official quant thread for cat08

[prakharc]

Originally Posted by dare2

one more with a bit higher difficulty level.

At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year, what position in line gives you the best chance of being the first duplicate birthday?

Is it 183 ??

[prakharc]

Originally Posted by dare2

one more with a bit higher difficulty level.

At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year, what position in line gives you the best chance of being the first duplicate birthday?

Okay one more solution..is the answer 41 ??

[shivam_01]

Originally Posted by dare2

@shivam.
post the answer with the solution.

Originally Posted by romy4allcse
4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5
soln: N = (1*3*5*7..........*97*99)/(2*4*6*...*100)
= (1/2)*(3/4)*(5/6)*............*(97/9*(99/100)
here we can see that lowest value is 0.5 and after that the value of the factors is increasing n reaches to 0.99
and the multiplication value gets lower and lower so it vl be less than 0.5 i.e 1/2 for sure
compute it for 3-4 factores
value goes down so as i m not able to find the exact soln
hence by seeing the variation i vl go with my "Tukka"
option(c)
wat is the ans.??

My take on the question
Multiply by the numerator on the top and denominator

so we get

100!/ 2^50*(50!)
after that i applied AM>=GM so finally in the range C
Option c

other way is
let a = 1*3*5*7....*99
using AM >= GM
(2500/50)^50 >= a
let b = 2*4*6*8.....*100
(2550/50)^50 >= b
now N= a/b = (2500/2550)^50 = (50/51)^50

N = (50/51)^50
so
1/N = (51/50)^50 so
= (1+1/50)^50 applying binomial theorm upto four terms and just fgetting the option 3 Option C

Not completely of my methods but think this willl do and i think shivani should get the answer and solution as she got this question ....
@shivai do u have the solution

[trustrajib]

Please try this..

A charity organisation which invited source people who have been divided in two groups G1 and G2. All the people from G1 will play a game with all the people from G2. The person who will loose will donate Rs.100 to the charity. How much will be the difference in the total donation to the charity if the no.of people had been 31 instead of 25?

[bishweshwar1234]

Originally Posted by trustrajib

Please try this..

A charity organisation which invited source people who have been divided in two groups G1 and G2. All the people from G1 will play a game with all the people from G2. The person who will loose will donate Rs.100 to the charity. How much will be the difference in the total donation to the charity if the no.of people had been 31 instead of 25?

I did not get it, on what conditions one will lose and also will the whole group will donate or the playing person will do.

[arbit_rageur]

N=1/2*(3/4)*......(99/100)

N= product from nk1 to 50 (2k-1)/2k= Product (1-1/2k)

Sum 1/2+1/4+...1/100=1/2*(Sum(1/1+....1/50))>1/2*ln(49)>3/2
Sum<2.

Product<(1-3/200)^50<1-3/4<1/4.

So I would say b) is the option

Originally Posted by romy4allcse

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

soln: N = (1*3*5*7..........*97*99)/(2*4*6*...*100)
= (1/2)*(3/4)*(5/6)*............*(97/9*(99/100)

here we can see that lowest value is 0.5 and after that the value of the factors is increasing n reaches to 0.99
and the multiplication value gets lower and lower so it vl be less than 0.5 i.e 1/2 for sure
compute it for 3-4 factores
value goes down so as i m not able to find the exact soln
hence by seeing the variation i vl go with my "Tukka"
option(c)

wat is the ans.??

[bishweshwar1234]

Originally Posted by shivam_01

Dude is the answer
volume of the cone is
= 1/3*pi*(40/3)^3 cubic units

Can you post the solution for this ?

Thanks
Bishweshwar

[MaskedMenace]

Originally Posted by bishweshwar1234

Can you post the solution for this ?

Thanks
Bishweshwar

Q) In a right circular cone,AB is a diameter of the base,V is the vertex and angle(AVB) = 90. Of all the right cylinders that can be inserted in the cone,the one with the greatest volume is L. If the radius of L is 20/3 cm,Find the volume of the cone.


1.) 100pi/3 (2) 500pi/3 (3) 1000pi/3 (4) 2000pi/3 (5)can't be determined

sol:- Let R and r be the radii of the cone and inscribed cylinder respectively.
Volume of cylinder (v) = pi*r^2*(R-r) = 4pi*(r/2)*(r/2)*(R-r)
As the sum of the 3 factors is constant (r/2 +r/2+R-r = R)
The product is max when they are equal i.e., when r/2 = R-r
=> r= 20/3(given) =>R = 10

so, volume of the cone = 1/3*pi*R^2*R = 1000pi/3

[bishweshwar1234]

Originally Posted by shivam_01

My take on the question
Multiply by the numerator on the top and denominator

so we get

100!/ 2^50*(50!)
after that i applied AM>=GM so finally in the range C
Option c

other way is
let a = 1*3*5*7....*99
using AM >= GM
(2500/50)^50 >= a
let b = 2*4*6*8.....*100
(2550/50)^50 >= b
now N= a/b = (2500/2550)^50 = (50/51)^50

N = (50/51)^50
so
1/N = (51/50)^50 so
= (1+1/50)^50 applying binomial theorm upto four terms and just fgetting the option 3 Option C

Not completely of my methods but think this willl do and i think shivani should get the answer and solution as she got this question ....
@shivai do u have the solution

@shivani : Please post a better approach to the problem.

Thanks
Bishweshwar

[bishweshwar1234]

Originally Posted by shivam_01

5)The number of points (x,y) satisfies i) 3y-4x=20,ii)x^2+y^2=<16
is/are?

x^2+y^2=<16 this is a circle whose radius is 4 and the line 3y-4x=20 intersects at one piont
Infact its is tangent so ans is 1 point ???

soln 2 :
(x-c)(x-b)/(x-a) = K
b+c>=2a
hence option 2 b=<a=<c

??

soln 2 .a cannot be in extreme ends since the denominator can become 0. Hence (b) is the ans.

soln3. (c) is the ans.