Official Quant thread for CAT 2013

@sameersapre23 said:
Aryan and Nandini are siblings. They participated in two different competition i.e. swimming and wrestling respectively. Aryan participated in swimming and Nandini in wrestling. Both were awarded rank between 1 to 50 (including 1 and 50). What is the probability that difference between their ranks is not more than 10 rank?9 / 2517 / 4040 / 1003 / 10
Total Cases = 50*50 = 2500.

For Rank 1 of Aryan, Nandini has 11 options.
For Rank 2 => Nandini has 12 options.
This will go on increasing till Aryan's Rank reaches 11.

For Aryan's rank between 12 to 40, we will have 21 options each for Nandini's Rank.

And now from 41 to 50, Nandini's Rank (options) will decrease from 20 to 11.

Total = 2*(11 + 12 + .... + 20) + 21 + 21*29 = 310 + 21 + 609 = 940.

So, Probability = 940/2500 = 9/25
@YouMadFellow said:
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Kya hua sirjee? :O
@YouMadFellow said:
..
KAA HUA..
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q: 67^73/ 100 , R?

@Faruq said:
A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
36 ?
x/8 = 5(7/8x/(n-1))
n - 1 = 35
n = 36

@Faruq said:
EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
37
@Faruq said:
A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
Suppose total toffees = 8x.

=> x = 5*7x/(x-1)
=> x - 1 = 35.

So, x = 36.
@viewpt said:
KAA HUA..-----------q: 67^73/ 100 , R?
N mod 100 = 67^73 mod 100 = 67*( 67^4) ^18 mod 100

Last two digits of 67 *( 89 ^2) ^18 = 67 *( 21)^18 = 67*(61) = 87 ?

There could be calculation mistakes

Arey sab sirf 36 likh rahe hai .. no approach 😐 .. Isliye ..
@viewpt said:
KAA HUA..-----------q: 67^73/ 100 , R?
87...
Q: Q: 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired?Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.

ans is not with me/ i don't have ans only/

@YouMadFellow said:
N mod 100 = 67^73 mod 100 = 67*( 67^4) ^18 mod 100 Last two digits of 67 *( 89 ^2) ^18 = 67 *( 21)^18 = 67*(61) = 87 ? There could be calculation mistakes Arey sab sirf 36 likh rahe hai .. no approach .. Isliye ..
In office...hence, cannot post the approach...pardon me..
@viewpt said:
KAA HUA..-----------q: 67^73/ 100 , R?
100 = 4*25

67^73 mod 4 = (-1)^73 = -1 = 3.

67^73 mod 25 = 17^73 mod 25 = 17^72*17 mod 25 = 14^36*17 mod 25 = (-4)^18*17 mod 25 = 4^18*17 mod 25 = 6^4*16*17 mod 25 = 1296*272 mod 25 = 21*22 = 462 mod 25 = 12.

So, 4p + 3 = 25q + 12
q = 3 and Remainder = 87.
@viewpt said:
KAA HUA..-----------q: 67^73/ 100 , R?
87 ???
@Faruq said:
A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
36
@viewpt said:
Q: Q: 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired?Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.ans is not with me/ i don't have ans only/
Already Discussed in the morning today..Check the solution by @scrabbler
http://www.pagalguy.com/posts/4770506
@YouMadFellow said:
N mod 100 = 67^73 mod 100 = 67*( 67^4) ^18 mod 100 Last two digits of 67 *( 89 ^2) ^18 = 67 *( 21)^18 = 67*(61) = 87 ?
Smart Fellow. ^^ :mg: :P
I opted the long method :splat: :|
@viewpt said:
Q: Q: 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired?Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.ans is not with me/ i don't have ans only/
8C4*4!* [ 1 - 1/1! + 1/2! - 1/3! + 1/4!] = 8C4*5 ways.

Now subtracting repeated cases.....
@viewpt said:
Q: Q: 4 couples of husband-wife go to a dance party where they are supposed to dance in pairs. They decide that no one will dance with his/her partner. In how many ways all of they can be paired?Remember that it is 21st century and a man can be paired with man and similarly a woman can also be paired with woman.ans is not with me/ i don't have ans only/
http://www.pagalguy.com/posts/4770506

regards
scrabbler

@Estallar12 said:
Smart Fellow. ^^ I opted the long method 8C2 - 4 = 24 pairs.
@scrabbler said:
in mein se kaun sa sahi hai?
what is wrong wid 8C2 - 4 = 24 pairs.
Working together A and B can complete a job in 5 days. If A work twice as fast as earlier and B work half as fast as earlier then it would take them 4 days to complete the work. find the number of days when A and B alone can complete the same job???????????????

@viewpt said:
KAA HUA..-----------q: 67^73/ 100 , R?
87 ?

E(100) = 40

67^73 mod 100 = 67^33 mod 100 = 67*(67^2)^16 mod 100 = 67*(89^2)^8 mod 100 = 67*(21^2)^4 mod 100 = 67*81*81 mod 100 = 87
@Faruq said:
Working together A and B can complete a job in 5 days. If A work twice as fast as earlier and B work half as fast as earlier then it would take them 4 days to complete the work. find the number of days when A and B alone can complete the same job???????????????
:wow:

One day's work by A = a
One day's work by B = b

(a + b) = 1/5
(2a + b/2) = 1/4 => 4a + b = 1/2

3a = 1/2 - 1/5 => a = 1/10 => b = 1/5 - 1/10 = 1/10

=> Each takes 10 days ?