# Thread: help with math word problem

1. ## help with math word problem

one elevator leaves first floor of a building at 9:00 a.m. It takes 5 seconds to travel between floors, stays 10 seconds at each floor and stops at each floor on its way to the top. Another elevator leaves the 12th (top) floor at 9:00 a.m. ; it takes 5 seconds to travel between floors, stays 15 seconds at each floor and stops at each floor on its way to the bottom. On what floor will both elevators be available at the same time?

2. ## Re: help with math word problem

I am guessing it would be somewhere in the middle, 6th or 7th floor but cannot prove it mathematically yet.

3. ## Re: help with math word problem

Let they meet on x th floor.
upward lift takes 5 + 10 = 15 s per floor
downward lift takes 5 + 15 = 20 s per floor
if they do not wait at each floor,
then time taken by upward lift = 5x
and time taken by downward lift = 5(12 – x)
5x = 5(12 – x)
x = 6 th floor

as they are waiting
then time taken by upward lift = (5 + 10)x = 15x
and time taken by downward lift = (5 + 15)(12 – x) = 20(12 – x)
15x = 20(12 – x)
35x = 240
x = 6.86 th floor
in between 6 and 7 th floor
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4. ## Re: help with math word problem

Hello, MimiMath3573!

One elevator (A) leaves first floor of a building at 9:00 a.m.
It takes 5 seconds to travel between floors, stays 10 seconds at each floor, and stops at each floor on its way to the top.
Another elevator (B) leaves the 12th (top) floor at 9:00 a.m.
It takes 5 seconds to travel between floors, stays 15 seconds at each floor, and stops at each floor on its way to the bottom.
On what floor will both elevators be available at the same time?

Elevator A takes: $5n + 10(n-1) \,=\,15n-10$ seconds to reach floor $n.$

Elevator B takes: $5(12-n) + 15(11-n) \,=\,225 - 20n$ seconds to reach floor $n.$

They meet when their times are equal:
. . $15n - 10 \:=\:225 - 20n \quad\Rightarrow\quad 35n \,=\,235 \quad\Rightarrow\quad n \,=\,\frac{235}{35} \,=\,6\tfrac{5}{7}$

Both elevators will be available on the 7th floor.

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Check

Elevator A takes: $t \:=\:15(7)-10 \:=\:95$ seconds to reach floor 7.
It stays there for 10 seconds ... until $t\,=\,105$ seconds.

Elevator B takes: $t\:=\:225 - 20(7) \:=\: 85$ seconds to reach floor 7.
It stays there for 15 seconds ... until $t\,=\,100.$

Therefore, both elevators are available on the 7th floor
. . from $t = 95$ to $t = 100.$