Official Quant thread for CAT 2013

MBA CAT 2024
8001
@vijay_chandola said:
How many positive integers less than a million can be formed with the digits 0, 7, 8 and 9?a) 4095b) 4096c) 3072d) 3075
4095.
all the six places can be filled by any of the four numbers. and since one of the number would be 0 with all six digits , we subtract that

hence 4^6-1= 4095

8002

How many digits from 100 to 999 including both doesnt contain 2,3,4,5??

8003
@Shray14 said:
How many digits from 100 to 999 including both doesnt contain 2,3,4,5??
I think bold underlined word should be numbers

abc,
a can be 1, 6, 7, 8, 9
b can be 0, 1, 6, 7, 8, 9
c can be 0, 1, 6, 7, 8, 9

So, 5*6*6 = 180 such numbers

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
a) 3/4
b) 3/8
c) 1/2
d) 2/3

8004
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
Bro INTERGER..

2/3??
8005
@Shray14 said:
How many digits from 100 to 999 including both doesnt contain 2,3,4,5??
Total digits = 900*3 = 2700

Digit 2 appears 280 times between 100 and 999 [ 20 times in each hundred, and extra 100 times from 200 to 299]
Similarly, other digits also appear 280 times

So, Result = 2700 - (280*4) = 1580 digits ?

If numbers instead of digits, then consider only 0,1,6,7,8,9
So, three digit numbers which contain only these digits = 6^3 - 6^2 = 180 numbers?
8006
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
3/8 ??
8007
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
3/8?
8008
@chillfactor said:

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
P(truth) = 3/4
P(6 with die) = 1/6

P(man reports 6) = P(6 with die)*(P(truth)) + P(not 6 with die)*P(false) = 1/6*3/4 + 5/6*1/4 = 8/24

P(actually 6) = (3/24)/(8/24) = 3/8 ?
8009
@Shray14 said:
apne kaise kiya
he said it is 6
total cases= 1/6*3/4 ( it is actually 6) + 5/6*1/4 ( it is not 6)
req prob= (1/6*3/4)/(1/6*3/4 + 5/6*1/4) = 3/8
8010
@Shray14 said:
How many digits from 100 to 999 including both doesnt contain 2,3,4,5??
_ _ _
Digits to use 10-4=6, 0 can't be there in first place so
5*6*6 = 180
8011
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
b)
(1/6 * 3/4)/(1/6*3/4 + 5/6*1/4)= 3/8

8012
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
(1/6)*(3/4)/[(1/6)*(3/4)]+5[(1/6)1/4]
3/8
8013
Let N = (1+10)⁰+(1+10)¹+(1+10)²+(1+10)³+....+(1+10)³⁴⁵
What is the digit sum of N?
1) 3
2) 6
3) 7
4) 9
5) None

No OA :), want to confirm. If you are posting only the answer, and not the approach, I will simply ignore the post 😃 Sorry in advance and Thanks in advance too :mg:
8014

@soham2208 said:
Let N = (1+10)⁰+(1+10)¹+(1+10)²+(1+10)³+....+(1+10)³⁴⁵What is the digit sum of N?1) 32) 63) 74) 95) NoneNo OA , want to confirm. If you are posting only the answer, and not the approach, I will simply ignore the post Sorry in advance and Thanks in advance too
Sum of digits of N=Nmod9
N=2^0+2^1...2^345
1(2^346 -1 )=2(8)^115-1=-2-1=-3=6
Sum of digits=6?
8015
@soham2208 said:
Let N = (1+10)⁰+(1+10)¹+(1+10)²+(1+10)³+....+(1+10)³⁴⁵What is the digit sum of N?1) 32) 63) 74) 95) NoneNo OA , want to confirm. If you are posting only the answer, and not the approach, I will simply ignore the post Sorry in advance and Thanks in advance too
N= 11^0 + 11^1 + so on till till 11^345

digit sum = N mod 9

so we get : 2^0 + 2^1 so on

E(9)=6


so after every 6 nos we get same result

342*(2^0+2^1+2^2+2^3+2^4+2^5) + 2^0+2^1+2^2 mod 9

we get 7


8016
@soham2208 said:
Let N = (1+10)⁰+(1+10)¹+(1+10)²+(1+10)³+....+(1+10)³⁴⁵What is the digit sum of N?1) 32) 63) 74) 95) NoneNo OA , want to confirm. If you are posting only the answer, and not the approach, I will simply ignore the post Sorry in advance and Thanks in advance too
1+11+11^2+11^3+....
Remainder by 9 follows a series : 1, 2, 4, 8, -2, -4 repeated till 11^342
Left is remainder with 1+2+4 = 7?
8017
1. A bag contains 6 white and 9 black balls. Three drawings of 1 ball each are made such that
(i) the balls are replaced before the next drawing (ii) balls are not replaced before the next drawing. Find the probability that all the black balls are drawn in each case.
8018
@Shray14 said:
Bro INTERGER..2/3??
@Brooklyn said:
3/8 ??
@maddy2807 said:
3/8?
@soham2208 said:
P(truth) = 3/4P(6 with die) = 1/6P(man reports 6) = P(6 with die)*(P(truth)) + P(not 6 with die)*P(false) = 1/6*3/4 + 5/6*1/4 = 8/24P(actually 6) = (3/24)/(8/24) = 3/8 ?
@maddy2807 said:
he said it is 6total cases= 1/6*3/4 ( it is actually 6) + 5/6*1/4 ( it is not 6)req prob= (1/6*3/4)/(1/6*3/4 + 5/6*1/4) = 3/8
@ScareCrow28 said:
b)(1/6 * 3/4)/(1/6*3/4 + 5/6*1/4)= 3/8
This question is solved example of NCERT and answer given is 3/8

But I'm getting 3/4

P(truth) = 3/4
P(6 with die) = 1/6
P(man reports 6) = P(6 with die)*(P(truth)) + P(not 6 with die)*P(false)*P(among 5 wrong possibilities he chose 6) = (1/6)*(3/4) + (5/6)*(1/4)*(1/5) = 4/24

P(actually 6) = 3/4

Alternatively

In 'abc', 'a' signifies the number that appears on the dice, 'b' is showing whether the person lied or told the truth (T for truth and L for lie) and 'c' signifies the number that the person said

Sample space:-
1T1, 2T2, 3T3, 4T4, 5T5, 6T6, 1L2, 1L3, 1L4, 1L5, 1L6, 2L1, 2L3, 2L4, 2L5, 2L6, 3L1, 3L2, 3L4, 3L5, 3L6, 4L1, 4L2, 4L3, 4L5, 4L6, 5L1, 5L2, 5L3, 5L4, 5L6, 6L1, 6L2, 6L3, 6L4, 6L5

P(1T1) = P(2T2) = P(3T3) = P(4T4) = P(5T5) = P(6T6)
and their sum = 3/4
So, P(1T1) = P(2T2) = P(3T3) = P(4T4) = P(5T5) = P(6T6) = 1/8

Also, probability of rest 30 cases is 1/4 and all are equiprobable. So, probability of each case will be
P = (1/4)/30 = 1/120

Required probability = (1/8)/{(1/8) + 5(1/120)} = 3/4
8019
@chillfactor : i was confused b/w 3/8 and 3/4 n chose 3/8 as i had reasd this q before ! :banghead:
8020
@SAHILVANSIL said:
1. A bag contains 6 white and 9 black balls. Three drawings of 1 ball each are made such that (i) the balls are replaced before the next drawing (ii) balls are not replaced before the next drawing. Find the probability that all the black balls are drawn in each case.
P(Replaced)=(9/15)^3
P(!Replaced)=(9*8*7)/(15*14*13)
P(For both)=[(9/15)^3]*(9*8*7)/(15*14*13)