A bike running at 80 km/h initially is slowed down to

I believe that the following reflects the intent of the problem:

nkmungila1 wrote:A bike running at 80 km/h initially is slowed down to 60 km/h as soon as the fuel indicator touches the half level mark. It keeps running at this speed till it runs out of fuel, thereby covering a total distance of 640 km in 10 hours. Traveling at the lower speed, the bike consumes 2 litres of fuel per hour. What is the capacity (in litres) of the fuel tank of the bike?

(A) 16
(B) 20
(C) 32
(D) 40
(E) 56

Average speed for the whole trip = d/t = 640/10 = 64 miles per hour.

This is a MIXTURE problem.
Two speeds (60mph and 80mph) are combined to form a mixture with an average speed of 64mph.
To determine how much time must be spent at each speed, we can use ALLIGATION.
Let S = the slower speed and F = the faster speed.

Step 1: Plot the 3 speeds on a number line, with S and F on the ends and the average speed for the whole trip in the middle.
S 60----------64------------80 F

Step 2: Calculate the distances between the values on the number line.
S 60----4----64----16-----80 F

Step 3: Determine the ratio of the two given speeds.
The ratio of S to F is equal to the RECIPROCAL of the values in red.
S:F = 14:6 = 8:2.

Implication:
Of the 10 hours of travel time, 8 hours are traveled at 60mph and 2 hours are traveled at 80mph.
Since 2 liters are consumed for every hour spent traveling at 60mph, the total amount of fuel consumed during the 8 hours traveled at 60mph = 2*8 = 16 liters.
Since the tank is half full when the bike begins to travel at 60mph, these 16 liters constitute half of the tank's capacity.
Thus, the capacity of the tank = 2*16 = 32 liters.

The correct answer is C.