Official Quant thread for CAT 2013

logrhythm · January 11, 2013, 7:33pm

@naga25french said:
@Logrhythm@pyashrajRight N = 444(upto 50 digits) except for the nth digit(where n ranges from 1-50). If N is divisible by 13 for nth digit, then find how many value can n take ?

did not understand the question

4444.... this is repeated 50 times? then what is nth here?

or is it 4444...upto (n-1) times and then the nth digit varies from 1 to 50?

pls help me understand. thanks

naga25french · January 11, 2013, 7:37pm

@Logrhythm said:
did not understand the question4444.... this is repeated 50 times? then what is nth here?or is it 4444...upto (n-1) times and then the nth digit varies from 1 to 50? pls help me understand. thanks

its something like this .. you have 50 digit number for which every digit except n th digit is 4 .. so find n if number is divisible by 13

logrhythm · January 11, 2013, 7:41pm

@naga25french said:
its something like this .. you have 50 digit number for which every digit except n th digit is 4 .. so find n if number is divisible by 13

sorry but still not able to get it....can you give one example? thanks

naga25french · January 11, 2013, 7:46pm

@Logrhythm said:
sorry but still not able to get it....can you give one example? thanks

sample numbers would be

n44444444(4 repeating 49 times)
4n4444444(4 repeating 48 times)
44x444444(4 repeating 47 times)

OA answer is 24

uditultimate · January 11, 2013, 7:47pm

any one post some fresh meal..

logrhythm · January 11, 2013, 7:48pm

@naga25french

1 digit - not possible

2 digit - not possible

3 dgit - 44x -> x can be 2

4 digit - 444x -> x can be 6

5 digit - 4444x -> x can be 7

is this is how this can be done. i am afraid this would be too arduous, any other approach?

grvsngh · January 11, 2013, 7:49pm

udit saini
82 82 82 82
iss se fresh nhi hai..

logrhythm · January 11, 2013, 7:49pm

@naga25french said:
sample numbers would ben44444444(4 repeating 49 times)4n4444444(4 repeating 48 times)44x444444(4 repeating 47 times)OA answer is 24

ok, how to solve it then?

pyashraj · January 11, 2013, 8:19pm

@naga25french

@Logrhythm

N = 444(upto 50 digits) except for the nth digit(where n ranges from 1-50). If N is divisible by 13 for nth digit, then find how many value can n take ?

Should Indeed be 24..

n=1, N=4, Rem[N/13] = 4

n=2, N=44, Rem[N/13]= 5

n=3, N=444, Rem[N/13]= 2, but if N=442, Its Divisible..

n=4, N=4444, Rem[N/13]=11, but if N=4446, Its Divisible..

n=5, N=44444, Rem[N/13]= 10, but if N=44447, Its Divisible..

n=6, N=444444, Rem[N/13]= 0..Hence N is itslf divisible..

n=7, N=4444444, Rem[N/13]= 4...and the cycle gets repeated..

Thus, for n= 1,2,3,4,5,6 ..there r 3 values at which this is satisfied..at n=3,4 n 5..

For, n=7,8,9,10,11,12..there r 3 values at which this is satifisfied..at n= 9,10 n 11..

.

For n = 43, 44, 45, 46, 47, 48..There r 3 values at which this is satisfied..at n= 45, 46 n 47..

For n=49 n 50..Its not satisfied..

Thus, a total of 3*8 = 24 values are there of n(1-50), at which the conditions r met..

Screwd the 1st tym..

pratskool · January 11, 2013, 9:08pm

@pirateiim478 said:
Last one :A fair coin is tossed 10 times . Find the probability that two heads do not occur consecutively ??(A) 1/(2^4)(B) 1/(2^3)(C) 1/(2^5)(D) noneGood night @rayus

1/2^4 ???

the arrangement can be _T_T_T_T_T

or T_T_T_T_T_

u can place either head or tail in the spaces, 5 spaces 2 arranegements hence, 2*(2^5) = 2^6

TOTAL no. of arrangements is 2^10...

hence 1/2^4

logrhythm · January 11, 2013, 9:23pm

@pratskool said:
1/2^4 ???the arrangement can be _T_T_T_T_Tor T_T_T_T_T_u can place either head or tail in the spaces, 5 spaces 2 arranegements hence, 2*(2^5) = 2^6TOTAL no. of arrangements is 2^10...hence 1/2^4

There is no restriction on the number of heads or tails that are appearing, pls refer to my post....i guess that is how it needs to be done.

pratskool · January 11, 2013, 9:26pm

@Logrhythm said:
There is no restriction on the number of heads or tails that are appearing, pls refer to my post....i guess that is how it needs to be done.

yeah.... realised it now.... arrangement can be like HTTTHT... also.....

i am out of sync of cat prep, need to warm up a little... pardon me

naga25french · January 12, 2013, 3:45am

@pyashraj 24 is right 😃 method is always tedious for these kind of problems 😃

jaituteja · January 12, 2013, 4:03am

@pratskool The question states that "Find the probability that two heads do not occur consecutively ??"

will we consider only two heads consecutively or three/four/five heads consecutively...

Since three and four already comprises of two heads consecutively case...

Please through some light...

rayus · January 12, 2013, 4:11am

@naga25french said:
Consider a regular polygon of n sides .The number of values of n for which the polygon will have angles whose degree values is an integer ?

( factors of 360 ) -2

= 22

cadmium · January 12, 2013, 4:27am

Simplify:

839478 * 625

jaituteja · January 12, 2013, 4:34am

@cadmium 839478 *625

(840000-522) * 625

840000*625 - 522*625

= 524673750

ravi.theja · January 12, 2013, 4:36am

@pirateiim478 said:
In how many ways 3 red balls and 4 blue balls can be arranged in 6 different boxes?

a + b + c + d + e + f = 3
so, 8C5 = 56 ways
a + b + c + d + e + f = 4
so, 9C5 = 126 ways

Total = 56*126 ways

ravi.theja · January 12, 2013, 4:40am

Find the last non-zero digit in 37!

x2maverickc · January 12, 2013, 4:51am

@vbhvgupta said:
Which of the followinf is not a perfect square100856, 325137, 945729? how to find quickly?

sure B isnt one as it ends with 7. Also the digital root of A is 2 😉 It shouldnt also be a perfect square 😛