[shanks4mba]

Originally Posted by KAC8L591

At least 4 head means
4H 2T = 6!/4!*2! =15ays
5H 1T = 6!/5! = 6 Ways
6H = 1 Ways

Total possible ways = 22
Req probability = 22/64 =11/32

then wats worng wit my approach?

[sachinverma]

Originally Posted by sachinverma

Points S ,T and U are selected on sides QR, PR, and PQ respectively of triangle PQR. The lines PS, QT, and RU meet at point Z. If area(PUZ) = 126, area(UQZ) = 63, and area(RTZ) = 24, the area of triangle PQR is ? (a) 324 (b) 351 (c) 360 (d) 364

Source:quantexpert(dot)in

Hello puys, no answer on this Q?

[shashank3012]

Hello people,

As expected,the official quant thread is running at a frenetic pace.There has been some server issues.If some of you may remember,the Introduce yourself thread was closed and a new one was opened for the very reason

So,people,there is a new official thread opened for quant discussions to encompass the discussions from July onwards

Please continue with the same.here is the link:

http://www.pagalguy.com/forum/quanti...ml#post1605026 (Official Quant thread for CAT 09-II [July 09 onwards])

Thread Closed.