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

[selebratinglife]

Originally Posted by implex

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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

Total distance any bus travels for a trip = 12kms
Since it takes 20 minutes its speed is 12/20km/min=3/5km/min
Since bus b starts 10 minutes after bus A, the distance that the bus A travels in 10 minutes is 3/5*10=6..So he should just miss the bus A for him to take maximum time for the journey..I'll go with option (5)

[Iota]

Originally Posted by vivekma

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However for X = 10,
Bus B takes 10/3 minutes to reach and bus A takes 50/3 minutes.
But B starts after 10 minutes so effective time when it reaches X =10 is 10+10/3 = 40/3 .

So passenger gets on Bus B and travels a distance of 10 Km in the anti clockwise direction.
.......

F(X) is 10 for X = 10. Since D is directly proportional to T , and speed is constant,

Option (4) is correct.

Oh i hope there are no mistakes! Correct me if i am wrong.

Please consider the following conditions:

If X = 9
the time taken
if board Bus A = 45/3 > 40/3
if board Bus B = 10 + 3/0.6 = 45/3 > 40/3

pls let me knw if my interpretations are wrong.

[killerkart19]

bus A travels 12 km in 20 mins, speed=12/20= .6 km/min
bus B travels at the same speed but leaves the terminus 10 mins late in the meanwhile the bus A has travelled .6*10=6 kms. so for the passenger to maximize his time for his journey he should just miss bus A , 6kms away from the terminus....

so i ll go with option (5) none...

[jumpfrog]

I'll go with option (1) ie 3.


00km |--|--|--||--|--|--| 6km
12km

at any time, the position of the buses are mirror images. So the furthest a bus can be is at the centre when both coincide.

[gunjan-ahuja]

Originally Posted by Iota

Quantitative Question # 061
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My Attempt :

The time when 'B' leaves the terminus 'A' reaches the 12/2 = 6 km which is also the half mark of the circular path(5 km).
from this point both the buses travel with equal speed and in opposite direction. So they meet in the halfway of their in between distance,
i.e (1+5)/2 = 3 Km from terminus in Bus 'B' route.
So The value X should be (12-3) = 9 to make the Time travelled by the person Maximum.
So Answer : 5 (none).

[gunjan-ahuja]

IOTA can u plz explain more how r u doing with options

[vivekma]

Originally Posted by Iota

Please consider the following conditions:

If X = 9
the time taken
if board Bus A = 45/3 > 40/3
if board Bus B = 10 + 3/0.6 = 45/3 > 40/3

pls let me knw if my interpretations are wrong.

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.

[vivekma]

I wonder who yesterday's quant devils are. Guys any idea?

[Noobieman1]

Hey all..first post here

This problem..if we find out where the 2 buses meet, it is at x=3 and x=9.
So, if he misses the bus at either pt, the time taken is (12+3)/.6 = 25 mins. At other points, the buses take lesser than 25 mins(try x<6 , x>6 or x=6..the prev posts have done this i guess). So should be x=3(available option)

[Iota]

Originally Posted by gunjan-ahuja

IOTA can u plz explain more how r u doing with options

The question was to find maximum f(X) for some value of X.
I didn't xplore the options but tried to find the value of X for Max f(x).
the approach was to find the distance when both the buses meet. that is when the time travelled will be maximum for that person. Otherwise the person would opt for the other bus (A/B)
e.g: Suppose that the distance is,
X= 3
then the time for Bus A = (3/0.6) = 5 mins. But for 'B' it is (12-3)/0.6 = 15 mins,
so if X= 3 the person will board bus A.
similarly for ne distance the person will board the bus which takes less time to travel that distance[assumption no one wants to travel for longer period unless s/he is in cruise liner ].
Now we hav to find the distance for which the Time will be Maximum which ever bus the person boards. so I tried to find the distance for which they will meet. and thats the answer.

Hope that i am clear,if otherwise pls let me knw. I will try to come up with some more xplanations.....