CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

gunjan-ahuja

thnx for the explanation IOTA

Iota

Quote:

Originally Posted by vivekma

When X =10, Distance travelled in Bus B is 10 Km which takes 10/.6 = 50/3 minutes > 45/3

I was trying to prove that B gets to X before A and hence person travels 10Km anti clockwise rather than 2 Km clockwise.

When X = 10, time if board Bus B = 10 + 1/0.6 = 35/3

tg8

Hi...
bt can u explain me my basic doubt hw the total distance is of 12 km when he is travelling 1km along the circular road of 10km Plz find the attatched image.

tg8

i am taking pt A as a terminus ....
is it wrong?

Iota

Quote:

Originally Posted by tg8

Hi...
bt can u explain me my basic doubt hw the total distance is of 12 km when he is travelling 1km along the circular road of 10km Plz find the attatched image.

Dude pls read the question once more carefully,it says as :

"Bus A leaves the terminus every 20 minutes, it travels a distance 1 km to a circular road of length 10 km and goes clockwise around the road"

that means that he travels 1 km in straight line (from terminus to start of circular path) and then the bus starts going along the circular path.
so the bus covers :
1 km (terminus to Start of circular path) + 10 Km (Circular path) + 1 Km (start of circular path to terminus) = 12

Hope that u r doubt is clear now. U cn also refer to the diagram attached jus b'4 u r post.

saurabhsmit2004

Quote:

Originally Posted by gunjan-ahuja

thnx for the explanation IOTA

Gunjan and iota its still not clear why we are concluding that value when A and B meet.
Rather X is the distance from the terminus at which the man will catch the bus.
Now, how can u say that since A and B meet at X=3; then for X=12-3=9 the time taken will be greatest.
Rather if X=10; this man walks 10 km and hence spends max time .
What do u say??

Varun Khullar

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Quantitative Question # 061
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Bus A leaves the terminus every 20 minutes, it travels a distance 1 km to a circular road of length 10 km and goes clockwise around the road, and then back along the same road to to the terminus (a total distance of 12 km). The journey takes 20 minutes and the bus travels at constant speed. Having reached the terminus it immediately repeats the journey. Bus B does the same except that it leaves the terminus 10 minutes after Bus A and travels the opposite way round the circular road. The time taken to pick up or set down passengers is negligible. A man wants to catch a bus a distance 0 < x < 12 km from the terminus (along the route of Bus A). Let f(x) the maximum time his journey can take. The value of x for which f(x) is a maximum is

(1) 3 (2) 5 (3) 8 (4) 10 (5) none

max time his journey can take .. is

let's see ..

1. he travels at 3/5 km per min
travel clockwise he takes 20 mins
okay the person shouldn't life on the straight one km stretch for max
have to catch bus B
the buses meet at 3km from terminal anti clockwise and 9 km from clockwise..
for max time of journey 9<x<11 should be the condition
to maximize time x-> 11 lower limit my answer is 5 .. none of these

Iota

Quote:

Originally Posted by saurabhsmit2004

Gunjan and iota its still not clear why we are concluding that value when A and B meet.
Rather X is the distance from the terminus at which the man will catch the bus.
Now, how can u say that since A and B meet at X=3; then for X=12-3=9 the time taken will be greatest.
Rather if X=10; this man walks 10 km and hence spends max time .
What do u say??

if X = 10 the man will board Bus 'B' whr the time will be
10 [Waiting Time for Bus B] + (1+1)/0.6 = 40/3 ==> this is not the maximum time as for X=9 , f(x) = 45/3 min

N.B: Why the person will board Bus B?==> [If I was the person I will always opt the option which will take lesser time for distamce X

]

arbit_rageur

My answer is 1)3.

The two buses together will span the entire distance twice in the time between the bus A going from the initial point to x=3 for the second time. Therefore suppose a man goes to some position at a particular time, the worst time it will take for the waiting at a particular point and for the journey together will be at x=3.

I assume the maximum time for journey includes time he spends waiting for the bus. Or else, I guess it will be x=6, but that seems a trivial solution.

randhir8

hi
i am new to the forum..
here is mine reply for the question no.61
some points on which the solution is based..
1.as the buses followed opposite route on the circular path,i.e road,thus the whole path from terminus to again back to terminus may be visualised as a circle, this makes reaching to the solution easy.
2. consider two points first, the terminus and the second one, the point at a distance of 6 km. from the terminus i.e. at the distance of 5 km on the circular path from its beginning.
3. at time 0,say the zero hour both the buses will be at the terminus, and then thereafter every 10 minutes will switch over their positions.
i mean to say that if we name these two point as terminus & mid-point,then at 10th
minute bus A will be at mid-point, while bus B will at the terminus
then at the 20th minute bus A will be at the terminus while bus B will at the mid-point, and so on...

the solution i.e. the X for which f(x) i.e. waiting time will be maximum can be formualised as follows:
x should be solution of
max [min (12-x,18-x)] , for x>6
or max [min (12-x,6-x)] , for x<6

3. be clear that x is the distance from the terminus measured clockwise.

thus the solution will be , x limits to 0 (i.e.zero)

and the max waiting time = (6-x).5/3 min.
and that equals to almost 6 minutes.

In summary max journey time corresponds to max waiting time
and so max f(x) is for x limits to 0,i.e. x--> 0 (x tends to zero).
so ans is the alternative (5)--none

in short it is the case of just missing the bus but not to worry the next bus for journey is atmost after 6 munites.