Carissa the chessplayer is a very, very slow walker. In fact, she walks at 100 meters per hour when walking uphill, 120 meters per hour when walking across flat ground, and 150 meters per hour when walking downhill. One day, Carissa walks across Boston from a café to a boba shop, and then takes the same route in reverse to return to the café. What was Carissa's average speed during the entire round trip?

Lilliam0216 Mar 9, 2024

#2**+1 **

Suppose that on the trip from the cafe to the boba shop a total of U metres is uphill, ( either in a single stretch or made up in bits), and similarly L metres is on the level, and D metres is downhill, then the time taken for the trip will be U/100 + L/120 + D/150.

On the return trip uphill becomes downhill and downhill becomes uphill so the time taken will be U/150 + L/120 + D/100.

Total time there and back will be U(1/100 + 1/150) + L(1/120 +1/120) + D(1/150 + 1/100) = (U + L + D)/60.

The total distance covered is 2(U + L + D) metres, so dividing distance by time, the average speed will be 120 m/hr.

Tiggsy Mar 10, 2024

#2**+1 **

Best Answer

Suppose that on the trip from the cafe to the boba shop a total of U metres is uphill, ( either in a single stretch or made up in bits), and similarly L metres is on the level, and D metres is downhill, then the time taken for the trip will be U/100 + L/120 + D/150.

On the return trip uphill becomes downhill and downhill becomes uphill so the time taken will be U/150 + L/120 + D/100.

Total time there and back will be U(1/100 + 1/150) + L(1/120 +1/120) + D(1/150 + 1/100) = (U + L + D)/60.

The total distance covered is 2(U + L + D) metres, so dividing distance by time, the average speed will be 120 m/hr.

Tiggsy Mar 10, 2024