Name: Monty
Who is asking: Other Question: This problem is from a ninth grade Math book: "Essentials of Mathematics" Russell V. Person. c. 1961 Second Pr. 1963 My neighbors daughter asked for my help with this homework problem in 1964. =============================== 35. A man and his wife walk up a moving escalator. The man walks twice as fast as his wife. When he arrives at the top, he has taken 28 steps. When she arrives at the top, she has taken 21 steps. How many steps are visible in the escalator at any one time. ============================== This is just for fun, it's not critical. Also, do they have problems this hard in ninth grade nowadays? Monty PS I doubt that you will need it but I'll send the solution if you do. Hi Monty, This is a great problem. Here is our solution. Just as they step on the escalator paint the first step red so that you can see how it progresses relative to how they progress. We then drew a diagram of the situation just as the man reaches the top. The man has walked up 28 steps so the red step is 28 steps behind him. The woman walks at half his speed so she is 14 steps above the red step. Thus the woman is still 14 steps from the top. We know that the woman only walks 21 steps to reach the top, and thus in the time she walks 7 more steps the escalator also goes up 7 steps. Hence the woman and the escalator are travelling at the same rate. Now look at the diagram again. In the time it took the red step to raise the distance labeled ?, the woman has climbed 14 steps. Since the escalator and the woman are travelling at the same rate, ? is 14. Hence you can see 28 + 14 = 42 steps. Is this how you solved the problem? We are not sure about the difficulty of grade nine problems, but we do teach a course to students who plan to be elementary school teachers. That course has problems that are somewhat similar. For example
Penny and Claude
