On a journey of 600 km, a train was delayed 1h and 30 min after having covered 1/4 of the way. To arrive on time at the destination, the engine driver had to increase the speed by 15km/hr. How long did the train travel for?

Question

anonymous · Answer

does that mean it reached 1/4 of its way in 1 hr 30 mins ?

anonymous · Answer

yes

anonymous · Answer

@ganeshie8

anonymous · Answer

@amistre64

anonymous · Answer

it would a bit less than 5 hr and 30 mins.

ganeshie8 · Answer

lets say the train was travelling at $v1$ kmph in the first quarter of journey

ganeshie8 · Answer

then, after 1/4th journey, it travels at $v1 + 15$ kmph

ganeshie8 · Answer

we can have these equations  :-

$\large v1(\frac{t}{4} + 1.5) = 150 ----------- (1)$


$\large (v1+15)(\frac{3t}{4} - 1.5) = 450 ----------- (2)$

ganeshie8 · Answer

two equations, two unknowns

ganeshie8 · Answer

you can solve them ?

raden · Answer

600/v = 150/v + 3/2 + 450/(v + 15)
simplify becomes :
(300 - v)/(2v) = 150/(v + 15)
300v = 4500 + 285v - v^2
v^2 - 15v - 4500 = 0
(v - 60)(v + 75) = 0
only v = 60 km/h which satifies
therefore :
t = d/v = 600/60 = 10 hours

dan815 · Answer

haha vmath why do u always ask these kinds of algebra questions

dan815 · Answer

ive seen u ask these kinds for months

whpalmer4 · Answer

The slip in ganeshie8's approach is that the first equation gives "distance credit" for that the hour and a half delay spent moving at $v_1$, but of course that isn't true, as the train isn't moving during the delay.  It should be $$v_1(\frac{t}{4}) + 0(1.5) = 150$$and combined with the second equation will give the correct answer.