Question

A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Answer 1

Speed of the boat in water =5 km/hr

Let the speed of the stream be x km/hr

So, the speed of the boat upstream will be (5-x) km / hr

So, the speed of the boat downstream is (5+x) k/hr

Time given to cover 40 km upstream = 3(time taken to cover downstream)

⇒40/ (5-x) km/hr = 3(5+x)

⇒1/(5-x)=3(5+x)

⇒5+x=15-3x

⇒x+3x=15-5

⇒4x=10

⇒X=10/4

⇒X=5/2

∴x=2.5 km/hr

Let the speed of the boat in still water be x km/h and speed of the stream is y km/h.

Therefore, speed of the boat while upstream is (x−y) km/h and speed of the boat while downstream is (x+y) km/h

As we know that speed=  

time

distance

​  

, therefore, time=  

speed

distance

​  

 Let the speed of the motorboat in still water and the speed of the stream are u km/h and v km/h, respectively.

Then, a motorboat speed in downstream =(u+v) km/h

and a motorboat speed in upstream = (u-v) km/h.

Motorboat has taken time to travel 30 km upstream,

t1=30u−vh

and motorboat has taken time to travel 28 km downstream,

t2=28u+vh

By first condition, a motorboat can travel 30 km upstream and 28 km downstream in 7 h i.e,  t1+t2=7h

⇒ 30u−v+28u+v=7 ...(i)

Now, motorboat has taken time to travel 21 km upstream and return i.e., t3=21u−v [for upstream]

and  t4=21u+v [for downstream]

By second condition,  t4+t3=5h

⇒ 21u+v+21u−v=5 ...(ii)

Let  x=1u+v and y=1u−v

Eqs. (i) and (ii) becomes 30x+28y=7 ...(iii)

and  21x+21y=5

⇒ x+y=521 ...(iv)

Now, multiplying in Eq. (iv) by 28 and then subtracting from Eq. (iii) , we get

30x+28y=728−x+28−y=14021−−−−−−−−−−−−−−

2x=7−203=21−203

⇒ 2x=13⇒x=16

On putting the value of x in Eq. (iv), we get

16+y=521

⇒ y=521−16=10−742=342⇒y=114

∴ x=1u+v=16⇒u+v=6 ...(v)

and  y=1u−v=114

⇒ u−v=14 ...(vi)

Now, adding Eqs. (v) and (vi), we get2u=20⇒u=10

On putting the value of u in Eq. (v), we get

10+v=6

⇒ v=−4

Hence, the speed of the motorboat in still water is 10 km/h and the speed of the stream 4 km/h.