Sets

A= {1,2,3,4,5.100}
B= {4,7,10,13,16,19.}
C= {2,4,6,8,10.}
B−C= {7,13,19, .97}
A∩ (B−C)= {7,13,19, .97}
sum of elements=832

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0

P, Q, R represents some students which play all three games. Hence no any option is correct.

n (C) = 73, n (T) = 65, n (C ∩ T) = x
n (C ∪ T) ≤ 100
⇒ n (C) + n (T) - n (C ∩ T) ≤ 100
⇒ x ≥ 38
n (C ∩ T) ≤ min (n (C), n (T) ⇒ x ≤ 65
⇒ 38 ≤ x ≤ 65

Let number of elements in T is R.
∴ 20R = 500 ⇒ R = 25
and 6R = 5N ⇒ N = 30

 max {n (A), n (B)} ≤ n (A U B) ≤ n (U)
⇒ 76 ≤ 76 + 63 - x ≤ 100
⇒ -63 ≤ -x ≤ -39
⇒ 63 ≥ x ≥ 39

Yes, the Bihar BEd CET Answer Key is released separately for all paper sets – A, B, C, and D. Candidates must download the key according to their question paper set.

A = {2,4,6,8, . . .50} ⇒ 25 element
A = {7,14,21, . . . .49} ⇒ 7 elements
A ∩ B = {14,28,42} = 3 elements
Required number of elements = 25 + 7 - 3 = 29

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

$\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$          

Another possible number is 9k + 5 forms are {104, .995}

$\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$  

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$  possible value of $\mathcal{l}$ = 5