Aptitude - Boats and Streams - Discussion

[(450-30x)+(450+30x)]/(225-x^2) = [900/(225-x^2)].

How did we get 30*30 = 900?

Are we taking 30 common or what?

Any shortcut method for this sum?

In a stream running at 2km/h, a motor boat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motorboat in still water? Please answer.

What happen to the 900 and the 2?

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!

See its simple to understand.

Given distance is 30km. Let assume speed of river is x.

Speed of boat is 15 once it goes downstream. So speed would be 15+x.

And upstream is 15-x.

Now what it told from going to 30 km time required would be 30/downstream and coming back speed would be 30/upstream.

So total time given here is 4.30 minutes which is 4*1/2 (1/2 here means 30 minutes which is half of 1 hr).

So adding both the times downstream and upstream would give total time which is 4*1/2.

i.e why its gives 30/(downstream speed)+30/(upstream) = 4*1/2.

And now you are smart enough to solve equations.

How we will getting 30 sir?

What is the difference between 4 sum and 8 sum having the same description?

Simple to find speed of stream to check option.

A) 4.
= 15+4/2 + 15-4/2.
= 4.25 hr so wrong.

B) 5.
= 15+5/2 + 15-5/2.
= 4.5 hr so right answer.