Answer: Option B. -> 3$3/4$ hours
Answer: (b)$3/5$ of usual speed will take $5/3$ of usual time.[Since, time & speed are inversely proportional]$5/3$ of usual time = usual time + $5/2$$2/3$ of usual time = $5/2$usual time = $5/2 × 3/2$= $15/4 = 3{3}/4$ hours. Using Rule 8,If an object travels certain distance with the speed of $A/B$ of its original speed and reaches its destination 't' hours before or after, then the taken time by object travelling at original speed isTime = $\text"A"/\text"(Difference of A and B)"$ × time (in hour)

Answer: Option D. -> 15 m/sec
Answer: (d)1 kmph = $5/18$ m/sec54 kmph = $5/18 × 54$= 15 m/sec.

Answer: Option C. -> 1 hour 12 minutes
Answer: (c)Time and speed are inversely proportional.Usual time × $7/6$ - usual time= 12 minutesUsual time × $1/6$ = 12 minutesUsual time = 72 minutes= 1 hour 12 minutesUsing Rule 8,Here, A = 6, B = 7, t = $12/60 = 1/5$ hrs.Usual time= $\text"A"/\text"Diff of A and B"$ × time= $6/{(7 - 6)} × 1/5 = 1{1}/5$ hrs.= 1 hrs. 12 minutes

Answer: Option D. -> 420 km
Answer: (d)Using Rule 1,Time = 10$1/2$ hours = $21/2$ hoursSpeed = 40 kmphDistance = Speed × Time= $40 × 21/2$ = 420 km

Answer: Option B. -> 23
Answer: (b)Using Rule 1,Speed = $150/25$ = 6 m/sec= $6 × 18/5 = 108/5$ = 21.6 kmph

Answer: Option C. -> 7 km
Answer: (c)If the required distance be x km, then$x/3 - x/4 = 1/2$${4x - 3x}/12 = 1/2$$x/12 = 1/2$ ⇒ x = 6 kmUsing Rule 9,Here $S_1 = 4, t_1 = x, S_2 = 3, t_2 = x + 1/2$$S_1t_1 = S_2t_2$$4 × x = 3(x + 1/2)$$4x - 3x = 3/2 x = 3/2$Distance= $4 × 3/2$ = 6 kms

Answer: Option A. -> 30
Answer: (a)Using Rule 1,Speed = $\text"Distance"/\text"Time"$= $200/24$ m/s$200/24$ m/s = $200/24 × 18/5$= 30 km/h [Since, x m/s = $18/5$ x km/h]

Answer: Option B. -> 4$5/8$ hours
Answer: (b)Speed of train = 60 kmphTime = 210 minutes= $210/60$ hours or $7/2$ hoursDistance covered= $60 × 7/2$ = 210 kmTime taken at 80 kmph= $210/80 = 21/8$ hours= 2$5/8$ hoursUsing Rule 9,Here, $S_1 = 60, t_1 = 210/60 hrs, S_2 = 80, t_2$ = ?$S_1t_1 = S_2t_2$60 × $210/60 = 80 × t_2$$t_2 = 21/8$ hrs = 2$5/8$ hrs

Answer: Option D. -> 2 hours
Answer: (d)Using Rule 1,The boy covers 20 km in 2.5 hours.Speed = $20/{2.5}$ = 8 km/hr.New speed = 16 km/hrTime = $32/16$ = 2 hours.

Answer: Option B. -> 55.44 minutes
Answer: (b)Using Rule 1,Distance covered on foot= 4 × 3$3/4$ km. = 15 km.Time taken on cycle= $\text"Distance"/\text"Speed" = 15/{16.5}$ hour= ${15 × 60}/{16.5}$ minutes= 54.55 minutes