Aptitude - Boats and Streams - Discussion
Discussion Forum : Boats and Streams - General Questions (Q.No. 4)
4.
A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
Answer: Option
Explanation:
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr.
 |
30 | + | 30 | = 4 | 1 |
| (15 + x) | (15 - x) | 2 |
 |
900 | = | 9 |
| 225 - x2 | 2 |
9x2 = 225
x2 = 25
x = 5 km/hr.
Video Explanation: https://youtu.be/lMFnNB3YQOo
Discussion:
63 comments Page 2 of 7.
Owen said:
1 decade ago
What confuses me is that I expected the time gained going down to be same as that lost going up.
Finding the velocities etc easy way to solve this, but I still find the thought difficult to get around that whatever the stream gives in time going down should be lost going back But its just not true!
Finding the velocities etc easy way to solve this, but I still find the thought difficult to get around that whatever the stream gives in time going down should be lost going back But its just not true!
Prabhat said:
9 years ago
Whoever is having confusion about how it got multiplied in place of addition in
(30/(15+x)) + (30/(15-x)) = 9/2 hrs.
It is by cross multiplication it will become,
(30 * (15-x))+(30 * (15+x)) / (15+x) * (15-x) = 9/2 hrs.
After simplifying it will become 900/15^2 - x^2 = 9/2.
(30/(15+x)) + (30/(15-x)) = 9/2 hrs.
It is by cross multiplication it will become,
(30 * (15-x))+(30 * (15+x)) / (15+x) * (15-x) = 9/2 hrs.
After simplifying it will become 900/15^2 - x^2 = 9/2.
Disha said:
7 years ago
@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.
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.
(3)
Tanu said:
6 years ago
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.
(1)
Sai said:
7 years ago
We know that.
Still in water + stream = downstream.
In question downstream distance and time is given, we get the downstream speed and still in water speed also given.
Then we can find the stream speed easily.
Still in water + stream = downstream.
In question downstream distance and time is given, we get the downstream speed and still in water speed also given.
Then we can find the stream speed easily.
(2)
Rohit chavan said:
1 decade ago
Total time=(distance of downstream/speed of downstream)+(distance of downstream/speed of downstream)
Therefore, (30/(15+x))+(30/(15-x)) = 9/2 hrs
Then find the value of x by simplifying the above equation.
Therefore, (30/(15+x))+(30/(15-x)) = 9/2 hrs
Then find the value of x by simplifying the above equation.
Kavitha said:
1 decade ago
The distance in both down&upstream is 30 as given;.
Here we have to find the speed. So distance/speed=time.
(30/15+x) + (30/15-x) = 4 1/2 = 9/2(2X4+1/2 = 9/2).
By solving it we get the answer.
Here we have to find the speed. So distance/speed=time.
(30/15+x) + (30/15-x) = 4 1/2 = 9/2(2X4+1/2 = 9/2).
By solving it we get the answer.
AdityaLath said:
10 years ago
Short trick:
D = z(x^2-y^2)/2x.
Where.
D = Distance.
Z = Time taken.
x = Speed of boat in still water.
y = Speed of stream.
30 = 9*(15^2-y^2)/2*2*15.
200 = 225-y^2.
y^2 = 25.
y = 5 answer.
D = z(x^2-y^2)/2x.
Where.
D = Distance.
Z = Time taken.
x = Speed of boat in still water.
y = Speed of stream.
30 = 9*(15^2-y^2)/2*2*15.
200 = 225-y^2.
y^2 = 25.
y = 5 answer.
Priya said:
8 years ago
The speed of the current is 5 then the speed of the boat in downstream =u+v =15+5 =20;.
The speed of the boat in upstream = u-v = 15-5 =10;.
Time =distance/speed.
4 1/2 = 30/20+30/10.
9/2=9/2.
The speed of the boat in upstream = u-v = 15-5 =10;.
Time =distance/speed.
4 1/2 = 30/20+30/10.
9/2=9/2.
Vinoth said:
1 decade ago
Simply follow formula when only distance the, time t and speed of boat x given.. and asked for speed of stream.... use in formula,
d^2/(x^2-y^2)=t ,
9/2=(225-y^2)/15, ==> y^2=25 , y=5
d^2/(x^2-y^2)=t ,
9/2=(225-y^2)/15, ==> y^2=25 , y=5
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