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shreyas123 · May 30, 2009, 03:14 AM

A fast train takes 3 hours less than the slow train for a journey of 600 km.If the speed of the slow train is 10km/he less than that of fast train.Find the speed of the two trains.

Unknown008 · May 30, 2009, 05:29 AM

For this type of question, you have to assign variables to your unknown values. Say, let

x be the speed of the fast train and
y be the speed of the slower train.

The time taken for the fast train to complete the journey is given by $\frac{600}{x}$

by the formula $Speed = \frac{Distance}{Time}$

The time for the slower train to complete the same journey will be $\frac{600}{y}$

Now, the time for the faster train is 3 hours less than that of the slower train, so,

$\frac{600}{x} + 3 = \frac{600}{y}$

Now, you're given that the speed y is 10km/h less than x, so you have another equation;

$y = x-10$

Now, replace that in your first equation and solve for x, the speed of the fast train. To fint the speed of the slow train, use your second equation and use the value of x you just obtained.

Hope it helped! :)

shreyas123 · May 30, 2009, 08:53 AM

Originally Posted by Unknown008

For this type of question, you have to assign variables to your unknown values. Say, let

x be the speed of the fast train and
y be the speed of the slower train.

The time taken for the fast train to complete the journey is given by $\frac{600}{x}$

by the formula $Speed = \frac{Distance}{Time}$

The time for the slower train to complete the same journey will be $\frac{600}{y}$

Now, the time for the faster train is 3 hours less than that of the slower train, so,

$\frac{600}{x} + 3 = \frac{600}{y}$

Now, you're given that the speed y is 10km/h less than x, so you have another equation;

$y = x-10$

Now, replace that in your first equation and solve for x, the speed of the fast train. To fint the speed of the slow train, use your second equation and use the value of x you just obtained.

Hope it helped! :)

I also did the same things till here..! :D
actually the problem is after substituting the vaue of y=x-10 in 1st equation...

Unknown008 · May 31, 2009, 07:20 AM

Ok, I'll do it further a little bit:

$\frac{600}{x}+3=\frac{600}{y}$

$\frac{600}{x}+3=\frac{600}{x-10}$

Multiply everything by x;

$\frac{600x}{x}+3x=\frac{600x}{x-10}$

$600+3x=\frac{600x}{x-10}$

Now by x-10;

$600(x-10)+3x(x-10)=\frac{600x(x-10)}{x-10}$

$600x-6000+3x^2-30x=600x$

Now, can you simplify and solve?