Aptitude - Boats and Streams - Discussion

Speed in still water = 15 km/hr.
Distance traveled in downstream = 30 km.
Time taken in upstream = 4 hr 30 min.
The speed of the stream = ?
Speed = Distance/time.
Distance = 30km.
Speed in upstream = (u-v)km/hr = (15-v)km/hr.
Speed in downstream = (u+v)km/hr = (15+v)km/hr.

Total time = (distance of downstream/speed of downstream) + (distance of upstream/speed of upstream).
30/(15 + x) + 30/(15 - x) = 4 1/2,
30/(15 + x) + 30/(15 - x) = 9/2,
30(15 - x) + 30(15 + x) = 9/2 * (15 - x)(15 + x),
900 = 9/2 * (225 + 15x - 15x - x^ 2),
900 = 9/2 * (225 - x ^ 2),
1800= 9(225 - x ^ 2),
1800 = 2025 - 9x ^ 2,
2025 - 1800 = 9x ^ 2,
225 = 9x ^ 2,
x ^ 2 = 225/9 = 25,
x = √(25) = 5.
Ans=>X=5.

(30/15+x) + (30/15-x) =4 (1/2).
By cross multiplying;

(30(15-x) + 30(15+x)) /(15+x) (15-x) = 9/2.
(450-30x+450+30x)/15^2 - x^2 = 9/2.
-30x and +30x get cancelled.

So,
900/15^2 -x^2 = 9/2.

Someone help me to understand the solution, please.

As we have given upstream speed=15 from option lets take 4 so for,

Downstream 15 + 4 = 19 and upstream 15 -4 = 11.

The distance is given so 30/19=1.58 and 30/11=2.72 it not satisfy condition so take next option
as 5 so similarly 15 +5 = 20 and 15 - 5 =10 so 30/20 = 1.5 and 30/10 = 4.

So 1.5 +4 = 4.5 and our time is also 4.5 so the answer is 4.5.

Motorboat goes down stream 30 kilometre and again returns to the starting point from a total time of 4 hours and 30 minutes if the speed of the stream is 5 kilometre per hour then find the speed of the motor boat in still water please solve this.

Speed in still water = 15 km/hr.
Distance traveled in downstream = 30 km.

Time taken in upstream = 4 hr 30 min.
The speed of the stream = ?

Speed = Distance/time.
Distance = 30km.

Speed in upstream = (u-v)km/hr = (15-v)km/hr.

Speed in downstream = (u+v)km/hr = (15+v)km/hr.

Total time = (distance of downstream/speed of downstream) + (distance of upstream/speed of upstream).

(30/(15+v))+(30/(15-v)) = 4 1/2.
(30(1/(15-v))+(1/(15+v)) = 4 1/2.
(30((15+v)+(15-v))/(15+v)(15-v))=9/2
(30(15+15+v-v)/(15^2-v^2))
30*30/225-v2=9/2
900/(225 - v^2) = 9/2
900 / (225 - v^2) = 9/2.
900 * 2 = 9(225 -v^2).
1800 = 2025 - 9v^2.
9v^2 = 2025 - 1800.
9v^2 = 225.
v^2 = 225/9.
v^2 = 25.
v = 5 Km/hr.

Thank you all for giving the answer.

30/(15+x)+30/(15-x)=4 1/2.

Don't multiply take LCM then it will be;

{ 30(15- x)+30(15+x)}/(15+x)(15-x) = 4 1/2.
450-30x+450+30x /(15+x) (15 - x) = 9/2.
-30x+30x will be cancelled then;
900 /15^2-x^2=9/2.
**(a+b)(a-b)=a^2-b^2 **

When cross multiplied we get;
2 * 900=9(15^2-x^2),
1800=2025-9x^2 so,
9x^2=2025-1800,
9x^2=225,
x^2=225/9,
x^2=25,
x = 5.

@All.

x=consider speed of boat , which is 15kmph.
d=30km.
t=4hr30 min.

Given in question that it comes back, means the boat has taken a lap .so it will be one time upstream and second downstream. That is why both x+y and x-y has been taken.
30/(x+y)+30/(x-y)=9/2.
x=5.

How 900 came? Explain.