### TIME AND DISTANCE

1. 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.

1. Explanation :

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

1. Explanation :

1. 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

1. Explanation :

1. 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

1. 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

1. 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

1. 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

1. 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.

1. 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

1. 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

1. 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.