Official Quant thread for CAT 2013

MBA CAT 2024
29145
@meow14 A)The integer sequence x1,x2,x3.....is such that x(n+2)=x(n+1)-x(n) for n>=1. Sum of first 1492 terms is 1985 and of frst 1985 terms is 1492. Find sum of first 2007 terms.
1)3477 2)986 3)493 4)1491 5)573
B)Find no. of terms common in series:
S1=1,3,6,9,10,15.......200 terms
S2=3,6,9,12,15..........200 terms
1)21 2)22 3)132 4)133 5)None of these


A) a b b-a -a -b -b+a a b
terms repeating after every 6
so sum of every 6 terms is zero..
now can be solved further...
29146
@Tusharrr said:
As x ranges over all real values, what is the maximum value of root [ (x^2−44x+288) × (−x2+100x−2304) ]?
{Wow, this is way outside CAT level. Olympiad se uthake laye ho kya? :splat:}

Anyway...
root [ (x^2−44x+288) × (−x2+100x−2304) ]

=root [ (x−36)(x−8) × (x−36)(64−x) ]
=(x-36)root[(x−8)(64−x)]
Now this is symmetric around 36, and the answer has to lie between 8 and 64 because beyond that the root part becomes imaginary. So there will be two solutions equidistant from 36 (say 36+k and 36-k)

Suppose one of the values we want is x = 36 + k. Then the function is
(x-36)root[(x−8)(64−x)]
=(k)root[(28-k)(k+28)]
=root[(k^2)*(784-k^2)]
which will be maximum when k^2 = 784 - k^2 => k^2 = 392 and the max value will also be root (392 * 392) = 392.

regards
scrabbler
29147
@meow14 said:
:HELP:A)The integer sequence x1,x2,x3.....is such that x(n+2)=x(n+1)-x(n) for n>=1. Sum of first 1492 terms is 1985 and of frst 1985 terms is 1492. Find sum of first 2007 terms.1)3477 2)986 3)493 4)1491 5)573B)Find no. of terms common in series:S1=1,3,6,10,15.......200 termsS2=3,6,9,12,15..........200 terms1)21 2)22 3)132 4)133 5)None of these
I get 23 😞

regards
scrabbler
29148
@Buck.up said:
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
8C3 = 56?

regards
scrabbler
29149
@Buck.up
a+b+c+d=7
b,c>1
a+b'+c'+d=5
5+4-1 c3
8c3=56 ways
29150
@Buck.up said:
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
8c3 = 56
29151

Determine the number of subsets of A={1,2, €Ś,10} whose sum of elements are greater than or equal to 28.

29152
@Tusharrr said:
Determine the number of subsets of A={1,2, €Ś,10} whose sum of elements are greater than or equal to 28.
Wild guess - 512?

regards
scrabbler
29153
@Buck.up said:
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
56
29154
@Buck.up said:
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
56.
29155
@scrabbler said:
I get 23 regardsscrabbler
@ChirpiBird said:
56.
@techgeek2050 said:
56
Plz isse elaborate kar do
29156
@Buck.up said:
Plz isse elaborate kar do
let me try to explain

we have to select 3 houses out of 10 houses..

for these 3 houses not to be adjacent, these have to be selected from in between of the 7 houses

7 houses have 8 gaps between them and these 3 houses can be placed in 8c3 ways

^here i have actually backtracked the problem

hope u get it
29157
@meow14 said:
B)Find no. of terms common in series:S1=1,3,6,10,15.......200 termsS2=3,6,9,12,15..........200 terms1)21 2)22 3)132 4)133 5)None of these
First series is summation (1, 1+2, 1+2+3 and so on). Now in this 2 out of every three will be divisible by 3 (can figure this n(n+1)/2 se, or just observe the pattern.)

Now the second series is multiples of 3 up to 600. So we need to find how many summation values are there up to 600 and there are 35 such (since summation 35 34 = 35 *34/2 = 595) out of which 12 will not be divisible by 3 and hence 35-12 = 23 34 - 12 = 22 common ones. Hence 22

Edited for stupid error :)

regards
scrabbler
29158
@Buck.up said:
Plz isse elaborate kar do
Out of 10, he has to choose 3 such that they are not adjacent. First place the 7 houses like this

_ x _x _x _x _x _x _x _

Now there are 8 spaces for 3 houses. When the 3 houses are kept in these spaces they'll always be separated by at least 1 house. so they can be chosen in 8C3 = 56 ways.


29159
@Buck.up said:
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
8C3

29160
@Logrhythm said:
let me try to explainwe have to select 3 houses out of 10 houses..for these 3 houses not to be adjacent, these have to be selected from in between of the 7 houses7 houses have 8 gaps between them and these 3 houses can be placed in 8c3 ways ^here i have actually backtracked the problem hope u get it
Very elegant and clear explanation, my explanation would involve a great deal more hand-waving and equation-mongering :splat:

regards
scrabbler
29161
@scrabbler said:
out of which 12 will not be divisible by 3

12? why?

29162

xxxxxx(A)xxxxxx(B)xxxxxx(C)xxxxxxx

p q r s

Lets assume thief will choose A,B,and C now rest p+q+r+s=7 now q and r cant be 0
so p+(q`+1)+(r`+1)+s=7
p+q`+r`+s=5
so non negative solution (5+4-1)C(4-1)=8C3
29163
@scrabbler said:
First series is summation (1, 1+2, 1+2+3 and so on). Now in this 2 out of every three will be divisible by 3 (can figure this n(n+1)/2 se, or just observe the pattern.)Now the second series is multiples of 3 up to 600. So we need to find how many summation values are there up to 600 and there are 35 such (since summation 35 = 35 *34/2 = 595) out of which 12 will not be divisible by 3 and hence 35-12 = 23 common ones. Hence None of These I guess, if I have not done something stupid?regardsscrabbler
but i got 22
29164
@meow14 said:
12? why?
Because if you see the original series we saw 2 out of 3 are divisible...so 1 out of 3 is not...so in 34, the 1st, 4th, 7th,....34th term (i.e. 12 terms) won't be divisible.

regards
scrabbler