**Explanation :**Aditya’s speed = 20 km/hr = {20 * 5/18} m/sec = 50/9 m/sec

Time taken to cover 400 m= { 400 * 9/50 } sec =72 sec = 1 12/60 min= 1 1/5 min.

**Explanation :**Speed = { 750/150} m/sec =5 m/sec = { 5 * 18/5 } km/hr =18km/hr

**Explanation :**

**Explanation :**Let the speed be x km/hr.

Then, distance covered in 1 hr. 40 min. i.e., 1 2/3 hrs = 5x/3 km

Remaining distance = { 24 – 5x/3 } km.

5x /3 = 5/7 *{ 24 - 5x/3 }

5x/3 = 5/7 { (72-5x) /3}

7x =72 –5x

12x = 72

x=6

Hence speed = 6 km/hr ={ 6 * 5/18 } m/sec = 5/3 m/sec

**Explanation :**

**Explanation :**Average speed = \(\frac { 2xy }{x+y}\) km/hr =\(\frac { 2*25*4 }{25+4}\) km/hr = \(\frac {200}{29}\) km/hr

Distance traveled in 5 hours 48 minutes =\(\frac {200}{29}\) *\(\frac {29}{5}\)=40 km

Distance of the post-office from the village ={ 40/2 } = 20 km

**Explanation :**Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then

\(\frac {x}{200}\) + \(\frac {x}{400}\) + \(\frac {x}{600}\) + \(\frac {x}{800}\) =\(\frac {4x}{y}\)

\(\frac {25x}{2500}\)=\(\frac {4x} {y}\)

y=\(\frac {2400*4}{25}\)=384

hence average speed =384 km/hr

**Explanation :**New speed =5/6 of the usual speed New time taken=6/5 of the usual time

So,( 6/5 of the usual time )-( usual time)=10 minutes.

=>1/5 of the usual time=10 minutes.

usual time=10 minutes

**Explanation :**Let the required distance be x km

Difference in the time taken at two speeds=1 min =1/2 hr

Hence x/5-x/6=1/5<=>6x-5x=6

x=6

Hence, the required distance is 6 km

**Explanation :**Suppose they meet x hours after 10 a.m. Then,

[Distance moved by first in x hrs] + [Distance moved by second in (x-1)hrs]=390.

65x + 35(x-1) = 390 => 100x = 425 => x = 17/4

So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.

**Explanation :**Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours

10x = 4 x 90 or x =36.

So, speed of goods train = 36kmph

**Explanation :**Relative speed of the policeman = (10-8) km/hr =2 km/hr.

Time taken by police man to cover 100m=\(\frac {100}{1000}\) x \(\frac{1}{2}\) =\(\frac{ 1}{20}\) hr.

In 1 hrs, the thief covers a distance of 8 x \(\frac {1}{20}\) km = \(\frac {2}{5}\) km = 400 m

**Explanation :**Let the distance be x km. Then,

( Time taken to walk x km) + (time taken to ride x km) =37 min.

( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.

But, the time taken to walk 2x km = 55 min. Time taken to ride 2x km = (74-55)min =19 min.