Time and Distances Questions in English

(C) Let the distance between the house and the school be $S \, km$.
Time taken to go to school $= \frac{S}{3} \, hours$.
Time taken to return from school $= \frac{S}{2} \, hours$.
The total time taken for the round trip is $5 \, hours$.
Therefore,$\frac{S}{3} + \frac{S}{2} = 5$.
Taking the least common multiple of $3$ and $2$,which is $6$,we get:
$\frac{2S + 3S}{6} = 5$.
$\frac{5S}{6} = 5$.
$5S = 30$.
$S = 6 \, km$.
Thus,the distance between his house and school is $6 \, km$.

(C) The formula for average speed when the distance covered is the same for two different speeds is given by $\frac{2v_1v_2}{v_1+v_2}$.
Here,$v_1 = 64 \, km/hr$ and $v_2 = 80 \, km/hr$.
Average speed $= \frac{2 \times 64 \times 80}{64+80} = \frac{2 \times 64 \times 80}{144}$.
Simplifying the expression: $\frac{10240}{144} = \frac{640}{9} \approx 71.11 \, km/hr$.

(C) Let the distance travelled on foot be $x \, km$.
Then the distance travelled by bicycle is $(61 - x) \, km$.
We know that $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
Time taken on foot = $\frac{x}{4} \, hours$.
Time taken on bicycle = $\frac{61 - x}{9} \, hours$.
Total time taken is $9 \, hours$,so:
$\frac{x}{4} + \frac{61 - x}{9} = 9$
Multiply the entire equation by $36$ (the $LCM$ of $4$ and $9$):
$9x + 4(61 - x) = 9 \times 36$
$9x + 244 - 4x = 324$
$5x = 324 - 244$
$5x = 80$
$x = \frac{80}{5} = 16 \, km$.
Thus,the distance travelled on foot is $16 \, km$.

(C) Total distance $= 6 \, km$.
Total time $= 45 \, minutes = \frac{45}{60} \, hr = 0.75 \, hr$.
First half distance $= 3 \, km$.
Time taken for first half $= \frac{2}{3} \times 45 \, minutes = 30 \, minutes = 0.5 \, hr$.
Remaining distance $= 6 \, km - 3 \, km = 3 \, km$.
Remaining time $= 45 \, minutes - 30 \, minutes = 15 \, minutes = \frac{15}{60} \, hr = 0.25 \, hr$.
Required speed for remaining distance $= \frac{\text{Remaining distance}}{\text{Remaining time}} = \frac{3 \, km}{0.25 \, hr} = 12 \, km/hr$.

(B) Average speed $= \frac{\text{Total distance travelled}}{\text{Total time taken}}$
Total distance travelled $= 10 \, km + 12 \, km = 22 \, km$
Time taken for the first part $= \frac{10 \, km}{12 \, km/hr} = \frac{5}{6} \, hr$
Time taken for the second part $= \frac{12 \, km}{10 \, km/hr} = \frac{6}{5} \, hr$
Total time taken $= \frac{5}{6} + \frac{6}{5} = \frac{25 + 36}{30} = \frac{61}{30} \, hr$
Average Speed $= \frac{22}{\frac{61}{30}} = \frac{22 \times 30}{61} = \frac{660}{61} \approx 10.82 \, km/hr$
Thus,the average speed is approximately $10.8 \, km/hr$.

(B) Let the distance for each of the three equal parts be $d \, km$.
Total distance $= 3d \, km$.
Time taken for each part is $t_1 = d/3$,$t_2 = d/4$,and $t_3 = d/5$ hours.
Total time $= 47 \, minutes = 47/60 \, hours$.
So,$d/3 + d/4 + d/5 = 47/60$.
Taking $LCM$ of $3, 4, 5$,which is $60$:
$(20d + 15d + 12d) / 60 = 47/60$.
$47d / 60 = 47/60$.
Therefore,$d = 1 \, km$.
Total distance $= 3d = 3 \times 1 = 3 \, km$.

(C) Distance travelled by the train from $A$ between $8\, a.m.$ and $9\, a.m.$ is $60\, km/hr \times 1\, hr = 60\, km$.
Remaining distance between the two trains at $9\, a.m.$ is $330\, km - 60\, km = 270\, km$.
Since the trains are moving towards each other,their relative speed is the sum of their individual speeds: $60\, km/hr + 75\, km/hr = 135\, km/hr$.
Time taken to meet after $9\, a.m.$ is $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{270\, km}{135\, km/hr} = 2\, hours$.
Therefore,the trains will meet at $9\, a.m. + 2\, hours = 11\, a.m.$

(A) Let the original duration of the flight be $t$ hours.
Let the original speed be $v = \frac{600}{t} \, km/hr$.
According to the problem,the new speed is $v' = v - 200 = \frac{600}{t} - 200$.
The new time is $t' = t + 0.5$ hours.
Since distance is constant $(600 \, km)$,we have $v' \times t' = 600$.
$(\frac{600}{t} - 200)(t + 0.5) = 600$.
$600 + \frac{300}{t} - 200t - 100 = 600$.
$\frac{300}{t} - 200t - 100 = 0$.
Divide by $100$: $\frac{3}{t} - 2t - 1 = 0$.
$3 - 2t^2 - t = 0 \implies 2t^2 + t - 3 = 0$.
Factoring the quadratic equation: $2t^2 + 3t - 2t - 3 = 0$.
$t(2t + 3) - 1(2t + 3) = 0$.
$(t - 1)(2t + 3) = 0$.
Since time cannot be negative,$t = 1$ hour.

(A) Let Abhay's speed be $x \, km/h$.
Time taken by Abhay to cover $30 \, km = \frac{30}{x} \, hours$.
Since Abhay takes $2 \, hours$ more than Sameer,the time taken by Sameer is $\left(\frac{30}{x} - 2\right) \, hours$.
If Abhay doubles his speed,his new speed is $2x \, km/h$.
Time taken by Abhay at double speed $= \frac{30}{2x} = \frac{15}{x} \, hours$.
According to the problem,this time is $1 \, hour$ less than Sameer's time:
$\frac{15}{x} = \left(\frac{30}{x} - 2\right) - 1$
$\frac{15}{x} = \frac{30}{x} - 3$
$3 = \frac{30}{x} - \frac{15}{x}$
$3 = \frac{15}{x}$
$x = \frac{15}{3} = 5 \, km/h$.
Thus,Abhay's speed is $5 \, km/h$.

(C) Let the length of the journey be $x \, km$.
The time taken at $40 \, km/hr$ is $t_1 = \frac{x}{40} \, hours$.
The time taken at $35 \, km/hr$ is $t_2 = \frac{x}{35} \, hours$.
Given that the difference in time is $15 \, minutes$,which is $\frac{15}{60} = \frac{1}{4} \, hours$.
Therefore,$\frac{x}{35} - \frac{x}{40} = \frac{1}{4}$.
Taking the least common multiple of $35$ and $40$,which is $280$:
$\frac{8x - 7x}{280} = \frac{1}{4}$.
$\frac{x}{280} = \frac{1}{4}$.
$x = \frac{280}{4} = 70 \, km$.
Thus,the length of the journey is $70 \, km$.

(A) The relative speed of $B$ with respect to $A$ is $(6 - 1) = 5$ rounds per hour.
This means $B$ gains $5$ full rounds over $A$ in one hour.
To meet for the first time,$B$ must gain exactly $1$ full round over $A$.
Time taken to gain $5$ rounds $= 60$ minutes.
Time taken to gain $1$ round $= \frac{60}{5} = 12$ minutes.
Since they start at $7.30 \, a.m.$,they will meet for the first time after $12$ minutes.
Therefore,the meeting time is $7.30 + 12 \text{ minutes} = 7.42 \, a.m.$

(B) Let the speed of car $P$ be $u \, km/hr$ and the speed of car $Q$ be $v \, km/hr$.
When the cars travel in opposite directions,their relative speed is $(u + v)$.
Since they meet after $1 \, hour$ over a distance of $120 \, km$,we have: $u + v = \frac{120}{1} = 120 \dots (1)$.
When the cars travel in the same direction,their relative speed is $(u - v)$.
Since they meet after $6 \, hours$ over a distance of $120 \, km$,we have: $u - v = \frac{120}{6} = 20 \dots (2)$.
Adding equation $(1)$ and $(2)$,we get: $(u + v) + (u - v) = 120 + 20$.
$2u = 140$,which implies $u = 70 \, km/hr$.
Thus,the speed of car $P$ is $70 \, km/hr$.

(A) The initial distance between the policeman and the thief is $200\, m$.
Relative speed of the policeman with respect to the thief $= 11\, km/h - 10\, km/h = 1\, km/h$.
To convert the relative speed into meters per minute,we use the conversion factor: $1\, km/h = \frac{1000\, m}{60\, min} = \frac{50}{3}\, m/min$.
Distance covered by the policeman relative to the thief in $6\, minutes = \text{Relative speed} \times \text{Time} = \frac{50}{3}\, m/min \times 6\, min = 100\, m$.
Therefore,the remaining distance between them after $6\, minutes = 200\, m - 100\, m = 100\, m$.

(A) Let the actual distance travelled be $x \, km$ and the time taken be $t \, hr$.
Since the time taken is the same in both scenarios,we can write:
$t = \frac{x}{10}$
Also,if he walks at $14 \, km/hr$,he covers $x + 20 \, km$ in the same time $t$:
$t = \frac{x + 20}{14}$
Equating the two expressions for $t$:
$\frac{x}{10} = \frac{x + 20}{14}$
$14x = 10(x + 20)$
$14x = 10x + 200$
$4x = 200$
$x = 50 \, km$
Thus,the actual distance travelled is $50 \, km$.

(C) Since $A$ and $B$ are moving in opposite directions, their relative speed is the sum of their individual speeds.
Relative speed $= 2 + 3 = 5\, \text{rounds per hour}$.
This means they cross each other $5$ times in every hour.
From $8\, a.m.$ to $9.30\, a.m.$, the total time elapsed is $1.5\, \text{hours}$.
Number of crossings $= \text{Relative speed} \times \text{Time} = 5 \times 1.5 = 7.5$.
Since they cross each other at discrete points, they will have completed $7$ full crossings before $9.30\, a.m.$

(C) Let the speed of the train be $v_t\, km/hr$ and the speed of the car be $v_c\, km/hr$.
Case $1$: $\frac{120}{v_t} + \frac{480}{v_c} = 8$ --- $(1)$
Case $2$: $\frac{200}{v_t} + \frac{400}{v_c} = 8 + \frac{20}{60} = 8 + \frac{1}{3} = \frac{25}{3}$ --- $(2)$
Divide $(1)$ by $40$: $\frac{3}{v_t} + \frac{12}{v_c} = \frac{1}{5}$ --- $(3)$
Divide $(2)$ by $200$: $\frac{1}{v_t} + \frac{2}{v_c} = \frac{25}{3 \times 200} = \frac{1}{24}$ --- $(4)$
Multiply $(4)$ by $3$: $\frac{3}{v_t} + \frac{6}{v_c} = \frac{3}{24} = \frac{1}{8}$ --- $(5)$
Subtract $(5)$ from $(3)$: $(\frac{12}{v_c} - \frac{6}{v_c}) = \frac{1}{5} - \frac{1}{8} \Rightarrow \frac{6}{v_c} = \frac{8-5}{40} = \frac{3}{40} \Rightarrow v_c = \frac{6 \times 40}{3} = 80\, km/hr$.
Substitute $v_c = 80$ in $(4)$: $\frac{1}{v_t} + \frac{2}{80} = \frac{1}{24} \Rightarrow \frac{1}{v_t} = \frac{1}{24} - \frac{1}{40} = \frac{5-3}{120} = \frac{2}{120} = \frac{1}{60} \Rightarrow v_t = 60\, km/hr$.
Ratio of speed of train to car $= v_t : v_c = 60 : 80 = 3 : 4$.

(C) The speed from $P$ to $Q$ is $v_1 = 40 \, km/hr$.
The return speed is increased by $50 \%$,so $v_2 = 40 + (0.50 \times 40) = 40 + 20 = 60 \, km/hr$.
The formula for average speed when distance is constant is $\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$.
Substituting the values: $\text{Average Speed} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \, km/hr$.

(C) Let the distance from the house to the school be $x \, km$.
Let the scheduled time to reach the school be $t \, hours$.
In the first case,the speed is $v_1 = 2.5 \, km/hr$ and the time taken is $t_1 = t + \frac{6}{60} = t + 0.1 \, hours$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$,we have $x = 2.5(t + 0.1)$.
In the second case,the speed is $v_2 = 2.5 + 1 = 3.5 \, km/hr$ and the time taken is $t_2 = t - \frac{6}{60} = t - 0.1 \, hours$.
Thus,$x = 3.5(t - 0.1)$.
Equating the two expressions for $x$: $2.5(t + 0.1) = 3.5(t - 0.1)$.
$2.5t + 0.25 = 3.5t - 0.35$.
$1.0t = 0.6$,so $t = 0.6 \, hours$.
Substituting $t$ back into the first equation: $x = 2.5(0.6 + 0.1) = 2.5 \times 0.7 = 1.75 \, km$.

(D) Let the actual speed of the car be $v \, km/hr$.
Given speed $= \frac{5}{7}v$.
Time taken $= 1 \, hour + 40 \, minutes + 48 \, seconds$.
Convert time into hours: $1 + \frac{40}{60} + \frac{48}{3600} = 1 + \frac{2}{3} + \frac{48}{3600} = 1 + \frac{2}{3} + \frac{1}{75} = \frac{75 + 50 + 1}{75} = \frac{126}{75} = 1.68 \, hours$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$:
$42 = (\frac{5}{7}v) \times 1.68$.
$42 = (\frac{5}{7}v) \times \frac{126}{75}$.
$42 = v \times \frac{5 \times 126}{7 \times 75} = v \times \frac{630}{525} = v \times 1.2$.
$v = \frac{42}{1.2} = 35 \, km/hr$.

(D) Total distance $= 600 + 800 + 500 + 100 = 2000 \, km$.
Total time taken $= \frac{600}{80} + \frac{800}{40} + \frac{500}{400} + \frac{100}{50}$.
$= 7.5 + 20 + 1.25 + 2 = 30.75 \, hours$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{2000}{30.75} = \frac{200000}{3075} = \frac{8000}{123} \, km/hr$.
$= 65 \frac{5}{123} \, km/hr$.

(B) Let the speed of $A$ be $v_A$ and the speed of $B$ be $v_B$ in $km/hr$.
Given: $v_A + v_B = 7$ and $v_A > v_B$.
The time taken by $A$ is $t_A = \frac{24}{v_A}$ and the time taken by $B$ is $t_B = \frac{24}{v_B}$.
Given: $t_A + t_B = 14$,so $\frac{24}{v_A} + \frac{24}{v_B} = 14$.
$\frac{24(v_A + v_B)}{v_A v_B} = 14$.
Substituting $v_A + v_B = 7$: $\frac{24 \times 7}{v_A v_B} = 14$.
$v_A v_B = \frac{24 \times 7}{14} = 12$.
We have $v_A + v_B = 7$ and $v_A v_B = 12$. The quadratic equation $x^2 - 7x + 12 = 0$ gives the speeds.
$(x - 4)(x - 3) = 0$,so $x = 4$ or $x = 3$.
Since $A$ is faster than $B$,$v_A = 4 \, km/hr$ and $v_B = 3 \, km/hr$.

(B) Let the usual speed be $v$ and the usual time taken be $t$.
Distance $d = v \times t$.
When the man walks at $\frac{6}{7}$ of his usual speed, the new speed is $v' = \frac{6}{7}v$.
The new time taken is $t' = \frac{d}{v'} = \frac{d}{\frac{6}{7}v} = \frac{7}{6} \times \frac{d}{v} = \frac{7}{6}t$.
Given that the man is $12 \, \text{minutes}$ late, the difference in time is $t' - t = 12 \, \text{minutes}$.
$\frac{7}{6}t - t = 12 \, \text{minutes}$.
$\frac{1}{6}t = 12 \, \text{minutes}$.
$t = 12 \times 6 = 72 \, \text{minutes}$.
$72 \, \text{minutes} = 1 \, \text{hour} \, 12 \, \text{minutes}$.

(C) Let the distance to point $A$ be $d$ km and the time taken to reach at $2\, p.m.$ be $t$ hours.
If speed is $10\, km/hr$,time taken is $t$,so $d = 10t$.
If speed is $15\, km/hr$,he reaches at $12\, noon$,which is $2$ hours earlier than $2\, p.m.$,so $d = 15(t - 2)$.
Equating the two expressions for $d$: $10t = 15(t - 2)$.
$10t = 15t - 30$.
$5t = 30$,so $t = 6$ hours.
The distance $d = 10 \times 6 = 60\, km$.
To reach at $1\, p.m.$,he needs to reach $1$ hour earlier than $2\, p.m.$
So,the required time is $t - 1 = 6 - 1 = 5$ hours.
Required speed $= \frac{\text{Distance}}{\text{Time}} = \frac{60\, km}{5\, hr} = 12\, km/hr$.

(C) Let the speed of the car be $x\, km/hr$ and the speed of the train be $1.5x\, km/hr$.
The distance between point $A$ and point $B$ is $75\, km$.
The time taken by the car to travel from $A$ to $B$ is $T_c = \frac{75}{x}\, \text{hours}$.
The time taken by the train to travel from $A$ to $B$ (excluding stops) is $T_t = \frac{75}{1.5x}\, \text{hours}$.
Since the train stops for $12.5\, \text{minutes}$ (which is $\frac{12.5}{60} = \frac{5}{24}\, \text{hours}$),the total time taken by the train is $T_t + \frac{5}{24}$.
Given that both reach at the same time,we have: $\frac{75}{x} = \frac{75}{1.5x} + \frac{5}{24}$.
Subtracting $\frac{75}{1.5x}$ from both sides: $\frac{75}{x} - \frac{75}{1.5x} = \frac{5}{24}$.
Factoring out $\frac{75}{x}$: $\frac{75}{x} (1 - \frac{1}{1.5}) = \frac{5}{24} \Rightarrow \frac{75}{x} (1 - \frac{2}{3}) = \frac{5}{24}$.
$\frac{75}{x} (\frac{1}{3}) = \frac{5}{24} \Rightarrow \frac{25}{x} = \frac{5}{24}$.
Solving for $x$: $x = \frac{25 \times 24}{5} = 5 \times 24 = 120\, km/hr$.

(B) The circumference of the track is $726 \, m = 0.726 \, km$.
Since they are walking in opposite directions,their relative speed is the sum of their individual speeds.
Relative speed $= 4.5 \, km/h + 3.75 \, km/h = 8.25 \, km/h$.
Time taken to meet for the first time $= \frac{\text{Distance}}{\text{Relative speed}} = \frac{0.726 \, km}{8.25 \, km/h}$.
Time in hours $= \frac{0.726}{8.25} \, h$.
To convert this into minutes,multiply by $60$:
Time in minutes $= \left( \frac{0.726}{8.25} \right) \times 60 \, min = 0.088 \times 60 \, min = 5.28 \, min$.

(B) Let the original speed of the person be $x \, \text{km/hr}$.
The distance covered at original speed in $30 \, \text{hours}$ is $D_1 = 30x$.
If he reduces his speed by $\frac{1}{15}^{th}$, the new speed becomes $x - \frac{1}{15}x = \frac{14}{15}x$.
The distance covered at the new speed in $30 \, \text{hours}$ is $D_2 = 30 \times \frac{14}{15}x = 28x$.
According to the problem, the difference in distance is $10 \, \text{km}$:
$D_1 - D_2 = 10$
$30x - 28x = 10$
$2x = 10$
$x = 5 \, \text{km/hr}$.

(C) Let the time taken by $A$ be $x$ hours.
Since $A$ takes $30 \text{ minutes}$ $(0.5 \text{ hours})$ more than $B$,the time taken by $B$ is $(x - 0.5)$ hours.
The ratio of speeds of $A$ and $B$ is $3:4$. Let the speeds be $3k$ and $4k$ respectively.
Since the distance is the same for both,we have:
$\text{Distance} = \text{Speed} \times \text{Time}$
$3k \times x = 4k \times (x - 0.5)$
Dividing both sides by $k$ $(k \neq 0)$:
$3x = 4(x - 0.5)$
$3x = 4x - 2$
$4x - 3x = 2$
$x = 2 \text{ hours}$.
Thus,the time taken by $A$ is $2 \text{ hours}$.

(D) Distance covered in the first $2\, hours\, 15\, minutes$ $(2.25\, hours)$ at $80\, km/h$ is:
$D_1 = 80 \times 2.25 = 180\, km.$
Remaining distance to be covered is:
$D_2 = 350 - 180 = 170\, km.$
Time taken to cover the remaining distance at $60\, km/h$ is:
$T_2 = \frac{170}{60} = \frac{17}{6}\, hours = 2\, hours\, 50\, minutes.$
Total time taken for the journey is:
$T_{total} = 2\, hours\, 15\, minutes + 2\, hours\, 50\, minutes = 5\, hours\, 5\, minutes.$
Anna started at $5:20\, a.m.$ Adding $5\, hours\, 5\, minutes$ to the start time:
$5:20 + 5:05 = 10:25\, a.m.$

(A) $1$. Calculate the actual travel time without stops: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{600 \, km}{100 \, km/hr} = 6 \, \text{hours}$.
$2$. Calculate the number of stops: The train stops every $75 \, km$. The number of stops in $600 \, km$ is $\frac{600}{75} = 8$. However, the train does not stop at the final destination, so the number of stops is $8 - 1 = 7$ stops.
$3$. Calculate the total halt time: $7 \, \text{stops} \times 3 \, \text{minutes/stop} = 21 \, \text{minutes}$.
$4$. Calculate the total time: $\text{Total time} = \text{Travel time} + \text{Halt time} = 6 \, \text{hours} + 21 \, \text{minutes} = 6 \, \text{hours} \, 21 \, \text{minutes}$.

(C) Step $1$: Convert the speed from $km/hr$ to $m/s$ by multiplying by $\frac{5}{18}$.
Speed $= 108 \times \frac{5}{18} \, m/s = 6 \times 5 \, m/s = 30 \, m/s$.
Step $2$: Calculate the distance using the formula $\text{Distance} = \text{Speed} \times \text{Time}$.
Distance $= 30 \, m/s \times 15 \, s = 450 \, m$.

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

(D) The perimeter of the square field is $4 \times \text{side} = 4 \times 35 \, m = 140 \, m$.
The speed of the boy is $9 \, km/hr$. To convert this into $m/s$,we multiply by $\frac{5}{18}$:
$\text{Speed} = 9 \times \frac{5}{18} \, m/s = 2.5 \, m/s$.
The time taken to run around the field is given by the formula:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{140 \, m}{2.5 \, m/s} = 56 \, \text{seconds}$.

(C) The average speed is calculated as $\text{Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.
For the first bus,$\text{Distance}_1 = 300 \, km$ and $\text{Time}_1 = 7.5 \, hours$.
$\text{Speed}_1 = \frac{300}{7.5} = 40 \, km/h$.
For the second bus,$\text{Distance}_2 = 450 \, km$ and $\text{Time}_2 = 9 \, hours$.
$\text{Speed}_2 = \frac{450}{9} = 50 \, km/h$.
The ratio of their average speeds is $\frac{\text{Speed}_1}{\text{Speed}_2} = \frac{40}{50} = \frac{4}{5}$ or $4:5$.

(C) In the first $2 \, hours$,the distance covered is $70 \times 2 = 140 \, km$.
In the next $2 \, hours$,the speed increases to $80 \, km/hr$,so the distance covered is $80 \times 2 = 160 \, km$.
Total distance covered in $4 \, hours = 140 + 160 = 300 \, km$.
Remaining distance to be covered $= 345 - 300 = 45 \, km$.
In the next interval,the speed increases to $90 \, km/hr$.
Time taken to cover the remaining $45 \, km$ at $90 \, km/hr = \frac{45}{90} = 0.5 \, hours$.
Total time taken $= 4 + 0.5 = 4.5 \, hours$ or $4\frac{1}{2} \, hours$.

(D) Initial speed of the train $= \frac{\text{Distance}}{\text{Time}} = \frac{10 \text{ km}}{12 \text{ min}} = \frac{10 \text{ km}}{12/60 \text{ hr}} = \frac{10 \times 60}{12} \text{ km/hr} = 50 \text{ km/hr}$.
If the speed is decreased by $5 \text{ km/hr}$,the new speed $= 50 - 5 = 45 \text{ km/hr}$.
Time taken to cover the same distance of $10 \text{ km}$ at the new speed:
Time $= \frac{\text{Distance}}{\text{New Speed}} = \frac{10}{45} \text{ hours}$.
Converting into minutes: $\frac{10}{45} \times 60 \text{ minutes} = \frac{10 \times 4}{3} \text{ minutes} = \frac{40}{3} \text{ minutes}$.
$\frac{40}{3} \text{ minutes} = 13 \frac{1}{3} \text{ minutes} = 13 \text{ minutes } + (\frac{1}{3} \times 60) \text{ seconds} = 13 \text{ minutes } 20 \text{ seconds}$.

(D) The speed of the man is $5\, km/hr$. To convert this into $m/s$,we multiply by $\frac{5}{18}$.
Speed $= 5 \times \frac{5}{18} = \frac{25}{18}\, m/s$.
The time taken to cross the bridge is $15\, minutes$. Converting this into seconds,we get $15 \times 60 = 900\, seconds$.
The length of the bridge is equal to the distance covered by the man in that time.
Distance $= \text{Speed} \times \text{Time}$.
Distance $= \frac{25}{18} \times 900$.
Distance $= 25 \times 50 = 1250\, meters$.

(A) Speed of the truck $= \frac{\text{Distance}}{\text{Time}} = \frac{550\, m}{60\, s} = \frac{55}{6}\, m/s$.
Speed of the bus $= \frac{\text{Distance}}{\text{Time}} = \frac{33\, km}{45\, min} = \frac{33 \times 1000\, m}{45 \times 60\, s} = \frac{33000}{2700}\, m/s = \frac{330}{27}\, m/s = \frac{110}{9}\, m/s$.
Ratio of the speed of the truck to the speed of the bus $= \frac{55/6}{110/9} = \frac{55}{6} \times \frac{9}{110} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.
Thus,the ratio is $3:4$.

(C) The distance travelled in each hour forms an arithmetic progression $(AP)$ where the first term $a = 35$ and the common difference $d = 2$.
The total distance travelled in $n = 12$ hours is the sum of the first $12$ terms of this $AP$.
The formula for the sum of the first $n$ terms of an $AP$ is $S_n = \frac{n}{2} \{2a + (n - 1)d\}$.
Substituting the given values:
$S_{12} = \frac{12}{2} \{2 \times 35 + (12 - 1) \times 2\}$
$S_{12} = 6 \{70 + 11 \times 2\}$
$S_{12} = 6 \{70 + 22\}$
$S_{12} = 6 \times 92$
$S_{12} = 552\, km$.

(D) The speed of the athlete is calculated using the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
Given: $\text{Distance} = 200 \text{ m}$, $\text{Time} = 24 \text{ s}$.
$\text{Speed} = \frac{200}{24} \text{ m/s} = \frac{25}{3} \text{ m/s}$.
To convert speed from $\text{m/s}$ to $\text{km/hr}$, multiply by $\frac{18}{5}$.
$\text{Speed} = \frac{25}{3} \times \frac{18}{5} \text{ km/hr} = 5 \times 6 \text{ km/hr} = 30 \text{ km/hr}$.

(C) Let the distance between the village and the post office be $d \, km$.
Time taken to go to the post office $= \frac{d}{25} \, hours$.
Time taken to return from the post office $= \frac{d}{4} \, hours$.
Total time taken $= 5 \, hours \, 48 \, minutes = 5 + \frac{48}{60} \, hours = 5 + \frac{4}{5} = \frac{29}{5} \, hours$.
According to the problem,$\frac{d}{25} + \frac{d}{4} = \frac{29}{5}$.
Taking the $LCM$ of $25$ and $4$,which is $100$,we get: $\frac{4d + 25d}{100} = \frac{29}{5}$.
$\frac{29d}{100} = \frac{29}{5}$.
$d = \frac{29}{5} \times \frac{100}{29} = 20 \, km$.
Thus,the distance of the post office from the village is $20 \, km$.

(B) Speed is defined as the ratio of distance traveled to the time taken.
Given: Distance $= 600 \, m$,Time $= 5 \, \text{minutes} = 5 \times 60 \, \text{seconds} = 300 \, \text{seconds}$.
Speed in $m/s = \frac{600 \, m}{300 \, s} = 2 \, m/s$.
To convert speed from $m/s$ to $km/h$,we multiply by $\frac{18}{5}$.
Speed in $km/h = 2 \times \frac{18}{5} = \frac{36}{5} = 7.2 \, km/h$.

(C) To compare the speeds,we must convert them to the same unit.
Convert $80 \, km/h$ to $m/s$ by multiplying by $\frac{5}{18}$:
$80 \times \frac{5}{18} = \frac{400}{18} = \frac{200}{9} \, m/s$.
Now,compare $\frac{200}{9} \, m/s$ with $10 \, m/s$:
Ratio $= \frac{200}{9} : 10 = \frac{200}{9} : \frac{90}{9} = 200 : 90 = 20 : 9$.

(B) Step $1$: Calculate the speed of Mohan.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{10.2 \, km}{3 \, h} = 3.4 \, km/h$.
Step $2$: Calculate the distance covered in $5 \, hours$.
Distance $= \text{Speed} \times \text{Time} = 3.4 \, km/h \times 5 \, h = 17 \, km$.

(A) First,convert the speed of the train from $km/h$ to $m/s$:
Speed $= 72 \times \frac{5}{18} = 20\, m/s$.
Next,calculate the total distance covered by the train in $25\, s$:
Total distance $= \text{Speed} \times \text{Time} = 20\, m/s \times 25\, s = 500\, m$.
The total distance covered while crossing a bridge is the sum of the length of the train and the length of the bridge:
Total distance $= \text{Length of train} + \text{Length of bridge}$.
$500\, m = 100\, m + \text{Length of bridge}$.
Length of bridge $= 500\, m - 100\, m = 400\, m$.

(B) Let the length of the train be $x \,m$.
The total distance covered by the train while passing the bridge is $(x + 150) \,m$.
The speed of the train is $60 \,km/h$. Converting this into $m/s$:
$60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \,m/s$.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$x + 150 = \frac{50}{3} \times 18$.
$x + 150 = 50 \times 6$.
$x + 150 = 300$.
$x = 300 - 150 = 150 \,m$.
Therefore,the length of the train is $150 \,m$.

(C) Given:
Speed of sound $(v)$ = $330 \text{ m/s}$
Time taken $(t)$ = $10 \text{ s}$
Distance $(d)$ = Speed $\times$ Time
$d = 330 \text{ m/s} \times 10 \text{ s} = 3300 \text{ meters}$
To convert meters into kilometers,divide by $1000$:
$d = 3300 / 1000 = 3.3 \text{ km}$
Therefore,the distance of the thundercloud is $3.3 \text{ km}$.

(B) Speed of the train $= 92.4 \, km/h$.
To convert speed into $m/s$,multiply by $\frac{5}{18}$:
Speed $= 92.4 \times \frac{5}{18} = \frac{462}{18} = \frac{77}{3} \, m/s$.
Time $= 10 \, minutes = 10 \times 60 = 600 \, seconds$.
Distance $= \text{Speed} \times \text{Time}$.
Distance $= \frac{77}{3} \times 600 = 77 \times 200 = 15400 \, m$.

(A) The distance of the sun from the earth is $143,400,000 \text{ km} = 1434 \times 10^5 \text{ km}$.
The time taken by light to travel from the sun to the earth is $7 \text{ minutes}$ and $58 \text{ seconds}$.
Converting time into seconds: $T = (7 \times 60) + 58 = 420 + 58 = 478 \text{ seconds}$.
The velocity of light $v$ is given by the formula: $v = \frac{\text{Distance}}{\text{Time}}$.
$v = \frac{1434 \times 10^5 \text{ km}}{478 \text{ s}} = 3 \times 10^5 \text{ km/s}$.
Thus,the velocity of light is $3 \times 10^5 \text{ km/s}$.

(C) The distance covered by the train is calculated using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
Given,$\text{Speed} = 48 \text{ km/h}$ and $\text{Time} = 50 \text{ minutes} = \frac{50}{60} \text{ hours}$.
$\text{Distance} = 48 \times \frac{50}{60} = 48 \times \frac{5}{6} = 8 \times 5 = 40 \text{ km}$.
Now,to cover the same distance of $40 \text{ km}$ in $40 \text{ minutes}$ (which is $\frac{40}{60} = \frac{2}{3} \text{ hours}$),the required speed is:
$\text{Required Speed} = \frac{\text{Distance}}{\text{New Time}} = \frac{40}{40/60} = \frac{40 \times 60}{40} = 60 \text{ km/h}$.
Therefore,the train must run at $60 \text{ km/h}$.

(A) The circumference of the wheel is $3 \frac{3}{4} \text{ m} = \frac{15}{4} \text{ m}$.
In $2$ seconds,the wheel makes $4$ revolutions.
Distance covered in $2$ seconds $= \text{Circumference} \times \text{Number of revolutions} = \frac{15}{4} \times 4 = 15 \text{ m}$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{15 \text{ m}}{2 \text{ s}} = 7.5 \text{ m/s}$.
To convert speed from $\text{m/s}$ to $\text{km/h}$,multiply by $\frac{18}{5}$.
Speed in $\text{km/h} = 7.5 \times \frac{18}{5} = 1.5 \times 18 = 27 \text{ km/h}$.