official quant thread for cat08

[prakharc]

An easy logical question:-

Once there was an immensely rich hotelier he had a whim and he builds a hotel with infinite rooms. Now there is a maths conference and infinite mathematician from all over the world attend this conference. All infinite mathematicians are accomodated in infinite rooms of that hotel.

1) One mathematician from America arrives late and goes to hotel manager for a room. How will the manager arrange a room for him and what's his room number??

2) Late as usual, another infinite mathematicians from India arrive a day later. How will the hotel manager arrange room for all Indian mathematician ??

[bishweshwar1234]

plz solve this

(1) Probability of a man taking one step towards his left is 2/5 and that on to his right is 3/5. what is the probability that at the end of a 9 such steps,he has moved a net distance of 1 step ?
(a) (5/6)4 (b) (6/25)4 (c) 126(6/25)5 (d) 120(6/25)4 (e) 126(6/25)4

[prakharc]

Originally Posted by bishweshwar1234

plz solve this

(1) Probability of a man taking one step towards his left is 2/5 and that on to his right is 3/5. what is the probability that at the end of a 9 such steps,he has moved a net distance of 1 step ?
(a) (5/6)4 (b) (6/25)4 (c) 126(6/25)5 (d) 120(6/25)4 (e) 126(6/25)4

Ans) e.... 126[(6/25)^4]

he'll be one step on either side if on one side he takes 4 step and other side 5 steps

So probability= 9C4 (2/5)^4 (3/5)^5 + 9C5 (2/5)^5 * (3/5)^4 = 9C4 (6/25)^4= 126(6/25)^4

[ankaj]

Originally Posted by prakharc

An easy logical question:-

Once there was an immensely rich hotelier he had a whim and he builds a hotel with infinite rooms. Now there is a maths conference and infinite mathematician from all over the world attend this conference. All infinite mathematicians are accomodated in infinite rooms of that hotel.

1) One mathematician from America arrives late and goes to hotel manager for a room. How will the manager arrange a room for him and what's his room number??

2) Late as usual, another infinite mathematicians from India arrive a day later. How will the hotel manager arrange room for all Indian mathematician ??

if we consider the hotel to be circular...not sure if that was you were looking for....

[prakharc]

Originally Posted by ankaj

if we consider the hotel to be circular...not sure if that was you were looking for....

No...hotel is straight....

though i think circular hotel with infinite radius is straight line only....

problem is...... if infinite rooms are occupied you cant ask a new person to go to infinite room that's actually no where...so how to allot him a room is the problem...

[bishweshwar1234]

Originally Posted by prakharc

Ans) e.... 126[(6/25)^4]

he'll be one step on either side if on one side he takes 4 step and other side 5 steps

So probability= 9C4 (2/5)^4 (3/5)^5 + 9C5 (2/5)^5 * (3/5)^4 = 9C4 (6/25)^4= 126(6/25)^4

One help I need, in probability how we can apply binomial probabilty. If you have any good link for probability, please send.

[prakharc]

Originally Posted by bishweshwar1234

One help I need, in probability how we can apply binomial probabilty. If you have any good link for probability, please send.

I don't have any link....remember it from my engineering times....

I can explain a bit though....

Let's say p is probability of success and q is probability of failure. And at each point probability remains same...i.e not affected by what happened earlier..for eg tossing a coin 1/2 is fixed etc.

then we can get probability by chosing particular term from binomial expansion of (p+q)^n

To choose that particular term:-
n=9 i.e. total number of events.....let's take moving left as success=p=2/5 and moving right as failure=q=3/5 ...now for one step left we need 5 success and 4 failures....so we need the term where power of p is 5 and q is 4 from binomial expansion
i.e. nC5 p^5 q^4 ...... and similarly if there are 4 success and 5 failures we choose the term where power of p is 4......

and then add these two to get resultant probability

[masoom]

Easy but a bit conceptual questions..

1. How many ordered pairs (p,q) are there such that the unit's digit of p^p and q^q are same...

provided p,q are natural numbers less than 10..

2. Given that a^5+b^5+c^5=91849... where a , b, c are distinct digits..

if 2a5b1c is divided by 11 what is the remainder..

[mrholmes]

Hi Puys,

this may be headache...but can anyone post sectional and overall cutoff marks for CAT 2000 to CAT 2007.
wanna check if i am anywhere near the elites .

this wud be a grt help...

also, was there ny paper in which the negative marks were 1/3


regards,
holmes

[HYgoldking]

2. Given that a^5+b^5+c^5=91849... where a , b, c are distinct digits..

if 2a5b1c is divided by 11 what is the remainder

9,8,2 is satisifying the condition.

(9+8+2)-(2+5+1) = 11

So ans is 0 ??