Answer: Option B. -> 8 hours
Answer: (b)Since the train runs at $7/11$ of its own speed,the time it takes is $11/7$ of its usual speed.Let the usual time taken be t hours.Then we can write, $11/7$ t = 22t = ${22 × 7}/11$ = 14 hoursHence, time saved= 22 - 14 = 8 hours

Answer: Option D. -> 40
Answer: (d)Using Rule 2,Total distance = 100 km.Total time = $50/50 + 40/40 + 10/20$= $1 + 1 + 1/2 = 5/2$ hoursAverage speed = ${100 × 2}/5$ = 40 kmph

Answer: Option D. -> 5$1/3$
Answer: (d)Using Rule 2,If a man travels different distances $d_1,d_2,d_3$,... and so on in different time $t_1,t_2,t_3$ respectively then,Average speed= $\text"total travelled distance"/\text"total time taken in travelling distance"$= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Answer: Option C. -> 120 kmph
Answer: (c)Using Rule 5,Here, the distances are equal.Average speed= $({2 × 100 × 150}/{100 + 150})$ kmph= ${2 × 100 × 150}/250$ = 120 kmph

Answer: Option B. -> 4 km
Answer: (b)Required distance of office from house = x km. (let)Time = $\text"Distance"/ \text"Speed"$According to the question,$x/5 - x/6 = {6 + 2}/60 = 2/15$${6x - 5x}/30 = 2/15$$x/30 = 2/15$$x = 2/15$ × 30 = 4 km. Using Rule 10,If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

Answer: Option C. -> 3 km/hour more
Answer: (c)Using Rule 2,If a man travels different distances $d_1,d_2,d_3$,... and so on in different time $t_1,t_2,t_3$ respectively then,Average speed= $\text"total travelled distance"/\text"total time taken in travelling distance"$= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Answer: Option D. -> 33$1/3$ km/hour
Answer: (d)
Let the total distance be x km.
Total time = ${x/3}/25 + {x/4}/30 +{{5x}/12}/50$
= $x/75 + x/120 + x/120$
= $x/75 + x/60 = {4x + 5x}/300 = {3x}/100$ hours
Average speed
= $\text"Total distance"/\text"Time taken"$
= $x/{{3x}/100} = 100/3 = 33{1}/3$ kmph
Using Rule 18,If a man covers $1/x$ part of Journey at u km/h,$1/y$ part at v km/h and $1/z$ part at w km/hr and so on, then his average speed for the whole journey will be $1/{1/{xu} + 1/{yv} + 1/{zw} +...}$

Answer: Option B. -> 14 hrs
Answer: (b)Using Rule 1,Distance = Speed × TimeSpeed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

Answer: Option C. -> 560 km.
Answer: (c)Distance = Speed × Time= (80 × 7) km. = 560 km.

Answer: Option C. -> 37.5
Answer: (c)Using Rule 5,Average speed of whole journey= $({2xy}/{x + y})$ kmph= ${2 × 50 × 30}/{50 + 30}$= ${2 × 50 × 30}/80$ = 37.5 kmph