Official Quant thread for CAT 2013

@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???????????????
5(a+b) = 4(2a+b/2)
a = b

=> Workin alone they complete in 10 days
3 different numbers A, B, C has 2, 3, 4 factors respectively. when A , B and C are multiplied a new number D is produced that has n number of factors. How many different values of n are possible ?
@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???????????????
A and B's efficiency be = x and y.

x + y = 1/5
And, 2x + y/2 = 1/4
=> 4x + y = 1/2

Subtracting (i) from (ii),
=> 3x = 1/2 - 1/5 = 3/10

So, x = 1/10 and y = 1/10.
Thus, both will take 10 days each.
@viewpt said:
in mein se kaun sa sahi hai?what is wrong wid 8C2 - 4 = 24 pairs.

Yeh bhi ho chuka hai :)

http://www.pagalguy.com/posts/4770614

regards
scrabbler
@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???????????????
1/a + 1/b = 1/5
1/2a + 2/b = 1/4

solving these 2 will get ....a = 10 , b = 10 ?
@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???????????????
10 days
@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???????????????
10 days both..
@Faruq said:
3 different numbers A, B, C has 2, 3, 4 factors respectively. when A , B and C are multiplied a new number D is produced that has n number of factors. How many different values of n are possible ?
7 values = 24, 16, 10, 18, 15, 7 and 12..? ( None same, 2 same, 3 same, 4 same )
A = a
B = b^2
C = x*y OR x^3

D = a*b^2*x*y OR a*b^2*x^3 ..

Edited ( Hastiness Hurts )..
@scrabbler said:
Sir, toh sahi kaun sa hai aur kyu?
@Faruq said:
3 different numbers A, B, C has 2, 3, 4 factors respectively. when A , B and C are multiplied a new number D is produced that has n number of factors. How many different values of n are possible ?
--- Snipped ---
@viewpt said:
Sir, toh sahi kaun sa hai aur kyu?
Sir waala dekho.

I am completing my method. Some cases need to be subtracted. :splat:
@Estallar12 said:
If A, B and C = 2^2, 2^3 and 2^4. => n = 10.Now, 6 common factors and 1 different => N = 7*2.If, 5 common and 2 same or 2 diff => 2 more cases.4 common and 3 diff OR 2,1. = 2 more.3 common and 2,2 OR 1,1,2 OR 1,1,1,1 => 2 more.2 common and 2,2,1 OR 4,1 OR 2,1,1,1 OR 3,1,1 OR 3,2 => 4 ways.Total = 12 ways ?
Isme to factors = 3, 4 and 5 hogae na?
@viewpt said:
Sir, toh sahi kaun sa hai aur kyu?
The latter link explains what is wrong with that approach of 8C2 - 4....read it carefully :)

regards
scrabbler

@ScareCrow28 said:
Isme to factors = 3, 4 and 5 hogae na?
Yeah. I misinterpreted the question! :splat:

Editing my method. :|
@Faruq said:
3 different numbers A, B, C has 2, 3, 4 factors respectively. when A , B and C are multiplied a new number D is produced that has n number of factors. How many different values of n are possible ?
A is of the form p
B is of the form (q^2)
C is of the form (r*s) | (r^3)

Possible forms of D = (p^4)*s | (p^3)*(r*s) | (p*q^2*r*s) | ( p*q^3 *s )| (p^2*q^2*s) | (p^6) | (p^3)*(r^3) | (p^4)*(q^2) | (p)*(q^5) | (p*(q^2)*(r^3)

So, possible values of n = 10, 16, 24, 18, 7, 15, 12 ?

@Faruq said:
The guards of two goods trains are in the respective last carriages of their trains. The two trains A and B of length 200 m and 300 m respectively, begin to enter a 500-metre long tunnel from opposite ends. The guard of one of the trains wants to pick up a flag from the other guard. How long after the instant when the two trains begin to enter the tunnel, does this exchange occur? (Given that speed of train A = 2 m/s, speed of train B = 3 m/s)
200 sec
@gautam22 said:
@Faruq for first it can be a, and second b^2 this can be d only possible factors for A therefore a total of 10 cases?????? @YouMadFellow sir kahan galat kar raha hoon?
Thoda aur elaborate karoge ? Mujhe kuch nahi chamka :splat: .. a,b,c,d me ulajh gaya solution me :splat:
@gautam22 said:
sir ye a,b,c,d bas denote karne ke liye ki ye different hain ek doosre se
See, there are 10 possible combinations possible, but if you actually find the number of factors for each of those 10, few possibilities would be repeated.

Check my solution: I have mentioned 10 possible combinations, but only 7 distinct values are coming up.
Total no. of people in a community =100

No. of people who like amy =83
No. of people who like bill =88
No. of people who like candy = 51
No. of people who like dany =37

1.Find the max no. of persons who like exactly one out of the four?
2.Find the max no. of persons who like exactly two out of four?
3.Find the min no. of person who like atleast three out of four?

Kindly share the approach.Suggest some good links for practising such questions.


Total no. of people in a community =100

No. of people who like amy =83
No. of people who like bill =88
No. of people who like candy = 51
No. of people who like dany =37

1.Find the max no. of persons who like exactly one out of the four?
2.Find the max no. of persons who like exactly two out of four?
3.Find the min no. of person who like atleast three out of four?

Kindly share the approach.Suggest some good links for practising such questions.