中二數學問題急！！！！

2009-03-05 6:25 am

2)Paul walks up a moving-up escalator.Find the number of steps in the
escalator and the speed of the escalator in steps per second
with the following conditions:
a)If he takes 2 steps per second , he will reach the top after taking 32
steps.
b)If he takes only 1 step per second , he will reach the top after
taking 20 steps.
Let x be the number of steps,and y steps per second be the speed
of the escalator.
請畀題解

回答 (1)

老爺子

2009-03-05 4:25 pm

✔ 最佳答案

Let x be the number of steps, and y steps per second be the speed of the escalator.

a) 2 steps per second:

Time taken
= 32/ 2 s
= 16s

Steps walked by Paul + Steps moved by escalator = Steps in the escalator
32 + 16y = x ...... (1)

b) 1 step per second:

Time taken
= 20/1
= 20 s

Steps walked by Paul + Steps moved by escalator = Steps in the escalator
20 + 20y = x ...... (2)

(1) = (2):
32 + 16y = 20 + 20y
4y = 12
y = 3

Put x = 3 into (2):
20 + 20(3) = x
x = 80

Ans: There are 80 steps in the escalator.
Ans: The speed of the escalator is 3 steps per second.
=

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中二數學問題 急！！！！

回答 (1)

中二數學問題急！！！！