Aptitude - Problems on Trains - Discussion

d1/d2=x1/x2;
d1=16x1;
d2=9x2;
9/16=(x1)2/(x2)2;
x1/x2=3/4;

Ishita Narula is correct

If distance between delhi and patna is x then, time taken in two conditions are :

1. 2x/60 and x/80+x/40
as eq. x/80+x/40= 2x/60+2

Simplified x=480.

L.C.M of 60,80 and 40 is 240..
so let be 240 is the distace .
then train with 60kmh will take 8 hrs to complete.
and trains with 80 kmh and 40 kmh will take 9 hrs to complete.

so the time difference is (9-8) =1.. so what is required is 4 hrs lap.

so multiply with 4 we will get 240*4 = 960 km ..that is the answer.

A train travelled from Delhi to Patna and back in a certain time at the rate of 60kmph. But if the train had travelled from Delhi to Patna at rate of rate 80Kmph. And back from Patna to Delhi at the rate of 40Kmph. It would take two hours Longer. Find the distance between Delhi and Patna?

Detailed solution please.

Its ok the formula is working but can anyone proove that formula!

For this question the direct formulas are:
(A's speed) : (B's speed) = root(b) : root(a)
Time at which they meet= root(a*b)

Derivation to the formulas:

Let X be the total distance, t be the time at which they meet
a,b are times at which they reach the destination after they meet.
s1, s2 are speeds of train1, train2 respectively.

then X = t(s1+s2) ...(1)
X = (t+a)s1 ...(2)
X = (t+b)s2 ...(3)

after equating (1) and (2)
t.s2 = a.s1 ...(4)
after equating (1) and (3)
t.s1 = b.s2 ...(5)

multiply (4) and (5)
t.t.a.b = s1.s2.a.b

" t = root(a*b) "

substitute t value in either (4) or (5)

then " s1/s2 = root(b)/root(a) "

Its an simple formula nd you all are making it so much complicated.

d1+d2= D;(total distance)

v1=d1/t1; =>v1=(D-d2)/t1----> EQ1

v2=d2/t2; => v2=(D-d1)/t2---->EQ2

d1 is distance traveled by first train to meet the 2nd, similarly d2 is distance traveled by second train to meet first train;

t1 is time taken by train 1 to meet the other
t2 is time taken by train 2 to meet first train

now d1+d2= D;(total distance)

according to given data

v1=(D-d1)/9;
v2=(D-d2)/16;


v1/v2=(16/9) *((D-d1)/(D-d2));

=(16/9) *((v2/t2)/(v1/t1))----->from EQ1 and EQ2

(v1/v2)^2 =(16/9)(t1/t2)

t1 and t2 are equal as trains start at same time and both meet after equal amounts of time so, memorize the above formula

Good Problem !!!Just memorize the FORMULA .