Official Quant thread for CAT 2013

MBA CAT 2022
2961
@maddy2807 said:
The inverse of the sum of the following series upto n terms can be written as3/4 + 5/36 + 7/144 + ...1) (n-1)^2/(n^2+2n)2) (n^2+2n)/(n-1)^23) (n^2+2n)/(n+1)^24) (n+1)^2/(n^2+2n)
3/4 + 5/36 + 7/144..
=-[(1/4-1/1) + (1/9-1/4) + (1/16-1/9)+...]
=-[-1+1/(n+1)^2]
=1-1/(n+1)^2
=(n^2+1+2n-1)/(n+1)^2
=(n^2+2n)/(n+1)^2

inverse = (n+1)^2/(n^2+2n)

4
2962
@milestogo3 @soumitrabengeri @maddy2807
proper question yehi h yaar, consecutive nhi given.. koi definite tareeka h nikalne ka i guess
i dnt hav answers to this, but i'll enquire about the question and get back to you asap
2963
@maddy2807 said:
835?
Yes thats right
@soham2208 said:
S = 4 + 11 + 22 + 37 + ...S = .......4 + 11 + 22 + ..... (Tn)Subtract and get TnTn = 4 + [7 + 11 + 15 + ....(n-1) terms] = 4 + S1S1 = 7 + 11 + 15 +.....S1 = .......7 + 11 + ....... (Gn)Subtract and get GnGn = 7 + [ 4 + 4 + ...(n-2) terms] = 7 + 4*(n-2) = 4n - 1S1 = Sigma(4n-1) = 2n(n+1) - n = 2n^2 + nTn = 2n^2 + n + 4S = Sigma(Tn) = (n*(n+1)*(2n+1)/3) + (n*(n+1)/2) + 4nFor n = 10, S = (10*11*21/3) + ( 10*11/2) + 40 = 770 + 55 + 40 = 865 ?PS: This is a general approach, there might be some calculation mistakes, also for this specific question, you can just do it manually as 10 terms is less enough
Thanks
2964

meanwhile help me with this
LCM of 2^6-1 and 2^9-1?
A.2^12 +27*(2^9) - 217
B.2^12 +63*(2^3) - 1
C.2^12 + 5*(2^9) - 1
D.2^12 + 9*(2^8) - 1

Also please show your method because i tried a lot but could not devise a way to find the lcm for numbers with powers?

2965
@rkshtsurana said:
The lengths of the sides of a triangle are integers, and its area is also an integer. One side is 21 and the perimeter is 48 . What is the shortest side of the triangle?p.s. - Currently No OA
21,17,10........SHORTEST SYD 10??????
2966

@realslimshady said:
meanwhile help me with thisLCM of 2^6-1 and 2^9-1? A.2^12 +27*(2^9) - 217B.2^12 +63*(2^3) - 1C.2^12 + 5*(2^9) - 1D.2^12 + 9*(2^8) - 1Also please show your method because i tried a lot but could not devise a way to find the lcm for numbers with powers?
option B?
2967
@rkshtsurana said:
The lengths of the sides of a triangle are integers, and its area is also an integer. One side is 21 and the perimeter is 48 . What is the shortest side of the triangle?p.s. - Currently No OA
x+y=27

taking x=10, y=17

area=root(24*3*(24-x)(24-y))
==>root(2^3*3^2*14*7)
==>root(2^4*3^2*7^2)
==>4*3*7=84



2968
@realslimshady said:
meanwhile help me with thisLCM of 2^6-1 and 2^9-1? A.2^12 +27*(2^9) - 217B.2^12 +63*(2^3) - 1C.2^12 + 5*(2^9) - 1D.2^12 + 9*(2^8) - 1Also please show your method because i tried a lot but could not devise a way to find the lcm for numbers with powers?

LCM=(2^6-1)*(2^9-1)/HCF
LCM=(2^6-1)*(2^9-1)/(2^3-1)
==>(2^3+1)(2^9-1)
==>2^12-2^3+2^9-1
==>2^12 +63*(2^3)-1

B

2969
@krum this is exactly the approach i tried, but i dont understand how you got hcf= 2^3 - 1
can u please explain this?
2970
@realslimshady said:
@krum this is exactly the approach i tried, but i dont understand how you got hcf= 2^3 - 1can u please explain this?
@krum understood it, u used a^n - B^n/ a-b.. how could i miss that :/
thanks :D
2971
@realslimshady said:
@krum this is exactly the approach i tried, but i dont understand how you got hcf= 2^3 - 1can u please explain this?
(2^6-1) = ((2^3)^2-1^2) = (2^3-1)(2^3+1)

(2^9-1) = ((2^3)^3-1^3) = (2^3-1)(2^6+1+2^3)

==> HCF = (2^3-1)

other questions of such type might have powers like (7^2001-1) & (7^2001+1)

in such cases remember that HCF of 2 consecutive even numbers =2
HCF of 2 consecutive odd numbers = 1
2972
@rkshtsurana said:
The lengths of the sides of a triangle are integers, and its area is also an integer. One side is 21 and the perimeter is 48 . What is the shortest side of the triangle?p.s. - Currently No OA
let x be shortest side
Other side be 27-x
Area =rt(24*3*(24-x)*(x-3))
x=10
Area is integer

2973
@milestogo3 said:
293768 431845725613 = INLETS
What is going on in this question ?
2974
@krum Thanks a lot :D
2975

what is the no. of diff combinations of two no.s such that sum of these two nos and their LCM is 323

2976
@Ashmukh said:
what is the no. of diff combinations of two no.s such that sum of these two nos and their LCM is 323
let numbers be ka,kb where a,b are coprime

ka+kb + lcm(ka,kb)=323 = 17.19
k(a+b+ab)=17.19
if k=1 then
a+b+ab=17.19
(a+1)(b+1) = 324 = 2.2.3.3.3.3
factors=15..so number of unordered solutions from here would be,
6 ways
if k=17
(a+1)(b+1) = 20 = 2.2.5
2 ways

if k=19
(a+1)(b+1) = 18 = 3.3.2
2 ways

10 solutions ?




2977
An ant starts from a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Find the probability that ant moves to the starting vertex on its tenth move.


he total number of two digit numbers in which the ten €™s digit is either identical, double, triple, quadruple, five times, six times, seven times or eight times the units digit is:
OPTIONS

1) 43
2) 21
3) 45
4) 22

2978
@sujamait almost there ans is 10
2979

what is the min value of 48cos x-20 sin x

2980
what is the min value of 48cos x-20 sin x