# @Skaushi

##### Sachin Kaushik 14 karma
###### CATOfficial Quant thread for CAT 2013
@ayushbhalotia 49: it comes out to be 24. I can't make out of proportional sign for 0....
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###### CATOfficial Quant Thread for CAT 2012 [Part 4]
Q2

scoring 60percent=losing 40 percent=100 marks

hence a+b+c+d=100
hence we have 103C3 solns..

but a,b,c cannot be >50

hence a'+b+c+d=49
we have 52C3 solns
hence 103C3-3*52C3=110551

Q4...The possibilities are
case1..all similar
only 9 cases are possible

case2 2 distinct digits present...there are sub cases to this
A>when none are 0

9C2(2*2*2*2-2)=504

B>when 0 is one of the digits

9C1(1*2*2*2-1)=63

hence total cases=504+72=576

Sam sir, can you explain a bit....a'+b+c+d=49 step...
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
Let abcd be the four digit number
Now, either first three digit will be consecutive or last three

Case-1: First three digit is consecutive
It is also divisible by 9
=> a+a+1+a+2+b=9k
=> 3a+b=9k-3
Now, maximum value of 3a+b=36
If 3a+b=6 (k=1)
=> (a,b)= (1,3)
If 3a+b=15 (k=2)
=> (a,b)= (4,3),(6,3)
If 3a+b=24 (k=3)
=> (a,b)= (7,3),(6,6)
Thus, total 5 cases

Case 2:- Last three digits are consecutive
=> a+3b= 9k-3
If a+3b=6
2 cases
If a+3b=15
2 cases
If a+3b=24
2 cases

Thus, total 11 cases

P.S:- Neetuji Pls check. I think i missed something

I think 14 cases will be there in total. Cases where b=0 (considering it is always the one's place digit) will be counted as 4 digit numbers as well.
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
28 = 7*4 or 7*2*2
24 = 8*3

from 28 factors , it has reduced to 24 factors.

Thus, power of 2 in first case is 3, which reduces to 2 in second case.
And power of 3 increases from 6 to 7.

Thus, 2n^2 = 2^3 * 3^6
3n^2 = 2^2 * 3^7

thus, 6n^2 = 2^3 * 3^7

Factors = 4*8 = 32

Vikram , could you elaborate a bit.. I have not been able to understand the approach...
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
yup..how did u get it

1 approach could be start with 5 variables, a,b,c,d,e in incrreasing order and now make eqns, eg highest 2 d+e=121 a+b=111 and so on....
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###### CATOfficial Quant Thread for CAT 2012 [part 2]
Ques 1)

X+Yif X=0 Y4X+3Y=221
DIVIDE BY 4 AND TAKE ONLY REMAINDERS
Y/4=1 (value of Y such that remainders=1 always)
so Y can take 1,5,9,...,65
so total no. of solution are 17.

Rajan, could you explain a bit bout division by 4 here...why we are doing it ?
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###### CATOfficial Quant Thread for CAT 2012 [part 2]
My take:b

10C2*2!*2!*9!

=>18*10!

Please thoda explain karo bhai...What stands for what ?
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###### CATOfficial Quant Thread for CAT 2012 [part 2]
Jatin_kaushik Says