hey abhi

wats the answer to the P&C; problem u posted earlier..............the boat one??

hi
that was a good solution.

anyway can u tell me hwo is R.S.Agarwal for quant

If u r preparing for CAT i would suggestyou to go for Arun Sharma only ..coz R.S Agarwal is for basic level esp for SSC type or Bank PO type xams......:)

And u dont have to go for so much basics if u r giving CAT this year.

hi
that was a good solution.

anyway can u tell me hwo is R.S.Agarwal for quant

I don't have much idea about RS Agarwal. But from what I have heard from others, it's as good as Arun Sharma. I would suggest, stick to one book for basics and later you can go through any number of questions from wherever you feel like.....

hi
that was a good solution.

anyway can u tell me hwo is R.S.Agarwal for quant

i hav a queston in number systems...
ders no solution 4 it in arun sharma...
find the no. of numbers between 100 to 400 which r divisible by either 2,3,5 and 7...


thanks in advanc.............

Numbers divisible by 2 are 102,104,.........,398
Total Number of numbers divisible by 2
398 = 102 + (n-1)2
n = 149
Numbers divisible by 3 are 105,111,.........,399 (excluding the even numbers)
Total Number of numbers divisible by 3 and not by 2 (as 2 has already been included)
399 = 105 + (n-1)3
n = 50
Numbers divisible by 5(and not by 2 and 3) are 115,125,145,155,175,185,205,215,235,245.........,395
n= 4 x 6 = 20
Numbers divisible by 7(and not by 2,3,5) are 119,133,161..............,385
n = 11

Ans = 149 + 50 + 24 + 12 = 230

Guys, let me know if the answer is incorrect. however, this is a very crude way of solving it. Plz post any other method for solving such questions.

i hav a queston in number systems...
ders no solution 4 it in arun sharma...
find the no. of numbers between 100 to 400 which r divisible by either 2,3,5 and 7...


thanks in advanc.............

one suggestion to all ..
Please post the question instead of page number etc., coz of most of us will be online @ office where we don't have book to see .. :)
flood the thread with questions ... we can have a healthy discussion ..

:snipersm:The crew of an 8 member rowing committee is to be chosen from 12 men, of which 3 can row on one side only and 2 can row on the other side only. Find the number of ways of arranging the crew with 4 member on each side.

a) 40,320 b) 30,240 c) 60,480 d) 10,080

Thanks !!! The answer is correct.But could you plz explain, how exactly you got '30' as the number of ways of arranging the variants..

cheers!!

hi ..Ans according to me is ( d )

Coz 30 ways of arranging the variants!
12C6 ways of picking the students and 6! x 6! the way in way inwhich we can arrange them
all together 30x (12C6) x 6! x 6!

Is my ans right dude?

but if the student just behind gets the same variant and so on , the whole row gets the same variant.

cheers
--------------
Get busy living or Get busy dying.

Dude.. the question is straight forward. if you read it carefully, you will understand it. Actually, students sitting in the first row and the one just behind him should get the same variant. For ex-- one of the arrangements can be....
1st row v1 v2 v3 v2 v1 v3
2nd row v1 v2 v3 v2 v1 v3

Hope this answers ur question...

but if the student just behind gets the same variant and so on , the whole row gets the same variant.

cheers
--------------
Get busy living or Get busy dying.

student sitting diagonally behind or behind in the same row (although this shud not b the case) ----not clear
can u pls clarify :)


cheers
-----------------
Get busy living or Get busy dying.

The student sitting just behind the one sitting in the first row.

student sitting diagonally behind or behind in the same row (although this shud not b the case) ----not clear
can u pls clarify :)


cheers
-----------------
Get busy living or Get busy dying.

Hi All,
Here is a question from permutation and combination. its question no 21, chapter 13. I am not getting the correct answer. May be the approach is incorrect.
Quest:
======
Three variants of CAT paper are to be given to 12 students. In how any ways can the students be placed in 2 rows of 6 each so that there should be no identical variants side by side and that the student sitting behind should get the same variant. Find the number of ways it can be done.:neutral:

a) 6!^2 b) 6 x 6! x 6! c) 6!^3 d) 30 x (12 C 6) x 6! x 6!

Pretty good intiative Abhijit...
Arun sharma is very nice book covering almost all the topics and concepts..
we can have a healthy discussion when ever we are stuck with anything..
Keep rocking...

Thanks ppl for the prompt reply and thanks abhi for posting the Q, but i think i was stuck on a diff Q. ...sorry for tat ..........will post it tommo......:)

and i really dont think discussing one chap at a time will work out coz diff persons will b solving diff chaps .
and discussing randomly we can parallely get an idea and fundas of other chaps as well.
wat say?


cheers
-------------
Get busy living or Get busy dying.

Good initiative.
Arun Sharma is a book with very good questions but very bad explanations.
This thread will help us explore the solutions in a better way.

Alos, LOD3 for profit and loss sucks, the problems are simply long with nothing much interesting.
So, it can be skipped ( i wasted one whole day on it grrrrr )

hey!
I have got the question.. . It goes like this

Solve the following inequalities
13. x^2-7x+12 < x-4|:grab:

a. x<2 b. x>4 c. 2<4 d.2<=x<=4


the answer is 'c' 2<4. If you substitute the values given in the option you can easily make out that only option 'c' satisfies the given inequality. there are many questions in quant where you can directly substitute the values and get the result... Cheers!!!1

Hi.. Here's another approach.. Instead of directly substituting which will be time consuming, we can play around with the equations to save some time

x^2-7x+12 < x-4
=(x-3)*(x-4)<|x-4
Makes it simple now..
If we substitute the same we get the answer

Cheerz

Count me in too ...
Good start..:D

hey!
I have got the question.. . It goes like this

Solve the following inequalities
13. x^2-7x+12 < x-4:grab:

a. x<2 b. x>4 c. 2<4 d.2<=x<=4


the answer is 'c' 2<4. If you substitute the values given in the option you can easily make out that only option 'c' satisfies the given inequality. there are many questions in quant where you can directly substitute the values and get the result... Cheers!!!1

Hi,
Could you plz post the question. And one more thing, it would be better if we could discuss one chapter at a time.What say ???

tats a nice suggestion and wud have done tat only but the prob is even im in office and dont really remem the exact equation ..........??:...but will definitely post the Q. tommo

cheers:
---------------
Get busy living or Get busy dying.

im having some probs in solving the inequalities chap of the book
can u give a detailed solution for Q.13 in tat chap for LOD1.

can u quote the question here only.so that ppl shud not go thru book while browsing..........as i m in office...nd cant knw which question is been talked abt.........................



-Chaitz
Just Another PAGAL
Just die for once to know how to live

hi abhijeet nd all,

its a gud thing to bring Arun sharma to r disscussion plate as most of us go thru it.....nd smewhr or the other finds some prob.....in solveing that..........

what i want to add is as for now we should just restrict our scope to LOD 1 and LOD 2 questions or AS,as LOD 3 are too much nd hardly question of such type r asked newhr.................

Chaits
Just Another PAGAL

Kudos!!! to u my frend for bringing up this thread ........

now itll b really helpful if u can solve one of doubts ....:)
im having some probs in solving the inequalities chap of the book
can u give a detailed solution for Q.13 in tat chap for LOD1.

cheers:

--------------
Get busy living or get busy dying.