The Official CAT 2007 Quant Thread

[harshavardhanab]

Originally Posted by esh.nil

Happy Solving....

1) How many numbers less than 10^5 have the sum of th digits as 10?

2) A survey about Tv viewership was conducted on 100 respondents. The results are,
93 liked sony,89 liked Zee, 81 liked Stratv, 75 liked zee cinema, 78 like Mtv.
1 did not like any of the above.
Find the minimum number of people liking all the the 5 channels.

3) A coin is tossed n times. if the probability of getting exactly 7 heads is less than the probability of getting exactly 4 heads, what is the maximum possible value of n??

For Q1.

We have to find non-negetive integral soln set to the equation,

x1 + x2 + x3 + x4 + x5 = 10 under the condition,

0<=x1,x2,x3,x4,x5<=9.

The solution to which is given by (10+5-1)C(10) - 5 = 996.

Is this correct esh ??

For Q3.

From problem, (7/2) *(n-7)/2 < 4/2 * (n-4)/2

which upon solving gives n<15. Max n = 14.

[esh.nil]

Originally Posted by harshavardhanab

For Q1.

We have to find non-negetive integral soln set to the equation,

x1 + x2 + x3 + x4 + x5 = 10 under the condition,

0<=x1,x2,x3,x4,x5<=9.

The solution to which is given by (10+5-1)C(10) - 5 = 996.

Is this correct esh ??

For Q3.

From problem, (7/2) *(n-7)/2 < 4/2 * (n-4)/2

which upon solving gives n<15. Max n = 14.

For Q1.. it is perfect... Can u explain on how u impose the condition
0<=x1,x2,x3,x4,x5<=9 on x1 + x2 + x3 + x4 + x5 = 10 ??
another soln:
Consider 10 identical balls, this must be divided into 5 grps which can be done using 4 lines, OOOOOOOOOO||||
the no of arrangements possible are 14!/(10!*4!)
out of these, 10 appears in 5 cases which is not acceptable, hence the reqd no of cases = 14!/(10!*4!)- 5 = 996...

For Q3. I dont have any clue how to do that?? also the answer was 11 i guess, but got no clue...

[esh.nil]

Originally Posted by harshavardhanab

For Q3.

From problem, (7/2) *(n-7)/2 < 4/2 * (n-4)/2

which upon solving gives n<15. Max n = 14.

@harsh.. i thought abt this and here goes my approach..
we have n tosses , so totally 2^n cases are possible,
given exactly 7 heads, means there are n-7 tails
so the total cases = n!/{7!*(n-7)!}
similarly, for exaclty four tails,
the number of cases = n!/{4!*(n-4)!}
given n!/{7!*(n-7)!}>n!/{4!*(n-4)!}
or, (n-4)(n-5)(n-6)>7.6.5
n>11....
In the qn the options that are given are 8,9,10,11,12...
so wiill 12 be the answer??? I am not sure...

[harshavardhanab]

Originally Posted by esh.nil

For Q1.. it is perfect... Can u explain on how u impose the condition
0<=x1,x2,x3,x4,x5<=9 on x1 + x2 + x3 + x4 + x5 = 10 ??
another soln:
Consider 10 identical balls, this must be divided into 5 grps which can be done using 4 lines, OOOOOOOOOO||||
the no of arrangements possible are 14!/(10!*4!)
out of these, 10 appears in 5 cases which is not acceptable, hence the reqd no of cases = 14!/(10!*4!)- 5 = 996...

For Q3. I dont have any clue how to do that?? also the answer was 11 i guess, but got no clue...

@esh, we need all nos less than 10^5, i.e., all nos less that or equal to 99999 [ 5 digits] whose digit sum is 10. Each digit can take any of values from 0 to 9 and their sum should be 10. Thats all...

After which i used polynomial multiplication and generating functions to solve the eqn.

Feel free to revert if it is not clear.

For 3rd qsn, I did consider coin to be a fair coin with P(head) = P(tail) = 1/2. If I toss it n times then probability of getting exactly 7 heads is probability of getting 7 heads and n-7 tails = 7/2*(n-7)/2

pls correct if i am wrong

[harshavardhanab]

Originally Posted by esh.nil

@harsh.. i thought abt this and here goes my approach..
we have n tosses , so totally 2^n cases are possible,
given exactly 7 heads, means there are n-7 tails
so the total cases = n!/{7!*(n-7)!}
similarly, for exaclty four tails,
the number of cases = n!/{4!*(n-4)!}
given n!/{7!*(n-7)!}>n!/{4!*(n-4)!}
or, (n-4)(n-5)(n-6)>7.6.5
n>11....
In the qn the options that are given are 8,9,10,11,12...
so wiill 12 be the answer??? I am not sure...

@esh, I did consider what u have posted i.e., nC7 < nC4. It will give only minimum value for n and not maximum value.

[harshavardhanab]

Originally Posted by esh.nil

For Q1.. it is perfect... Can u explain on how u impose the condition
0<=x1,x2,x3,x4,x5<=9 on x1 + x2 + x3 + x4 + x5 = 10 ??
another soln:
Consider 10 identical balls, this must be divided into 5 grps which can be done using 4 lines, OOOOOOOOOO||||
the no of arrangements possible are 14!/(10!*4!)
out of these, 10 appears in 5 cases which is not acceptable, hence the reqd no of cases = 14!/(10!*4!)- 5 = 996...

For Q3. I dont have any clue how to do that?? also the answer was 11 i guess, but got no clue...

@esh... For Q3, I got ans as 11. The eqns are fine, just did calculation mistake. Damn me.

(7/2) * (n-7)/2 < 4/2 *(n-4)/2

7n - 49 < 4n - 16 ==> n<11.

[gautamgomzi]

Q. What is the rightmost digit preceding the zeroes in the value of 20^53 ?
a. 2 b. 8 c. 1 d. 4

Q. What is the remainder when 2(8!) – 21(6!) + 14 (13!) ?
a. 1 b. 7! C. 8! D. 9!

Q. How many integer value of x and y are there such that 4x + 7y = 3 , while mod x < 500 and mod y <500 ?
a. 144 b. 141 c. 143 d. 142


**************GAUTAM*************

[nsrikanth]

Originally Posted by gautamgomzi

Q. What is the rightmost digit preceding the zeroes in the value of 20^53 ?
a. 2 b. 8 c. 1 d. 4

Q. What is the remainder when 2(8!) – 21(6!) + 14 (13!) ?
a. 1 b. 7! C. 8! D. 9!

1.(a) is correct answer.

20^53 = 2^53 * 10^53

Units digit for 2^53 is 2

didn't get second question?

[esh.nil]

Originally Posted by harshavardhanab

@esh, I did consider what u have posted i.e., nC7 < nC4. It will give only minimum value for n and not maximum value.

@harsh.. It is indeed true that 12 gives the minimum possible answer,
but i have some doubt in ur approach,
when there are exaclty 7 heads, then the probability of it's occurance is,
n!/{7!*(n-7)!}/2^n
but how it is 7/2*(n-7)/2??? the probability also looks to be greater than 1.. are u sure about this??

"i used polynomial multiplication and generating functions to solve the eqn." Can u pls comment on this? what is this method, it would be gr8 if u could give a brief desc of this or a link pointing to this..
thanks in advance buddy..

[gautamgomzi]

Q. What is the remainder when 2 (8 ! ) – 21 ( 6 ! ) + 14 (13 ! ) ?
a. 1 b. 7 ! C. 8 ! D. 9 !