A deer is standing 80m in from the west end of a tunnel. The

If the answer is in fact 180 meters, then I think I know one way to solve this problem. It is helpful to create a diagram of where the deer is when the train approaches that sort of looks like this:

E____________________D______________W________T2_______________T1

Where E represents the east end of the tunnel, D represents the deer's position in the tunnel, and W represents the west end of the tunnel. T1 represents the train when the deer first sees it approach. T2 represents the train’s position when the deer arrives at W, the western entrance of the tunnel.

We are told that the deer arrives at the west end of the tunnel when the train is 20 meters away from the west end of the tunnel. Further, we are told that if the deer runs towards the east end of the tunnel, the deer would be hit by the train at the east end of the tunnel.

To make things simpler, let the distance between T2 and T1 equal X, and let the distance between D and E equal Z.

We are trying to find the total length of the tunnel. From the above information, we know that

Total length = ED + DW

Total length= Z + 80

Let y equal the rate at which the deer runs. We know that the train is moving 10 times faster, so let the train’s rate equal 10y.

In the first situation, where the deer runs towards the train, set up a rate time distance table like the one below. Since the deer is running towards the train, ADD rates and distances, and leave time the same.

Rate Time Distance (meters)

Deer y t 80

Train 10y t X

Total 11y t 80+X


Plug in a simple number for y. Let y= 1 m/s. Solve for X and t.

Rate Time Distance

Deer 1 80 sec 80 m

Train 10 80 800 m

Total 11 80 880 m

In the second situation, the deer runs AWAY from the train. So, subtract the deer’s rate from the train’s rate, and subtract the deer’s distance from the train’s distance. Note that in this situation, the train must cover the additional distance of 20 meters, plus the distance from the deer’s original position to the western tunnel entrance (80 meters), plus the distance from the deer’s original position to the eastern tunnel entrance (Z). We know from the first scenario that X= 800 meters when the Deer’s rate= 1 m/s.

Rate Time Distance

Deer 1 t2 Z

Train 10 t2 Z+80+20+X

Total 9 t2 100+800

Solve for t2:

9 (t2) = 900

t2 = 100

Then plug in t2 to either the deer or the train’s rate equation.

Deer: 1 (t2) = Z
100 = Z

Add 80 meters to Z to find the total length of the tunnel

Total length= Z+80 = 180

You're welcome. I'm not entirely happy with my solution, as it is very long and probably too complicated to complete in under two minutes. I think there is probably a simpler solution. Hopefully someone else will chime in with it!

mainhoon,
that was great!!!!!!!!:)
but im not able to understand the second step....can you please explian it in detail.....

Bunuel
thankyou so much for the detailed explanation....
your explanation helped me to understand another typical quant question too thankyou once again...

No it is not a second method.. Basically 1 is the case when deer runs towards the train .. You use that to figure out how far us the train from the tunnel.. That is 800+20. the second is use the case when the deer is running away from the train.. So that becomes
(800+20+D)/x=(D-80)/10x
This is just distance/time.. Now you get 180 for D



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Posted from my mobile device

mainhoon,
im sorry its the second step...not second method...

I hope it is clear now...

Posted from my mobile device

Sorry to ask.
But is this a GMAT type question?
More importantly is this just a 600-700 level question?

T ------T' --W ---- D ----W'----E
If the deer goes eastward when the train is at T', the deer must be at W'.
WD = WD' = 80.
T'D' = 20 + 80 + 80 = 180.

Train catches Deer at W => "Catch up" : same distant, different time and rate. [when Train is at T', Dear is at W']
Deer: Rate = a, time = t, distant = at
Train: Rate = 10 a, time = t - 180/10a, distant = at
at = 10a (t - 180/10a) => at = 20

WE = 80 + 80 + 20 = 180.

Answer: D

Hello

I have a doubt in this question. After the point where we find that the distance the train covered is 800m, I tried to use the relative speed concept.

In my understanding, the relative speed would be 9x. The distance to be covered would be 900(800+20+80).

After this deduction, I am unable to apply the concept.

Could you please help in the application?

Thanks and Regards

Using Ratios, it can be done quite simply:


........................................................|...........................||_______________________|_________________________________||.....................
Train..................................................|........20m........TunnelW.............80m............Deer.............................................TunnelE................

First Case:
Speed of Deer:Speed of Train = 1:10
Deer covers 80 m, so in same time, train covers 800 m.

........................................................|...........................||_______________________|_________________________________||.....................
Train.....................800m.....................|........20m...........TunnelW.............80m............Deer.........................X................TunnelE................

Second Case:
Deer covers, say X m of the tunnel.
In same time, train covers 800+20+80+X meters i.e. 900+X meters.

X/(900+X) = 1/10 (Ratio of their speeds)
X = 100 meters

So length of tunnel = 80 + 100 = 180 meters

In the first case, the train would have traveled 800m towards the deer, in the time the deer takes to travel 80m and exit the tunnel. So the distance from where the train is sighted to the east end of the tunnel is 800+20+80+y, where y is the portion of the tunnel to the east of where the deer is standing

So from the second case we have, by equating time (800+20+80+y)/10 = y/1
The numerator in the LHS is the distance traveled by the train in the second case and the RHS numerator is the distance traveled by deer in the second case.
We get y=100. The distance of the tunnel =80+y=180