Official Quant thread for CAT 2013

MBA CAT 2024
17006
Rem(123123123.....300 digits)/504?
17007
Rem of 7777.....(2n+1)times/1232
17008
@Calvin4ever said:
Anjali and Bhagwat fired 45 shots each.Total of 66 bullets hit the target and the remaining bullets missed it. How many times does Anjali hit the target if it is known that the number of hits per one miss shown by Anjali is twice that of Bhagwat ?@Logrhythm
A(h) + B(h) = 66
A+B = 66
A(m) + B(m) = 24
A(t) + B(t) = 90

solve karlo bhai...waise aese type ke ques options se kara karo..
17009
@sunbabaphu here.
17010
@TootaHuaDil said:
Rem(123123123.....300 digits)/504?
504 = 8*9*7

123%8 = 3
(1+2+3)*100%9 = 600%9 = 6
any digit repeated 6 times is div by 7...
123123%7 = 0
123123123123%7 = 0
123123123123123123%7 = 0
hence, 123*6k%7 = 0

so remainder is of the form -> 8x+3 = 9y+6 = 7z....
17011
@TootaHuaDil said:
Rem of 7777.....(2n+1)times/1232
let n = 1
1232 = 16*77
777%16 = 9
777%77 = 7

so remainder -> 16x+9 = 77y+7...
17012
@TootaHuaDil said:
Rem(123123123.....300 digits)/504?
3?
17013
@bs0409 said:
z=1^2-2^2 +3^2.............-2998^2+2999^2a) 2999000b)5998000c)4498500 d)2249250
1^2-2^2 +3^2.............-2998^2+2999^2 = 1+5+9+13+....+(2(2999)-1)
total terms in this series = 1500
so sum = 1500/2*5998 = 4498500
17014
@Pratishruti said:
3?
No OA is 483
17015
@TootaHuaDil said:
Rem(123123123.....300 digits)/504?

8a+3=9b+6=7c
72d+51=7c

remainder = 483


@TootaHuaDil said:
Rem of 7777.....(2n+1)times/1232

1232=2*2*2*2*7*11.
777777.....(2n+1)time can be written as
77777777.....00000+77777
now 77777777.....00000 is divisible by 2*2*2*2*7*11
so you get the remainder by 77777(mod 1232)= 161(mod1232).
17016
Try this -

Find the sum of all positive integers less than 300 and relatively prime to 150.
17017
2 coins are going to be flipped
>It is guaranteed that at least one out of the two flips will turn up heads

What is the probability that both coins will come up heads?
17018
@naveenkrs said:
Try this -Find the sum of all positive integers less than 300 and relatively prime to 150.
300(1-1/2)(1-1/5)(1-1/3)
=> 80 ?
17019
@veertamizhan said:
2 coins are going to be flipped
>It is guaranteed that at least one out of the two flips will turn up heads

What is the probability that both coins will come up heads?
1/2 ?
17020
@gs4890 said:
300(1-1/2)(1-1/5)(1-1/3)
=> 80 ?
bhai , i guess 80 wil be the count of such nos. SUM shud be 150 x 80 = 12000...
correct me if m wrong..
17021
@karan20 said:
bhai , i guess 80 wil be the count of such nos. SUM shud be 150 x 80 = 12000...correct me if m wrong..
17022
@TootaHuaDil said:
A palindromic number with even number of digits in base 16 will always be divisible by?(a)2(b)17(c)15(d)7
(abba)16 = 4096a + 256b + 16b + a = 4097a + 272b = 17 (241a + 16b ) ==.> so,divisble by 17
17023
@naveenkrs said:
Try this -Find the sum of all positive integers less than 300 and relatively prime to 150.
300 = 5^2 * 3 * 2^2

E ( 300 ) = 300 * 4/5 * 2/3* 1/2 = 80 .

Sum of all co-primes less than 150 = 150/2 * 80 = 6000??
17024
@karan20 said:
bhai , i guess 80 wil be the count of such nos. SUM shud be 150 x 80 = 12000...correct me if m wrong..
@gs4890 said:
bhai if its less than 300 then its 300/2 * 80 ........but here its asked for less than 150 so shud it be 150/2 * 80 ?? correct me if m wrng :(
17025

@naveenkrs said:
Try this -Find the sum of all positive integers less than 300 and relatively prime to 150.
whats the OA, know only how to calculate less than 150