Official Quant thread for CAT 2013

@blowcat13 said:
A jar full of milk contains 40% of water. a part of this milk is replaced by another containing 19%water and now the percentage of water is found to be 26%. The quantity of milk replaced is??A)1/3 B)2/3 C)2/5 D)3/5weak at alligations explain in detail plzzz
OA: C) 2/3

let em be a,b,c,d,e,f,g ;
a= b+c+d+e+f+g+4;
b+12=b+c+d+e+f+g+1;
11=c+d+e+f+g;
solved.

@blowcat13 said:
@dragster OA is 2/3 bhai..
milk content replaced with respect to overall solution must be 2/5 , n that to milk content only - it shall be 2/3..
Q: 17^23^31^41 div 59 R?
@rahulmittal2211 said:
There are seven children standing in line, not all of whom have the same number of cakes with them. if the forst child distributes his cakes to the remaining six children such that he doubles their respective no of cakes, then he will be left with 4 cakes. instead, if the second child takes two cakes from each of the remaining six children, then he will be left with three cakes less than the number of cakes that the first child initially had. what is the total no of cakes that are there with the third child to seventh child?A. 11 B. 14 C. 12 D. 15 someone plz explain the solution for this...
10
Q: in how many ways a regular hexagon can be painted with a black color ?
@viewpt its a time material question and according to it answer is 11. dnt knw how?
@viewpt said:
OA: C) 2/3
By the rule of alligation, we have:
Strength of first jar Strength of 2nd jar
40 19
26
7 14
So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
Required quantity replaced = 2/3

@viewpt said:
Q: in how many ways a regular hexagon can be painted with a black color ?
is it 63 ways?
@viewpt said:
Q: 17^23^31^41 div 59 R?
by remainder theorem;
59 and 17 are coprime..
59 is prime , f59= 58;
for 23^31^41,
divide by 58 , now f58= 28
for 31^41
divide by 28 , remainder=19 ;
so we can reduce it as 23^19 / 58 (intermediate)

23*(7)^9 / 58

23*7^2 /58 = 25

so, 17^25 / 59

17*(-6)^12 /59

17*(-8) /59

51*17 /59

41..

@viewpt said:
Q: in how many ways a regular hexagon can be painted with a black color ?
2^6-1=63
@naveenkrs
@naveenkrs said:
In triangle ABC, D, E, and F are the trisection points of AB, BC, and CA nearer A,B,C, respectively. Let BF and AE meet at J. Let CD and AE meet at K and CD and BF meet at L. Find 1) BJ : JF2) AJ : JE3) DK : KL : LC4) EJ : JK : KA5) FL : LJ : JB
i got the following ans .. wats the OA ??
1) 3:4
2) 6:1
3) 1:3:3
4) 1:3:3
5) 1:3:3
@swapnil4ever2u said:
@naveenkrs i got the following ans .. wats the OA ??1) 3:42) 6:13) 1:3:34) 1:3:35) 1:3:3
right answer
@viewpt said:
Q: 17^23^31^41 div 59 R?
e(59) is 58..
23^31^41%58
e(58) is 28
31^41%28
e(28) is 12
41%12 = 5
31^5%28 = 3^5%28 = 19
23^19%58 = 25..
17^25%59 = 41..

bhai kahin agar galti na hui ho toh...
@vishcat said:
@naveenkrs no wrong answer
whats the OA?
@Logrhythm said:
it wld be all primes (except 2) 5^2-4^2)and 4*primes so 24+9 = 33 such numbers...By "uniquely" i understand only in one way...
bhai wat abt numbers like
9(not prime)=> 5^2-4^2 only one way..
36 (4*9(not prime)) can also be expressed in only one way ..
@swapnil4ever2u said:
bhai wat abt numbers like 9(not prime)=> 5^2-4^2 only one way..36 (4*9(not prime)) can also be expressed in only one way ..
bhai.. these are squares and can be expressed as 3^2-0^2 = 9
6^2-0^2 = 36...
@Logrhythm fir to 1 bhi hona chahiye 1^2 - 0^2
There are 4 people Erick, Helena, Johny and David and they have 9 seats available to sit on. Now Johny and Helena don €™t like each other and that €™s why they always need to have one blank seat in between them and others can sit on any seat but no one can sit on the blank seat between Johny and Helena.

First, the seats are arranged in a linear fashion and then the number of different linear arrangements possible are found to be M.Then the seats are arranged in circular fashion after which the number of different arrangements possible are found to be N. Then M + N is found to be
OPTIONS

1) 270
2) 450
3) 480
4) None of these
@viewpt said:
Q: 17^23^31^41 div 59 R?
41?