Two trains left from two stations P and Q towards station Q and statio

Given:
1. Two trains left from two stations P and Q towards station Q and station P respectively.
2. 3 hours after they met, they were 675 Km apart.
3. First train arrived at its destination 16 hours after their meeting and the second train arrived at its destination 25 hours after their meeting.

Asked: How long did it take for the first train to make the whole trip?

1. Two trains left from two stations P and Q towards station Q and station P respectively.
Let the speed of the trains be v1 & v2 respectively

2. 3 hours after they met, they were 675 Km apart.
Since they were running in opposite direction.
3(v1 + v2) = 675
v1 + v2 = 225 (1)

3. First train arrived at its destination 16 hours after their meeting and the second train arrived at its destination 25 hours after their meeting.
Let us assume that the trains met after time t
Distance between P & Q = (v1+v2)t = 225t
Distance covered by first train = v1*t
Distance covered by first train = (225-v1)*t

Distance travelled by first train after the meeting = 225t - v1*t = (225-v1)t = 16v1
(225-v1)t = 16v1

Distance travelled by first train after the meeting = 225t - (225-v1)*t = v1*t = 25v2 = 25*(225-v1)
v1*t = 25*(225-v1)

t = 16v1/(225-v1) = 25(225-v1)/v1
16v1^2 = 25(225-v1)^2
4v1 = 5(225-v1)
9v1 = 5*225
v1= 125

100t = 16*125
t = 4*5 = 20 hours

Time taken by first train = 20 + 16 = 36 hours

IMO C

If we let r = the speed of the first train and s = the speed of the second train, we can create the equation:

3r + 3s = 675

r + s = 225

s = 225 - r

Now, if we let d = the distance between stations P and Q and t = the number of hours it takes for the two trains to meet, then we can create the equations:

r(t + 16) = d → Eq. 1

and

s(t + 25) = d → Eq. 2

and

(r + s)t = d → Eq. 3

Since r + s = 225, then Eq. 3 becomes:

225t = d

Substituting 225t for d in Eq. 1, we have:

r(t + 16) = 225t

r = 225t/(t + 16)

Substituting 225t for d in Eq. 2, we have:

s(t + 25) = 225t

s = 225t / (t + 25)

Finally, since r + s = 225, we have:

225t / (t + 16) + 225t / (t + 25) = 225

t / (t + 16) + t / (t + 25) = 1

Multiplying the above equation by (t + 16)(t + 25), we have:

t(t + 25) + t(t + 16) = (t + 16)(t + 25)

t^2 + 25t + t^2 + 16t = t^2 + 25t + 16t + 400

t^2 = 400

t = 20

Since the two trains met 20 hours after they started and it takes the first train 16 more hours to reach its destination, it takes the first train a total of 20 + 16 = 36 hours to reach its destination.

Answer: C

Can this be done in 2 minutes?

How were you able to take the square roots of only R.H.S ?

Can you please explain the highlighted portion.
Is there some kind of typo?

This is what I am wondering too... There's got to be a faster way perhaps when we can reverse plug in?

[quote="nick1816"]Speed of first train = \(v_1\), Speed of second train = \(v_2\)

Time taken by first train to arrive at its destination after their meeting = \(t_1\) = 16 hrs

Time taken by second train to arrive at its destination after their meeting = \(t_2\) = 25 hrs

\(\frac{v_1}{v_2}\) = \(\sqrt{\frac{t_2}{t_1}}\)

\(\frac{v_1}{v_2}\) = \(\sqrt{\frac{25}{16}}\)

\(\frac{v_1}{v_2}\) = \(\frac{5}{4}\)

Distance travelled by second train after meeting is same as distance travelled by first train before meeting. Hence time taken by first train to travel that distance = 25 *(4/5) = 20 hrs

Time taken by first train to make the whole trip = 20+16 = 36 hrs



Hi Nick,

Could you please explain how we can equate the speed and square roots of times like that? Is there some formula or rule that I am missing?..

Thank you!

ababych D4kshGargas

Here is the proof of the formula, i used.

P................R............................Q


Suppose both trains meet at point R.

Speed of train A started from P = \(v_1\)

Speed of train B started from Q = \(v_2\)

Ratio of PR and RQ must be equal to ratio of their speed.

\(\frac{PR}{RQ} = \frac{v_1}{v_2}\)........(1)

After meeting, train A has to cover RQ to reach to the destination.
Time taken, \(t_1\), by train A to cover RQ = \(\frac{RQ}{v_1}\)

After meeting, train B has to cover PR to reach to the destination.

Time taken, \(t_2\), by train A to cover PR = \(\frac{PR}{v_2}\)

\(\frac{t_1}{t_2} = (\frac{RQ}{v_1})÷(\frac{PR}{v_2}) \)

\(\frac{t_1}{t_2} = (\frac{RQ}{PR})*(\frac{v_2}{v_1}) \)

From equation (1)

\(\frac{t_1}{t_2} = (\frac{v_2}{v_1})*(\frac{v_2}{v_1}) \)

\(\frac{t_1}{t_2} = (\frac{v_2}{v_1})^2 \)

\(\sqrt{\frac{t_1}{t_2}} = (\frac{v_2}{v_1}) \)

effatara


Thank you for this! Seems like such a straight forward approach, but somehow wasn't able to get there on my own.

Elaborating the highlighted concept :-

Let a be the number of hours after P and Q met. p and q be the speeds.

Distance covered by P in a hours = a*p
This distance is covered by Q in 25 hrs after they met = 25*q

Therefore \(a*p = 25*q\) => \(p = 25*\frac{q}{a}\)

Similarly,
Distance covered by Q in a hours = a*q
This distance is covered by P in 16 hrs after they met = 16*p

Therefore \(a*q = 16*p\)

=> \(a*q = 16*25*\frac{q}{a}\)

=> \(a^2 = 16*25\) => a = 5*4 = 20 hrs

Therefore time taken by P = 20+16 = 36 hrs -- Answer C

Let trains be P & Q, at point A & B , meeting at point M
Speed of P = SP, Speed of Q = SQ


1. Distance traveled after meeting = 675 (in 3hrs)
-> (SPx3) + (SQx3) = 675
-> SP+SQ = 225 .................................(1)

2. Train P took 16 hrs to reach destination (M to B) & Train Q took 25 hrs to reach destination (M to A)
-> Let time 't' be the time took by P&Q to reach meeting point M
-> Now, equating distance after meeting point M
-> Distance (MA) = t x SP = 25 x SQ ..............(2)
-> Distance (MB) = t x SQ = 16 x SP ..............(3)

Solving above eqn 2 & 3 , we get : SP/SQ = 4/5 ...........(4)

Solving eqn 1 & 4 , we get : SP =125 & SQ=100

Use above in either eqn 2 , we get t=20

So total time taken by P = t+16 = 20+16 = 36

OMG! This logic seems so simple to understand when someone walks you through it but is going to take some time (some practice) for me to inculcate.

Hello nick1816, is this some kind of a property?

Hi,

Nick has explained it in a subsequent post. Also, "effatara" has explained the same with a walkthrough.

I think, it's not GMAT question!

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