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 D

Explanation:

Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr

So we know from question that it took 4(1/2)hrs to travel back to same point.

So,
\begin{aligned}
\frac{30}{15+x} - \frac{30}{15-x} = 4\frac{1}{2} \\
=> \frac{900}{225 - x^2} = \frac{9}{2} \\
=> 9x^2 = 225 \\
=> x = 5 km/hr
\end{aligned}

Answer: Option C

Explanation:

Let the distance is x km
Rate downstream = 5 + 1 = 6 kmph
Rate upstream = 5 - 1 = 4 kmph
then
x/6 + x/4 = 1 [because distance/speed = time]
=> 2x + 3x = 12
=> x = 12/5 = 2.4 km

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 A

Explanation:

First of all, we know that
speed of current = 1/2(speed downstream - speed upstream) [important]

So we need to calculate speed downstream and speed upstream first.

Speed = Distance / Time [important]

\begin{aligned}
\text {Speed upstream =}\\ (\frac{15}{3\frac{3}{4}}) km/hr \\
= 15 \times \frac{4}{15} = 4 km/hr \\

\text{Speed Downstream = }
(\frac{5}{2\frac{1}{2}}) km/hr \\
= 5 \times \frac{2}{5} = 2 km/hr \\
\text {So speed of current = } \frac{1}{2}(4-2) \\
= 1 km/hr
\end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{Speed Upstream} = \frac{3}{\frac{20}{60}} = 9 km/hr \\
\text{Speed Downstream} = \frac{3}{\frac{18}{60}} = 10 km/hr \\
\text{Rate of current will be} \\
\frac{10-9}{2} = \frac{1}{2} km/hr
\end{aligned}