The speed of a motorboat in calm water is 75% greater than the velocity of the river current. If the boat takes a total of 28 hours to travel a specific distance downstream and then return upstream to the starting point, how long would it take the boat to cover that same distance in calm water?

The correct answer is 9³/₇ hours.

Step 1: Define the speeds of the boat and river current
Let the velocity of the river current be $v$.
The speed of the boat in calm water, denoted by $b$, is 75% greater than the velocity of the river current:
$b=v+0.75v=1.75v=\frac{7}{4}v$

Step 2: Find downstream and upstream speeds
When going downstream, the river current assists the boat:
$v_{\text{down}}=b+v=\frac{7}{4}v+v=\frac{11}{4}v$
When going upstream, the river current opposes the boat:
$v_{\text{up}}=b-v=\frac{7}{4}v-v=\frac{3}{4}v$

Step 3: Set up equation for total journey time
Let $d$ be the distance traveled each way.
Time taken downstream:
$t_{\text{down}}=\frac{d}{v_{\text{down}}}=\frac{d}{\frac{11}{4}v}=\frac{4d}{11v}$
Time taken upstream:
$t_{\text{up}}=\frac{d}{v_{\text{up}}}=\frac{d}{\frac{3}{4}v}=\frac{4d}{3v}$
The total time for downstream and upstream trips is 28 hours:
$t_{\text{down}}+t_{\text{up}}=28$
$\frac{4d}{11v}+\frac{4d}{3v}=28$

Step 4: Solve for the distance-to-velocity ratio
Factor out $\frac{4d}{v}$:
$\frac{4d}{v}\left(\frac{1}{11}+\frac{1}{3}\right)=28$
$\frac{4d}{v}\left(\frac{3+11}{33}\right)=28$
$\frac{4d}{v}\left(\frac{14}{33}\right)=28$
$\frac{56d}{33v}=28$
$\frac{d}{v}=28\times \frac{33}{56}=\frac{33}{2}$

Step 5: Calculate time taken in calm water
The time required to cover distance $d$ in calm water is:
$t_{\text{calm}}=\frac{d}{b}=\frac{d}{\frac{7}{4}v}=\frac{4}{7}\times \frac{d}{v}$
Substituting $\frac{d}{v}=\frac{33}{2}$ into the equation:
$t_{\text{calm}}=\frac{4}{7}\times \frac{33}{2}=\frac{66}{7}=9\frac{3}{7}$ hours.

Therefore, the boat will take 9³/₇ hours to cover the same distance in calm water.