Quant Boosters  Sagar Gupta, CAT Quant  99.2 Percentile  Set 3

Author : Sagar Gupta  MBA student at Symbiosis Institute of Operations Management. 99.2 Percentile in CAT 2015 (Quant)
From an 'x' litre of solution of alcohol and water, 10% of the solution is removed and that amount of water is added back. Now, 9.09% of solution is removed and again water is added in the same amount. Again 8.33% of solution is removed and water is added in the same quantity. If initially, the mixture had alcohol and water in the ratio of 2:1, and has 1:1 alcohol and water afterwards, what can be the value of 'x' ?
a) 2 litre
b) 3 litre
c) 4 litre
d) Not dependent on xSuppose the original total is x, then alcohol as a fraction of total (stepbystep) becomes x * 2/3 * 9/10 * 10/11 * 11/12 = 1/2 and hence alcohol : water will become 1 : 1 always
sqrt (x+2) + sqrt (x2)= sqrt(4x3)
The number of solutions for this equation are :
a) 0
b) 1
c) 2
d) more than 2x+2 + x2 + 2 sqrt ( x^2  4 ) = 4x 3
2x  3 =2 sqrt ( x^2  4)
4x^2 + 9  12x = 4x^2  16
25 = 12x
x=25/12N = 57^99 + 55^99. What is the remainder when N is divided by 224?
Euler[224] =96
99 mod 96 = 3
57^3 + 55^3 mod 224
[57+55 ] ( 57^2 + 55^2  57*55 ) mod 224
112 * 3 mod 224 = 112Or Use binomial
(56 + 1)^99 + (56  1)^99
Remainder will be 56 * 99 * 2 or 112(2 * 49 + 1) or 112Find the remainder when 987698769876.... 400 digits is divided by 31
Take [9876] as one digit, so total 100 digits
Euler(31) = 30, so just take 100mod30=10, ie last ten digits
Which is 9876 (1 + 10^4 + 10^8 + .... + 10^36) mod 31
(GP sum inside bracket) mod 31 = 13
9876mod 31=18
So 18 * 13 mod 31=17Sum of 5 terms in an increasing GP is 1031. All the terms are integers. Find the sum of first and the last term of this GP.
a + ar + ar^2 + ar^3 + ar^4 = 1031
a [ 1 + r + r^2 + r^3 + r^4 ] = 1031
a [ r^5 1 ] = 1031 [ r 1 ]
rearrange RHS to express 1031 in the form of r^5 1
Look for an integer whose fifth power is more than 1031
3a [ r^5  1 ] = 3093 [ r1 ]
5^5 = 3125
3a [ r^5  1 ] = [ 3125  32 ] [ r1 ]
3a [ r^5 1 ] = 32 [3125/32 1 ] [ r1 ]
r = 5/2 = 2.5
3a = 32 ( r1 )
3a = 32 (1.5)
a=16The no of positive integers n such that (n^6 + 206) is divisible by (n^2 + 2) is ?
n^6 + 206 = k ( n^2 + 2 )
Rearrange :
n^6 + 8 = k (n^2 + 2)  198
(n^2 + 2 ) ( n^4 + 4  2n^2 ) = k (n^2 + 2 )  198
Divide throughout by n^2 + 2
n^4 + 4  2n^2 = k  [ 198 / n^2 + 2 ]
198 / n^2 + 2 must be an integer
n=1 , n = 2 , n=3 , n=4 , n=8 , n=14 > Hence 6 ValuesX takes 9 and Y takes D days to complete a job working alone. They work on alternate days. If they take exactly the same time irrespective of who starts, how many positive integer values are possible for D ?
2m + 18m/d = 18
m + 9m/d = 9
d = 9m / 9m
m=6 : 54/3 =18
m=8 : 72 / 1 = 72Find the sum of all values of x , that satisfy (4x^2 + 15x + 17 ) / ( x^2 + 4x+12) = (5x^2 + 16x+18 ) / ( 2x^2 + 5x + 13 )
m/n = m+p/n+p
mn + mp = mn + np
m = n
4x^2 + 15x + 17 = x^2 + 4x + 12
3x^2 + 11x + 5 = 0
Sum of roots = 11/3The values of the numbers 2^2004 and 5^2004 are written one after another. How many digits are there in all?
(1) 4008
(2) 2003
(3) 2004
(4) None of theseNo. of digits = logN + 1 ... here it's 2004(log10) + 1=2005
Or by pattern : 2^1 5^1 = 25 : 2 digits
2^2 5^2 = 425 : 3 digits
2^3 5^3 = 8125 : 4 digits
2^4 5^4 = 16625 : 5 digits
2^5 5^5 = 323125 : 6 digits
..
2^2004 5^2004 : 2005 digitsHow many integer solutions exist for the equation 5x  3y = 140 such that x and y are of opposite signs?
a) 8
b) 9
c) 10
d) 115x  3y = 140
x=25 , x=22 , x=19 , x=16 , x=13 , x=10 , x=7 , x=4 , x=1
9 solutions