Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec
Distance = (35 x 4) m = 140 m.
Time taken = 140 x (2/5) sec= 56 sec
If the speed of the faster horse be and that of slower horse be then
and
Now, you can go through options.
The speed of slower horse is 20km/h
Since, 20+30=50
and
Relative speed of the thief and policeman = (11 ? 10) km/hr = 1 km/hr
Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m
Therefore, Distance between the thief and policeman = (200 ? 100) m = 100 m.
Speed = = 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr
Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.
Time taken in walking both ways = 7 hours 45 minutes --------(i)
Time taken in walking one way and riding back= 6 hours 15 minutes-------(ii)
By equation (ii)*2 -(i), we have
Time taken to man ride both ways, = 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes
speed = (5x5/18)m/sec
= 25/18 m/sec.
Distance covered in 15 minutes = (25/18 x 15 x 60)m
= 1250 m.
Total time taken = (160/64 + 160/8)hrs
= 9/2 hrs.
Average speed = (320 x 2/9) km.hr
= 71.11 km/hr.
Suppose they meet x hrs after 8 a.m
then,
[Distance moved by first in x hrs] + [Distance moved by second in (x-1) hrs] = 330.
Therefore, 60x + 75(x-1) = 330.
=> x=3.
So,they meet at (8+3) i.e, 11a.m.
Time taken by Akash = 4 h
Time taken by Prakash = 3.5 h
For your convenience take the product of times taken by both as a distance.
Then the distance = 14km
Since, Akash covers half of the distance in 2 hours(i.e at 8 am)
Now, the rest half (i.e 7 km) will be coverd by both prakash and akash
Time taken by them = 7/7.5 = 56 min
Thus , they will cross each other at 8 : 56am.
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