Let distance be $D$ km.
Time at 10 km/hr = $\frac{D}{10}$, Time at 15 km/hr = $\frac{D}{15}$.
Time difference = 2 hours (2 PM - 12 PM = 2 hours)
$\frac{D}{10} - \frac{D}{15} = 2 \Rightarrow \frac{3D-2D}{30} = 2 \Rightarrow D=60$ km.
Time at 15 km/hr = $\frac{60}{15} = 4$ hours (arrives at 12 PM, started at 8 AM)
Time at 10 km/hr = $\frac{60}{10} = 6$ hours (arrives at 2 PM, started at 8 AM)
Thus, to reach at 1 PM, time required = 5 hours.
Required speed = $\frac{60}{5} = 12$ km/hr.

Speed = 72 km/h = $72 \times \frac{5}{18} = 20$ m/s. Distance covered in 25 seconds = $20 \times 25 = 500$ m. Length of bridge = $500 - 100 = 400$ m.

The thief will cover 15 km between 1 PM and 2 PM. Relative speed = 18-15 = 3 km/h. Time = $\frac{15}{3} = 5$ hours. After 2 PM, 5 hours = 7 PM. (Note: Solution in book matches answer key).

In half an hour, the bus covered distance = $70 \times 0.5 = 35$ km. Remaining distance = 410 - 35 = 375 km. Relative speed = 70+80 = 150 km/h. Time = 375/150 = 2.5 hours. Distance covered by bus = $70 \times (0.5 + 2.5) = 70 \times 3 = 210$ km.

Required speed = $\frac{45 \times 6}{5} = 54$ km/h

Let the speed of the fast train = $x$ km/hr
Speed of the slow train = $(x-10)$ km/hr
According to the question: $\frac{900}{x-10} - \frac{900}{x} = 3$
$\frac{900(x - (x-10))}{x^2 - 10x} = 3$
$9000 = 3(x^2 - 10x)$
$3000 = x^2 - 10x$
$x^2 - 10x - 3000 = 0$
$x^2 - 60x + 50x - 3000 = 0$
$x(x-60) + 50(x-60) = 0$
$(x-60)(x+50) = 0$
$x = 60$ (Speed cannot be negative.)
Thus, the speed of the fast train = 60 km/hr

Average speed = $\frac{2xy}{x+y}$
$x = 10, y = 15$
Average speed = $\frac{2 \times 10 \times 15}{10+15} = \frac{300}{25} = 12$ km/hr.

Speed 1 = 2.5, Speed 2 = 3.5. Time diff = 6 + 6 = 12 mins = 12/60 hours. Distance = rac{S_1 imes S_2}{S_2 - S_1} imes ext{time diff} = rac{2.5 imes 3.5}{3.5 - 2.5} imes rac{12}{60} = rac{8.75}{1} imes rac{1}{5} = 1.75 km = 1 3/4 km.

Total distance = 225 + 375 = 600 meters
Speed = $90 \times \frac{5}{18} = 25$ meters/second
Time = $\frac{600}{25} = 24$ seconds.

Relative speed = 21 - 15 = 6 meters/minute. Time = 114 / 6 = 19 minutes.

Speed = 180 / 6 = 30 m/s. In km/h = 30 imes rac{18}{5} = 108 km/h.

Average speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{67 + 97}{1.5 + 2.5} = \frac{164}{4} = 41$ km/h.

A to C (uphill) and C to A (downhill). Let AB = x, BC = y. x/10 + y/12 = 6; y/15 + x/12 = 5. Solving this, the time taken on BC is 2 hours.

Dog speed = (4 * 120) / 3 = 160 cm/s = 1.6 m/s. Cat speed = (3 * 80) / 2 = 120 cm/s = 1.2 m/s. Relative speed = 0.4 m/s. Time = 100 / 0.4 = 250 sec = 4 minutes 10 sec.

Train speed = $\frac{250}{12}$ m/s = $\frac{250}{12} \times \frac{18}{5} = 75$ km/h.

15 min = 1/4 hour, 20 min = 1/3 hour, remaining 10 min = 1/6 hour. Distance = $x(1/4) + 2x(1/3) + x(1/6) = 39$. $\frac{3x+8x+2x}{12} = 39 \Rightarrow x = 36$ km/h.

40% distance = 120 km. Time taken = 2 hours. Remaining distance = 180 km. Remaining time = 2 hours. New speed = 180/2 = 90 km/h. Old speed = 120/2 = 60 km/h. Increase = 90 - 60 = 30 km/h.

Distance according to first speed = $50 imes 2.5 = 125$ km. Distance according to second speed = $70 imes 1.5 = 105$ km. Total distance = 230 km. Total time = 4 hours. Average speed = 230/4 = 57.5 km/h.

Let the distance of the journey = $x$ km.
$\frac{x}{40} - \frac{x}{50} = \frac{30}{60} = \frac{1}{2}$
$\frac{5x - 4x}{200} = \frac{1}{2}$
$\frac{x}{200} = \frac{1}{2} \Rightarrow x = 100$ km.

Distance between cities = 300 km. Time taken to meet = 300 / (60+90) = 300/150 = 2 hours. Distance from 'K' city = 2 * 60 = 120 km.