Official Quant thread for CAT 2013

MBA CAT 2024
81
@rkshtsurana said:
Set S = { a1 , a2 , a3.....} where elements of set S are only the combination of powers of 2 and 3 like a1 = 2 a2 = 3 a3= 4 a4 = 6 a5 = 9 a6 = 12 and so on..Find 1/a1 + 1/a2 + 1/a3 ............a) 1b) 2c) 3d) 4
2.

Try adding up till first 6-7 numbers..will be 1. something..cannot exceed 2.

PS: Please check the question!
82
@billuisback its correct sir..u may be missing something..check my approach once
83
@rkshtsurana said:
s1 = 1/2 + 1/2^2 + 1/ 2^3 ........ = 1s2 = 1/3 + 1/3^2 .......= 1/2combination = ( 1+ s1)(1+ s2) - 1 2 * 3/2 - 1 = 2
yaar..
What about term like 6,12,18,...??are these cover in above two series...??
84

Which of the following numbers is always divisible by n^7-n?


1) 13
2) 9
3) 7
4) 5
85
@rkshtsurana said:
@billuisback its correct sir..u may be missing something..check my approach once
Bhia..you have written a4=6, 6 kaha se aayega? :P

86
@soumitrabengeri said:
Which of the following numbers is always divisible by n^7-n?1) 132) 93) 74) 5
7. Simple Formula based hai... since 7 will always be prime, so 7.
87
@surajsrivastav said:
yaar..What about term like 6,12,18,...??are these cover in above two series...??
ha multiply hga na see it like dis
( 1 + { 1/2 + 1/4 + 1/8.....}) * ( 1 + [1/3 + 1/9 + 1/27...]) - 1

so jab 1/3 ..1/2 se multiply hga it ll give 1/6 and so on
that last -1 is for when 1 is multiplied by 1 and that is nt needed in r series
88
@rkshtsurana said:
ha multiply hga na see it like dis( 1 + { 1/2 + 1/4 + 1/8.....}) * ( 1 + [1/3 + 1/9 + 1/27...]) - 1so jab 1/3 ..1/2 se multiply hga it ll give 1/6 and so onthat last -1 is for when 1 is multiplied by 1 and that is nt needed in r series
got it... :-)
89
@billuisback sir i said combination of power of 2 and 3.. it mean a general term of element of set S is ( 2^a * 3^b)
90
@soumitrabengeri said:
Which of the following numbers is always divisible by n^7-n?1) 132) 93) 74) 5
Fermat theorem
n^7 = n mod 7 7 is prime
91
@rkshtsurana said:
@billuisback sir i said combination of power of 2 and 3.. it mean a general term of element of set S is ( 2^a * 3^b)
okies. And yaar sir to mat hi bolo..Office mat banao plz!

Waise answer to sahi hai :P
92
@krum said:
8c4(x^2+y)^4*(1/y+1/x^2)^4=>8c4*(4c0x^8+4c1x^6y+4c2x^4y^2+4c3x^2y^3+4c4y)(4c01/x^8+4c1*1/x^6*1/y+4c2*1/x^4*1/y^2+4c3*1/x^2*1/y^3+4c4*1/y)const term=>8c4(4c0^2+4c1^2+4c2^2+4c3^2+4c4^2)=>8*7*6*5/24*(1+16+36+16+1)=>4900
how are you writing this-
8c4(x^2+y)^4*(1/y+1/x^2)^4
93
@fedbite said:
forming different pairsas (a+b+c+d)^8a^2*b^2*c^2*d^2 similarly every term below 8!/2!^4+8!/3!^2*2+8!/4!^2*2=4900 (Ans.)
i couldn't get that, please elaborate a bit
94
@krum said:
i have paired 2 terms , only this pair will leave a constant term as u can see
okay and why this 8C4?
95
@invincible2910 said:
okay and why this 8C4?
what if it were (x+1/x)^8
to find constant term powers of x and 1/x should be same
this happens at 4th term ie. 8c4*(x)^(8-4)*(1/x)^4

similarly here take (x^2+y) as 'a' and (1/y+1/x^2) as 'b' in the expansion of (a+b)^n
=>((x^2+y) + (1/y+1/x^2))^8 , 4th term will be
=>8c4(x^2+y)^4*(1/y+1/x^2)^4
96
@krum said:
what if it were (x+1/x)^8 to find constant term powers of x and 1/x should be samethis happens at 4th term ie. 8c4*(x)^(8-4)*(1/x)^4similarly here take (x^2+y) as 'a' and (1/y+1/x^2) as 'b' in the expansion of (a+b)^n=>((x^2+y) + (1/y+1/x^2))^8 , 4th term will be =>8c4(x^2+y)^4*(1/y+1/x^2)^4
yes, i got it now, thank you! :)
97

this was a qs. posted on last thread by one of the guys here..


if C(n,0)*C(n,n-2) + C(n,1)*C(n,n-3) + C(n,2)*C(n,n-4)............C(n,n-2)*C(n,0) =C(x,y)

given n = 100. find the values of (x,y)

banging my head over this for past 45 mins.. (cold and thinking is a brutal combination btw)
98
@invincible2910 said:
i couldn't get that, please elaborate a bit
multinomial coefficients
99

ohh Lord!!

100

ratio of area of triangle FOE to area of triangle AOD??