Official Quant thread for CAT 2013

MBA CAT 2024
7421
@kingsleyx said:
P is a prime number greater than 37. Then the largest number that will always divide (P-1)(P+1) is ______
for all such prime numbers, two continuous even numbers will always leave at least 3*2^3

24
7422
@Brooklyn said:
yar question is : q+r to be divisible by 11 !! not n
check my solution, you will know why n is also divisible by 11, if (q + r) is divisible by 11
7423
@jain4444 said:
Let n be a 5-digit number, and let q and r be the quotient and remainder, respectively, when n is divided by 100. For how many values of n is q + r divisible by 11?
Q will be of three digit(abc) and r will be of 2(de) q+r should be of 11k type
100a+10b+c+10d+e = 11k = mod 11 =0
a-b+c-d+e=0
This is the property for divisibility by 11
abcde mod 11=0 means all 5 digit numbers divisible by 11 = 8181
7424
@Cat.Aspirant123 said:
P and Q are 120 km apart. A starts from P towards Q at 6 a.m. B starts from Q towards P at 11a.m. A is 50% faster than B. They cross each other at 8 p.m. A €™s speed (in km/hr) is _____.
120-5(1.5)x/(2.5)x=9
120-7.5x=22.5x
120=30x
x=4
A speed is 6km/hr
7425
@arumugadas said:
jain bhai throw some light on this
@Brooklyn said:
@jain4444sir logic i thnk to this problem:q can range b/w 100 to 9999r can range b/w 0 to 99now to get q+r divisble by 11 we need either both q = r=11k typeor (11K+5, 11+6), (11K+2, 11K+9) so on??

100*q + r

least value for which (q + r) is divisible by 11 is 10010 and 10010/11 = 910

largest value for the same is 99990 and 99990/11 = 9090

in between 9090 and 910 all the values will be possible (including both)

so , number of total values will be = 9090 - 910 + 1 = 8181
7426
@arumugadas said:
the answer sheet of 5 eng students can be checked by one of the 9 prof what is the probability that all the 5 answwer sheets are checked by exactly two professors
9C2*30/(9C1+9C2*30+9C3*150+9C4*240+9C5*120)
7427
@Cat.Aspirant123 said:
If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
1025?
7428
For a regular nonagon ABCDEFGHI which of the following holds true.

AF = AB + AC
AF > AB + AC
AF AF = 2AB+(AC/2)
7429
7430
7431
@ashishsoul said:
Kya karna kya chah rahe ho bhai ...?
7432
@Torque024 af=ab+ac
7433
@Torque024 said:
For a regular nonagon ABCDEFGHI which of the following holds true.AF = AB + ACAF > AB + ACAF
By looking at the figure..
i guess
AF> AB + AC
7434
@Torque024 said:
For a regular nonagon ABCDEFGHI which of the following holds true.AF = AB + ACAF > AB + ACAF
taking regular side=a
AF=asin80/sin20=2.87a
AC=asin140/sin20=1.87a
then AC+AB=1.87a+a=2.87a=AF.......hence option 1?
7435
@htomar
@19rsb http://www.qbyte.org/puzzles/p091s.html ye lo answerr bhai ab mat puchna kaise hua bas cat ke liye prepare karte karte samne aya tha mere :P
7436
If T1 = 1 , and Tn = (Tn-1+ 1)/Tn-1 , find T75

a) 9-12
b) 12-15
c) 15-18
d)None of the above
Mine is coming 1.98, but OA is B) . Anyone solve this out.
Tn-1= T(n-1)
7437
@Torque024 said:
If T1 = 1 , and Tn = (Tn-1+ 1)/Tn-1 , find T75a) 9-12b) 12-15c) 15-18d)None of the above
Mine is coming 1.98, but OA is B) . Anyone solve this out.
Tn-1= T(n-1)
Getting ratio of fibbonaci terms (golden ratio)
so answer should be 1.618
7438
@techsurge said:
Getting ratio of fibbonaci terms (golden ratio)so answer should be 1.618
I proceeded this way,
T75=T74+1/T74 = (2(T73)+1)/(T73+1) = 3(T72)+2/(2(T72)+1)
= 74(T1)+73/(73(T1)+1)= 74+73/73+1 = 74+73/74 = 1+(73/74)=1.98xyz
7439
Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?
7440
@Torque024 said:
I proceeded this way,
T75=T74+1/T74 = (2(T73)+1)/(T73+1) = 3(T72)+2/(2(T72)+1)
= 74(T1)+73/(73(T1)+1)= 74+73/73+1 = 74+73/74 = 1+(73/74)=1.98xyz
f(1) =1
f(2)=2/1
f(3)=3/2
f(4) =5/3
f(5) =8/5
f(6) 13/8
As these are nothing but fibonnaci terms
The ratio of n+1 th fibbonaci term by nth term can be written as 1.618 for large n
(if i am correct)