Official Quant thread for CAT 2013

MBA CAT 2024
14563
@ishu1991 said:
@TootaHuaDil(100-2)5 + (100-1)5+(100+1)5 + (100+2)598^5 + 102^5 = 2(100^5 + 5C2 100^3 (2)^2 +5C4 100 (2)^4)99^5 + 101^5 = 2((100^5 + 5C2 100^3 (1)^2 +5C4 100 (1)^4)so the sum would be:4 100^5 + 10^7(2^3+2) + 1000 (2^4+1) finally the prob condenses torem(17000 /400) == 200
yes...right :)
14564
@TootaHuaDil 0 ?
14565
@TootaHuaDil said:
What is the remainder when 98^5 + 99^5 + 101^5 + 102^5 is divided by 400? (1)0 (2) 100 (3) 200 (4) 300 (5) none of these
200?
14566

Do this........

1234123412341234.......................(400 digits) mod 909=?

14567
@techgeek2050 said:
@x2maverickc nope. this question was in xat 2013 so i dunno the answer. the options given are 1.) 26 2.)25 3.)24 4.)23 5.) 22


Options make everything look so simple.
I mean if B is a point inside circle and C on the circle with BC=2.
then OC=root50
hence OB=root50 - 2
the square of which is 50+4-4_/50 which comes a lil over 54-28=26
14568
@x2maverickc said:
Options make everything look so simple.I mean if B is a point inside circle and C on the circle with BC=2.then OC=root50hence OB=root50 - 2the square of which is 50+4-4_/50 which comes a lil over 54-28=26
o,b,c arent given colinear..then how u got OB?
14569
@bs0409 said:
Do this........1234123412341234.......................(400 digits) mod 909=?
909=9*101
N mod 9 = 1(using sum of digits)
N mod 101=57(using alternate sum of digits)

now by chinese remainder thm.
N mod 909=764
14570
@x2maverickc is it necessary that OBC will lie in a straight line?
14571
@bs0409 said:
Do this........1234123412341234.......................(400 digits) mod 909=?
1234 * 400 / 4mod 909

=>1234*100 mod 909

=> 685 will be answer.




14572
@TootaHuaDil said:
yes...right
a^n + b^n + c^n + d ^n... is divisible by a+ +b+ c+ d ... when n is odd..isn't this true??
14573
@anuragu79 said:
o,b,c arent given colinear..then how u got OB?
EDIT: take back my words.
God save me.
14574
@anuragu79 said:
a^n + b^n + c^n + d ^n... is divisible by a+ +b+ c+ d ... when n is odd..isn't this true??
It is true only when a,b,c and d are in ARITHMETIC PROGRESSION and n is odd.
14575
There are 10 friends, each individual has some message. How many minimum telephone calls will be needed, so that evry1 has all messages?
14576
@x2maverickc said:
EDIT: take back my words. God save me.
then solution?
14577
@techgeek2050 said:
The radius of a circle with center o is root 50 cm. A and C are two points on the circle ,and B is a point inside the circle . The length of AB is 6 cm, and the length of BC is 2 cm . The angle ABC is a right angle . find the square of the distance OB
Got it somewhere. Posting the solution. Solved by Kamal lohia Sir.
clearly required distance OB is hypotenuse of right triangle shown in Red, whose square equals (196 + 64)/10 = 26
14578
@TootaHuaDil thanks bro!!
14579

And can someone explain this solution?? 😞

14580
@nagpal9 said:
There are 10 friends, each individual has some message. How many minimum telephone calls will be needed, so that evry1 has all messages?
55?
14581
@x2maverickc said:
909=9*101
N mod 9 = 1(using sum of digits)
N mod 101=57(using alternate sum of digits)

now by chinese remainder thm.
N mod 909=764
_/\_ Bhai, can you explain chinese remiander theorem.?
N mod k.,
we segregate k into 2 coprimes a and b., then.?
14582
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?