Official Quant thread for CAT 2013

@vbhvgupta said:
@joyjitpal Find the no of no between 200 and 300 , both incuded , which are not divisible by 2,3,4,5.How to go for it?
300*1/2*2/3*4/5 equals to 80

210*1/2*2/3*4/5 equals to 56



80-56 is 24

now add 2 for 203 and 207



so 24 plus 2 equals to 26


@krum


@joyjitpal My method is different but the logic is same.

At 8am, the difference between hour hand and minute hand is 240 degree. Now we need the time when the difference between them is 180 degree. So both hands move at a relative speed of 5.5 degree.

Time when both hands will be at a distance of 180 degree will be => (240-180)/5.5

=> 60/5.5 => 120/11

time will be 8am + 120/11

Similarly , for 2 pm , the difference is 60 degree between both the hands,

2:00-2:10 ( approx) = 60 degree
2:10-2:41( approx) = 180 degree

Therefore, time will be (60 + 180)/5.5 => 480/11

Note: the minute hand first cover 60 degree approx. ( a bit more) to coincide with the hour hand first and then together they will have to maintain a distance of 180 degree at the relative speed.

time will be 2pm + 480/11

now difference in time will be (2pm + 480/11) - (8am + 120/11)

6hr + 360/11

6hr Answer.... ( 360/11 means they return to the same point of time...)

Hope it helps..!!!
@vbhvgupta said:
@joyjitpal Find the no of no between 200 and 300 , both incuded , which are not divisible by 2,3,4,5.How to go for it?
Is it 26??
This should done with simple set theory...
300*1/2*2/3*4/5 - 199*1/2*2/3*4/5 = 26.93
Hence 26 numbers
@vbhvgupta said:
@joyjitpal Find the no of no between 200 and 300 , both incuded , which are not divisible by 2,3,4,5.How to go for it?
26?

Q1

A number n^2 has 11 factors less than n. find in how many ways can n^3 be written as product of 2 distinct nos ?

@vbhvgupta said:
Q1
3^32mod50
=(243)^6*9mod50
=49^3*9mod50
=41

41/50 = .82

x=2

@vbhvgupta e) none of these..... is it correct???
@vbhvgupta e) none of these..... is it correct???
@Zedai said:
A number n^2 has 11 factors less than n. find in how many ways can n^3 be written as product of 2 distinct nos ?
n^2 has 23 factors
so n=a^11

n^3=a^33

so 17 ways
@vbhvgupta said:
Q1
2?
@vbhvgupta e) none of these..... is it correct???

Birbal says to akhbar "I am thrice as old as you were when i was as old as you are" . If the sum of their present ages is 80 years,then how many years ago was birbal twice as old as akhbar ?


@vbhvgupta said:
Q1
x=2
@jaituteja said:
@vbhvgupta e) none of these..... is it correct???
ANs is 2
@nole said:
Birbal says to akhbar "I am thrice as old as you were when i was as old as you are" . If the sum of their present ages is 80 years,then how many years ago was birbal twice as old as akhbar ?
b=3(a-(b-a))
b=-3b+6a
2b=3a

a+b=80

==> a=32, b=48

16 years ago
@Zedai said:
A number n^2 has 11 factors less than n. find in how many ways can n^3 be written as product of 2 distinct nos ?
@krum bhai, after seeing you getting the right answer. i think the question is wrongly printed in the book.
i was banging my head against the wall thinking over the situation from half an hour!
OA is 17.
@vbhvgupta i got it.. thanks..!!!
Find the last non-zero digit of 54!
@bs0409 said:
Find the last non-zero digit of 54!
54 = 5*10+4

R(2^10 * R(10!) * R(4!))

= R(4*8*4)

=8

@Zedai what was printed in the book