Official Quant thread for CAT 2013

MBA CAT 2024
19146
@saurabh9gupta said:
Hi.Will this method work for longer numbers?

Yes it can be adapted, even 7 or 8 digit can be done to within 1% accuracy with practice.

regards
scrabbler
19147
@DeAdLy said:
Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?
let's say the prob he know the ans is x and y denotes the prob of him not knowing the ans.
x+y=1

in case of x----> prob he gets it correct=1
in case of y----> prob he gets it correct=1/4

thus x + y/4=2/3

solve the two equations x=5/9.

OA?
19148
@IIMAIM 111/145 ?
19149

@amitranjan2311 said:
@IIMAIM 111/145 ?
adding 3% to numerator and denominator
114/150
adding 1/3rd of numerator and denominator
152/200=76

OA=76.5
19150
@IIMAIM 111/132
adding 10% to numerator and denominator it will be 122/145
again adding 3 % to numerator and denominator it will be 125/150
adding 1/3rd of numerator and denominator
166/200 =83
OA 0.84
is this correct approach or we can do it with some other way?
19151
@IIMAIM said:
adding 3% to numerator and denominator114/150adding 1/3rd of numerator and denominator 152/200=76OA=76.5
Also, with practice one can see ways of enhancing it. For example here after 114/150 just multiply by 2 and get 228/300 = 0.76.

regards
scrabbler
19152
@amitranjan2311 said:
@IIMAIM111/132adding 1% to numerator and denominator it will be 122/145 again adding 3 % to numerator and denominator it will be 125/150 adding 1/3rd of numerator and denominator 166/200 =83OA 0.84 is this correct approach or we can do it with some other way?
Do minimum effort. Get the denominator as something whose table you know. For example here if you know table of 13, make 111/132 as 109/130 and just divide to get 0.839. The whole point is to get a reasonably accurate answer fast.

regards
scrabbler
19153
@amitranjan2311 said:
@IIMAIM111/132adding 1% to numerator and denominator it will be 122/145 again adding 3 % to numerator and denominator it will be 125/150 adding 1/3rd of numerator and denominator 166/200 =83OA 0.84 is this correct approach or we can do it with some other way?
yea correct way adding 10% in 1st step you mistakenly wrote as 1% edit it

19154
@gnehagarg said:
there is number which is minimum is 500 and maximum number is 999500*2=1000 and 999*2=1998I am checking numbers between 1000 and 1998 having 22,12,20,22,24,26,28,32,42,52,62,72,82,92=>1410230240250260270280290214*91200,1202-----------------------------------129850 munbers126+50=176
can you explain why you took these numbers(2,12,20,22,24,26,28,32,42,52,62,72,82,92) as reference and also 102,302,402 when multiplied by 2 wont give a four digit number.. right?
19155
@iLoveTorres

There can be any number which on muliplication by 2 gives 1002,1012,1020,1022,1024,1026 and soon...this 2 is 1002 12 is 1012

Minimum number is 500 500*2=1000
Maximum number is 999 999*2=1998

I am calculating number with one two,2 two 3 two from 1000 to 1998
19156
@gnehagarg said:
@iLoveTorresThere can be any number which on muliplication by 2 gives 1002,1012,1020,1022,1024,1026 and soon...this 2 is 1002 12 is 1012Minimum number is 500 500*2=1000Maximum number is 999 999*2=1998I am calculating number with one two,2 two 3 two from 1000 to 1998
whats the fault when you do it this way

case1) keep the thousand's place as 1 and the hundred's place as 2.. now the tens place can be filled in 10 ways and the units place can be filled in 5 ways so total 50 ways

case2) now keep the tens place 2 and thousand's place as 1.. hundred's place can be filled in 10 ways and units place in 5 ways again giving 50 ways

case 3) keep the units place as 2 and thousand's place as 1.. now you can fill the hundred's place in 10 ways and also the tens place.. so totally 100

in total 200 ways
19157
@iLoveTorres said:
whats the fault when you do it this waycase1) keep the thousand's place as 1 and the hundred's place as 2.. now the tens place can be filled in 10 ways and the units place can be filled in 5 ways so total 50 wayscase2) now keep the tens place 2 and thousand's place as 1.. hundred's place can be filled in 10 ways and units place in 5 ways again giving 50 wayscase 3) keep the units place as 2 and thousand's place as 1.. now you can fill the hundred's place in 10 ways and also the tens place.. so totally 100in total 200 ways
You can do this, but you need to minus the cases where the number contains three 2s or three 2s as they are double/triple counted.

regards
scrabbler
19158
@scrabbler said:
You can do this, but you need to minus the cases where the number contains three 2s or three 2s as they are double/triple counted.regardsscrabbler
isnt 1222 a valid 4 digit number?
19159
@iLoveTorres said:
isnt 1222 a valid 4 digit number?
Of course it is....but you have counted it 3 times in your enumeration as it is of the form 1 2 2 _ and 1 2 _ 2 and 1 _ 2 2.

See, you have 200 cases which have at least one 2. Out of those there are 5 + 10 + 10 =25 numbers with two 2s (1 2 2 _ and 1 2 _ 2 and 1 _ 2 2) and 1 (1222) with three 2s. So use Venn diagram logic => 200 - 25 + 1 = 176 numbers total.

Hope it helps...

regards
scrabbler
19160
@scrabbler said:
Of course it is....but you have counted it 3 times in your enumeration as it is of the form 1 2 2 _ and 1 2 _ 2 and 1 _ 2 2.See, you have 200 cases which have at least one 2. Out of those there are 5 + 10 + 10 =25 numbers with two 2s (1 2 2 _ and 1 2 _ 2 and 1 _ 2 2) and 1 (1222) with three 2s. So use Venn diagram logic => 200 - 25 + 1 = 176 numbers total.Hope it helps...regardsscrabbler
awesome. thanks for your valuable help bro. so this way we can extend it to all the permutation logic? like say a previous question on the forum was between 1000 and 10000 how many numbers are there with sum of the digits 9. so here i can basically adapt it as for numbers from 1001 to 9999 sum of the digits is 9 as finding all the multiples of nine and then subtracting all the redundant cases?
19161
@iLoveTorres said:
awesome. thanks for your valuable help bro. so this way we can extend it to all the permutation logic? like say a previous question on the forum was between 1000 and 10000 how many numbers are there with sum of the digits 9. so here i can basically adapt it as for numbers from 1001 to 9999 sum of the digits is 9 as finding all the multiples of nine and then subtracting all the redundant cases?
Well, that might work but it would be damn long. There are better ways....

How I did it was, I considered that there are 4 digits _ _ _ _ So I have to fill in a total of 9 units (identical!) into these 4 places. But the first place cannot be empty (as first digit cannot be 0) so put one there and it reduces to a problem of distributing 8 units into 4 places and this can be done (using theory of partitioning) in (8+3)C3 or 11C3 or 165 ways.

regards
scrabbler
19162

Find the Number of ways of forming a 4 digit number using 1,2,3 in every number generated ?

19163
@scrabbler said:
Well, that might work but it would be damn long. There are better ways....How I did it was, I considered that there are 4 digits _ _ _ _ So I have to fill in a total of 9 units (identical!) into these 4 places. But the first place cannot be empty (as first digit cannot be 0) so put one there and it reduces to a problem of distributing 8 units into 4 places and this can be done (using theory of partitioning) in (8+3)C3 or 11C3 or 165 ways. regardsscrabbler

why have you considered the number of digits as 9? 0-9=10 right? now subtracting the case of 0 you get 9 possibilities

19164
@IIMAIM said:
Find the Number of ways of forming a 4 digit number using 1,2,3 ?
3^4?
19165

no @iLoveTorres