; m + 120 \; m = 360 \; m$.
Time taken $t = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{360 \; m}{10 \; m/s} = 36 \; sec$.

(B) First, convert the speed of the train from $\text{km/hr}$ to $\text{m/s}$:
Speed $= 54 \times \frac{5}{18} = 15 \; \text{m/s}$.
Let the length of the train be $L_t$ and the length of the platform be $L_p$.
When the train passes a man standing on the platform, it covers its own length in $20 \; \text{seconds}$:
$L_t = \text{Speed} \times \text{Time} = 15 \times 20 = 300 \; \text{m}$.
When the train passes the platform, it covers the sum of its own length and the platform's length in $36 \; \text{seconds}$:
$L_t + L_p = \text{Speed} \times \text{Time} = 15 \times 36 = 540 \; \text{m}$.
Substituting the value of $L_t$:
$300 + L_p = 540$
$L_p = 540 - 300 = 240 \; \text{m}$.
Thus, the length of the platform is $240 \; \text{m}$.

(C) Let the length of the bridge be $x \; m$.
Total distance covered by the train to cross the bridge = (Length of train + Length of bridge) = $(130 + x) \; m$.
Speed of the train = $45 \; km/hr$.
Convert speed to $m/s$: $45 \times \frac{5}{18} = \frac{225}{18} = 12.5 \; m/s$.
Time taken = $30 \; s$.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$130 + x = 12.5 \times 30$.
$130 + x = 375$.
$x = 375 - 130 = 245 \; m$.
Thus,the length of the bridge is $245 \; m$.

(C) The relative speed of the faster train with respect to the slower train is $72 - 54 = 18 \; km/h$.
Converting this speed into $m/s$: $18 \times \frac{5}{18} = 5 \; m/s$.
The total distance to be covered by the faster train to completely cross the second train is the sum of their lengths: $100 + 120 = 220 \; m$.
Time taken = $\frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{220}{5} = 44 \; s$.

(B) Speed of the train $= 30 \; km/hr = 30 \times \frac{5}{18} \; m/s = \frac{25}{3} \; m/s$.
Time taken by the train to pass a standing man $= \frac{\text{Length of train}}{\text{Speed of train}} = \frac{100 \; m}{\frac{25}{3} \; m/s}$.
Time $= \frac{100 \times 3}{25} = 12 \; sec$.

(B) Speed of the train $= 120 \; km/hr = 120 \times \frac{5}{18} \; m/s = \frac{100}{3} \; m/s$.
Total distance to be covered to cross the platform $=$ Length of the train $+$ Length of the platform.
Total distance $= 120 \; m + 130 \; m = 250 \; m$.
Time taken $= \frac{\text{Total distance}}{\text{Speed of the train}}$.
Time taken $= \frac{250}{\frac{100}{3}} = \frac{250 \times 3}{100} = \frac{750}{100} = 7.5 \; sec$.

(C) Step $1$: Convert the speed of the train from $km/hr$ to $m/s$.
Speed $= 135 \times \frac{5}{18} = 7.5 \times 5 = 37.5 \; m/s$.
Step $2$: Let the length of the train be $l$ meters. Since the length of the platform is equal to the length of the train,the length of the platform is also $l$ meters.
Step $3$: When a train crosses a platform,the total distance covered is the sum of the length of the train and the length of the platform.
Total distance $= l + l = 2l$.
Step $4$: Use the formula: $\text{Speed} = \frac{\text{Total Distance}}{\text{Time}}$.
$37.5 = \frac{2l}{40}$.
Step $5$: Solve for $l$.
$2l = 37.5 \times 40$.
$2l = 1500$.
$l = 750 \; m$.
Therefore,the length of the train is $750 \; m$.

(C) Relative speed of the faster train with respect to the slower train $= 45 - 36 = 9 \; km/hr$.
Converting to $m/s$: $9 \times \frac{5}{18} = 2.5 \; m/s$.
Total distance to be covered to cross each other $=$ Sum of lengths of both trains $= 250 + 200 = 450 \; m$.
Time taken to cross each other $= \frac{\text{Total distance}}{\text{Relative speed}} = \frac{450}{2.5} = 180 \; sec$.

(D) Speed of the first train $= 60 \; km/hr = 60 \times \frac{5}{18} \; m/s = \frac{50}{3} \; m/s$.
Let the speed of the second train be $x \; m/s$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds: $\text{Relative speed} = (\frac{50}{3} + x) \; m/s$.
The total distance covered to cross each other is the sum of their lengths: $\text{Total distance} = 120 \; m + 130 \; m = 250 \; m$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$:
$\frac{50}{3} + x = \frac{250}{6}$.
$\frac{50}{3} + x = \frac{125}{3}$.
$x = \frac{125}{3} - \frac{50}{3} = \frac{75}{3} = 25 \; m/s$.
Converting the speed back to $km/hr$: $25 \times \frac{18}{5} = 90 \; km/hr$.