CRE 4 - Boats & Streams

Speed downstream = 60 + 20 = 80 kmph

Speed upstream = 60 - 20 = 40 kmph

Hence, option (d).

Speed of stream = 2 – 1 = 1 km/h

Hence, option (c).

Man’s speed in still water = (10 + 5)/2 = 7.5 km/h

Hence, option (c).

Speed upstream = (40 ÷ 8) = 5 km/h

Speed downstream = (36 ÷ 6) = 6 km/h

Speed of boat in standing water = (5 + 6)/2 = 5.5 km/h

Hence, option (c).

Man's speed = 6 km/hr

River’s speed = 1.2 km/hr

Speed while going downstream = (b + r) = (6 + 1.2) = 7.2

Speed while going upstream = (b - r) = (6 - 1.2) = 4.8

(x/7.2) + (x/4.8) = 10 

x = 28.8

Hence, option (b).

Speed of the boatman upstream = $\frac{1}{\frac{20}{60}}$ = 3 km/hr.

Speed of the boatman downstream = $\frac{1}{\frac{18}{60}}$ = 10/3 km/hr.

Rate of current = $\frac{downstream speed - upstream speed}{2}$ = (10/3 - 3)/2 = 1/6 km/hr.

Hence, option (b).

S_down = 15 + S_s

T = $\frac{D}{S_{down}}=\frac{200}{15+S_s}$

10 = $\frac{200}{15+S_s}$

15 + S_s = 200/10 = 20

S_s = 5 kmph

T = $\frac{D}{S_{up}}=\frac{200}{15-5}$

Hence, 20.

The boats together travel 10 + 5 = 15 km in 60 minutes.

∴ In one minute they can travel 1/4 km

∴ They are 1/2 km apart, two minutes before they collide.

Hence, option (d).

Let the speed of the boat be Vb and that of the stream be V_w.

Now, the time taken by the boat in going upstream = $\frac{10}{V_b-V_w}$ = 2

⇒ V_b - V_w = 5 ... (i)

And the time taken by the boat in going downstream = $\frac{10}{\frac{V_b}{2}+V_w}$ = 2

⇒ V_b/2 + V_w = 5 ... (ii) 

Adding (i) and (ii),

(3V_b)/2 = 10

V_b = 20/3 

V_w = 5 - (10/3) = 5/3 kmph

The speed of the water flow is 5/3 kmph.

Hence, option (a).

Let P km/hr be the speed of boat P, and Q km/hr be the speed of boat Q. 

Let s km/hr be the rate of flow of stream. 

We have,

P - s = 80/10 = 8 and P + s = 80/5 = 16

Solving, we get P = 12 and s = 4

Again, Q - s = 60/5 = 12

Thus, s = 4 and Q = 16.

Thus, speed of boat Q in still water is 16 km/hr.

Hence, option (b).

It is clear that the water flow is from A to B.

Distance travelled by boat starting from A = 80 kms.

Distance travelled by boat starting from B = 40 kms.

Since time travelled is same for both the boats:

$\frac{80}{V_b+V_w}=\frac{40}{V_b-V_w}$

⇒ V_b = 3V_w

⇒ V_b : V_w = 3 : 1

Hence, option (c).