Official Quant thread for CAT 2013

MBA CAT 2024
13203
@TootaHuaDil said:
4. A certain number 'C' when divided by N1 it leaves a remainder of 13 and when it is divided by N2 it leaves a remainder of 1, where N1 and N2 are the positive integers. Then the value of N1+N2 is, if N1/N2=5/4(a) 36(b) 27(c)54(d) can't be determined uniquely
Is it 27??
13%15 = 15
13%12 = 1
N1+N2 = 15+12 = 27
13204

@audiq7 said:
22?
@bs0409 said:
b)22N=14D=34
@joyjitpal said:
22 N=nD +143N=mD + 8adding them 4N=(m+n)D +22so rem 22
Two values of divisors will come 17 and 34 and thus remainder will be 5 or 22 and thus Can;t be determined
13205
A,B,C were playing a game. At the begning A and B together had 100% more money then C and B,C had 300% more money then A. By the end of the game A, B has 100% more then C and A had 12.5% less money then B,C. Finally A gained Rs 800 in the game.
Find the percentage change of the money of B
1.40
2.30
3.57.14
4.42.8
13206

Number of 5 digit numbers that contain 5 only once is .....

13207
@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!
since 18! is highest so we consider that and so..3 zeros..
13208
@Logrhythm said:
Is it 27??13%15 = 1513%12 = 1N1+N2 = 15+12 = 27
Nop...N1 can be 15, 20, 30 and 60 and correspondingly N2 can be 12, 16, 24 and 48
13209
@TootaHuaDil said:
4. A certain number 'C' when divided by N1 it leaves a remainder of 13 and when it is divided by N2 it leaves a remainder of 1, where N1 and N2 are the positive integers. Then the value of N1+N2 is, if N1/N2=5/4(a) 36(b) 27(c)54(d) can't be determined uniquely
Is the answer (b)?
Let N1 = 5x and N2 = 4x, where x=3
For any no N of form 20x + 13
N/5x = 13
N/4x = 1
etc..
13210
@bs0409
@Logrhythm 2/3 ??
base conversion give (.33333333....)base 10 = (.1555555555....)base6
closest is 2/3=.1666666666
13211
@TootaHuaDil said:
2. A number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34. Such a least possible number is:(a) 175(b) 75(c) 680(d) 580
75..option to question
13212
@sails yes ..correct
13213
@pirateiim478 said:
Number of 5 digit numbers that contain 5 only once is .....
Not sure but is it 9^4 + 9^3*8*4

wrong sound kar raha hai.
13214
@TootaHuaDil said:
1. When a number 'N' is divided by a proper divisor 'D' then it leaves a remainder of 14 and if the thrice of that number i.e. 3N is divided by the same divisor D, the remainder comes out to be 8. Again if the 4 times of the same number i.e. '4N' is divided by D the remainder will be: (a) 35(b) 22(c) 5(d) can't be determined
n=dk+14
3n=dz+8
4n+d(k+z) +22..so 22
13215
@bs0409 @somnathbhatta - No. it is 0.2 or 1/5

it can be done like this: 0.33333.... is 1/3 in base 10
let this be 0.xyz.... in base 6, this can be written as x/6+y/36+z/216+.... = 1/3
by observation we can see that at x = 2 and y=z=...=0 - LHS=RHS
hence 1/3 in base 10 is 1/5 in base 6
13216
@TootaHuaDil said:
4. A certain number 'C' when divided by N1 it leaves a remainder of 13 and when it is divided by N2 it leaves a remainder of 1, where N1 and N2 are the positive integers. Then the value of N1+N2 is, if N1/N2=5/4(a) 36(b) 27(c)54(d) can't be determined uniquely
36....N1=20..n2=16..
13217
@pirateiim478 said:
Number of 5 digit numbers that contain 5 only once is .....
9^4 + 4*8*9^3 = 41*9^3 = 29889
13218
@sails said:
n=dk+143n=dz+84n+d(k+z) +22..so 22
n=dq+14
3n=3(dq+14)
=3dq+42
=(3dq+34)+8
So 34 must be divisible by divisor...Divisor will be factor of 34. It can be 1,2, 17 or 34...But divisor cannot be smaller than remainder so we are left with two values of d...17 and 34...For 17 remainder for 4n=4(17q+14) ...remainder=5 and for d=34 remainder = 22
13219
@pirateiim478 said:
Number of 5 digit numbers that contain 5 only once is .....
9^4 + 9^3*8*4
13220
@Logrhythm not convinced with the solution. base conversion is far more accurate. i don't know if this is correct. i won't say ur approach is wrong. but i would stick to base conversion.. what say @bs0409 ?
13221
@TootaHuaDil said:
n=dq+143n=3(dq+14)=3dq+42=(3dq+34)+8So 34 must be divisible by divisor...Divisor will be factor of 34. It can be 1,2, 17 or 34...But divisor cannot be smaller than remainder so we are left with two values of d...17 and 34...For 17 remainder for 4n=4(17q+14) ...remainder=5 and for d=34 remainder = 22
the answer s 22 right?
13222
@sails said:
36....N1=20..n2=16..
n1q1+13=n2q2+1
n1(4q2-5q1)=60
n1= all the factors of 60 such that n1>13