Official Quant thread for CAT 2013

MBA CAT 2024
9921
@anilapex said:
ThanksEuler method for finding remainder needs less calculation than the method given in Quant (Arun Sharma)What's Euler number & how to find remainder by this method?
Suppose a number N has prime factors p,q,r,...
N = p^a * q^b * r^c * ...

Euler number of N
E(N) = N * (1 - 1/p) * (1 - 1/q) * (1 - 1/r) * ...

For example,
N = 6 = 2 * 3
E(6) = 6 * (1 - 1/2) * (1 - 1/3) = 6 * 1/2 * 2/3 = 2

N = 7
E(7) = 7 * (1 - 1/7) = 7 * 6/7 = 6

http://www.pagalguy.com/news/remainders-reloaded-euler-fermat-wilsons-theorems-cat-2011-a-19067
9922
@anilapex said:
Find the remainder when 54^124 is divided by 17?a. 4 b. 5 c. 13 d. 15
a. 4

54^124 mod 17
= 54^12 mod 17
= 3^12 mod 17
= 10 * 10 * 10 * 10 mod 17
= -15 * -15 mod 17
= 225 mod 17
= 4 mod 17
9923
@maddy2807 said:
270?
Yes it is 270.
9924
A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?
1> 1/8
2> 1/16
3> 1/32
4> None of these
9925
@anilapex said:
Find the remainder when 54^124 is divided by 17?a. 4 b. 5 c. 13 d. 15
4?
9926
@hiteshkhurana82 said:
A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?1> 1/82> 1/163> 1/324> None of these
4. none of these ?

_ = 1
_H_ = 10C1 = 10
_H_H_ => a + b' + c = 7 => 9C2 = 36
_H_H_H_ => 8C3 = 56
_H_H_H_H_ => 7C4 = 35
_H_H_H_H_H_ => 6C5 = 6
Total = 144

Sample space = 1024

So 9/64
9927
@anilapex said:
ThanksEuler method for finding remainder needs less calculation than the method given in Quant (Arun Sharma)What's Euler number & how to find remainder by this method?
It is very easy..

If N is a prime number, Euler number of N = E(N) = N-1
It N is non-prime, then N will be of the from N = a^p*b^q*c^r where a,b,c are the prime factors of N. In this case, E(N) = N*(1 - 1/a)(1-1/b)(1-1/c)
9928
@hiteshkhurana82 said:
A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?1> 1/82> 1/163> 1/324> None of these
when no head = 1

when 1 head = 10C1

when 2 heads
_ H1 _ H2 _

a + b + c = 7
=> 9C2 = 36

when 3 heads
_H1 _ H2 _ H3 _

a + b + c + d = 5
=> 8C3 = 56

when 4 heads
_ H1 _ H2 _ H3 _ H4 _

a + b + c + d + e = 3
=> 7C4 = 35

when 5 heads
_H1 _ H2 _ H3 _ H4 _ H5 _

a + b + c + d + e + f = 1
=> 6C5 = 6

sum of all = 144
total cases = 2^10

so , prob. = 144/2^10

alternate method

when coin is tossed 1 time = 2 cases
when 2 times = 3 cases (TT , HT , TH)
when 3 times = 5 cases (TTT , THT , HTT , TTH , HTH)

so , apply fibonacci here

2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144

so , prob. = 144/2^10
9929
@hiteshkhurana82 said:
A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?1> 1/82> 1/163> 1/324> None of these
According to me : 2 heads do not occur consecutively implies we have to consider atleast 2 heads and at max 5 heads

2 heads :
_x_x
a+b+c = 7
So, 9C2 = 36

3 heads :
_x_x_x_
8C3 = 56

4 heads :
_x_x_x_x_
7C4 = 35

5 heads :
_x_x_x_x_x_
6C5 = 6

Total = 128 = 2^7

Prob = 2^7/2^10 = 1/2^3 ??
9930
@grkkrg
@jain4444 Sir, What is the difference between :
1. What is the probability that 2 heads do not occur....
AND
2. What is the probability that no head occurs consecutively... ???
Does it mean that in 1st question we have to take a minimum of 2 heads in 10 throws ??
9931
@hiteshkhurana82 said:
A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?1> 1/82> 1/163> 1/324> None of these
Case 1 : All tails (T) -> 1 way
Case 2: 1 head, 9 tails -> 10 ways
Case 3: 2 head, 8 tails -> _H_H_ => (g1 + g2 + g3 = 8, where g2 >=1) => 9C2 = 36
Case 4: 3 head, 7 tails -> _H_H_H_ => (g1 +g2 + g3 + g4 = 7) => 8C3 = 56
Case 5: 4 head, 6 tails -> _H_H_H_H_ => 7C4 = 35
Case 6: 5 head, 5 tail -> _H_H_H_H_H_ => 6C5 = 6

Total fav cases = (1 + 10 + 36 + 56 + 35 + 6) = 144
Prob = 144/1024

None of these ?
9932
@ScareCrow28 said:
@grkkrg@jain4444 Sir, What is the difference between :1. What is the probability that 2 heads do not occur....AND2. What is the probability that no head occurs consecutively... ???Does it mean that in 1st question we have to take a minimum of 2 heads in 10 throws ??
I think both mean the same.
9933
@grkkrg said:
I think both mean the same.
Zada tez banne k chakkar me galat karva baitha
9934
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
9935
@jain4444 said:
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
6 hours?

(1/2 + 2/2 + 3) x = 9
x = 2

So 3x = 6 hours
9936
@jain4444 said:
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
6 hours?
9937
@jain4444 said:
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
6 hours?
9938

@jain4444 6 hrs ?
9939
@jain4444 said:
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
Lengths = x, 2x, 3x

Burn rate = r

First candle burns out in (x/r), Second in (2x/r), Third in (3x/r)

(x/r) + (3x/2r) + (2x/r) = 9 =>9x/2r = 9 => (x/r) = 2

3x/r = 6 hours ?
9940
@jain4444 said:
There are 3 candles with their lengths in the ratio 1 : 2 : 3 (Every other dimension is the same for all the candles). They are lit in such a way that when the second candle has been lit, the first candle had been reduced to half its original length & when the third candle is lit, the second candle is half its original length. The total time taken for all the candles to totally burn out is 9 hours. Assume that the candles are lit in increasing order lengths. In how much time does the longest candle completely burn?
6 hours??