Aptitude - Boats and Streams - Discussion

The speed of the boat in still water = Vb.
The speed of stream = Vs.

Question asked -- ratio, Vb : Vs = ?..

Vb + Vs = Vr1 (resultant speed downstream), and time taken to cover a distance "d" with speed Vr1 is 4 hrs.

Similarly,
Vb - Vs = Vr2 (resultant speed upstream), and time is taken to cover the same distance with speed Vr2 is 8 hrs, 48min.
By adding both the equations , we get
2Vb =Vr1 + Vr2.
Vb = (Vr1 + Vr2) 1/2.

lets find Vr1, which is distance d / time taken during upstream i.e Vr1 = d/4,
Similarly, Vr2 = d/(8 + 4/5),
= 5d/44.

in 60 min it is 1 hr so, in 1 min it is 1/60 hrs and in 48 min it is 1/60 * 48 = 4/5 hrs }
Putting Vr1 and Vr2 values in eq 3, we get -
Vb = 2d/11.

Putting value of Vb in eq 1 we'll get;
Vs = 3d/44.

Therefore, Vb/Vs = 8/3 Ans.

Given,
Time (upstream)=8 hour 48 min=(8*60+48) = 528 min.
Time (downstream)= 4 hour = (4*60)min = 240 min.

Let distance be x.
we know speed =distance/time.
so, the speed of the boat (upstream) = distance/time.
= x/528.
speed of the boat (downstream) = distance/time.
= x/240.

We know that;
speed of boat in still water = (downstream+upstream)/2.
= (x/240 + x/528)/2.
= x/330.

We know that
speed of stream/current water.
= (downstream-upstream)/2.
So we have to find the ratio of the speed of the boat in still water by the speed of the stream/current =
(x/330)/(x/528)=8/3 or 8:3
Therefore, ratio = 8:3 Answer.

Given explanation is quite difficult to understand. Let me explain you this in a simpler manner.

Distance is same during upstream and Downstream and Distance is equal to : Speed/Time.

So, time in downstream = 4 hours,
Time in upstream = 44/5 hours (8 hours 48mins).
Now let's assume x as the distance.
A as the downstream speed and B as the upstream speed.

Now write both (as the distance formula),

44/5 = x/B ---> equation 1.
4 = x/A ---> equation 2.

Solving these two equations we get,

44/20 = a/b ---> equation 3.

By equation 3, we can say that a = 44b/20

And we want to calculate u : v.
i.e 1/2(a + b) : 1/2(a - b),
i.e (a + b) : (a - b),
i.e 44b/20 + b : 44b/20 - b,
i.e 64b/20 : 24b/20,
i.e 64b : 24b,
i.e 8 : 3.

That's It. I know it's a bit lengthy But you would be able to calculate 3/4 of these steps orally.

Speed of the boat upstream = Speed of boat - Speed of stream.

Speed of stream = Speed of boat - Speed of the boat upstream--------- (1).

Speed of the boat downstream = Speed of boat + Speed of stream.

Speed of stream = Speed of the boat downstream - Speed of boat-------- (2).

Now, equating (1) & (2),

Speed of boat-speed of boat upstream = Speed of boat downstream - Speed of boat.

2 (Speed of boat) = Speed upstream + Speed downstream.

Speed of boat = (Speed upstream + Speed downstream)/2.

If, Speed upstream = x and Speed downstream = y then,

Speed of boat = (x+y)/2.

Similarly, making speed of the boat as subject in eq (1) & (2) and then equating we get,

Speed of stream = (Speed downstream - Speed upstream)/2.

= (y-x) /2.

Let,

Upstream speed = X kmph.
Downstream spead = Y kmph.

So,
We know that distance is the same.
And we know, D=SXT (speedxtime)
D= X(8+48/60) upstream case--->eq1
D=Y(4)---> eq 2
Now, equate both.

4Y = X(8+4/5).
4Y= X(44/5).
Y=X(11/5).
Y/X = 11/5.

So now relate as Y=11, X=5.
Now speed of water(S.W) = downstrean-upstream/2.
= Y-X/2.
=11-5/2 = 6/2 = 3.

Now, speed of boat = downstream - speed of the water.
= Y- (S.W).
= 11-3 = 8.

Now ratio of both as given in question,
Ratio of boat speed to water speed = 8/3.
Ratio = 8:3.
Hope you'll get it well.

Speed=distance/time
therefore distance=speed*time.
let consider speed at upstream is "x" and at downstream is "y", then distance covered (upstream in 8 hrs 48 min)= x*8*4/5

here 8hours and 48minutes=8*48/60=8*4/5

and distance covererd (downstream in 4hours=y*4

to relate both these equation
from question ,we know that distance covered are same

therefore x*8*4/5 = y*4

y=11/(5*x)
we know that speed are different in down stream and upstream.
if both are in same direction
take difference(x-y)
if they are in opposite
take sum(x+y)

required ratio =(x+y)/2 : (x-y)/2 // here 2 is to take average.
then answer is 8:3

Let the man's rate upstream be U kmph and that downstream be V kmph.

Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.

[U*8*48/60] = [V*4] [BECAUSE 1 MINUTE=60SECONDS]
[U*8*4/5] = [4V]
[U*44/5] = [4V]
44/5 U = 4V

THEN, USE THE FORMULA-----If the speed of the boat in still water is U km/hr and the stream of the boat is V km/hr then.

speed downstream = (u+v)km/hr
speed up stream = (u+v)km/hr

so, our U= 44/5 COME AND V= 4 COME

NOW, [U+V]/[U-V]
[44/5+4] / [44/5-4]
[44+20] / [44-20]
64/24,
8/3,
8:3 ANS

We all know this basic rule of finding ratio which is:
x/y = a/b.

Further when we solve this, we get:
x+y/x-y = a+b/a-b .... equation (1).

Solution:

Let the boat's speed upstream be x & downstream be y.

Therefore,
According to the formula, Distance = Speed*Time.
We get for upstream, D= x*44/5.
& for downstream, D= x*4.

Now, in the question, it is given that the distance covered is same while going upstream and downstream.

Therefore,
x*44/5 = 4y.
x/y = 11/5 (satisfies the equation: x/y = a/b).

Now put this value of x/y in equation (1).

x+y/x-y = 11+5/11-5.
x+y/x-y = 16/6.
x+y/x-y = 8/3.

8:3 is the Answer.

Let as assume,

The speed of boat is : b km/hr.

The speed of the water is : w km/hr.

In downstream the relative speed of the boat is: d = (b + w) km/hr.

In upstream the relative speed of the boat is : u = (b - w) km/hr.

According to the question,

The distance covered in upstream = (b + w) *4 km.

The distance covered in downstream = (b - w) * (8 + 48/60) km.

As the distance covered by boat is same in both direction, Hence.

(b + w) *4 = (b - w) *8.8.

(b*8.8) - (b*4) = (w*4 + w*8.8).

b*4.8 = w*12.8.

b/w = 8/3.

Or b:w = 8:3.

By using the formula we have;
Speed upstream = u-v.
Speed downstream= u+v.
Where u&v is speed of the boat in still water and speed of stream respectively.
Given that distance covered is equal so;

By the time distance formula;
Speed *time= distance.
Both the values put according to formula.
(u-v) *8*48/60=(u+v) 4.
(u-v) 8*4/5=(u+v) 4.
44/5u-44/5v= 4u+4v.
44/5u- 4u = 44/5v+4v.
u[44/5-4] = v[44/5+4].
24u = 64v.
u/v = 64/24.
u/v = 8/3.

So, the ratio between the speed of the boat and speed of the water current is;
8:3.