At full throttle, boat travels 16 miles per gallon
--> Total distance = 16*10 = 160 miles (for 10 gallons)
--> Total time boat can travel = 160/20 = 8 hours

Let the time taken for upstream = \(t_1\) & for downstream = \(t_2\)
--> \(t_1 + t_2 = 8\)

Speed during upstream = 20 - 5 = 15 mph & during downstream = 20 + 5 = 25 mph

Let the distance traveled in upstream and downstream = d each
--> \(\frac{d}{15} + \frac{d}{25} = 8\)
--> \(\frac{(5d + 3d)}{75} = 8\)
--> \(\frac{8d}{75} = 8\)
--> \(d = 75\) miles

IMO Option C

milegae of boat ; 16*10 ; 160 miles
so at 20 mph total time ; 160/20 ; 8 hrs
speed of boat upstream ; 20-5 ; 15mph
speed of boat downstream ; 20+5 ; 25 mph
total time 8 hrs ;
d/15+d/25 = 8
solve for d = 75 miles
IMO C

We see that, with the current, the boat travels 20 + 5 = 25 mph, and against the current, the boat travels 20 - 5 = 15 mph. However, the boat can travel 10 x 16 / 20 = 8 hours on a 10-gallon gasoline tank, no matter if there is a current or not. Therefore, if we let d = the distance the boat can travel upstream (or downstream), we can create the equation:

d/15 + d/25 = 8

Multiplying the equation by 150, we have:

10d + 6d = 1200

16d = 1200

d = 75

Answer: C
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This is a quicker approach based on the premise that there is a proportionate decrease/increase in MPG with a decrease/increase in speed.

When speed is 20 mph (still water speed), the MPG is 16.
When speed is 15 mph (upstream speed), the MPG is (16/20)*15 = 12.
When speed is 25 mph (downstream speed), the MPG is (16/20)*25 = 20.

Let 'x' miles be the distance of each leg of the return trip that the boat can make on a full tank of 10 gallons.
Fuel needed for the upstream trip = x/12 gallons
Fuel needed for the downstream trip = x/20 gallons
x/12 + x/20 = 10.....> x=75 miles

ANS: C