Ankurb's posts
Ankurb replied to Official Quant Thread for CAT 2010
solve for any one variable..
1/(4-p) + 3p/(7p-3) = 1
9 - 12p + 4p^2 = 0
p = 3/2
r = 5/3
q = 2/5
pqr = 1
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Ankurb replied to Official Quant thread for CAT 09-III (September...
what u r saying is true but you can simplify it further, to apply Euler:-
2^1929 mod 4*33 =
2^1927 mod 33 = now apply euler
E(33) = 20
2^1927 mod 33 = 2^7 mod 33 = 29
remainder = 29*4 = 116
Ankurb replied to Official Quant thread for CAT 09-III (September...
take point (x,y) anywhere on line y = 3 or on y = 1,
area will always be = 0.5 * base * height = 0.5 * 5*1 = 5/2
Ankurb replied to Official Quant thread for CAT 09-III (September...
8^643 mod 4 = 0
8^643 mod 3 = -1
8^643 mod 11 = 6
11p + 6 = 3q-1
q = 11p + 7)/3,==>p = 1
17 + 33r = 4t,==>r = 3
remainder = -16
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Ankurb replied to Official Quant thread for CAT 09-III (September...
option B)
Cauchy Equation:-
f(x+y) =f(x)f(y), ==>
f(x) = e^ax
f1(x) = ae^ax =af(x)
.
.
fn(x) = a^nf(x), ==> k = a^n
given condition:-
g(x) = (f(x) - 1)/x = (e^ax - 1)/x
as limit(x-->0)g(x) = T(a constant), ==>
T = a
==>
k = T^n
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Ankurb replied to Official Quant thread for CAT 09-III (September...
option b)30
N = a^p b^q c^r
condition:-
(N^(p+1)(q+1)(r+1)/2)/N = N
==>
(p+1)(q+1)(r+1) = 4
N will either be of the form a^3 or a.b
a^3--- 1 number
a.b-- 29 numbers
total = 30
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Ankurb replied to Official Quant thread for CAT 09-III (September...
select 1 ball from first bag, 2 balls from 2nd bag and so on.
if balls wer identical, reading wud be = 5500 gms
if 1st bag have lighter balls , reading = 5500 - 10 = 5490 gms
if 2nd bag have lighter balls , reading = 5500 - 20 = 5480 gms
.
.
Ankurb replied to [JMET 2010] Notification, How to apply and rela...
binomial:-
(1+0.05)^1.2 = 1+0.05*1.2 = 1.06(approx)
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Ankurb replied to Official Quant thread for CAT 09-III (September...
did it this way...
with no condition:-
9*7*5*3
with a pair consisting two boys:-
3*7*5*3
9*7*5*3 - 3*7*5*3 = 6*7*5*3 = 630
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Ankurb replied to Official Quant thread for CAT 09-III (September...
a+b = 8 - 2c
2c^2 = 25 -(a^2+b^2)
using above two:-
2c^2 = 25 -((8-2c)^2 - 2ab)
2c^2 = 25 -64 - 4c^2 + 32c + 2ab
6c^2 = 2ab + 32c - 39
also:-
ab <=(4 - c)^2
for max:-
6c^2 = 2(4 - c)^2 + 32c - 39
6c^2 - 24c + 21/2
c = 1/2 or 7/2
c(max) = 7/2
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