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Hi all,

CAT material for giveaway. I took the CAT in 2006 and have got quite a bunch of mock CAT question papers( mostly from TIME and IMS) to give away. No price, no costs, no hidden charges, just come over to my place and collect it. I'm presently in Bombay till 6th June, 2008 so please try to establish contact with me before that.

Location: Colaba, Bombay

contact: hfoblues{at}gmail{dot}com

CAT material for giveaway. I took the CAT in 2006 and have got quite a bunch of mock CAT question papers( mostly from TIME and IMS) to give away. No price, no costs, no hidden charges, just come over to my place and collect it. I'm presently in Bombay till 6th June, 2008 so please try to establish contact with me before that.

Location: Colaba, Bombay

contact: hfoblues{at}gmail{dot}com

Question: 174ABCD is a rectangle. O is the midpoint of CD. BO meets AC at M. E is a point outside the rectangle such that AE = BE and <(AEB) = 90 degrees. If BE = BC = 2, then area of AEBM is

(a) 6-(12)^1/2 (b) 2+3^1/2 (c) 8/3+(3)^1/2 (d) 2+(32^1/2)/3

I am getting the answer as 2 + (32)^1/2/3 Option d

AE=BE=BC=2 (withAB= 2*root(2)) => CO=root(2) Now tri MOC and tri AMC are similair. with oc/AB=1/2..Hence height of the base AB of tri = 4/3

so required area = 2*2 /2 +2root(2)*4/3*1/2

=2 + 4 root(2) /3 = 2 + root(32)/3 which is option d

option d for me too... methd same as hanuman's

Posting after a long time

My answers are:

A> 2 Only statement II is true. there can be a case where both unni and vasu may not be involved.

B> since wagle has no conditions/restrictions on him, his selection gives the maximum number of combinations. hence 4

C> only the 3rd statement is true... as can be deduced from question A. hence option 4 is correct

CORRECTION....

FOR QUESTION (B)

i now go with saleem's answer , while wagle gives only 7 combinations... S&T; gives 8 combinations.

My answers are:

A> 2 Only statement II is true. there can be a case where both unni and vasu may not be involved.

B> since wagle has no conditions/restrictions on him, his selection gives the maximum number of combinations. hence 4

C> only the 3rd statement is true... as can be deduced from question A. hence option 4 is correct

All i can say is that CAT never ceases to surprise me......

------------------------------------------------------------Quant Question # 120------------------------------------------------------------Question:A right circular cone has base radius 1. The vertex is V. C is a point on the circumference of the base. The distance VC is 3. A particle travels from C around the cone and back by the shortest route. Its minimum distance from V is(a) 5/4 (b) 2 (c) 5/3 (d) 3/2answer is 3/2Hence option (d) is the correct choice.:)

Me too getting answer as 3/2...a sitter after a long time

done and thanks

The meet scheduled for 2day (July 2nd) is cancelled.

hey its a good thing the meet was cancelled/postponed. i couldn't have made it yesterday and i really didn't want to miss it.

By the way, for us first timers in the meet could somebody please send us a copy of/ link to the paper to be discussed, CAV7 so we can also contribute something to the discusions

Cheers

santosh_s Saysplez explain how you ppl got the equation b=2a+19

Look at the relation given in the problem

XY = YB + BC + CZ = WZ = WD + DA + AX

BC = 19, PQ = 87,

Considering XY=WZ=b and AX=YB=ZC=DW=a,

We get the relation...

I would say this is a typical CAT 2 marker, although a bit on the easier side.

What say?

Cheers

3 minutes for me, i.e. a reading speed of around 210wpm.

I guess i've really got to pull up my socks considering that the junta here has clocked an average time of 2 mins.

I guess i've really got to pull up my socks considering that the junta here has clocked an average time of 2 mins.

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