# Official Quant Thread for CAT 2011 [Part 8]Quantitative Report

Please continue here with all the quant queries and their discussions.
:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...
28 28
=> S = 1 + (4/7) + (9/7^2) + ...
=> S/7 = 1/7 + (4/7^2) + (9/7^3) + ...
Subtract them we get
=> 6S/7 = 1 + 3/7 + 5/49 + ...
=> 6S/49 = 1/7 + 3/49 + 5/343 + ...
Subtract them again
=> 36S/49 = 1 + 2/7 + 2/49 + ...
=> 36S/49 = 1 + (2/7) / (6/7)
=> S = 49/27
4 4
S= 1+(4/7)+(9/(7^2))+(16/(7^3))+.....
S/7 = 1/7 +(4/(7^2))+(9/(7^3))+.....
6S /7 = 1+3/7+5/7^2+.......
Now this series is in A.P. & G.P.
S = ab/1-r+dab/(1-r)^2
here a = first number in A.P.
B= first number in G.P.
1/1-1/7+2/(1-1/7)^2 = 70/18
so sm is: 7/6*70/18
2 2
find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....
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1 1
Solve and find the pair of non negative integers (x, y) such that
(x+ y - 5)^2 = 9xy.
Doing Second: log(4)a-b = log(4) (_/ a-_/b)^2+log(4)4
log(base 4)a-b = log(base 4) 4*(_/ a-_/b)^2
a-b = 4*(_/ a-_/b)^2
taking sqaure root
_/ (a-b) = 2_/a-2_/b
iI think it should be 1 only keeping option in mind
(12^5/3) ^11 % 3
obviously this will be 0 .
8^48/16^18 = 2^144/ 2^72 : remainder will again 0 so my take 0
remaining whatever it is divided into
S:G:M = 4: 80: 60 = 1:20:15
Now let profit y then 80% of the same will be distributed and S will get y/36*80/100+20/100Y
G = 20/36*80/100y
M=15/36*80/100y
so the final ratio will be 800:1600:1200
8:16:12 = 2:4:3
2x 4x 3x
5x = 2200
...
p(x) = 26/52+6/52 =32/52
p(y) = 13/52 +13/52+2/52 =28/52
p(z) = 13/52+3/52+3/52 = 19/52
so p(x)> p(y) > p(z)
option A
P.S.: Pata nahi me tash k patto ko thik se samaj payi hu ya nai, I always got confused betn "Ace" and "club" or "jack"
x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)
2.5x+2.5y = 1+3xy
let x=0 y =1/2.5
y=0 X= 1/2.5
I think soln can be more than 1
10x+y is the number
x*y+x^2 =52
x*y+y^2 =117
y^2-x^2 = 65
y+x = 13
y-x= 5
aading 2y = 18 y= 9
x= 4
so value is 49
5^7^9 = 5k
so it is 4k+1
3*3*3*3*3 /41 remainder 38
The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?
10^n-1 is obviously multiple of 9 and 11
so 4707/9 =523 (sum will be 9*n where n is no. by which 9 is repeated)
so value is 9999999999999999............523 times
P.S.:- jain saha...
The remainder is 1.
E(11) = 10
25^anything has unit digit 5
5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1
hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..
5^25^125^3125 /11..
thanks all...
I have not applied Euler in that part...I just calculated the remainder of 7^9 with 16....Then I represented 7^9 as (16K + 7), then I segregated common multiple from numerator and denominator. After that I applied Euler of 8....
What is the O.A.??
(3^5^7^9)/41
3 and 41 are coprimes.
=> E(41) = 40
5^7^9/40 => 40 = 5*8
=> 5^7^9/8 => E(8 ) = 4
=> 7^9/4 = 3^9/4 => -1 or 3 remainder
=> 5^3/8 = 125/8 => 5 remainder
5^7^9 => 5x = 8y+5 => 5 remainder
3^5/41 = 3*81/41 = 120/41 = 38 remainder
No problem.
1 1
but dude u cant apply euler in bold part...5 and 40 are nt co prime
E(41) = 40
We need to find 5^7^9 % 40
E(40) = 16
7^9%16 = 7
5^(16K+7) % 40 = 5 * 5^(16K+6)%8
Now, E(8 ) = 4
Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38
The remainder is 38
i am getting 38 ... oa?
C(41)=40
checking for 5^7^9 mod 40
C(40) = 4
7^9 mod 4= 3
so... 5^3 mod 40=5
and 3^5 mod 41 = 38