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

[siyar]

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



wel i think the answer is 10 kms(provided we cannot take any decimel values here!)..
if the man is at a distance of 10kms..he has to wait for 10 mins for bus B to arrive..and..den bus b will travel for extra 10kms..
so..that is d maximum journey time!

[Aarav]

The bus takes 20 mins for 12 km or 5/3 min/km. For x <= 1, the best strategy is to wait up to 10 minutes for a returning bus, so f(x) = 10 + 5x/3. Similarly for 11 <= x <= 12, f(x) = f(12-x).
For 1 < x <= 6, the worst case is that he just misses the right bus, so that the wrong bus comes 10 minutes later and the right bus 10 minutes after that. So he can wait 10 minutes and then travel 12-x km for a total time of 10 + 5(12 - x)/3 = 30 - 5x/3 or he can wait 20 minutes and then travel x km for a total time of 20 + 5x/3. The first is better for x >= 3. So, summarising:


f(x) = 10 + 5x/3 for 0 ≤ x <= 1
20 + 5x/3 for 1 < x <= 3
30 - 5x/3 for 3 < x <= 6
10 + 5x/3 for 6 < x <= 9
40 - 5x/3 for 9 < x < 11
30 - 5x/3 for x ≥ 11

The maximum value of 25 minutes occurs at x = 3 and x = 9.

[randhir8]

hi
i were at the wrong side in my previous mail becuse of the wrong interpretation of the q. no. 61,as i took it as man is waiting at the terminus while any bus (A or B) at the distance x.
1.for any canculating person jouney distance in the track should not be greater than 12+6=18 km.
2. max journey time -> max journey distance, comes in the case of just missing the optimum directional bus,i.e for x<6 it is the bus B (anticlock wala)
and for X>6 it is the bus A (clockwise wala)
3. for x >6,
for x=11 minimum j.distance has to be min (1+12,6+11)=13
explanation--with bus A one have to travel at the most 13 km & with the bus B it would be 17 km,case is that one just misses the bus A
similarly for x=10,min j.d.=min(2+12,6+10)=14
for x=9,min j.d.=min (3+12,6+9)=15

for x=8,min j.d.=min(4+12.6+=14, and so on

4.For X<6
for x=5,min j.d.=min(6+7,5+12)=13--- always the first expression whithin the parentheses is for bus A & the second is for bus B

similarly for x=4, min j.d.=14
for x=3, min j.d.=15--Min(6+9,3+12)=15
and thereafter goes on decreasing.

thus maximum journey time corresponds to x=3 or x=9. and max j.d=15,max j.t.=15.5/3=25 min.

corresponding minimum goes with x=6.

[tg8]

hey thanks IOTA .... i got it later ....
bt thanx a lot dude...

[randhir8]

hi aarav
i am just impressed with your formulation for solution of the prob. 61
however the answer as found is right ,but i think some refinement on the part of whole system of equation is needed, just consider the case---
let x= 0.9 & x= 1.1, obviosly in both the cases the same bus ride would be advisable

now calculate the corresponding journey time
for x=0.9 it equals to 10+4.5/3 =11.5
and for x=1.1, it gives 20 + 5.5/3=21.8

similar may be on the other extreme,i.e.x=11

probably i have not annoyed you!
thanks...

[madadi]

Originally Posted by vivekma

madadi, How will he spend the 10 minutes traveling when the bus is at the terminus? I agree with ur latter half of the logic though.

Hey vivekma.... just have a nice look at the question... it talks about total time taken 'during the journey' n not while 'travelling'...! when you stand at a bus depot, n wait for a bus, let's say for 10 mins, as is in this case, n you take 30 mins travelling in the bus, so the total time taken for your journey would be 30+10= 40 min... n not 30 mins alone...
hope this clears your doubt...

[vikas_mba_iim]

hi u all !
I am a new member of pg!
My solution is :
assuming that he does not spends time waiting for bus.
maximum time will be if the passenger gets bus B just as he steps down the bus A.
so,
the speed of the bus is 36km/hr.
so,
time required by bus A to reach x=time required by bus B to reach that point
=> x/36=10/60+(12-x)/36
=> x=9
so,according to me x=9

[astute10]

hey my method is more of visulisation

If you consider whole route its basically a circle with radius. Centre is the terminus

Q says where should the man stand where it takes max time for either of the bus to reach.

in first 10min A will travel 12/2=6 KM from and will 5 km alon the circle.

In next 5 min both B and A will travel 3 km and meet finally ie 9 km from terminus in clockwise direction ie the route of the bus

So we can see 9 km from the termius is the last point where both reach therefore ans is 9

[bleach it out]

This problem was from old QQAD files (2006), problems unused, and it took me some time to write a good solution. It requires some effort to look for Invariant i.e. a quantity that doesn't change.

The key word was "any successive k" -> Let us denote our vertices as
A1, A2, A3, ..., A12 and let -1 be at A1. Let the 4 successive vertices be A1, A4, A7, A10.
The product of the numbers on these vertices is -1 to start with (if we take -1 at A1), and if we are changing the sign of each of A1, A4, A7, A10 then also the product should remain -1. But when we say that we want to shift
-1 to an adjacent vertex, then product on nodes A1, A4, A7, A10 will be 1, which is not possible.

Similarly, we can argue for k = 6 -> Take successive vertices as A1, A3, A5, A7, A9, A11.

For k = 3, we need even number of touches in 10 nodes and odd number of touches in 2 nodes to make our case happen. But, this means in all we require even number of touches, or in other words we should perform our operation of any 3 successive changes even number of times which in effect means we are back at k = 6, which we proved is not possible.

sorry for being late...but m not getting that "key word" thing!!
any successive k should mean a1,a2,a3...y r u taking a1,a4,a7...

plz clarify...

[bleach it out]

my interpretaion of question no. 61,
if we consider the entire route to be a circular one and the terminus as a point somewhere in the circumference
speed of bus A can be calculated by (12/20)= 0.6km/min
and bus B leaves teminus 10 mins after A..that means when B is about to start A must have covered (.6*10)= 6kms...since there is one more condition that "it travels a distance 1 km to a circular road of length 10 km and goes clockwise around the road"
therefore total distance of A from the terminus when B is about to start is 4kms.
assuming same speed of both A and B...the most suited option is (1)
as the distance of terminus from both A and B would be 3kms
and distance between A and B being 4kms.

m i right?