__Boats and Streams Practice Problems – Set 3__**1. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:**

a) 1 km/hr

b) 2 km/hr

c) 1.5
km/hr

d) 2.5
km/hr

e) 3.5
km/hr

**Answer:**

**A)**

**Explanation:**

Assume that he moves 4 km downstream in x hours

Then, speed downstream = distance/time =4/x km/hr

speed upstream = 3/x km/hr

He rows to a place 48 km distant and comes back in 14 hours

⇒48/(4/x)+48/(3/x)=14 ⇒12x+16x=14 ⇒6x+8x=7 ⇒14x=7 ⇒x=1/2

Hence, speed downstream =4/x=4(1/2) = 8 km/hr

speed upstream = 3/x=3(1/2) = 6 km/hr

**2. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place ?**

a) 2.4km

b) 3 km

c) 2.5 km

d) 3.6 km

e) 4.2 km

**Answer:**

**A)**

**Explanation:**

Speed downstream = (5 + 1) kmph = 6 kmph.

Speed upstream = (5 − 1) kmph = 4 kmph.

Let the required distance be x km.

Then, x/6 + x/4 = 1

=> 2x + 3x = 12 => x = 12/5 = 2.4km

**3. In a stream running at 2 kmph, a motor boat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motor boat in still water.**

a) 22kmph

b) 20kmph

c) 24kmph

d) 23kmph

e) 21kmph

**Answer:**

**A)**

**Explanation:**

Let the speed of the motorboat in still water be x kmph. Then,

Speed downstream = (x + 2) kmph,

Speed upstream = (x − 2) kmph

∴ 10/(x –
2) + 10/(x + 2) = 55/66

⇒ 1/(x – 2) + 1/(x + 2) = 11/120

⇒ [(x + 2) + (x – 2)]/(x^2 – 4) = 11/120

⇒ 11x^2 – 240x – 44 = 0

⇒ (x − 22)(11x + 2) = 0 ⇒ x = 22

∴ Speed of the motorboat in still water = 22
kmph

**4. There are 3 point A, B and C in a straight line such that point B is equidistant from points A and C. A boat can travel from point A to C downstream in 12 hours and from B to A upstream in 8 hours. Find the ratio of boat in still water to speed of stream.**

a) 9 : 2

b) 8 : 3

c) 7 : 1

d) 4 : 1

e) 7 : 3

**Answer:**

**C)**

**Explanation:**

Let speed in still water = x km/hr, of current = y km/hr

Downstream speed = (x+y) km/hr

Upstream speed = (x – y) km/hr

Let AC = 2p km. So AB = BC = p km.

So, 2p/(x+y) = 12 ……(i) and p/(x-y) = 8 ……(ii)

Divide both equations, and solve

x/y = 7/1

**5. A boat can row 18 km downstream and back in 8 hours. If the speed of boat is increased to twice its previous speed, it can row same distance downstream and back in 3.2 hours. Find the speed of boat in still water.**

a) 9 km/hr

b) 5 km/hr

c) 4 km/hr

d) 8 km/hr

e) 6 km/hr

**Answer:**

**E)**

**Explanation:**

Let speed of boat = x km/hr and that of stream = y km/hr

So, 18/(x+y) + 18/(x-y) = 8 …….(i)

when speed of boat becomes 2x km/hr:

18/(2x+y) + 18/(2x-y) = 3.2 …….(ii)

On solving both the equations we get, x= 6 km/hr

**6. A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in 6 hrs and 30 minutes. The speed of the boat in still water is:**

a) 10 km/h

b) 4 km/h

c) 14 km/h

d) 6 km/h

e) None of
these

**Answer:**

**A)**

**Explanation:**

Let the speed of the boat in still water be a kmph

Let the speed of the current be b kmph

Therefore, Boat will travel downstream in (a + b) kmph and upstream in (a
– b) kmph

=> 28/(a + b) + 24/(a – b) = 6 ……(i)

And 21/(a + b) + 30/(a – b) = 6 ½ = 13/2 ……(ii)

Multiply equation (i) with 3 and equation (ii) with 4

84/(a + b) + 72/(a – b) = 18 …. (iii) and 84/(a + b) + 72/(a – b) = 26 …..
(iv)

On subtracting (iv) from (iii) we get

-48/(z – b) = -8 => (a – b) = 6 …. (v)

On substituting (v) in (i) then we get

28/(a + b) + 24/6 = 6 => (a + b) = 14 …. (vi)

Adding (v) and (vi) we get a = 10 then b = 4

**7. Speed of a speed-boat when moving in the direction perpendicular to the direction of the current is 25 km/h, speed of the current is 5 km/h. So, the speed of the boat against the current will be:**

a) 25 km/h

b) 20 km/h

c) 15 km/h

d) 13 km/h

e) None of these

**Answer:**

**C)**

**Explanation:**

Speed of boat = 25 – 5 = 20 km/h

∴ Speed of boat against the current

= 20 – 5 = 15 km/h

**8. A ship 55 km from the shore springs a leak which admits 2 tonnes of water in 6 minutes, 80 tonnes would suffice to sink her, but the pumps can throw out 12 tonnes an hour. The average rate of sailing that she may just reach the shore as she begins to sink is:**

a) 9.17
km/h

b) 0.97
km/h

c) 55 km/h

d) 5.5 km/h

e) None of
these

**Answer:**

**D)**

**Explanation:**

In 1 h water entered into shop = (20 – 12) = 8 tonnes

Now, it will take 10 hrs to allow to enter 80 tonnes of water into ship

and in this time ship has to cover 55 km of distance.

Hence, required speed = 5.5 km/h.

**9. The ratio of speeds of a motor boat to that of the current of water is 36 : 5. The motor boat, goes along with the current in 5 h 10 min. Find the time to come back of motor boat.**

a) 5 h 50
min

b) 6 h

c) 6 h 50
min

d) 12 h 10
min

e) None of
these

**Answer:**

**C)**

**Explanation:**

Let speed of boat in downstream=(36x+5x)=41x

i.e. distance=41x *(31/6) km

Now, speed of boat in upstream=(36x-5x)

i.e Required time=(41x * 31/6) *1/(31x)=41/6 hr=6 hr 50 min

**10. A boat goes 6 km in an hour in still water. It takes thrice as much time in covering the same distance against the current. Speed of the current is :**

a) 2 km/hr

b) 3 km/hr

c) 4 km/hr

d) 5 km/hr

e) None of
above

**Answer:**

**C)**

**Explanation:**

x=6 kmph

Speed against current (x-y)=6/3=2 kmph.

i.e. Speed of current=6-y=2 y=4 kmph

**11. A man can row three-quarters of a kilometer against the stream in 45/4 minutes and return in 15/2 minutes. The speed of the man in still water is:**

a) 2 km/hr

b) 3 km/hr

c) 4 km/hr

d) 5 km/hr

e) None of
these

**Answer:**

**D)**

**Explanation:**

Speed against the current =3/4 X 4/45 X 60= 4 kmph

Speed along the current=3/4 X 15/2 X 60= 6 kmph

Speed of man in still water=1/2 (6+4)= 5 kmph.