Official Quant thread for CAT 2013

MBA CAT 2024
29686
@ScareCrow28 said:
I am not sure of the range values here in the question, I am assuming it to be excluding the extremities. Again in the second bold part it may be open brackets..so 0 factors in that range. Not sure though.
haan wahi....
no one has answered yet on Bodhi_vriksha..
mujhe to question mein kuch galat lag rha h
29687
@Subhashdec2 said:
since it is a general question i think we can take any value of n and solvelets take n=22^2013factors in the range of (1,2013)=2^0,2^1.....2^10=11 factorsin the range of [2^2012,2^2013]=2 factos only11:2no OA is matching here
yes exactly n can be a prime number or a composite number in each case oa will vary, isnt????
29688
@heylady said:
what is N=13?
when did i say N=13??
29689
@abhishek.2011 said:
yes exactly n can be a prime number or a composite number in each case oa will vary, isnt????
still whatever it is.... it is a general question so we should have the flexibility to chose N whatever we want.


aage pta ni yaar....
ans aane do kisika tabhi pta chalega
29690
@abhishek.2011 said:
yes exactly n can be a prime number or a composite number in each case oa will vary, isnt????
Actually for prime nos...(N²º¹², N²º¹³) would have 0 factors..so can't take that may be
I tried to check it for N=6 but its getting a bit lengthy
29691
@Subhashdec2 said:
when did i say N=13??
srry wrote is by mistake....I was supposed to write if....though nothing is mentioned explicitly in the question lets try with different funda,,
29692
@ScareCrow28 said:
Actually for prime nos...(N²º¹², N²º¹³) would have 0 factors..so can't take that may be I tried to check it for N=6 but its getting a bit lengthy
ya 4 prime factors r zero but lets say k=n^2013 n is prime but k is not???? as this 1 is a general question so there should be a provision to take n what we want or should have specified??
29693
@ScareCrow28 said:
Actually for prime nos...(N²º¹², N²º¹³) would have 0 factors..so can't take that may be I tried to check it for N=6 but its getting a bit lengthy
take n=4 den bhai
tab b 10:1 aayega....:P
29694
4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
29695
@Subhashdec2 said:
take n=4 den bhaitab b 10:1 aayega....
Arey..n=4 to woi baat hogai na 😛 I meant to say composite nos.
29696
@abhishek.2011 said:
ya 4 prime factors r zero but lets say k=n^2013 n is prime but k is not???? as this 1 is a general question so there should be a provision to take n what we want or should have specified??
Yeah, Just wanted to confirm that only. If it goes unsolved here, then there should be some flaw in the question OR we all are missing something. I would leave this Q for now.. :)
29697
@abhishek.2011 said:
Find the least number of all equal weight pieces which can be cut from three cakes of weights; 484g, 308g and 253g.A. 90B. 95C. 100D. 105
95 ?

hcf(484,308,253) = 11
484/11 + 308/11 + 253/11 = 44 + 28 + 23 = 95
29698
@Buck.up said:
4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
432?
29699
@ScareCrow28 said:
Arey..n=4 to woi baat hogai na I meant to say composite nos.
bhai 4 composite hee to h...

tab 4^2012 to 4^2013 mein ek factor aa jaega 2^4025 wala...:D
29700
@Buck.up said:
4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
Arrange 4 men in 3! ways = 6 ways
Now arrange 4 women..I am trying it manually.. = 2*2 = 4 ways
So total = 6*4 = 24 ways?

Tried once again
4 women can be arranged in only 2 ways

So total of 12 ways ..!! ?? :splat:
29701
@Buck.up said:
4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
12 ways??
29702
@Subhashdec2 said:
432?

@ScareCrow28 said:
Arrange 4 men in 3! ways = 6 waysNow arrange 4 women..I am trying it manually.. = 2*2 = 4 waysSo total = 6*4 = 24 ways?
Yes. I think. This is right.Women arrangement should be 4 ways ? But see chill saar solution( Post no #256). Even spectramind got same 12.
http://www.pagalguy.com/forums/quantitative-ability-and-di/permutations-and-combinations-t-4826/p-91145?page=13
Are we doing some wrong ?
29703
@Buck.up said:
4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
last to last years cl proc moc question

3!*2*2=24
29704
@Buck.up said:
Yes. I think. This is right.Women arrangement should be 4 ways ? But see chill saar solution( Post no #256). Even spectramind got same 12.http://www.pagalguy.com/forums/quantitative-ability-and-di/permutations-and-combinations-t-4826/p-91145?page=13Are we doing some wrong ?
No I got my mistake It should be 12 ways only..women can be arranged in 2 ways..I checked once again. Please check and confirm
29705
@ScareCrow28 said:
Arrange 4 men in 3! ways = 6 waysNow arrange 4 women..I am trying it manually.. = 2*2 = 4 waysSo total = 6*4 = 24 ways?
bhai men ka arrangement 3!*2 ways mein nahi hoga?