Official Quant thread for CAT 2013

MBA CAT 2024
28605
@albiesriram said:
The question is asking for value of m and you are quoting two digit numbers??
No it is asking for the constant before pi.....whether its 2 digit or 1 digit and intially the radius is 15 so it has to be a 2 digit number in any case....
28606
@Tusharrr said:
At time t=0 s, the radius of a circle is equal to 15 cm. The radius of the circle increases at a rate of 0.5 cm/s. The rate of change of area at t=20 s is equal to mπ cm^2/s, where m is a positive integer. What is the value of m?
A(at time t)= pie*Rt^2
dA/dt = 2pie* Rt * dR/dt = 2pie * Rt * 0.5
At, t=20s

dA/dt = 2pie* 25 * 0.5 = 25*pie = mn

m = 25 .. ( If that little thing after m is "pie" )
28607
@heylady said:
No it is asking for the constant before pi.....whether its 2 digit or 1 digit and intially the radius is 15 so it has to be a 2 digit number in any case....
My bad it looked like mn to me.. I thought answer would be a two digit number..
28608
@ScareCrow28 said:
A(at time t)= pie*Rt^2 dA/dt = 2pie* Rt * dR/dt = 2pie * Rt * 0.5 At, t=20sdA/dt = 2pie* 25 * 0.5 = 25*pie = mnm can be anything like, 25 or 25*2 or 25*11 or 25*22 ?? Any condition on n?? ..
Its not n its pi I suppose...
28609
@ScareCrow28 said:
A(at time t)= pie*Rt^2 dA/dt = 2pie* Rt * dR/dt = 2pie * Rt * 0.5 At, t=20sdA/dt = 2pie* 25 * 0.5 = 25*pie = mnm = 25 .. ( If that little thing after m is "pie" )
same aa raha hai mera bhi ans..
28610
@heylady said:
Its not n its pi I suppose...
dR/dt=1/2
so r= t/2+c
at t=0 r=15
hence c=15
So
R= t/2+15
A= pi*R^2
dA/dt= 2*pi*R*dR/dt
R at t=20= 25....
m=25....
28611

Q1

28612

Q2

28613

Q 3

28614
@vbhvgupta said:
Q1
Already done
28615
@ScareCrow28 said:
Already done
bhai yad nahi mujhe.
28616
@ScareCrow28 said:
Already done
spam kar raha hai idhar
28617
@DeAdLy said:
spam kar raha hai idhar
This doesn't come under spam! QA Thread pe aayega tab pata chalega na 😛 Spam to tu kara h
28618
@vbhvgupta said:
Q2
tukka maar raha hun a
28619
@ScareCrow28 said:
This doesn't come under spam! QA Thread pe aayega tab pata chalega na Spam to tu kara h
haan ab to hamare post spam hi lagenge...bada aadmi ban gaya tu to ab
28620
@ScareCrow28 said:
Already done
@vbhvgupta said:
bhai yad nahi mujhe.
@DeAdLy said:
spam kar raha hai idhar
@ScareCrow28 said:
This doesn't come under spam! QA Thread pe aayega tab pata chalega na Spam to tu kara h
@DeAdLy said:
tukka maar raha hun a
@DeAdLy said:
haan ab to hamare post spam hi lagenge...bada aadmi ban gaya tu to ab
Guys PM's can be used for usual chit-chat.

Find the sum of areas of all possible triangles that can be formed by joining the vertices of a cube of dimension 1 unit.
28621
@chillfactor said:

Find the sum of areas of all possible triangles that can be formed by joining the vertices of a cube of dimension 1 unit.
Total 56 triangles

6*4 triangles of 0.5 sq.units 12

24 triangles of 0.707 sq units

8 triangles of 0707 each =5.67

8 triangles of 1.22 sq units calc error.
24 triangles of 1.22 =29.28

46.95 sq units?
28622
@vbhvgupta said:
Q3
ans 3
A...product of roots shud be one...
28623
@rkshtsurana said:
ans 3 A...product of roots shud be one...
if there is a case say roots are 5/3 , 5/3 , 3/5 . this satisfies the condition, however product is still not 1... also if product is 1 does not guarantee that the condition is satisfied... i have been trying to solve this problem for the pas half an hour ...
28624
A bug starts at one vertex of a cube and moves along the edges of the cube according to the
following rule. At each vertex the bug will choose to travel along one of the three edges
emanating from that vertex. Each edge has equal probability of being chosen, and all choices are
independent. What is the probability that after seven moves the bug will have visited every
vertex exactly once?

(A) 1/2187
(B) 1/729
(C) 2/243
(D) 1/81
(E) 5/243