Official Quant thread for CAT 2013

Circles with radii 1, 2, and 3 are mutually externally tangent. What is the area of the triangle determined by the points of tangency?
(A) 3/5 (B) 4/5 (C) 1 (D) 6/5 (E) 4/3

@sauravd2001 said:
last five terms of 41^234
five terms ..yaar 2 ya 3 toh chlta yrr...last two 61 hnge
@rkshtsurana tere wale ka kya 6/5 hai?
@sauravd2001 said:
@rkshtsurana tere wale ka kya 6/5 hai?
yup..approach bata

minimum and maximum value of 4^(sinx)+4^(cosx)??

@rkshtsurana said:
Circles with radii 1, 2, and 3 are mutually externally tangent. What is the area of the triangle determined by the points of tangency?(A) 3/5 (B) 4/5 (C) 1 (D) 6/5 (E) 4/3
gm
@rkshtsurana bhaimaine toh kuch golmal karke nikala hai maine bade triangle ka nikal pahle area uska aya 6 ..uske age main tujhe diagram se btata hun ruk
@rkshtsurana bhai @krum bhai jaise hi kiya hai
@astute99 bhai option do
@astute99 max vale toh 4^(1+root3)/2
aur min valeu toh 4^(-root2)
@astute99 said:
minimum and maximum value of 4^(sinx)+4^(cosx)??
max at pi/4 => 4^(1/root2)+4^(1/root2)=>2^(1+root2)
min at 5pi/4 =>4^(-1/root2)+4^(-1/root2)=>2^(1-root2)
@gautam22 said:
sir thoda ez chote vale triangles mein 1/2 *product of sides *sin of angle between dem laga do
wahi kia hai
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@rkshtsurana said:
First player will win with 100% chance when square exactly circumscribes the coin..that is if side of square is dbut side of tile is l50 -50 k liye l-d ka square ka area half ho puri tile(side l ) ke(l-d)^2 = l^2 /2 l = (2 + rt2)d
even if the square circumscribes the coin, we need not have that it falls completely in 1 square. It may happen that the centre of the coin be the corner of a square.in that case it lies
in 4 squres.


am i missing smthng?
@wovfactorAPS said:
even if the square circumscribes the coin, we need not have that it falls completely in 1 square. It may happen that the centre of the coin be the corner of a square.in that case it lies in 4 squres.am i missing smthng?
YA THAT TOO RYT
@rkshtsurana said:
A coin of diameter d is thrown randmly on a floor tiled with squares of side l. Two players bet that the coin will land on exactly one, respectively, more than one, square. What relation should l and d satisfy for the game to be fair? @jain4444 - welcome!!!!
For the game to be fair, P(each of them to win) = 1/2
Number of tiles = N

The center of the coin can lie inside the square of (L - d) inside the original square of side L

So, P(lies inside one square) = N*(L-d)^2 / N*L^2 = p1

p1 should be 1/2 => (L-d)/L = +/- (1/root(2)) .. clearly L > d for it to be fair.
=> L = (2 + root(2))*d

EDIT: Didn't notice that you already gave the solution :)
@krum
@krum said:
max at pi/4 => 4^(1/root2)+4^(1/root2)=>2^(1+root2)min at 5pi/4 =>4^(-1/root2)+4^(-1/root2)=>2^(1-root2)
krum bhai kaise karte hai aise questions ko and pi/4 and 5pi/4 pe max and min kyu nikaala, plz zara basic se samjhao ise???????
@astute99 said:
@krum krum bhai kaise karte hai aise questions ko and pi/4 and 5pi/4 pe max and min kyu nikaala, plz zara basic se samjhao ise???????
differentiate kro...u ll get pie/4 pe maximum...periodic function he ..so min at 5pie/4
Solve x(x+1)(x+3)(x+4)+2 = 0 for x.
@rkshtsurana bhai yeh toh samjh mein aya ki pue/4 pe max n min kaise hoga
4^(1/root2)+4^(1/root2)=2^(1+root2)?????yeh samjha