A train is on its way east was delayed 1 hr when it was 560 km west of New York City. By increasing its normal rate 10 km/hr, the train arrived at New York City on schedule. What is the train's normal rate? Solutions and answer please, I'll fan and give a medal :) thank you!

Question

A train is on its way east was delayed 1 hr when it was 560 km west of New York City. By increasing its normal rate 10 km/hr, the train arrived at New York City on schedule.  What is the train's normal rate? Solutions and answer please, I'll fan and give a medal :) thank you!

anonymous · Answer

@PaulaLovesSchool13

anonymous · Answer

@igreen @gurv

anonymous · Answer

@Joel_the_boss

anonymous · Answer

460km/hr i think

akashdeepdeb · Answer

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akashdeepdeb · Answer

the trick is, usually, the train moves at the normal speed. 

let x be the normal rate or speed of the train. 
let d be the distance it covers before, stopping for 1 hour, 560 km before NYC. 
so usually, the time taken  t must be, 
$$t = \frac{560 + d}{x}$$

But in the question, after traveling properly for d kilometers, the train stops for 1 hour and then speeds up to reach at time. so, again:
$$t = \frac{d}{x} + 1 + \frac{560}{x+10}$$

t is the same for both, so:

$$\frac{560 + d}{x} = \frac{d}{x} + 1 + \frac{560}{x+10}$$

Solve this. 

NOTE:

t is just a variable to make things simpler. d and x are required to get the answer.

phi · Answer

the hardest part (probably) is writing down equations.

one way is to say:
the train is traveling at x km/hr  and will take t hours to go 560 km
using rate*time = distance,  we can write

x t = 560

next, it says if we speed up (by going 10 km/h faster  or x+10 )  we will "make up" an hour. that means  instead of t hours it will take t-1 hours
again use rate * time = distance

(x+10)(t-1) = 560

both equations equal 560, so we can say
x t = (x+10)(t-1) 
x t =  xt - x +10t - 10
add -xt to both sides and we get
0 = -x +10t -10
x+10 = 10t
or 
$$ t = \frac{x+10}{10} $$
let's use that in the first equation 
x t = 560
$$ x  \frac{x+10}{10} = 560 $$
$$ x(x+10)=  5600 \
x^2 +10x = 5600 \
x^2 +10x -5600 = 0 $$
where x is the *old speed* (or *normal* speed) i.e. what we want to find

it looks like we should use the quadratic formula to solve for x.

anonymous · Answer

Thank you, I'll try both of it, and I'll give the medal to the correct one. I'll fan both of you :)

akashdeepdeb · Answer

What @phi did gives us the same equation. Both are correct.

anonymous · Answer

Yah I just finished :) anyhow, I can't give medals to both so I'll just choose later. :) Thank you so much ^_^