Official Quant thread for CAT 2013

MBA CAT 2024
27228
@vbhvgupta said:
approach
total = 100(let)
P=10
Q=20
R=20
Others=50

after increase
total=115
P=12
Q=22
R=22

so others =59

50 --------- 59 18 %

27229

27

27230
@vbhvgupta said:
40
D. 2500 x=20%
27231
@vbhvgupta said:
24
18%
27232
@vbhvgupta said:
27
c?
27233
@jain4444 said:
Let ϕ(n) be the Euler Phi Function. If 1 ≤ n ≤ 1000, what is the smallest integer value of n that minimizes ϕ(n)/n ?
n=10?
27234

If 4 dice are rolled once, the number of ways of getting the sum atleast 10 is...


a - 870
b - 950
c - 1050
d - 1170

27235
@iLoveTorres said:
n=10?
ye wala kaise kiya??
27236
Find the largest possible real value of
root[(x−20)(y− x)] + root[(140−y)(20−x)] + root[(x−y)(y−140)]

subject to −40 ≤ x ≤ 100 and −20 ≤ y ≤ 200.
27237
@iLoveTorres said:
c?
a
27238
@Dexian said:
ye wala kaise kiya??
look i may be wrong too par mera logic kya tha toh for phi(n)/n to be min phi(n) min and n max ( but it should be a small number as well so we kno phi(prime) = prime-1. Now find two small enough prime numbers whose euler should be small.. 2,3 i.e E(6)=2 E(10)=4 E(14)=6 E(22)=11 E(26)=12. if you observe min value E(10)/10 hi dega.. as the prime number increases its Euler increases. like for E(35)=24 E(77)=60 so dono E(n) and n should be small enough so that the difference between the two has an effect on minimizing the value.

Hope samajh aaya
27239
There are 5 red, 2 green and 2 blue flags. Make a signal of 5 flags using them noting that repeating a colour is allowed.What will be the total number of possible flags if different arrangements are different signals?
OPTIONS

1) 141
2) 121
3) 127
4) 240

pls explain me..i dont understand dis solution
27240
@jain4444 said:
Find the largest possible real value of root[(x−20)(y− x)] + root[(140−y)(20−x)] + root[(x−y)(y−140)]subject to −40 ≤ x ≤ 100 and −20 ≤ y ≤ 200.
60 ?
27241
@pavimai said:
There are 5 red, 2 green and 2 blue flags. Make a signal of 5 flags using them noting that repeating a colour is allowed.What will be the total number of possible flags if different arrangements are different signals?OPTIONS1) 141 2) 121 3) 127 4) 240 pls explain me..i dont understand dis solution
when all the 5 are red you have 5!/5!=1 arranement
when 4 are red and remaining one is either green or blue you have 5!/4!*2=10
when 3 are red 2 are same (either green or blue) = (5!/3!*2!)*2!=20
when 3 are red and 2 are different 5!/3!=20
when 2 are red other 2 of one kind and 1 of the remaining kind (5!/2!*2!)*2=60
when 1 red and 2 each of the remaining kind 5!/2!*2!=30

Hence 1)141
27242
@iLoveTorres said:
look i may be wrong too par mera logic kya tha toh for phi(n)/n to be min phi(n) min and n max ( but it should be a small number as well so we kno phi(prime) = prime-1. Now find two small enough prime numbers whose euler should be small.. 2,3 i.e E(6)=2 E(10)=4 E(14)=6 E(22)=11 E(26)=12. if you observe min value E(10)/10 hi dega.. as the prime number increases its Euler increases. like for E(35)=24 E(77)=60 so dono E(n) and n should be small enough so that the difference between the two has an effect on minimizing the value.Hope samajh aaya
let n = 9. Then gcd(9, 3) = gcd(9, 6) = 3 and gcd(9, 9) = 9. The other six numbers in the range 1 ≤ k ≤ 9, that is, 1, 2, 4, 5, 7 and 8, are relatively prime to 9. Therefore, φ(9) = 6. As another example, φ(1) = 1 since gcd(1, 1) = 1.
http://en.wikipedia.org/wiki/Euler%27s_totient_function
27243

pls suggest me some ebooks for CAT preperation....thanks in advance


27244
@iLoveTorres said:
look i may be wrong too par mera logic kya tha toh for phi(n)/n to be min phi(n) min and n max ( but it should be a small number as well so we kno phi(prime) = prime-1. Now find two small enough prime numbers whose euler should be small.. 2,3 i.e E(6)=2 E(10)=4 E(14)=6 E(22)=11 E(26)=12. if you observe min value E(10)/10 hi dega.. as the prime number increases its Euler increases. like for E(35)=24 E(77)=60 so dono E(n) and n should be small enough so that the difference between the two has an effect on minimizing the value.Hope samajh aaya
its 210

E(n) = n*(1 - 1/a)*(1 - 1/b)........

"a" and "b" are prime factors of "n"

we will get minimum of E(n) when we would have n = a*b*c......

i mean we should not have any extra power of any prime factor

so , 2*3*5*7 = 210
27245
Find the sum of all the real values of €˜x €™ that satisfy the equation:


243^log81(x) -2x = 2^log16(x)+2 -8

PS: log x and the no is base , very difficult to type log
27246
@pavimai 141 is it

27247
@jain4444 pls suggest me some ebooks for CAT preperation....