[vipulcat]

Originally Posted by vipulcat

hi iam doin it in this way plz let me know what is wrong in it????????

1sector can be painyed in 3 ways ,2nd in 2 ways ,3rd in 2ways ,4th in 2wys and 5th in 1 way....so iam getting ans 24 ........

sry i forgot to tell the question number it is 19 ..........
plz let me know

[dante rage]

Originally Posted by inava

Hey,

Thanks for your response.

Even when i add the area of the exposed triangle, i'm getting the incorrect answer.

According to the solutions given,

the area of the pyramid = area of the three triangles + area of base.

which is coming out to be 54 + 18*3^(1/2).

I'm unable to understand why the formula that i have applied is not giving me the correct Surface Area.

suppose the area of the cube is A so what you need to do is subtract the area of the three triangles and add the area of the base of the pyramid...ie

A -3X(area of one triangle) + area of the base of the pyramid

3 x(area of triangle) is the curved surface are of the pyramid....

[amithbabu]

Originally Posted by Neo007

VA : 4 (a disaster again )

QA : 45

DI :48

OA: 97

disappointed to see VA not contributing any thing

can some body suggest what kind of question should i attempt to clear cut off at least

concentrate on Rcs...improve ur reading speed...

[tejthekiran]

Originally Posted by naga25french

few puys pmed me reg tennis set ..so thought of posting here..

26 ,28,29 are very easy application of funda i posted already ..

let me discuss only 25 and 27

25)43 is the winner.. who is not runner


lets see 43 road to finals

43 + 22 = 65 --> round 1

22(43)+11 = 33 --> round 2 .. see without upset 22 would have faced 11 but as 43 upset 22 , it played 11 in round 2.. this is the only funda which will be difficult .. if u get hold of it , then its cake walk

11(43) + 6 = 17 --> round 3

6(43)+3 = 9-->QF

3(43)+2 = 5--> semis

43 and 1 in the finals...

so we can conclude that - one team in the option should have ultimately lost to 2 or 3 in some round .. lets check which one it is ?

option 1 - 44 opponents are 21,12,5,4

option 2 - 45 opponents are 20,13,4

option 3 - 46 opponents are 19,14,3

option 4 - 36 opponents are 29,4

option 5 - 37 opponents are 28,5,4..

so we get option 3 as answer ..

it hardly take a min to find this

27) 23 vs 46 ...

so our aim is to find the opponent first

23 - 42,10,7,2

46 - 19,14,3

so definitely 23 and 46 met in semis .. as 2+3 = 5

so now check the option ..

as 1 is the answer let me check from option 5


option 5 -

34 - 31,2 and 56 - 9 ,24,8,1 which is final line up and impossible as from 2 23 is the team in semis

option 3 -

5 - 4 and 51 - 14,19,3 which is final line up and impossible as from 3 46 is the team in semis

option 1 :

9 - 56 , 24,8,1 and 13-52,20,4 .. so there is possibility of semifinal matches between both teams

so option 1

hope this post clears the doubts

a big post for jus two question

hope u guys get it !

concept post : http://www.pagalguy.com/forum/cat-an...ml#post1713338 (AIMCAT 1003 Scores,Discussions(Please don't open if you have not taken the test))

Hi thanks for the reply, but a small doubt
for option 1
i.e seed 44 opponents would be 21(65),12(32),5(16),4(,1(4). thus when 4 players were left in contention 43 and 1 might have been the contenders and 1 would have won to move to round of two where it played with 43. so can 44 be answer