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 B

Explanation:

Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph

Answer: Option C

Explanation:

Please remember,
If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)

=> Rate in still water = 1/2(20+10) = 15 kmph
=> Rate of current = 1/2(20-10) = 5 kmph

Answer: Option D

Explanation:

Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph

Answer: Option B

Explanation:

Friends first we should analyse quickly that what we need to calculate and what values we require to get it.

So here we need to get speed of current, for that we will need speed downstream and speed upstream, because we know
Speed of current = 1/2(a-b) [important]

Let the speed upstream = x kmph
Then speed downstream is = 3x kmph [as per question]

\begin{aligned}
\text{speed in still water = } \frac{1}{2}(a+b) \\
=> \frac{1}{2}(3x+x) \\
=> 2x \\
\text{ as per question we know, }\\
2x = 9\frac{1}{3} \\
=> 2x = \frac{28}{3} => x = \frac{14}{3} \\
\end{aligned}
So,
Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr.
Speed of the current \begin{aligned} =\frac{1}{2}[14 - \frac{14}{3}]\\
= \frac{14}{3}
= 4 \frac{2}{3} kmph \end{aligned}

Answer: Option A

Explanation:

Let speed downstream = x kmph
Then Speed upstream = 2x kmph

So ratio will be,
(2x+x)/2 : (2x-x)/2
=> 3x/2 : x/2 => 3:1