I have a fiendish math question with trains...?

Question

A train leaves a station at noon. 2 hours later, a second train leaves the same station on the same track going 50 mph faster. At 3 o'clock, train two is 10 miles behind train one. How fast is each train going?

(Explain if possible)

Thank you.

Answer 1

Let s = the speed of the first train
Let s + 50 = the speed of the second train.

Compute the distance of each train:
The first train travels for 3 hours at a rate of s
d1 = 3s

The second train travels for 1 hour at a rate of s + 50:
d2 = s + 50

We know that d1 is 10 miles ahead of d2:
d1 = d2 + 10

Substituting everything in:
3s = s + 50 + 10
3s = s + 60
2s = 60
s = 30

So the first train travels at 30 mph
the second train travels at 80 mph

After three hours, the first train has gone 90 miles, the second has gone 80 miles (and is therefore 10 miles behind).

This confirms our answers:
Speed of the first train = 30 mph
Speed of the second train = 80 mph

Answer 2

Algebra...basically write down all the formulas you know first...

Given: Y = X + 50 mph

Train A goes X miles/hr for 3 hours
Train B goes Y miles/hr for 1 hour

Distance train A went : X * 3...that's X mph for 3 hours
Distance train B went: Y * 1 ---or--- X * 3 - 10...that's Y mph for one hour or, by the given information, 10 miles less than X went.
so we know... Y *1 = X * 3 - 10
we also know... Y = X + 50 therefore X = Y - 50
substitute X = Y -50 in first equation to get to only one variable...
Y = (Y - 50) * 3 - 10
Y = 3Y - 150 - 10
-2Y = -160
Y = 80
Now since Y = X + 50 then X must equal 30 mph.

Check your work...train A goes 30 mph for three hours (90 miles) and train B goes 80 mph for 1 hour (80 miles). 90 miles - 10 miles = 80 miles so it checks out.

Answer 3

Let x be the speed of train1
so in 3 hours it has travelled 3x miles
in 1 hour train2 will have travelled x + 50 miles and be 10 miles behind train1
So
3x-10 =x+50
3x - x = 50 + 10
2x = 60
x = 30 = speed of train1
speed of train2 = 30 + 50 = 80

Check
3 x 30 = 90 miles
80 + 10 = 90

Answer 4

Let the speed of the first train be x miles per hour. So it travels 3x miles distance in the time allowed. The speed of the second train is x+50 miles per hour, and it travels exactly x+50 miles distance since it only has one hour of travel. So the difference between 3x and x+50 is 10.

So you have 3x - (x+50) = 10

That is, 3x - x - 50 = 10

or, 2x = 60

x = 30

This works because three hours at 30 mph is ten more than one hour at 80 mph.

Answer 5

t = time after noon in hours
v = speed of first train in mph
x1, x2 = their positions in miles

Train 1:
x1 = v * t

Train 2:
x2 = (v + 50) * (t - 2)

At three o'clock:
t = 3 and x1 = x2 + 10

So
v * t = (v + 50) * (t - 2) + 10
v * 3 = (v + 50) + 10 = v + 60
v * 2 = 60
v = 30 mph
and train 2 goes v + 50 = 80 mph

Answer 6

You can set up equations in the speed and distance of each train from the station at any given time.
V2 = V1 + 50.
D1 = V1 x 3. (where first train is at 3 PM.)
D2 = D1 - 10.
D2 = V2 x 1.
Now you have four equations in four unknowns, and can grind out the algebra to get a solution.

Answer 7

Sorry, can't help you. All I know is they better divert train 2, or it will be crashing into Train 1 soon! LOL!