official quant thread for cat08

[parmindersaluja]

Originally Posted by IIM maniac

@parmindersaluja,Can you please comment on this?

yup, since in this ques no other condition has been given, so its safe to take the extreme cases directly in order to save time....however, there may be ques where a small addition of a condition such as x>=0, might change the approach..
regards

[stalwart_itsme]

Originally Posted by parmindersaluja

because |x+y| + |x-y| = 4, can take values such as (4,0), (0,4), (3,1), (1,3), (2,2)...and their negative counterparts...
now, solving for (x,y) for the extreme cases (4,0), (0,4), (-4,0) and (0,-4) will give u values of x and y which will form a square encompassing all other values.

P.S. if u still have any doubt, dont hesitate to ask.

dude .... i think these values (4,0), (0,4), (3,1), (1,3) does not satify the equation...
its out of range...correct me if i m wrong....

[IIM maniac]

Originally Posted by parmindersaluja

yup, since in this ques no other condition has been given, so its safe to take the extreme cases directly in order to save time....however, there may be ques where a small addition of a condition such as x>=0, might change the approach..
regards

okie, thanks a lot for the explainaion, now it is clear to me.Regards.

[parmindersaluja]

Originally Posted by stalwart_itsme

dude .... i think these values (4,0), (0,4), (3,1), (1,3) does not satify the equation...
its out of range...correct me if i m wrong....

by (4,0) here i mean that x+y=4 and x-y=0.....now these two will give a pair of (x,y) and similarly the others

wat say???

[IIM maniac]

Originally Posted by parmindersaluja

by (4,0) here i mean that x+y=4 and x-y=0.....now these two will give a pair of (x,y) and similarly the others

wat say???

Exactly, i am agree with parmindersaluja.only thing is we should look for the case that covers all possible paris of co-ordinates.correct me if i am wrong.

[stalwart_itsme]

Puys solve this sitter.....

In how many possible ways can two distinct numbers be chosen randomly from the set of first fifteen natural numbers such that the sum of the two chosen numbers is divisible by three but not by four.
1. 24 2. 25 3. 26 4. 27 5. 28

[parmindersaluja]

Originally Posted by stalwart_itsme

Puys solve this sitter.....

In how many possible ways can two distinct numbers be chosen randomly from the set of first fifteen natural numbers such that the sum of the two chosen numbers is divisible by three but not by four.
1. 24 2. 25 3. 26 4. 27 5. 28

the minimum sum of the given numbers is 3 and maximum sum is 29.
Now we can have
3---possible in 1 way
6---2 ways
9---4 ways
12--5 ways
15--7 ways
18--5 ways
21--4 ways
24--2 ways
27--1 way

out of this eliminate the case of 12 and 24 as they are divisible by 4, so 24 is the answer...

is their any shorcut for this????except for the fact that the series will be a palindrome.

[IIM maniac]

Originally Posted by parmindersaluja

the minimum sum of the given numbers is 3 and maximum sum is 29.
Now we can have
3---possible in 1 way
6---2 ways
9---4 ways
12--5 ways
15--7 ways
18--5 ways
21--4 ways
24--2 ways
27--1 way

out of this eliminate the case of 12 and 24 as they are divisible by 4, so 24 is the answer...

is their any shorcut for this????except for the fact that the series will be a palindrome.

I think it is possible in 27 ways excluding 12 & 24 as they are divisible by 4.
3 in 1 way
6 in 2 ways
9 in 4 ways
12 excluded
15 in 7 ways

18 in 6 ways (15,3)(14,4)(13,5)(12,6)(11,7)(10,eight)
21 in 5 ways (15,6)(14,7)(13, eight) (12,9)(11,10)
24 excluded
27 in 2 way (15,12)(14,13)

[parmindersaluja]

yup, thnx for correcting me....

[stalwart_itsme]

Originally Posted by IIM maniac

I think it is possible in 27 ways ........

Yup.... its correct.....
Enjoy solving this too...

#)If the sum of the product of three positive real numbers taking two at a time is 11, then
what is the maximum possible value of the product of these three numbers?
1. 11/3
2 6^(3/2).
3 (11/3)^(3/2)
4 (11/2)^(4/3)
5. None of these

#)In how many ways 3 identical books may be distributed among 5 students such that no student
gets more than 2 books?
1. 125 2. 64 3. 48 4. 40 5. 30

Puys post some more questions...