Official Quant thread for CAT 2013

MBA CAT 2024
13163
@somnathbhatta said:
please provide a less calculation intensive method for the attached question. don't know the answer
From A to B=4C2*6C2=90
From B to C=6C1*2C1+1=13

Total no. of ways=1170
13164
@somnathbhatta said:
please provide a less calculation intensive method for the attached question. don't know the answer
1. 90 2. 1170

No of paths possible is given by (total no of lines)!/(Verticle line count)!(horizantal line count)!
Note: Count only horizantal, vertical lines and add them up for total no of lines

1. Here P is shortest path as it is diagonal of small rectangle in the path. Lets denote ends as X and Y.
No of paths from A to X is 4!/2!2! = 6
Diagonal = 1 path
No. of paths from Y to B is 6!/2!4! =15
Total =6*1*15 =90 ways

2.||ar to 1st one 90*13 =1170 ways
13165
@ravi.theja said:
How many numbers less than 1000 are there such that such that when divided by 2, 3, 5, 7 and 11 they leave remainders 1, 1, 2, 0 and 8 respectively.a) 1b) 2c) 3d) 4
Is it only 1, ie, 217??
Since the number when divided by 7 leaves a remainder of 0, hence the very first number till here wld be 7 only
LCM(2,3,5,7)x + 7 = 11y + 8
210x + 7 = 11y + 8
this gives only one number
13166
@pirateiim478 for the 2 .. isn't the answer 630 .. 90*7 .
i could only find 7 shortest paths from B to C.
13167
@ravi.theja said:
How many numbers less than 1000 are there such that such that when divided by 2, 3, 5, 7 and 11 they leave remainders 1, 1, 2, 0 and 8 respectively.a) 1b) 2c) 3d) 4
By applying CRT any number of this form shld be of the form 2310k+217

So, for less than 1000, there is only one number.


How many zeros are at the end of 13!+14!+15!+16!+17!+18!
13168
@ravi.theja said:
How many numbers less than 1000 are there such that such that when divided by 2, 3, 5, 7 and 11 they leave remainders 1, 1, 2, 0 and 8 respectively.a) 1b) 2c) 3d) 4
with 6 remainder is 1, with 5 remainder is 2, with 7 it is 0 and 11 it is 8...
1st such no. fulfilling 1st 2 conditions is, 37, next comes 67, next 97,... and so on...
no. is of the form 30k + 7 .. now for remainder 0 with 7 , 1st such no. is 217, which leaves remainder 8 with 11, 2nd is 427 which leaves 2, 3rd is 637 which leaves 10, 4th 1 is 847, which leaves 0, and next 1 is more than 1000...
so only 1 number exists that is 217.
13169
@pirateiim478 said:
1. 90 2. 1170No of paths possible is given by (total no of lines)!/(Verticle line count)!(horizantal line count)!Note: Count only horizantal, vertical lines and add them up for total no of lines1. Here P is shortest path as it is diagonal of small rectangle in the path. Lets denote ends as X and Y.No of paths from A to X is 4!/2!2! = 6Diagonal = 1 pathNo. of paths from Y to B is 6!/2!4! =15Total =6*1*15 =90 ways2.||ar to 1st one 90*13 =1170 ways
2.||ar to 1st one 90*13 =1170 ways
how is it 13??


13170
@ravi.theja said:
How many numbers less than 1000 are there such that such that when divided by 2, 3, 5, 7 and 11 they leave remainders 1, 1, 2, 0 and 8 respectively.a) 1b) 2c) 3d) 4
a) 1.
217

what is OA?
13171
@bs0409 said:
By applying CRT any number of this form shld be of the form 2310k+217So, for less than 1000, there is only one number.How many zeros are at the end of 13!+14!+15!+16!+17!+18!
13! (1 + 14 + 15 .. 18) = 13!*something ending with 1...
we are concerned only with 13 now..
hence 2 zeros?
13172
@bs0409 said:
By applying CRT any number of this form shld be of the form 2310k+217So, for less than 1000, there is only one number.How many zeros are at the end of 13!+14!+15!+16!+17!+18!
3 hoga kya ?
13173
@pratskool said:
13! (1 + 14 + 15 .. 18) = 13!*something ending with 1...we are concerned only with 13 now..hence 2 zeros?
13! (1 + 14 + 14*15 .. 14*15*16*17*18)
13174
@pratskool said:
13! (1 + 14 + 15 .. 18) = 13!*something ending with 1...we are concerned only with 13 now..hence 2 zeros?
It shld be 13!(1+14+14*15+14*15*16+14*15*16*17+14*15*16*17*18)
13175
@somnathbhatta said:
@pirateiim478 for the 2 .. isn't the answer 630 .. 90*7 .i could only find 7 shortest paths from B to C.
@joyjitpal said:
2.||ar to 1st one 90*13 =1170 wayshow is it 13??
It is 90*13
Here rectangle with D is restricting Neelam. So wee need to calculate no of ways from B to Z(Let Z be upper right end of smaller rectangle with D) and then from Z to C So it will be 6!/1!5! * 2!/1!1! => 6*2 =12. But there can path covering only outer path which will not be included in this So 12+1 =13 ways. Total 90*13=1170 ways
13176
@bs0409 said:
It shld be 13!(1+14+14*15+14*15*16+14*15*16*17+14*15*16*17*18)
yeah 1 more 0 getting from the term (as it ends with 5 and there are extra 2's in 13!) inside the bracket hence 3,.... loosing my head since the results....
13177
@nikemen said:
a) 1. 217what is OA?
OA : 1 😃 number is 217 !! perfect
13178
@pirateiim478 sorry dude .. but i'm still not getting the count .. i still count only 7 shortest paths. can u upload a pic of ur rough sheet .. that'll be helpful . thnx in advance
13179

P is a natural number. 2P is having 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?


Options :35,40,45,48,NOT
13180
@pirateiim478 said:
P is a natural number. 2P is having 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?Options :35,40,45,48,NOT

Should be 35??
P = 2^5*3^3

2P = 2^6*3^3 -> 28 divisors
3P = 2^5*3^4 -> 30 divisors
=> 6P = 2^6*3^4 -> 35 divisors
13181
@pirateiim478 said:
P is a natural number. 2P is having 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?Options :35,40,45,48,NOT
P=3^3*2^5

6P=3^4*2^6

6P has 35 divisors

13182
@somnathbhatta said:
@pirateiim478 sorry dude .. but i'm still not getting the count .. i still count only 7 shortest paths. can u upload a pic of ur rough sheet .. that'll be helpful . thnx in advance
Find attached the image. Here rectangle with D is restricting the path of Neelam. SO we need to apply formulas 2 times as shown and then add the outer path

P.S: Ignore my bad handwriting :D