CAA: Christian Anime Alliance

PostPosted: **Sat Jan 20, 2007 7:13 am**

So, I got my PSAT scores back and seeing that I have some work to do I decided to go to the site and see what I got wrong. A good bit of the maths kick my butt. Plus, if I go on to take the real SATs and miss out on a scholarship because of some stupid math questions, I might cry.

Anyways the site has explanations for the questions, and I was looking over some of them. I don't understand this one at all. Maybe I'm slow... :eh:

Mathematics: Question 19

A train 200 meters long was traveling at a constant rate of 20 meters per second through a tunnel. It took 1 minute and 10 seconds from the time that the front of the train entered the tunnel until the time that the back of the train left the tunnel. What is the length of the tunnel, in meters?

1. 1,000
2. 1,200
3. 1,400
4. 1,600
5. 2,000

Hide Answer Explanation

Choice (B) is correct. The train traveled for 1 minute and 10 seconds (which is 70 seconds) at 20 meters per second. Using the formula rate x time = distance, you can see that the train traveled 20 x 70 = 1,400 meters. This period of traveling began when the front of the train entered the tunnel and ended when the back of the train left the tunnel, at which point the front of the train was 200 feet beyond the tunnel (since 200 meters is the length of the train). Thus, the train traveled the length of the tunnel plus 200 meters. Therefore, (length of tunnel) + 200 = 1,400. It follows that the length of the tunnel is 1,200 meters.

This makes no sense to me. I was able to understand to the 1,400 m. But then afterwords they lose me so bad. I can't understand why I was supposed to subtract 200...

I hate maths... :bang:

PostPosted: **Sat Jan 20, 2007 7:24 am**

Consider it like this:
The train is 200 m long.
At t=0, the head of the train enters the tunnel.
At t=70 s, the rear of the train exits the tunnel.
The train is known to be travelling at 20 m/s.

The key to solving the problem is straightforward:
If the *head* of the train travelled 1400 m in the given time period, it follows that the *rear* of the train is 200 m behind it. After 70 s though the rear of the train is just exiting the tunnel. Therefore, the tunnel length must be:

70 s * 20 m/s -200m=1200 m

PostPosted: **Sat Jan 20, 2007 7:27 am**

Oh...okay. I get it now! I wasn't considering the fact that I was only counting the front of the train. It makes a bit more sense now.

Thanks!

CAA: Christian Anime Alliance

PSAT Math sucks...

PSAT Math sucks...