CGL 2012 Tier 2 Preparations and Results

Staff Selection Commission (SSC) SSC - CGL
2215
@JS-M said:
ohh.......my bad.... then it equilateral triangle
ACUTE ANGLED
2216
@cbiofficer my bad!!

book=80
pen =100
2217
@mickym said:
Acute ANGLED
Tan C will be = 2
Tan A tan B= 3
=> all angles ACUTE
logic??
2218
@sandeepv882 said:
but bhai ye tan A + tan B+ tan C=6 ko satisfy nhi karta
@sandeepv882 A+B = 180 - C..apply tan both the sides and try solving that way
2219
@Brooklyn said:
@cbiofficer my bad!!
book=80
pen =100
2220
@anurag1701 said:
@cbiofficer 6 : pi ??
correct
2221
@cbiofficer said:
made dumb calc error :banghead: !!!
2222
@Brooklyn said:
made dumb calc error !!!
ye yaha toh chl jayga bt xam me nai
2223
@mickym said:
Acute ANGLED
Tan C will be = 2
Tan A tan B= 3
=> all angles ACUTE
i din get it.. acute h wo pata kaise chalta h?
2224

kiske pass 2009 ka cglpaper ka link hai..

2225
. An empty fuel tank of a car was filled with A type of petrol. When the tank was one-third empty, it was filled with B type of petrol. Again, when the tank was one-third empty, it was filled with A type of petrol. Again, when the tank was one-third empty it was filled with B type of petrol. At this time what was the percentage of B type of petrol in the tank?
1) 51 %
2) 48 %
3) 49 %
4) 50 %

n plz explain hw to solve dis....
2226

If A=sin^2B+cos^4B, for any value of B, then the value of A is



2227
@mrudalmodi said:
i din get it.. acute h wo pata kaise chalta h?
@Brooklyn
See bro
Tan A Tan B= 3

Okk now in a triangle both A & B can't be >90 ( obtuse)
So let's Say A > 90 = 120
which can be written like
Tan 120 = tan (180-60)
= - tan 60 = -rt3
B has to be acute
but result is -ve in this case
So to be positive both A & B hav to b Acute
Similarly

tanA. tanB. tanC = tanA+tanB+tanC

=6

if an angle is obtuse, then tanA. tanB.tanC is -ive.

But 6 is +ve

so actule angled.



2228
@sh.arora10 said:
. An empty fuel tank of a car was filled with A type of petrol. When the tank was one-third empty, it was filled with B type of petrol. Again, when the tank was one-third empty, it was filled with A type of petrol. Again, when the tank was one-third empty it was filled with B type of petrol. At this time what was the percentage of B type of petrol in the tank?
1) 51 %
2) 48 %
3) 49 %
4) 50 %
n plz explain hw to solve dis....
2???
2229
@gudda1122 said:
2???
explain....
2230
@mickym
@mickym said:
See bro
Tan A Tan B= 3
Okk now in a triangle both A & B can't be >90 ( obtuse)
So let's Say A > 90 = 120
which can be written like
Tan 120 = tan (180-60)
= - tan 60 = -rt3
B has to be acute
but result is -ve in this case
So to be positive both A & B hav to b Acute
Similarly

tanA. tanB. tanC = tanA+tanB+tanC

=6

if an angle is obtuse, then tanA. tanB.tanC is -ive.

But 6 is +ve

so actule angled.

thanks a lot bro
2231
@sh.arora10 said:
. An empty fuel tank of a car was filled with A type of petrol. When the tank was one-third empty, it was filled with B type of petrol. Again, when the tank was one-third empty, it was filled with A type of petrol. Again, when the tank was one-third empty it was filled with B type of petrol. At this time what was the percentage of B type of petrol in the tank?
1) 51 %
2) 48 %
3) 49 %
4) 50 %
n plz explain hw to solve dis....
48% is it????

2232
@sh.arora10 said: If A=sin^2B+cos^4B, for any value of B, then the value of A is
13/16 ????????
2233
@mickym said:
48% is it????
explain..
2234
@sh.arora10 said:
explain..
Yr assume Volume = 99
now after 1st pass
A B
66 33

2nd pass
A B
44 22

3rd pass
A B
77* 2/3 22*2/3 +33
Reqd ratio= (22*2/3 +33)/ 99= 48.14% approx