Official Quant thread for CAT 2013

MBA CAT 2024
12923
@Zedai said:
@anytomdickandhary Sir, please help me out to solve this question using equations:3 different dies were rolled. find the no. of ways the summation of no will be 12?my approach:x+y+z=12(6-x')+(6-y')+(6-z')=12x'+y'+z'=6no of non negative values: 8C2=28 where am i going wrong with the concept sir?
is the answer 25?

and the problem with your solution is that you have nowhere ensured that value of x', y' and z' cant be 6.. so you have to remove 3 answers in which x, y and z = 6
12924
@bs0409 said:
yes...2nd section=70%ile but it is not shocking since I did not study second section and ought to be punished..........How many four-digit numbers are there with less than 6 different prime factors?(1) 1224 (2) 8476 (3) 9000 (4) 7613 (5) none of these
a similar fate here......94.29 with a respectable 97 in quant and a pathetic 81 in VA.............................btw a great signature and as far as this question goes I am blank.....
12925
@bs0409 said:
yes...2nd section=70%ile but it is not shocking since I did not study second section and ought to be punished..........How many four-digit numbers are there with less than 6 different prime factors?(1) 1224 (2) 8476 (3) 9000 (4) 7613 (5) none of these
The Smallest number with six different prime number is 2*3*5*7*11*13=30030 which is a 5 digit number so every 4 digit number will have less than 6 diff prime number

9*10*10*10= 9000
12926
@cadmium said:
Can anyone please explain me in detail the concept behind this problem? The least number which when divided by 35,45 and 55 leaves a remainder 3 but when divided by 9 leaves no remainder, isA. 1683B. 1677C. 2523D. 3363
options are all wrong. please check. and the best way to solve such a ques will be to check options individually..
12927
@pirateiim478 said:
@krum can you explain how you got 1/7!!
sorry, power cut :splat:

x+2x+3x+4x+5x+6x=1

==> x=1/21


so 3x=1/7
12928
@rayus said:
options are all wrong. please check. and the best way to solve such a ques will be to check options individually..
on second thought.. question is wrong.. for any number to be divisible by 45, then number has to be divisible by 9.. so the (number + 3) can never be divisible by 9 .. so there wud never be any number which would satisfy the conditions given in the question
12929
Triplets Consisting of three different numbers formed from number 1 to 10
HOW MANY SUCH TRIPLETS ARE FORMED THAT SUM OF THE TRIPLETS IS DIVISIBLE BY 3

12930

@rayus

Sorry sir.

My mistake in typing.


Question is:


The least number which when divided by 5,6,7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A. 1683

B.1677

C. 2523

D. 3363


Please elaborate how to solve this kind of problems through options.

Thanks.

12931
@cadmium said:
Can anyone please explain me in detail the concept behind this problem? The least number which when divided by 35,45 and 55 leaves a remainder 3 but when divided by 9 leaves no remainder, isA. 1683B. 1677C. 2523D. 3363
Lcm(35,45,55) = 3465

so number is of the form 3465k+3

acc to ques

(3465k+3)mod9==0

but 3465mod9=0

so no such number is possible
12932
@cadmium said:
@rayus Sorry sir. My mistake in typing.Question is:The least number which when divided by 5,6,7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:A. 1683B.1677C. 2523D. 3363Please elaborate how to solve this kind of problems through options.Thanks.
Lcm(5,6,7,8) = 840

so number is of the form 840k+3

acc to ques

(840k+3)mod9==0

so 840kmod9=6

for k=2

1683



12933
@krum said:
Lcm(35,45,55) = 3465so number is if the form 3465k+3acc to ques(3465k+3)mod9==0but 3465mod9=0so no such number is possible
sorry sir..
It was my mistake in typing. I apologise.
I have corrected myself and re-written the question in my post above.
Please tell me how to solve this kind of problems.
12934
@krum said:
Lcm(5,6,7,8) = 840so number is of the form 840k+3acc to ques(840k+3)mod9==0so 840kmod9=6for k=21683
sir, what is 'mod9' ?
12935

@cadmium said:
@rayus Sorry sir. My mistake in typing.Question is:The least number which when divided by 5,6,7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:A. 1683B.1677C. 2523D. 3363Please elaborate how to solve this kind of problems through options.Thanks.
In given options only one number satisfying the condition i.e 1683 is divisible by 9 and also leaving remainder 3 when divisible by 5,6,7,8.

Another Method
Take LCM of 5,6,7,8 it will be 840 . Now 840 is divisible by 5,6,7,8. To get the remainder add 3 So 843 will leave remainder 3 with 5,6,7,8.

To make it divisible by 9
843 + 840k mod 9
K=1
So ans 1683

For more Detail
Go to this link
http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536/784613
12936
@Amrofa said:
Triplets Consisting of three different numbers formed from number 1 to 10HOW MANY SUCH TRIPLETS ARE FORMED THAT SUM OF THE TRIPLETS IS DIVISIBLE BY 3
24?
12937
@Amrofa said:
In given options only one number satisfying the condition i.e 1683 is divisible by 9 and also leaving remainder 3 when divisible by 5,6,7,8.Another MethodTake LCM of 5,6,7,8 it will be 840 . Now 840 is divisible by 5,6,7,8. To get the remainder add 3 So 843 will leave remainder 3 with 5,6,7,8.To make it divisible by 9843 + 840k/9K=1So ans 1683For more DetailGo to this linkwww.pagalguy.com/forums/quanti...

thanks a ton, bro.

12938
@Zedai said:
@anytomdickandhary Sir, please help me out to solve this question using equations:3 different dies were rolled. find the no. of ways the summation of no will be 12?my approach:x+y+z=12(6-x')+(6-y')+(6-z')=12x'+y'+z'=6no of non negative values: 8C2=28 where am i going wrong with the concept sir?
you have to realize that x, y and z cannot take values = 0

so when you replace x = 6-x' and dont put a constraint that x'

ATDH.

12939
@Amrofa said:
Triplets Consisting of three different numbers formed from number 1 to 10HOW MANY SUCH TRIPLETS ARE FORMED THAT SUM OF THE TRIPLETS IS DIVISIBLE BY 3
42?

@cadmium mod9 means remainder when divided by 9
12940
@krum said:
42?
Correct

Additional Question

How many of the triplets formed are such that the sum of number is divisible by 9 and they do not have a 9 in them ?????
12941
@Amrofa said:
Triplets Consisting of three different numbers formed from number 1 to 10HOW MANY SUCH TRIPLETS ARE FORMED THAT SUM OF THE TRIPLETS IS DIVISIBLE BY 3
triplets of form 3k from 1 to 10 (3,6,9)-->1 way
3k+1 form (1,4,7,10) -->4c3= 4 ways
3k+2 form (2,5,8) = 1 way
with one of form 3k, one 3k+1 and one 3k+2 => 3*4*3 =36 ways
Adding all of them 42 ways
12942
@Amrofa said:
CorrectAdditional QuestionHow many of the triplets formed are such that the sum of number is divisible by 9 and they do not have a 9 in them ?????
triplets as in 111,222... or like the last question?