A train leaves Kyoto for Tokyo travelling 240 miles/hour at 12 noon. Ten minutes later another train leaves Tokyo for Kyoto travelling at 160 miles/hour. If Tokyo and Kyoto are 300 miles apart, what time will the trains pass each other?
This is in the Mat pre guide - rates and work problem set - Pg 34. I am not sure how to solve these kinds of problems. Please help!
Thanks
Jay
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- sjayva
- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
first of all, whenever you do a word problem, you should always check the feasibility of your answers. in other words, look at the answer choices, mentally check them against the rest of the problem, and then see whether they make sense.
in this problem, there's no way the trains can meet at 5:00pm; that's five hours after the first train leaves. a glance at the numbers will show that the faster train - going 300 mi at 240 mph - could make the whole trip by itself in a little over an hour (because it goes 240 of those 300 miles in the first hour alone); similarly, the slower train could make the entire trip by itself in about two hours.
in any case, here's the best way to set up the problem:
let t be the amount of time traveled by the train that leaves first. then the later train travels for (t - 1/6) hours.
try your best to read the following chart. it won't format properly in the forum, so i'll add dots to try to make it more comprehensible.
TRAIN ..... rate ... time ... distance
fast one ... 240 ... t ... 240t
slow one ... 160 ... t - 1/6 --- 160(t - 1/6)
the key fact is that the two trains cover the 300 miles together, so their distances add to 300 miles.
240t + 160(t - 1/6) = 300
400t - 160/6 = 300
400t - 80/3 = 300
400t = 980/3
t = 980/1200
= 98/120
= 49/60
so that's 49/60 hours, or 49 minutes
therefore, the trains meet each other at 12:49pm.
(note: if you let 't' stand for the _other_ train's time, you'll get 39/60; this isn't as nice, because then you still have to go through some motions to figure out the time)
in this problem, there's no way the trains can meet at 5:00pm; that's five hours after the first train leaves. a glance at the numbers will show that the faster train - going 300 mi at 240 mph - could make the whole trip by itself in a little over an hour (because it goes 240 of those 300 miles in the first hour alone); similarly, the slower train could make the entire trip by itself in about two hours.
in any case, here's the best way to set up the problem:
let t be the amount of time traveled by the train that leaves first. then the later train travels for (t - 1/6) hours.
try your best to read the following chart. it won't format properly in the forum, so i'll add dots to try to make it more comprehensible.
TRAIN ..... rate ... time ... distance
fast one ... 240 ... t ... 240t
slow one ... 160 ... t - 1/6 --- 160(t - 1/6)
the key fact is that the two trains cover the 300 miles together, so their distances add to 300 miles.
240t + 160(t - 1/6) = 300
400t - 160/6 = 300
400t - 80/3 = 300
400t = 980/3
t = 980/1200
= 98/120
= 49/60
so that's 49/60 hours, or 49 minutes
therefore, the trains meet each other at 12:49pm.
(note: if you let 't' stand for the _other_ train's time, you'll get 39/60; this isn't as nice, because then you still have to go through some motions to figure out the time)
3 posts Page 1 of 1