Time and Distances Questions in English

(D) Let the total distance of the journey be $2d$.
The time taken to cover the first half $(d)$ at $30 \text{ km/h}$ is $t_1 = \frac{d}{30}$.
The time taken to cover the second half $(d)$ at $20 \text{ km/h}$ is $t_2 = \frac{d}{20}$.
Average speed is defined as the total distance divided by the total time:
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2} = \frac{2d}{\frac{d}{30} + \frac{d}{20}}$.
Factoring out $d$ from the denominator:
$\text{Average speed} = \frac{2}{\frac{1}{30} + \frac{1}{20}} = \frac{2}{\frac{2+3}{60}} = \frac{2 \times 60}{5} = \frac{120}{5} = 24 \text{ km/h}$.
Since the calculated average speed is $24 \text{ km/h}$,which is not among the options $25, 28, 32$,the correct choice is $D$.

(B) Let the distance covered be $d$.
Time taken to travel by bus,$t_1 = \frac{d}{16}$.
Time taken to return by cycle,$t_2 = \frac{d}{9}$.
Total distance covered = $d + d = 2d$.
Total time taken = $t_1 + t_2 = \frac{d}{16} + \frac{d}{9} = \frac{9d + 16d}{144} = \frac{25d}{144}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{25d}{144}} = \frac{2 \times 144}{25} = \frac{288}{25} = 11.52 \text{ km/h}$.
Alternatively,using the formula for average speed when distance is constant: $\text{Average speed} = \frac{2s_1s_2}{s_1 + s_2} = \frac{2 \times 16 \times 9}{16 + 9} = \frac{288}{25} = 11.52 \text{ km/h}$.

(C) Let the speed of the car from $A$ to $B$ be $v_1 = 64 \, km/h$ and the return speed from $B$ to $A$ be $v_2 = x \, km/h$.
The formula for average speed for a round trip is given by $v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}$.
Given that $v_{avg} = 56 \, km/h$,we substitute the values:
$56 = \frac{2 \times 64 \times x}{64 + x}$
Divide both sides by $8$:
$7 = \frac{2 \times 8 \times x}{64 + x} = \frac{16x}{64 + x}$
Cross-multiply:
$7(64 + x) = 16x$
$448 + 7x = 16x$
$16x - 7x = 448$
$9x = 448$
$x = \frac{448}{9} \approx 49.78 \, km/h$.
Thus,the return speed is $49.78 \, km/h$.

(B) Let the distance between $A$ and $B$ be $d \text{ km}$.
Time taken for the onward journey from $A$ to $B$ is $t_1 = \frac{d}{10} \text{ hours}$.
Time taken for the return journey from $B$ to $A$ is $t_2 = \frac{d}{8} \text{ hours}$.
The total time taken is $t_1 + t_2 = 4 \frac{1}{2} = \frac{9}{2} \text{ hours}$.
So,$\frac{d}{10} + \frac{d}{8} = \frac{9}{2}$.
Taking the least common multiple of $10$ and $8$,which is $40$,we get $\frac{4d + 5d}{40} = \frac{9}{2}$.
$\frac{9d}{40} = \frac{9}{2}$.
$d = \frac{9}{2} \times \frac{40}{9} = 20 \text{ km}$.
The total distance covered during the entire journey is the sum of the onward and return distances,which is $d + d = 2d$.
Total distance $= 2 \times 20 = 40 \text{ km}$.

(C) The formula for average speed when the distance covered is the same for two different speeds is given by $\text{Average Speed} = \frac{2s_1s_2}{s_1 + s_2}$.
Here,$s_1 = 64 \text{ km/h}$ and $s_2 = 80 \text{ km/h}$.
Substituting the values into the formula:
$\text{Average Speed} = \frac{2 \times 64 \times 80}{64 + 80}$
$\text{Average Speed} = \frac{10240}{144}$
$\text{Average Speed} = 71.11 \text{ km/h}$.

(D) Let the total distance of the journey be $D\, km$.
Time taken to cover the first half of the distance $(D/2)$ at $50\, km/h$ is $t_1 = \frac{D/2}{50} = \frac{D}{100}\, hours$.
Time taken to cover the remaining half of the distance $(D/2)$ at $70\, km/h$ is $t_2 = \frac{D/2}{70} = \frac{D}{140}\, hours$.
The total time taken is $t_1 + t_2 = 6\, hours$.
So,$\frac{D}{100} + \frac{D}{140} = 6$.
Taking the least common multiple of $100$ and $140$,which is $700$:
$\frac{7D + 5D}{700} = 6$
$\frac{12D}{700} = 6$
$12D = 4200$
$D = \frac{4200}{12} = 350\, km$.
Wait,re-evaluating the formula provided in the prompt: The average speed for equal distances is $v_{avg} = \frac{2s_1s_2}{s_1+s_2}$.
Total distance $D = v_{avg} \times T = \left(\frac{2 \times 50 \times 70}{50+70}\right) \times 6 = \left(\frac{7000}{120}\right) \times 6 = \frac{7000}{20} = 350\, km$.
Since $350$ is not among the options,the correct answer is $D$ (none of these).

(A) Let Rakesh's speed be $v_1$ and Suresh's speed be $v_2$.
Let the time taken by them after meeting be $T_1 = 9 \, \text{hours}$ and $T_2 = 16 \, \text{hours}$ respectively.
The formula for the ratio of speeds when two people meet and then complete their journeys in times $T_1$ and $T_2$ is given by $\frac{v_1}{v_2} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the given values: $\frac{16}{v_2} = \sqrt{\frac{16}{9}}$.
$\frac{16}{v_2} = \frac{4}{3}$.
$4 \times v_2 = 16 \times 3$.
$4 \times v_2 = 48$.
$v_2 = 12 \, \text{km/h}$.
Thus,Suresh's speed is $12 \, \text{km/h}$.

(C) The average speed is defined as the total distance traveled divided by the total time taken.
Total distance $= 225 \ km + 370 \ km = 595 \ km$.
Total time $= 3.5 \ hours + 5 \ hours = 8.5 \ hours$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{595}{8.5} \ km/h$.
Average speed $= 70 \ km/h$.

(B) Average speed is defined as the total distance traveled divided by the total time taken.
Total distance $D = 6 + 8 + 32 = 46 \, km$.
Time taken for walking $t_1 = \frac{6}{1.5} = 4 \, hours$.
Time taken for running $t_2 = \frac{8}{2} = 4 \, hours$.
Time taken for bus travel $t_3 = \frac{32}{8} = 4 \, hours$.
Total time $T = t_1 + t_2 + t_3 = 4 + 4 + 4 = 12 \, hours$.
Average speed $= \frac{D}{T} = \frac{46}{12} = \frac{23}{6} = 3 \frac{5}{6} \, km/h$.

(D) The formula for average speed is $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.
First,convert all time intervals into hours:
$T_1 = 30 \text{ min} = 0.5 \text{ h}$
$T_2 = 45 \text{ min} = 0.75 \text{ h}$
$T_3 = 2 \text{ h}$
Next,calculate the distance traveled in each interval $(D = \text{speed} \times \text{time})$:
$D_1 = 40 \times 0.5 = 20 \text{ km}$
$D_2 = 60 \times 0.75 = 45 \text{ km}$
$D_3 = 70 \times 2 = 140 \text{ km}$
Total Distance $= 20 + 45 + 140 = 205 \text{ km}$
Total Time $= 0.5 + 0.75 + 2 = 3.25 \text{ h}$
$\text{Average Speed} = \frac{205}{3.25} \approx 63.07 \text{ km/h}$.
Since $63.07$ is not among the options,the correct answer is $D$.

(B) Let the usual speed be $S$ and the usual time be $T$.
Since distance is constant,$D = S \times T$.
When the speed becomes $\frac{3}{4}S$,the new time taken is $T + 20$.
Thus,$S \times T = \frac{3}{4}S \times (T + 20)$.
Dividing both sides by $S$,we get $T = \frac{3}{4}(T + 20)$.
$4T = 3T + 60$.
$T = 60 \text{ minutes}$.
Therefore,the usual time is $60 \text{ minutes}$.

(A) Let the distance be $d \, km$.
Time taken by the first man $= \frac{d}{4} \, hours$.
Time taken by the second man $= \frac{d}{3} \, hours$.
According to the problem,the first man arrives half an hour $(0.5 \, hours)$ before the second man.
Therefore,$\frac{d}{3} - \frac{d}{4} = 0.5$.
Taking the common denominator,we get $\frac{4d - 3d}{12} = 0.5$.
$\frac{d}{12} = 0.5$.
$d = 0.5 \times 12 = 6 \, km$.
Thus,the distance is $6 \, km$.

(C) Let the distance between $A$ and $B$ be $d \, km$.
Time taken by the first car to cover the distance $d$ is $t_1 = \frac{d}{20} \, hours$.
Time taken by the second car to cover the distance $d$ is $t_2 = \frac{d}{30} \, hours$.
The second car starts $1.5 \, hours$ later and reaches $2.5 \, hours$ earlier,so the difference in time is $t_1 - t_2 = 1.5 + 2.5 = 4 \, hours$.
Substituting the values: $\frac{d}{20} - \frac{d}{30} = 4$.
Taking the $LCM$ of $20$ and $30$,which is $60$: $\frac{3d - 2d}{60} = 4$.
$\frac{d}{60} = 4$.
$d = 240 \, km$.
Thus,the distance from $A$ to $B$ is $240 \, km$.

(D) Let the distance between Tilak Nagar and Moti Nagar be $d \, km$.
In the first case,the total time taken is $T_1 = \frac{d}{3.5} + \frac{d}{3.5} = \frac{2d}{3.5} = \frac{4d}{7} \, hours$.
In the second case,the total time taken is $T_2 = \frac{d}{3} + \frac{d}{4} = \frac{4d + 3d}{12} = \frac{7d}{12} \, hours$.
Given that the difference in time is $10 \, minutes = \frac{10}{60} = \frac{1}{6} \, hours$.
So,$T_2 - T_1 = \frac{1}{6}$.
$\frac{7d}{12} - \frac{4d}{7} = \frac{1}{6}$.
Taking $LCM$ of $12$ and $7$ as $84$:
$\frac{49d - 48d}{84} = \frac{1}{6}$.
$\frac{d}{84} = \frac{1}{6}$.
$d = \frac{84}{6} = 14 \, km$.
Since $14 \, km$ is not among the options,the correct answer is $D$ (none of these).

(B) Let the original speed of the train be $s \, \text{km/h}$.
The time taken at this speed is $T_1 = 8 \, \text{hours}$.
The distance $D$ covered is $D = \text{speed} \times \text{time} = 8s$.
If the speed is increased by $5 \, \text{km/h}$, the new speed is $(s + 5) \, \text{km/h}$.
The new time taken is $6 \, \text{hours} \, 40 \, \text{minutes} = 6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3} \, \text{hours}$.
Since the distance remains the same, we equate the two expressions for distance:
$8s = (s + 5) \times \frac{20}{3}$
Multiply both sides by $3$:
$24s = 20(s + 5)$
$24s = 20s + 100$
$4s = 100$
$s = 25 \, \text{km/h}$.
Thus, the lowest speed of the train is $25 \, \text{km/h}$.

(C) Let the speed without stoppage be $s_{1} = 42 \, km/h$ and the speed with stoppage be $s_{2} = 28 \, km/h$.
The formula for stoppage time per hour is given by: $\text{Stoppage time} = \frac{s_{1} - s_{2}}{s_{1}} \times 60 \, \text{minutes}$.
Substituting the values: $\text{Stoppage time} = \frac{42 - 28}{42} \times 60 \, \text{minutes}$.
$= \frac{14}{42} \times 60 \, \text{minutes}$.
$= \frac{1}{3} \times 60 \, \text{minutes} = 20 \, \text{minutes}$.
Therefore, the person stops for $20 \, \text{minutes per hour}$.

(B) First,convert the speed of the train from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$:
Speed $= 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \, m/s$.
Let the length of the train be $L \, m$.
When a train passes a platform,the total distance covered is the sum of the length of the train and the length of the platform.
Distance $= L + 130 \, m$.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$L + 130 = \frac{50}{3} \times 15$.
$L + 130 = 50 \times 5$.
$L + 130 = 250$.
$L = 250 - 130 = 120 \, m$.

(A) The speed of the train is given as $54 \, km/h$.
To convert the speed from $km/h$ to $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 54 \times \frac{5}{18} = 3 \times 5 = 15 \, m/s$.
The time taken to cross the telegraph post is $9 \, seconds$.
Since the train covers its own length while passing a stationary point (telegraph post),the length of the train is calculated as:
$\text{Length} = \text{Speed} \times \text{Time}$
$\text{Length} = 15 \, m/s \times 9 \, s = 135 \, m$.
Therefore,the length of the train is $135 \, meters$.

(B) To pass a telegraph post,the train must cover a distance equal to its own length.
Given:
Length of the train $(L)$ = $135\, m$
Speed of the train $(v)$ = $54\, km/h$
First,convert the speed from $km/h$ to $m/s$:
$v = 54 \times \frac{5}{18} = 3 \times 5 = 15\, m/s$
Time taken $(t)$ = $\frac{\text{Distance}}{\text{Speed}} = \frac{L}{v}$
$t = \frac{135}{15} = 9\, seconds$
Therefore,the train will pass the telegraph post in $9\, seconds$.

(C) The speed of the train is calculated by dividing the length of the train by the time taken to cross the man.
Speed $= \frac{\text{Length of the train}}{\text{Time taken}}$
Speed $= \frac{160 \ m}{18 \ s} = \frac{80}{9} \ m/s$.
To convert the speed from $m/s$ to $km/h$,multiply by $\frac{18}{5}$.
Speed in $km/h = \frac{80}{9} \times \frac{18}{5} \ km/h$.
Speed $= 16 \times 2 = 32 \ km/h$.

(A) To cross 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 $= 280\,m + 220\,m = 500\,m$.
The speed of the train is $60\,km/h$. Converting this to $m/s$:
Speed $= 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3}\,m/s$.
Time taken $= \frac{\text{Total distance}}{\text{Speed}} = \frac{500}{50/3} = \frac{500 \times 3}{50} = 10 \times 3 = 30\,seconds$.

(B) To find the speed of the train,we use the formula:
Speed = (Total distance covered) / (Time taken)
When a train passes a platform,the total distance covered is the sum of the length of the train and the length of the platform.
Total distance = $50\, m + 100\, m = 150\, m$.
Time taken = $10\, s$.
Speed = $150\, m / 10\, s = 15\, m/s$.
Therefore,the correct option is $B$.

(D) Length of the first train,$L_1 = 300 \text{ m}$.
Length of the second train,$L_2 = 200 \text{ m}$.
Total distance to be covered to cross each other,$D = L_1 + L_2 = 300 + 200 = 500 \text{ m}$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed,$S_{rel} = 90 \text{ km/h} + 60 \text{ km/h} = 150 \text{ km/h}$.
Convert the relative speed into $\text{m/s}$ by multiplying by $\frac{5}{18}$:
$S_{rel} = 150 \times \frac{5}{18} = \frac{750}{18} = \frac{125}{3} \text{ m/s}$.
Time taken to cross each other,$T = \frac{D}{S_{rel}} = \frac{500}{125/3} = \frac{500 \times 3}{125} = 4 \times 3 = 12 \text{ seconds}$.

(C) Let the speed of each train be $x \, m/s$.
Given: Length of each train $L_1 = L_2 = 135 \, m$ and speed $s_1 = s_2 = x \, m/s$.
Since the trains are moving in opposite directions,their relative speed is $s_1 + s_2 = x + x = 2x \, m/s$.
The total distance covered to cross each other is the sum of their lengths: $D = 135 + 135 = 270 \, m$.
Using the formula: $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$
$18 = \frac{270}{2x}$
$2x = \frac{270}{18} = 15 \, m/s$
$x = 7.5 \, m/s$.
To convert the speed from $m/s$ to $km/h$,multiply by $\frac{18}{5}$:
$\text{Speed} = 7.5 \times \frac{18}{5} = 1.5 \times 18 = 27 \, km/h$.

(C) The length of the train is $L = 150 \, m$.
The speed of the train is $v_t = 95 \, km/h$.
The speed of the man is $v_m = 5 \, km/h$.
Since both are moving in the same direction,the relative speed is $v_{rel} = v_t - v_m = 95 - 5 = 90 \, km/h$.
Convert the relative speed into $m/s$: $v_{rel} = 90 \times \frac{5}{18} = 25 \, m/s$.
The time taken to pass the man is $t = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{150}{25} = 6 \, seconds$.

(B) Length of the train $(L_1)$ = $100 \, m$.
Speed of the man $(s_2)$ = $6 \, km/h$.
Time taken $(t)$ = $3 \frac{3}{5} \, s = \frac{18}{5} \, s$.
Let the speed of the train be $x \, km/h$.
Since the man and the train are moving in opposite directions,their relative speed is $(x + 6) \, km/h$.
Converting the relative speed to $m/s$: $(x + 6) \times \frac{5}{18} \, m/s$.
Using the formula: $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$.
$\frac{18}{5} = \frac{100}{(x + 6) \times \frac{5}{18}}$.
$\frac{18}{5} = \frac{100 \times 18}{(x + 6) \times 5}$.
$18 = \frac{100 \times 18}{x + 6}$.
$1 = \frac{100}{x + 6}$.
$x + 6 = 100$.
$x = 94 \, km/h$.

(C) Relative speed of the two trains moving in the same direction is the difference of their speeds.
Relative speed $= (50 - 30) \, km/h = 20 \, km/h$.
To convert the speed from $km/h$ to $m/s$,multiply by $\frac{5}{18}$.
Relative speed $= 20 \times \frac{5}{18} = \frac{100}{18} = \frac{50}{9} \, m/s$.
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 $= \text{Relative speed} \times \text{Time}$.
Distance $= \frac{50}{9} \, m/s \times 18 \, s = 50 \times 2 = 100 \, m$.
Therefore,the length of the faster train is $100 \, m$.

(A) Let the speeds of the two trains be $v_1$ and $v_2$ (where $v_1 > v_2$) in $m/s$.
Total length of the trains $L = 130 + 110 = 240 \,m$.
When moving in the same direction,relative speed is $(v_1 - v_2)$. Time taken is $60 \,s$.
$v_1 - v_2 = \frac{240}{60} = 4 \,m/s$ --- (Equation $1$)
When moving in opposite directions,relative speed is $(v_1 + v_2)$. Time taken is $3 \,s$.
$v_1 + v_2 = \frac{240}{3} = 80 \,m/s$ --- (Equation $2$)
Adding Equation $1$ and Equation $2$:
$2v_1 = 84 \implies v_1 = 42 \,m/s$.
Subtracting Equation $1$ from Equation $2$:
$2v_2 = 76 \implies v_2 = 38 \,m/s$.
Thus,the speeds are $42 \,m/s$ and $38 \,m/s$.

(C) Let the speed of the faster train be $v_1$ and the speed of the slower train be $v_2$ in $m/s$.
Total length of both trains $= 90 + 90 = 180\, m$.
When moving in the same direction,relative speed $= v_1 - v_2$.
Time taken $= \frac{\text{Total distance}}{\text{Relative speed}} \implies 18 = \frac{180}{v_1 - v_2} \implies v_1 - v_2 = 10\, (i)$.
When moving in opposite directions,relative speed $= v_1 + v_2$.
Time taken $= \frac{\text{Total distance}}{\text{Relative speed}} \implies 9 = \frac{180}{v_1 + v_2} \implies v_1 + v_2 = 20\, (ii)$.
Adding equations $(i)$ and $(ii)$:
$2v_1 = 30 \implies v_1 = 15\, m/s$.
Subtracting $(i)$ from $(ii)$:
$2v_2 = 10 \implies v_2 = 5\, m/s$.
Thus,the speeds are $15\, m/s$ and $5\, m/s$.

$(A)$ The first train travels for $1.5$ hours ($1 \frac{1}{2}$ hours) before the second train starts.
Distance covered by the first train in $1.5$ hours $= 60 \times 1.5 = 90 \, km$.
Relative speed of the second train with respect to the first train $= 75 - 60 = 15 \, km/h$.
Time taken by the second train to catch up with the first train $= \frac{\text{Distance}}{\text{Relative Speed}} = \frac{90}{15} = 6$ hours.
Meeting time $= 6.30 \, am + 6$ hours $= 12.30 \, pm$.
Distance from the station $= 75 \, km/h \times 6$ hours $= 450 \, km$.

(D) $1$. The distance between stations $A$ and $B$ is $100 \, km$.
$2$. The train from $A$ starts at $7 \, am$ and travels at $20 \, km/h$. By $8 \, am$,it has traveled for $1 \, hour$,covering $20 \, km \times 1 \, h = 20 \, km$.
$3$. At $8 \, am$,the remaining distance between the two trains is $100 \, km - 20 \, km = 80 \, km$.
$4$. Since the trains are moving towards each other,their relative speed is $20 \, km/h + 25 \, km/h = 45 \, km/h$.
$5$. The time taken to meet after $8 \, am$ is $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{80 \, km}{45 \, km/h} = \frac{16}{9} \, hours = 1 \, hour \, 46 \, minutes \, 40 \, seconds$.
$6$. Therefore,the meeting time is $8:00 \, am + 1 \, hour \, 46 \, minutes \, 40 \, seconds = 9:46:40 \, am$.
$7$. Since $9:46:40 \, am$ is not among the options,the correct choice is $D$.

(A) $1$. The first train starts at $9\, am$ and travels for $2$ hours until the second train starts at $11\, am$.
$2$. Distance covered by the first train in $2$ hours $= 50\, km/h \times 2\, h = 100\, km$.
$3$. Remaining distance between the two trains at $11\, am = 210\, km - 100\, km = 110\, km$.
$4$. Since the trains are moving towards each other,their relative speed $= 50\, km/h + 60\, km/h = 110\, km/h$.
$5$. Time taken to meet after $11\, am = \frac{\text{Remaining Distance}}{\text{Relative Speed}} = \frac{110\, km}{110\, km/h} = 1\, hour$.
$6$. The trains meet at $12\, pm$ (noon).
$7$. Total distance from station $A$ at the meeting point = (Distance covered by first train in $2$ hours) + (Distance covered by first train in $1$ hour after $11\, am$) = $100\, km + (50\, km/h \times 1\, h) = 150\, km$.

(B) Let the time taken by the trains to meet be $t$ hours.
Distance travelled by the first train $(d_1)$ = $60t$.
Distance travelled by the second train $(d_2)$ = $40t$.
According to the problem, the difference in distance is $20\, km$:
$60t - 40t = 20$
$20t = 20$
$t = 1\, \text{hour}$.
The total distance between Mumbai and Pune is the sum of the distances covered by both trains:
Total Distance = $d_1 + d_2 = 60t + 40t = 100t$.
Substituting $t = 1$, we get:
Total Distance = $100 \times 1 = 100\, km$.

(B) The total distance covered is $3 + 3 + 3 + 3 = 12 \, km$.
The time taken for each stretch is calculated using the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$.
Time $t_1 = \frac{3}{10} \, h$,$t_2 = \frac{3}{20} \, h$,$t_3 = \frac{3}{30} \, h$,and $t_4 = \frac{3}{60} \, h$.
Total time $T = \frac{3}{10} + \frac{3}{20} + \frac{3}{30} + \frac{3}{60} = \frac{18 + 9 + 6 + 3}{60} = \frac{36}{60} = \frac{3}{5} \, h$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{12}{3/5} = \frac{12 \times 5}{3} = 20 \, km/h$.

(D) Let the distance from $P$ to $R$ be $x \text{ km}$.
Since both men start at the same time and meet at $R$,the time taken by $A$ to reach $R$ is equal to the time taken by $B$ to reach $Q$ and then return to $R$.
Time taken by $A$ to cover distance $PR = x \text{ km}$ at $3 \text{ km/h}$ is $t_A = \frac{x}{3} \text{ hours}$.
Time taken by $B$ to cover distance $PQ = 21 \text{ km}$ and then $QR = (21 - x) \text{ km}$ at $4 \text{ km/h}$ is $t_B = \frac{21 + (21 - x)}{4} = \frac{42 - x}{4} \text{ hours}$.
Equating the times: $\frac{x}{3} = \frac{42 - x}{4}$.
Cross-multiplying: $4x = 3(42 - x)$.
$4x = 126 - 3x$.
$7x = 126$.
$x = \frac{126}{7} = 18 \text{ km}$.

(A) Let the speed of the boy be $v_b$ and the speed of the car be $v_c$.
Since the time taken $t$ is the same for both,we use the formula $\text{speed} = \frac{\text{distance}}{\text{time}}$,which implies $\text{time} = \frac{\text{distance}}{\text{speed}}$.
Given that the time taken is equal:
$\frac{12}{v_b} = \frac{36}{v_c}$
Rearranging the terms to find the ratio of speeds $\frac{v_b}{v_c}$:
$\frac{v_b}{v_c} = \frac{12}{36}$
$\frac{v_b}{v_c} = \frac{1}{3}$
Therefore,the ratio of the speeds of the boy and the car is $1:3$.

(B) Let the speed from $A$ to $B$ be $v_1 = 64 \text{ km/h}$ and the return speed from $B$ to $A$ be $v_2 = x \text{ km/h}$.
The formula for average speed when the distance is the same for both legs of the journey is given by:
$\text{Average Speed} = \frac{2 v_1 v_2}{v_1 + v_2}$
Given that the average speed is $56 \text{ km/h}$,we have:
$56 = \frac{2 \times 64 \times x}{64 + x}$
Multiply both sides by $(64 + x)$:
$56(64 + x) = 128x$
$3584 + 56x = 128x$
Subtract $56x$ from both sides:
$3584 = 128x - 56x$
$3584 = 72x$
Solve for $x$:
$x = \frac{3584}{72} \approx 49.77 \text{ km/h}$

(B) The speed of the bus excluding stoppages is $54 \, km/h$.
This means the bus covers $54 \, km$ in $60$ minutes.
The speed of the bus including stoppages is $45 \, km/h$.
This means the bus covers $45 \, km$ in $60$ minutes.
The distance lost due to stoppages in one hour is $54 \, km - 45 \, km = 9 \, km$.
The time taken to cover this distance of $9 \, km$ at the original speed of $54 \, km/h$ is:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{9 \, km}{54 \, km/h} = \frac{1}{6} \, \text{hours}$.
Converting this into minutes: $\frac{1}{6} \times 60 \, \text{minutes} = 10 \, \text{minutes}$.
Therefore, the bus stops for $10$ minutes per hour.

(A) Let the time taken for them to meet be $t$ hours.
In time $t$,the first boy writes $200t$ lines.
The second boy writes $150t$ lines.
The total number of lines written by both is $817$.
So,$200t + 150t = 817$.
$350t = 817$.
$t = \frac{817}{350} \approx 2.334$ hours.
The number of lines written by the first boy is $200 \times \frac{817}{350} = \frac{817 \times 4}{7} = \frac{3268}{7} \approx 466.85$.
Since the first boy has completed $466$ lines and is currently working on the $467^{th}$ line,they will meet on the $467^{th}$ line.

(A) The length of the bridge is $1 \, km$.
The length of the train is half the length of the bridge,which is $0.5 \, km$.
To pass the bridge completely,the train must cover a total distance equal to the sum of the length of the bridge and the length of the train.
Total distance $= 1 \, km + 0.5 \, km = 1.5 \, km$.
The time taken is $2 \, minutes$,which is equal to $\frac{2}{60} \, hours = \frac{1}{30} \, hours$.
Speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{1.5 \, km}{1/30 \, h} = 1.5 \times 30 \, km/h = 45 \, km/h$.

(D) Total distance $= 80 \text{ km}$.
Total time $= 10 \text{ hours}$.
Half of the journey $= 40 \text{ km}$.
Time taken for the first half $= \frac{3}{5} \times 10 = 6 \text{ hours}$.
Remaining distance $= 80 - 40 = 40 \text{ km}$.
Remaining time $= 10 - 6 = 4 \text{ hours}$.
Required speed for the remaining distance $= \frac{\text{Remaining Distance}}{\text{Remaining Time}} = \frac{40}{4} = 10 \text{ km/h}$.

(A) Let the side of the square field be $a \, km$.
The distance covered by Amit from $A$ to $C$ along the boundary is $AB + BC = a + a = 2a$.
Time taken is $0.5 \, h$ and speed is $8 \, km/h$.
Distance = $\text{Speed} \times \text{Time} = 8 \times 0.5 = 4 \, km$.
Therefore,$2a = 4 \, km$,which gives $a = 2 \, km$.
The area of the square field is $a^2 = 2^2 = 4 \, km^2$.

(A) Let the speed of the train be $x\, km/h$.
Since the man is walking in the same direction,the relative speed is $(x - 6)\, km/h$.
Converting to $m/s$,the relative speed is $(x - 6) \times \frac{5}{18}\, m/s$.
The train covers its own length $(100\, m)$ in $5\, s$:
$(x - 6) \times \frac{5}{18} \times 5 = 100$
$(x - 6) \times \frac{25}{18} = 100$
$x - 6 = 100 \times \frac{18}{25} = 72$
$x = 78\, km/h$.
Now,let the speed of the car be $y\, km/h$. The relative speed is $(78 - y)\, km/h$.
The train passes the car in $6\, s$:
$(78 - y) \times \frac{5}{18} \times 6 = 100$
$(78 - y) \times \frac{5}{3} = 100$
$78 - y = 100 \times \frac{3}{5} = 60$
$y = 78 - 60 = 18\, km/h$.

(A) $1$. Time taken by the motorcyclist to travel from Mumbai to Pune: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{192 \, km}{32 \, km/h} = 6 \, hours$.
$2$. The car starts $2.5 \, hours$ after the motorcyclist and reaches $0.5 \, hours$ earlier than the motorcyclist.
$3$. Total time taken by the car: $6 \, hours - 2.5 \, hours - 0.5 \, hours = 3 \, hours$.
$4$. Speed of the car: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{192 \, km}{3 \, hours} = 64 \, km/h$.
$5$. Ratio of the speed of the motorcycle to the speed of the car: $\frac{32}{64} = \frac{1}{2}$,which is $1:2$.

(C) Let the length of the other train be $x \, m$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds:
Relative speed $= (62 + 40) \, km/h = 102 \, km/h$.
Convert the relative speed into $m/s$ by multiplying by $\frac{5}{18}$:
Relative speed $= 102 \times \frac{5}{18} = \frac{510}{18} = \frac{85}{3} \, m/s$.
The total distance covered when crossing each other is the sum of the lengths of the two trains,which is $(250 + x) \, m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$250 + x = \left(\frac{85}{3}\right) \times 18$
$250 + x = 85 \times 6$
$250 + x = 510$
$x = 510 - 250 = 260 \, m$.
Thus,the length of the other train is $260 \, m$.

(B) Let the length of the train be $x \, \text{meters}$ and its speed be $y \, \text{meters/second}$.
When the train passes a pole, the distance covered is equal to its own length:
$\frac{x}{y} = 15 \implies y = \frac{x}{15} \quad \dots(1)$
When the train passes a $100 \, \text{meter}$ long platform, the total distance covered is $(x + 100) \, \text{meters}$:
$\frac{x + 100}{y} = 25 \implies x + 100 = 25y \quad \dots(2)$
Substitute the value of $y$ from equation $(1)$ into equation $(2)$:
$x + 100 = 25 \times \left( \frac{x}{15} \right)$
$x + 100 = \frac{5x}{3}$
$3x + 300 = 5x$
$2x = 300$
$x = 150 \, \text{meters}$.
Thus, the length of the train is $150 \, \text{meters}$.

(B) The total distance to be covered to pass each other is the sum of the lengths of both trains: $D = 150 \, m + 100 \, m = 250 \, m$.
Time taken is $t = 10 \, s$.
Relative speed $v_{rel} = \frac{D}{t} = \frac{250 \, m}{10 \, s} = 25 \, m/s$.
To convert the relative speed into $km/h$,multiply by $\frac{18}{5}$: $v_{rel} = 25 \times \frac{18}{5} = 90 \, km/h$.
Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds: $v_{rel} = v_1 + v_2$.
Given $v_1 = 30 \, km/h$,we have $30 + v_2 = 90$.
Therefore,$v_2 = 90 - 30 = 60 \, km/h$.

(C) Total distance $= 12 \, km$.
Total time $= 45 \, minutes = \frac{45}{60} \, hours = 0.75 \, hours$.
Distance covered $= \frac{3}{4} \times 12 \, km = 9 \, km$.
Time taken $= \frac{2}{3} \times 45 \, minutes = 30 \, minutes = 0.5 \, hours$.
Remaining distance $= 12 \, km - 9 \, km = 3 \, km$.
Remaining time $= 45 \, minutes - 30 \, minutes = 15 \, minutes = \frac{15}{60} \, hours = 0.25 \, hours$.
Required speed $= \frac{\text{Remaining distance}}{\text{Remaining time}} = \frac{3 \, km}{0.25 \, h} = 12 \, km/h$.

(A) Let the speed of the train be $x \text{ km/h}$.
Since the man is moving against the train,the relative speed is $(x + 6) \text{ km/h}$.
To convert the relative speed into $\text{m/s}$,we multiply by $\frac{5}{18}$:
Relative speed $= (x + 6) \times \frac{5}{18} \text{ m/s}$.
The distance covered by the train to pass the man is equal to its own length,which is $110 \text{ meters}$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$110 = (x + 6) \times \frac{5}{18} \times 6$.
$110 = (x + 6) \times \frac{30}{18}$.
$110 = (x + 6) \times \frac{5}{3}$.
$x + 6 = 110 \times \frac{3}{5}$.
$x + 6 = 22 \times 3$.
$x + 6 = 66$.
$x = 66 - 6 = 60 \text{ km/h}$.
Thus,the speed of the train is $60 \text{ km/h}$.

(D) Let the speed of the train be $x \, km/h$.
When the train moves against the man,the relative speed is $(x + 6) \, km/h$.
Converting to $m/s$: $(x + 6) \times \frac{5}{18} \, m/s$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$,we have:
$110 = (x + 6) \times \frac{5}{18} \times 6$
$110 = (x + 6) \times \frac{5}{3}$
$x + 6 = \frac{110 \times 3}{5} = 66$
$x = 60 \, km/h$.
When the train moves in the same direction as the man,the relative speed is $(60 - 6) \, km/h = 54 \, km/h$.
Converting to $m/s$: $54 \times \frac{5}{18} = 15 \, m/s$.
Time taken to cross the second man $= \frac{\text{Length of train}}{\text{Relative speed}} = \frac{110}{15} = \frac{22}{3} = 7 \frac{1}{3} \, s$.