Question

plzzzzzzzzz send me the answer ​

Answer 1

Question :- Fast train takes 3 hrs less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/hr less than the fast train, find the speed of the two trains.

Solution :-

Let the speed of slow train be x km/hr.

So, according to the condition, speed of fast train = x + 10 km/hr

(since, the slow train has less speed by 10)

Now, we know that time taken = Distance/speed

So, time taken for slow train = 600/x hrs.

And, time taken for fast train = 600/(x + 10)

But it is given that fast train takes 3 hrs less than slow train. Hence, time taken by fast train = time taken by slow train - 3

= (600/x) - 3

= (600 - 3x)/x

So, by equating the two, we get

$\frac{600 - 3x}{x} = \frac{600}{x + 10}\\  \\By \: cross \: multiplying,\\\\(600 - 3x)(x + 10) = 600x\\\\= 600x + 6000 - 3x^{2} - 30x = 600x\\\\= -3x^{2} - 30x + 6000 + 600x - 600x = 0\\\\= -3x^{2} - 30x + 6000 = 0\\\\= -3(x^{2} + 10x - 2000) = 0\\\\= x^{2} + 10x - 2000 = 0/3\\\\= x^{2} + 10x - 2000 = 0$

Now splitting the middle term, we get

$x^{2} + 50x - 40x - 2000 = 0\\\\x(x + 50) - 40(x + 50) = 0\\\\(x - 40)(x + 50) = 0$

Now,

(x - 40) = 0/(x + 50)

x - 40 = 0

x = 40

Similarly, x = -50

But, xpeed can not be negative, hence,

x = 40

So, speed of slow train = 40 km/h

and speed of fast train = x + 10 = 40 + 10 = 50 km/h