link to previous thread :-

http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html

:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...

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link to previous thread :-

http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html

:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...

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=> 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

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

find the infinite sum of the foll seuence

1+(4/7)+(9/(7^2))+(16/(7^3))+.....

1+(4/7)+(9/(7^2))+(16/(7^3))+.....

members. They are involved in three jobs: teaching, training and research. Each faculty member working

with TT School of Management has to be involved in at least one of the three job ...

1

Solve and find the pair of non negative integers (x, y) such that

(x+ y - 5)^2 = 9xy.

(x+ y - 5)^2 = 9xy.

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

obviously this will be 0 .

8^48/16^18 = 2^144/ 2^72 : remainder will again 0 so my take 0

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(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 - (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

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

so it is 4k+1

3*3*3*3*3 /41 remainder 38

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...

E(11) = 10

25^anything has unit digit 5

5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1

5^25^125^3125 /11..

thanks all...

What is the O.A.??

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

but dude u cant apply euler in bold part...5 and 40 are nt co prime

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

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