Question 296126
Samuel left the house at noon heading to David's store, 60 miles away. David left his store at 2 pm, heading to Samuel's house and traveling at twice Samuel's speed. The two men met at 3 p.m. How fast was David traveling?


Since the journey from David's store to Samuel's house is 60 miles, then the sum of the distances traveled by the 2 will total 60, or, 


Samuel's distance covered + David's distance covered = 60 miles


Let Samuel's speed be S, then David's speed will be 2S, since David is traveling at twice Samuel's speed


Then Samuel's distance traveled = S(3), since it took him 3 hours to cover the distance to the meeting point, and


David's distance covered = 2S(1), since it took David 1 hour to cover the distance to the meeting point


Therefore, we'll have: 3S + 2S = 60


5S = 60


S = {{{60/5 = 12}}}


Now, since S, or Samuel's speed is 12 mph, then David's speed = {{{highlight_green(24)}}}mph (2S, or 2*12).