CGL 2012 Tier 2 Preparations and Results

Staff Selection Commission (SSC) SSC - CGL
761
@007pagal said: initially 4b^3 - 8b^ + 4b after 1st diff: 12b^2 - 16b + 4 after 2nd: 24b - 16
yeah it's correct

Yaar i did by simply equating all three
i.e. for max V
b = 2-2b
3b = 2
b= 2/3...................
762
@cbiofficer said: in tan^2 A= 2tan^2 B + 1 then cos^2 A + sin^2 B =?
1/rt2
763
@mickym said:
yeah it's correct
Yaar i did by simply equating all three
i.e. for max V
b = 2-2b
3b = 2
b= 2/3...................
yar what u are doing is that u are trying to make a cube from he box,which is clear as u have equated all sides, and u are getting b=2/3 according to which vol is 8/27
and according to my claculation i am getting a cuboid of sides 1/3 , 4/3 , 4/3 which gives vol 16/27
i am getting greater volume bro
764
@007pagal said: @mickym said:
yeah it's correct
Yaar i did by simply equating all three
i.e. for max V
b = 2-2b
3b = 2
b= 2/3................... yar what u are doing is that u are trying to make a cube from he box,which is clear as u have equated all sides, and u are getting b=2/3 according to which vol is 8/27 and according to my claculation i am getting a cuboid of sides 1/3 , 4/3 , 4/3 which gives vol 16/27 i am getting greater volume bro
yes bro u r correct.................

765
@Vineeth555 said:
see bro, 24k^2 + 201 needs to be written in terms of 2k +3.. so i took it as 6 ( 2k+3)(2k-3) +255.. now as 2k+3 is a divisor of the first term in the summation, it must also divide 255.. now, 255 = 3X 5 X 17 num of factors of 255= (1+1)(1+1)(1+1) = 8 ie 1,3, 5, 15,17, 51,....255 among thses, 1 and 3 wont satisfy the 2k+3 condition..so 6 numbers.. hope, this is right
Bhai how did u find the factors i e 1,3, 5, 15,17, 51,....255???
766

tan 2# tan 4# = 1

find tan3#
767
@sumit1234567890 said: many a student/ are frustrated/ because of unemployment
find error
is frustrated
768

sin # - cos # = 7/13


find sin # + cos #

if 0
769
@mickym said: sin # - cos # = 7/13
find sin # + cos #
if 0
17/13
770
@mickym said:
ans is 31 bro
1001 = 11*13*7
now let the nos on each face be
a b c d e f
so (a+b)(c+d)(e+f)= 1001
so sum of nos= 11+13+7= 31
Bhai how come"so (a+b)(c+d)(e+f)= 1001"??or kindly explain this
771
@sscjindabad said:
Bhai how come"so (a+b)(c+d)(e+f)= 1001"??or kindly explain this
bhai at each vertex 3 edges are meeting
& we r multiplying the nos on these 3 edges
total 8 vertices
so (a+b)(c+d)(e+f) = (ac+ad+bc+bd)(e+f) = ace+acf+ade+adf+bce+bdf+bde+bdf
772
@cbiofficer said:
17/13
773
@sscjindabad
tan 2# tan 4# = 1
find tan3#

774

regular tetrahedron each edge 6cm.. find volume and area..?

775
@mickym said: @sscjindabad
tan 2# tan 4# = 1
find tan3#
1
776
regular tetrahedron each edge 6cm.. find volume and area..?
777

ABCD is parallelogram P Q R S are mid pints on AB BC CD AD respect...

find PQ+QR+RS+SP
778

diagnols of a rhombus are in the ratio 1:rt 3

find angles of rhombus?
779

speed of A to B=33.3 kmph

speed of B to C = 50kmph
speed of C to A = 55 kmph
find average speed?
780
@cbiofficer said: regular tetrahedron each edge 6cm.. find volume and area..?
Vol = 18rt2
Area= 36rt3