PROBLEMS:
1. In one hour a boat goes 11 km long the stream and 5 km against the stream. The speed of the boat in still water is?
Sol:
Speed in still water = ½ ( 11+5) km/hr
= 8 kmph
2.A man can row 18 kmph in still water. It takes him thrice as long as row up as to row down the river. find the rate of stream.
Sol:
Let man's rate up stream be xkmph
then, in still water =1/2[3x+x]=2x kmph
so, 2x= 18, x=9
rate upstream =9kmph
rate downstream =27 kmph
rate of stream = ½ [27-9]
= 9kmph
3.A man can row 71/2kmph in still watre . if in a river running at 1.5 km an hour, if takes him 50 min to row to place and back. how far off is the place?
Sol:
speed down stream =7.5+1.5=9kmph
speed upstream =7.5-1.5=6kmph
let the required distence x km. then ,
x/9+x/6=50/60 = 2x+3x= 5/6*18
5x=15, x=3
Hence, the required distence is 3 km
4.A man can row 3 quarters of a km aganist the stream is 111/4 min. the speed of the man in still water is?
Sol:
rate upstream = 750/625 m/sec =10/9 m/sec
rate downstream =750/450 m/sec = 5/3 m/sec
rate in still water =1/2[10/9+5/3] = 25/18 m/sec
= 25/18*18/5=5 kmph
5.A boat can travel with a speed of 13 kmph in still water. if the speed of stream is 4 kmph,find the time taken by the boat to go 68 km downstream?
Sol: Speed down stream = 13+4= 17 kmph
time taken to travel 68km downstream =68/17 hrs
= 4 hrs
6.A boat takes 90 min less to travel 36 miles downstream then to travel the same distance upstream. if the speed of the boat in still water is 10 mph .
the speed of the stream is :
Sol:
Let the speed of the stream be x mph .
then, speed downstream = [10+x]mph
speed upstream =[10-x] mph
36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2]
(x+50)(x-2) =0
x=2 kmph
7.At his usual rowing rate, Rahul 12 miles down stream in a certain river in 6 hrs less than it takes him to travel the same distance upstream. but if he could double his usual rowing rate for his 24 miles round the down stream 12 miles would then
take only one hour less than the up stream 12 miles. what is the speed of the
current in miles per hours?
Sol:
Let the speed in still water be x mph and the speed of the current be y mph.
then, speed upstream = (x-y)
speed downstream =(x+y)
12/(x-y) - 12/(x+y) = 6
6(x2 - y2) m= 2xy => x2 - y2 =4y -(1)
and 12/(2x-y) - 12/(2x+y) =1 => 4x2 - y2 = 24y
x2= ( 24y + y2)/4 -->(2)
from 1 and 2 we have
4y+ y2 =( 24y+y2)/4
y=8/3 mph
y= 22/3 mph
8.There is a road beside a river. two friends started from a place A, moved to a
temple situated at another place B and then returned to A again. one of them
moves on a cycle at a speed of 12 kmph, while the other sails on a boat at a
speed of 10 kmph . if the river flows at the speedof 4 kmph , which of the
two friends will return to place A first?
Sol:
Clearly, The cyclist moves both ways at a speed of 12 kmph
so, average speed of the cyclist = 12 kmph
the boat sailor moves downstream = (10+4) = 14 kmph
upstream =(10-4) = 6 kmph
So, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph
=42/5 kmph =8.4 kmph
Since, the average speed of the cyclist is greater, he will return to A first.
9.A boat takes 19 hrs for travelling downstream from point A to point B. and
coming back to a point C midway between A and B. if the velocity of the steam is 4 kmph. and the speed of the boat in still water is 14 kmph. what is the distance between A and B?
Sol:
speed downstream =14+4 =18 kmph
speed upstream = 14 -4 = 10 kmph
let the distence between A and B be x km. then,
x/18 + (x/2)/10 = 19
x/18 + x/20 =19
19x/180 =19 =>x = 180km
Hence, the distence between A and B bw 180 km
