Asked by Liana
At 1 pm two trains, traveling toward each other leave from towns that are 312 km apart. One train averages at most 82km/h and the other at most 74km/h. What is the earliest possible time for them to meet?
Please help me solve this. I'm not exactly sure how to set up the equation or inequality.
Please help me solve this. I'm not exactly sure how to set up the equation or inequality.
Answers
Answered by
Reiny
For these kind of questions you have to ask yourself,
which of my quantities are equal so I can make an equation?
When the meet, will they not have travelled the same time? So the times are equal
let the distance covered by the faster train be x km
then the distance covered by the slower train is 312-x km
time taken by faster train = x/82
time taken by slower train = (312-x)/72
so ..
x/82 = (312-x)/72
crossmultiply and simplify to get
x =166.13 km
so time is 166.13/82 = 2.026 hours
which of my quantities are equal so I can make an equation?
When the meet, will they not have travelled the same time? So the times are equal
let the distance covered by the faster train be x km
then the distance covered by the slower train is 312-x km
time taken by faster train = x/82
time taken by slower train = (312-x)/72
so ..
x/82 = (312-x)/72
crossmultiply and simplify to get
x =166.13 km
so time is 166.13/82 = 2.026 hours
Answered by
joseph
Simplify please...thnks
X/50=(312-X)60
X/50=(312-X)60
Answered by
..
i dont get it
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