XAT Prep Group

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About this group
This is a prep group thought of, by me and my friends. All those who want to participate are free to join. But be beware no nuisance etc will be tolerated. There is a specific we are going to follow, it you feel it suits you your most welcom...
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** MBA is hyped but post MBA life is over hyped **
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all the best guys!!!!! we will meet after exam to discuss..

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All the best puys 😃

Where are the birds?
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All the best for XAT...

Cat-94.86|XAT13 - 99.33|IIFT-51.75|| Calls-iift,xl-bm n hr,spjain,ximb||Rejects-xl,spjain,iift||Converted XIMB
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All the best to all of you for XAT.. I am sure there are many would be juniors reading this...
XLRI School of Business and Human Resources BM-2012-14 Batch Member- PGDT'11, UDT'10, KDT'10
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all the best puys.....

i hope the last episode (XATTAK....saw this name somewhere on pg ....looked Apt ) of the season ends well ....

Guns -- Check
Ammunitions -- Check
Artillery -- Check

IT'S TIME TO FIRE........
Something that doesnt Kill u ...Simply makes u STRONGER ...!!!
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all d best for xat puys!! we will meet at new thread of xat discussion and analysis after xam. 😃

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@thundercoder said:
@YouMadFellow said: try to karo Area = 65
Well, I am not sure about the answer, but I did this :D


May be the destination is not for me, but the journey is..http://www.pagalguy.com/forums/chit-chat/shoutbox-t-83788/p-3594016/r-4343099
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@thundercoder said:
Area = 65 Take (0,0) and radius 8 for C1, (13,0) and radius 3 for C2Then equation of direct common tangentsL1 : 1.588x-y-15.011 = 0L2 : 1.588x-y+15.011 = 0Equation of transverse common tangentsL3 : 0.4166x - y - 8.665 = 0L4 : 0.4166x - y + 8.665 = 0Then intersection points are A(5.419,6.408), B(11.813,3.744)C(11.813,-3.744)D(5.419,-6.408), Area of quadrilateral for ABCD = 65.Ask if you need details at any step..question hai yeh to. Ab bacchon ko tang mat karo aur so jao
wow ma, accurate %.4f , aapka dimag to chacha choudhari se bhi tej chalta hai
back 2 shirking :O ZZZZ
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@YouMadFellow said: try to karo

Area = 65

Take (0,0) and radius 8 for C1, (13,0) and radius 3 for C2

Then equation of direct common tangents
L1 : 1.588x-y-15.011 = 0
L2 : 1.588x-y+15.011 = 0

Equation of transverse common tangents
L3 : 0.4166x - y - 8.665 = 0
L4 : 0.4166x - y + 8.665 = 0

Then intersection points are
A(5.419,6.408),
B(11.813,3.744)
C(11.813,-3.744)
D(5.419,-6.408),

Area of quadrilateral for ABCD = 65.

Ask if you need details at any step..question hai yeh to.
Ab bacchon ko tang mat karo aur so jao
I am Titanium.
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