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Sample Question Paper - 3, Mathematics (041), Class- XII, Session: 2021-22, TERM II, Time Allowed: 2 hours, , Maximum Marks: 40, , General Instructions:, 1. This question paper contains three sections – A, B and C. Each part is compulsory., 2. Section - A has 6 short answer type (SA1) questions of 2 marks each., 3. Section – B has 4 short answer type (SA2) questions of 3 marks each., 4. Section - C has 4 long answer-type questions (LA) of 4 marks each., 5. There is an internal choice in some of the questions., 6. Q 14 is a case-based problem having 2 sub-parts of 2 marks each., Section A, 1., , Evaluate: ∫, , cos x, sin, , 2, , [2], , dx, , x+4 sin x+5, , OR, a, , If ∫0, 2., , π/2, −, 3, sin x dx, √x dx = 2a ∫0, , a+1, , , find the value of integral ∫a, , Find the general solution of (1 + x2 ) dy + 2xydx =, , xdx, , [2], , cot xdx(x ≠ 0), –, √3, , 3., , If a⃗, b,⃗ c ⃗ are three mutually perpendicular unit vectors, then prove that | a⃗ + b ⃗ + c ⃗ | =, , 4., , Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, -1,, , [2], [2], , 2), B (5, 2, 4) and C(-1, -1, 6)., 5., , The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. Find, , [2], , the probability that out of 5 such bulbs at least one will fuse after 6 months of use., 6., , In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10%, , [2], , of the girls study in class XII. What is the probability that a student chosen randomly studies, in class XII given that the chosen student is a girl?, Section B, (3x+5), , [3], , 7., , Evaluate: ∫, , 8., , It is given that the rate at which some bacteria multiply is proportional to the instantaneous, , 3, , 2, , ( x − x +x−1), , dx, , ., , [3], , number present. If the original number of bacteria doubles in two hours, in how many hours, will it be five times?, OR, Solve the differential equation:, 9., 10., , dy, dx, , - y = xex, →, , , then show that b ⃗ = c .⃗, , [3], , Find the equation of line passing through points A (0,6,-9) and B(-3,-6,3). If D is the foot of, , [3], , If a⃗, b,⃗ c ⃗ are vectors such that a⃗ ⋅ b ⃗ =, , ⃗, a⃗ ⋅ c ,⃗ a⃗ × b = a⃗ × c ,⃗ a⃗ ≠, , 0, , perpendicular drawn from the point C (7,4,-1) on the line AB, then find the coordinates of, point D and equation of line CD., , Winning Star Institute, , winningstar.in, , 7200037940
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OR, Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to, ^, the vector 3^, i + 5^, j − 6k, , Section C, ∣, ∣cot x+ cot, , 3, , x∣, ∣, , [4], , 11., , Evaluate the integral: ∫, , 12., , Find the area bounded by the circle x2 + y2 = 16 and the line √3y = x in the first quadrant,, , 1+ cot, , 3, , dx, , x, , –, , [4], , using integration., OR, Find the area of the region in the first quadrant enclosed by x-axis, line x =, curve x, , 2, , 13., , + y, , 2, , –, √3y, , and the given, , = 4, , Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4), and perpendicular to, , [4], , the plane x - 2y + 4z = 10., CASE-BASED/DATA-BASED, 14., , Elpis Limited is a company that produces electric bulbs. The quality of their bulbs is really, very good. The customers are well satisfied and it has been as well recommended brand in the, market. The probability that a bulb produced by Elpis Limited will fuse after 150 days of use is, 0.05., , Find the probability that out of 5 such bulbs, i. No bulb will fuse after 150 days of use., ii. Not more than one will fuse after 150 days of use., , Winning Star Institute, , winningstar.in, , 7200037940, , [4]
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Solution, MATHEMATICS 041, Class 12 - Mathematics, Section A, cos x, , 1. Let I = ∫, , sin, , 2, , dx, , x+4 sin x+5, , Also let sin x = t then cos x dx = dt, So, I, , dx, , = ∫, , 2, , t +4t+5, dt, , = ∫, , 2, , 2, , 2, , t +2t(2)+(2 ) −(2 ) +5, dt, , = ∫, , 2, , (t+2 ) +1, , Again, Let (t + 2) = u then dt = du, dt, , I = ∫, , 2, , u +1, , = tan-1 (u) + c [Since, ∫, , dx = tan-1 x + c], , dt, 2, , u +1, , I = tan-1 (t + 2) + c, I = tan-1 (sin x + 2) + c, OR, We have,, ∫, , a, , 0, , −, √x dx =, , Let I, , 3, , π/2, , = ∫, , 0, , π/2, , I = ∫, , 2, , 3/2, , [x, , sin, , =, , 1, 4, , 1, 4, , 1, 3, , a, , It is given that ∫0, ⇒, , 2, 3, , 3/2, , a, , 3/2, , ⇒ a, , , then, , = 2a (, , 1, , dx =, , [(−3 cos, , [0 − (−3 +, , ..........(i), , 3/2, , a, , 3, , xdx, , 4, , I =, , 2, , =, , 0, , 3 sin x−sin 3x, , 0, , ⇒, , 3, , a, , ], , π, 2, , 1, , +, , 3, , 1, , )] =, , 4, , ∫, , π/2, , 0, , 2, 1, , [3 −, , 4, , 3π, , cos, , 3, , −, √x dx = 2a ∫, , (3 sin x − sin 3x)dx =, ) − (−3 +, , ] =, , π/2, , sin, , 0, , 2, 3, , 2, 3, 3, , 1, 3, , 1, 4, , [−3 cos x +, , 1, 3, , π/2, , cos 3x], , 0, , )], , . . . . . (ii), xdx, , ), 3, , = 2a ⇒ a, , 2, , = 4a, , 2, , ⇒ a (a − 4) = 0, , ⇒ a = 0, 4 [Using(i) and (ii) ], , When a = 4, we get, ∫, , a+1, , a, , xdx = ∫, , 5, , 4, , 5, , 2, , xdx = [, , x, , ], , 2, , =, , 4, , 25, 2, , −, , 16, , =, , 2, , 9, 2, , When a = 0, we get, ∫, , a+1, , a, , xdx = ∫, , 1, , 0, , a+1, , Hence, ∫a, , xdx =, , 9, , 2. It is given that (1 + x, dy, dx, , +, , 2xy, 2, , (1+x ), , =, , x, , ], , 2, , or,, , 2, 2, , ⇒, , 1, , 2, , xdx = [, , =, , 0, , 1, 2, , − 0 =, , 1, 2, , 1, 2, , ) dy + 2xydx = cotxdx, , cot x, 2, , 1+x, , This is equation in the form of, ∫, , 2x, , dy, dx, , + py = Q, , (where p =, , 2x, 2, , (1+x ), , and Q, , =, , cot x, 2, , 1+x, , ), , dx, 2, , (1+x2 ), , log(1+x ), 2, Now, I.F. = e∫ pdx = e, = e, = 1+ x, Thus, the solution of the given differential equation is given by the relation:, , y(I. F) = ∫ (Q × I. F)dx + C, 2, , ⇒ y ⋅ (1 + x ) = ∫ [, , cot x, 2, , 1+x, , 2, , ⋅ (1 + x )] dx + C, , 2, , ⇒ y ⋅ (1 + x ) = ∫ cot xdx + C, 2, , ⇒ y (1 + x ) = log | sin x| + C, , Therefore, the required general solution of the given differential equation is, 2, , y (1 + x ) = log | sin x| + C, , 3. Given that a, b, c are mutually prependicular vectors,, , Winning Star Institute, , winningstar.in, , 7200037940
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So,, ⃗, ⃗, a⃗ ⋅ b = b ⋅ c ⃗ = c ⃗ ⋅ a⃗ = 0, , ....(1), Also, a,b and c are unit vectors, so, ⃗, |a⃗| = |b| = |c |⃗ = 1, 2, ⃗, ⃗, ⃗ 2, |a⃗ + b + c |⃗, = (a⃗ + b + c ), , ⃗ 2, ⃗ 2 + 2a⃗. b ⃗ + 2b.⃗ c ⃗ + 2c .⃗ a⃗, + (b) + (c ), , 2, , = (a⃗), , ∣⃗, + |b∣, , 2, , = |a⃗|, , 2, , = (1), , 2, , + |c |⃗, 2, , + (1), , ⃗, , 2, , ⃗, , 2, , |a⃗ + b + c |⃗, |a⃗ + b + c |⃗, , 2, , + 2(0) + 2(0) + 2(0), 2, , + (1), , [from (1)], , + 0, , = 1+ 1+ 1, = 3, , –, ⃗, |a⃗ + b + c |⃗ = √3, , 4. Let A, B, C be the three points in the plane. D is the foot of the perpendicular drawn from a point P to the, →, , →, , →, , plane. PD is the required distance to be determined, which is the projection of AP on AB × AC, →, , →, , →, , Hence, PD = the dot product of AP with the unit along AB × AC, →, , So, AP, and, , ^, ^, ^, = 3 i + 6 j + 7k, , ∣, ∣, AB × AC = ∣, →, , ^, k∣, ∣, ^, ^, ^, 2 ∣ = 12 i − 16 j + 12k, , ^, i, , ^, j, , 2, ∣, ∣ −4, , 3, , →, , →, , →, , ∣, 4∣, , 0, , Unit vector along AB × AC, Hence, =, , =, , ^, ^, ^, PD = (3 i + 6 j + 7k) ⋅, , ^, ^, ^, 3 i −4 j +3 k, √34, ^, ^, ^, 3 i −4 j +3 k, √34, , 3 √34, 17, , 5. Let X represent the number of bulbs that will fuse after 6 months of use in an experiment of 5 trials. The, trials are Bernoulli trials., It is given that, p = 0.05, ∴ q = 1 −p = 1 − 0.05 = 0.95, X has a binomial distribution with n = 5 and p = 0.05, ∴, , P(X = x) = nCxqn−x px, where x = 1,2,...n, , = 5Cx(0.95)5−x (0.05)x, P(at least one) = P(X ≥ 1), 1-P(X < 1), =1 − P(X = 0), =1 − 5C0(0.95)5 × (0.05)0, =1 − 1 × (0.95)5, =1 − (0.95)5, 6. Let E denotes the event that student chosen randomly studies in class XII, F denotes the event that randomly, chosen student is girl., P (E|F) = ?, 430, , P (F ) =, , 1000, , = 0.43, , P (E ∩ F ) =, P (, =, , E, F, , 43, 1000, , = 0.043, , P(E∩F), , ) =, , 0.043, 0.43, , P(F), , = 0.1, , Section B, 7. Let the given integral be, I, , = ∫, , 3x+5, , Now by partial fractions putting,, , Winning Star Institute, , 2, , 2, , (x −x +x−1), , dx, , 3x+5, 3, , 2, , (x −x +x−1), , =, , A, x−1, , +, , Bx+C, 2, , (x +1), , winningstar.in, , 7200037940
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A(x2 + 1) + (Bx + C)(x - 1) = 3x + 5, Putting x - 1 = 0,, X=1, A(2) + B(0) = 3 + 5 = 8, A=4, By equating the coefficient of x2 and constant term, A + B = 0, 4+B=0, B = -4, A-C=5, 4-C=5, C = -1, From equation (1),we get,, 3x+5, , ∫, , 4, , =, , 2, , (x−1)(x +1), 3x+5, 2, , (x−1)(x +1), , +, , x−1, , −4x−1, 2, , (x +1), 1, , dx = 4 ∫, 4, , = 4 log(x − 1) −, , 2, , x−1, 2, , log(x, , 2log(x2, , dx − 4 ∫, , 1, 2, , (x +1), , + 1) − tan, , −1, , dx − ∫, , 1, 2, , (x +1), , dx, , x+ c, , tan-1x, , = 4log(x - 1) + 1) +c, 8. Let the original count of bacteria be N0 and at any time t the count of bacteria be N. We have,, dN, , ∝ N, , dt, , ⇒, ⇒, , dN, , = λN, , dt, dN, , = λdt, , N, , ⇒ ∫, , , where λ is a constant, , 1, , dN = λ ∫ dt, , N, , .....(i), , ⇒ log N = λt + C, , We have, N = N0 at t = 0. Putting t = 0 and N = N0 in (i), we get,, ∴ q log N0 = 0 + C ⇒ C = log N0, , Putting C = log N0 in (i), we have,, log N = λt + log N0, N, , ⇒ log(, , N0, , ) = λt, , ...(ii), , It is given that the original number of bacteria doubles in 2 hrs., That is when t = 2 hours, N = 2 N0. Put t = 2 and N = 2N0 in (ii), we have,, log(, , 2N0, N0, , ) = 2λ ⇒ λ =, , Putting λ =, log(, , N, , 1, 2, , ) = (, , N0, , 2, , ⇒ t =, , log 2, , log 2, 1, 2, , 1, 2, , log 2, , in (ii), we have,, , log 2) t, , log(, , N, N0, , ), , ...(iii), , Suppose the count of bacteria becomes 5 times i.e. 5 N0 in t1 hours. Putting t = t1 and N =5N0 (iii), we have,, t1 =, , 2, log 2, , log(, , 5N0, N0, , ) =, , 2, log 2, , (log 5) =, , 2 log 5, log 2, , hours., , OR, The given differential equation is,, dy, dx, , - y = xex, , It is a linear differential equation. Comparing it with,, dy, dx, , + Py = Q, , P = -1, Q = xex, I.F. =, , ∫ pdx, , e, , − ∫ dx, , = e, , = e-x, Solution of the equation is given by,, y × (I.F.) = ∫ Q × (I.F.) dx + c, , Winning Star Institute, , winningstar.in, , 7200037940
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Now, CD⊥ AB, ∴ a1a2 + b1b2 + c1c2 = 0, Where, a1 = −λ − 7, b1 = −4λ + 2, c1 =, and a2 = -1, b2 = -4, c2 = 4 [DR's of line AB], , 4λ − 8, , [DR's of line CD], , ⇒ (−λ − 7)(−1) + (−4λ + 2)(−4) + (4λ − 8)4 = 0, ⇒, , λ + 7 + 16λ − 8 + 16λ − 32 = 0, , ⇒, , 33λ − 33 = 0, , λ = 1, , On putting λ = 1 in Eq. (ii), we get required foot of perpendicular,, D = (-1,2,5), Also, we have to find equation of line CD, where, C(7,4,-1) and D(-1,2,-5)., ∴ Required equation of line is, x−7, , y−4, , =, , −1−7, , x−7, , ⇒, , x−7, , ⇒, , −2, y−4, , =, , 4, , [using Eq. (i)], , −5+1, , y−4, , =, , −8, , z+1, , =, , 2−4, , 1, , z+1, , =, , −4, z+1, , =, , [dividing denominator by -2], , 2, , OR, ^, ^, ^, n⃗ = 3 i + 5 j − 6k, −, −−−−−−−−, −, −, −, |(n⃗ )| = √9 + 16 + 144 = √70, n⃗, , ^ =, n, , =, , ∣n⃗ ∣, ∣ ∣, , ^, i +, , 3, √70, , ^, j −, , 5, √70, , 6, √70, , ^, k, , ^ = 7, r .⃗ n, r. (, , 3, √70, , ^, i +, , ^, j −, , 5, √70, , 6, √70, , ^, k) = 7, , Section C, 11. Let the given integral be,, 3, , I = ∫ (, , cot x+cot, 3, , 1+cot, , 2, , cot x(1+cot, , = ∫ [, , = ∫ (, , 3, , 1+cot, cot x csc, 3, , 1+cot, , 2, , x, , x, , ) dx, , x), , ] dx, , x, , x, , x, , ) dx, , Putting cot x = t, ⇒, , -cosec2x dx = dt, , ⇒, , cosec2x dx = -dt, , ∴ I = −∫, = −∫, , tdt, 3, , 1+t, tdt, 2, , (1+t)(t −t+1), t, , Using partial fraction let, , 2, , =, , (1+t)(t −t+1), , A, t+1, , +, , Bt+C, 2, , t −t+1, , 2, , t, , ⇒, , 2, , (1+t)(t −t+1), , ⇒, , A(t −t+1)+(Bt+C)(t+1), , =, , 2, , (t+1)(t −t+1), , t = A(t2 - t + 1) + Bt2 + Bt + Ct + C, , t = (A + B)t2 + (B + C - A)t + A + C, Equating Coefficients of like terms, A + B = 0 ...(i), B + C - A = 1 ...(ii), A + C = 0 ...(iii), Solving (i), (ii) and (iii), we get, ⇒, , A = −, B =, , 1, 3, , 1, 3, , Winning Star Institute, , winningstar.in, , 7200037940
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1, , C =, , 3, t, , ∴, , = −, , 2, , (1+t)(t −t+1), t, , ⇒, , t, , 1, , ∴ I = − [−, 1, , = +, 1, , =, , 3, , t+1, , −, , 1, 6, , 1, , 6, , 3(t+1), , ∫ (, , ∫, , 1, , +, , t+1, 1, , −, , t+1, , dt, , ∫, , 3, , dt, , ∫, , dt, , ∫, , 3, , 3(t+1), , = −, , 2, , (1+t)(t −t+1), , 1, , +, , 1, , = −, , 2, , (1+t)(t −t+1), , ⇒, , 1, 3(t+1), , 2t−1, 2, , t −t+1, , (2t−1)dt, , 1, 6, , [, , [, , 2t+2, 2, , t −t+1, , 2, , t −t+1, , 2t−1, 2, , 2, , ], , 2t−1+3, , t −t+1, , 1, , (t −t+1), , ), , 2, , t −t+1, , ) dt −, , −, , 2, , 6, , +, , ∫ (, , 6, , 1, , +, , t+1, , (, , 3, , ], 1, , ) dt +, , 1, , 2, , t −t+1, , ], , dt, , ∫, , 2, , dt, , ∫, , 2, , 1, , 2, , t −t+, , 4, , −, , 1, 4, , +1, , dt, , ∫, , 2, , 1, , (t−, , 2, , √3, , ) +(, , 2, , ), , 2, , let t2 - t + 1 = p, ⇒ (2t - 1)dt = dp, 1, , ∴ I =, , dt, , ∫, , 3, , −, , t+1, , dp, , 1, , ∫, , 6, , p, , −, , 1, , dt, , ∫, , 2, , (t−, , 1, , 2, , 1, , =, , 3, , log |t + 1| −, , 1, 6, , log |p| −, , 1, 2, , ×, , 2, , ), 2, , tan, , √3, , 2, , √3, , ) +(, , 2, , −1, , t−, , (, , 1, 2, , ) + C, , √3, 2, , 1, , =, , 3, 1, , =, , 3, , log |t + 1| −, , 1, 6, , log |p| −, , log | cot x + 1| −, , 1, 6, , 1, √3, , tan, , −1, , (, , 2t−1, √3, , ) + C, , 2, log∣, ∣cot x − cot x + 1∣, ∣−, , 1, √3, , tan, , −1, , (, , 2 cot x−1, √3, , ) + C, , 12. According to the question ,, Given equation of circle is x2 + y2 = 16 ...(i), Equation of line given is ,, –, √3y = x, 1, , ⇒ y =, , √3, , ...(ii), represents a line passing through the origin., , x, , To find the point of intersection of circle and line ,, substitute eq. (ii) in eq.(i) , we get, 2, , 2, , x, , x, , +, 2, , 2, , 3 x +x, , = 16, , 3, , 2, ⇒4x, ⇒, ⇒, , = 16, , 3, , = 48, , x2 = 12, –, , x= ±2√3, –, , When x= 2√3, then y, , =, , 2 √3, √3, , = 2, , Required area (In first quadrant) = ( Area under the line y =, –, , 1, √3, , –, , x from x = 0 to 2√3) + (Area under the circle, , from x = 2√3 to x=4 ), = ∫, , 2 √3, , 0, , =, , 1, √3, , 1, √3, 2, , [, , x, , 2, , xdx + ∫, , 2 √3, , ], 0, , 4, , 2 √3, , +[, , x, 2, , −, −−−−, −, √16 − x2 dx, , −, −−−−, −, √16 − x2 +, , Winning Star Institute, , 4, , 2, , (4), 2, , sin, , −1, , (, , x, 4, , )], 2 √3, , winningstar.in, , 7200037940
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– 2, −1, [(2√3) − 0] + [0 + 8 sin, (1) −, , 1, , =, , 2 √3, , –, = 2√3 + 8 (, , π, 2, , 2 √3, , )−, , × 2 − 8 sin, , 2, , –, –, = 2√3 + 4π − 2√3 − 8 (, = 4π −, , π, 3, , −1, , 2 √3, 2, , −, −−−−, −, −1, √16 − 12 − 8 sin, (, , 2 √3, 4, , )], , √3, , (, , 2, , ), , ), , 8π, 3, , 12π−8π, , =, , 3, 4π, , =, , 3, , sq units., OR, , The graphical representation of given line and curve, , Area OAB = Area ΔOCA + Area ACB, Area of △OC A, 2, , √3, , =, , 2, , x, 2, , 2, , ∫, , √3, , −, −−−, −, √4 − x2 +, , 4, 2, , −1, , = [0 + 2 × si n, , = [π −, , = [, , π, 3, , × base × height, , sq. units ...(1), , Area of AC B =, = [, , 2, , –, × √3 × 1, , 1, , =, , 1, , =, , −, , √3, , 2π, , −, , 2, √3, 2, , ], , 3, , −, −−−, −, √4 − x2 dx, , sin, , −1 x, , (1) −, , 2, √3, 2, , 2, , ], √3, , −−−, −, −1, √4 − 3 − 2sin, (, , √3, 2, , )], , ], , ...(2), , Area required = area of OAC + Area of ACB, √3, , =, , 2, , +, , π, 3, , −, , √3, 2, , =, , π, 3, , sq. units, , 13. The equation of the plane passing through (2, 1, -1) is, a(x - 2) + b(y - 1) + c(z + 1) = 0 ...... (i), Since, this passes through (-1, 3, 4), ∴ a(−1 − 2) + b(3 − 1) + c(4 + 1) = 0, , ..... (ii), Since, the plane (i) is perpendicular to the plane x - 2y + 4z = 10., ⇒ −3a + 2b + 5c = 0, , ∴ 1⋅a − 2⋅b+ 4⋅c = 0, , ........ (iii), On solving Eqs. (ii) and (iii), we get, ⇒ a − 2b + 4c = 0, , a, 8+10, , =, , −b, −17, , =, , c, 4, , = λ, , ⇒ a = 18λ, b = 17λ, λ = 4λ, , From Eq. (i),, 18λ(x − 2) + 17λ(y − 1) + 4λ(z + 1) = 0, ⇒ 18x − 36 + 17y − 17 + 4z + 4 = 0, ⇒ 18x + 17y + 4z − 49 = 0, ∴, , 18x + 17y + 4z = 49, , CASE-BASED/DATA-BASED, 14. Let p = Probability of a success and q = Probability of a failure, p = P (a bulb will fuse after 150 days) = 0.05 and q = 1 – 0.05 = 0.95, n = 5 and P (X = r) = C (n, r) prqn-r, , Winning Star Institute, , winningstar.in, , 7200037940
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i. No bulb is fused, r = 0, P (X = 0) = C (5, 0), (0.05)0 (0.95)5, , = (, , 19, 20, , 5, , ), , 5, , = (0.95), , ii. Not more than one fused bulb, P (not more than one fused bulb) = P (X = 0) + P (X = 1), = (, = (, = (, = (, , 19, 20, 19, 20, 19, 20, 19, 20, , 5, , ), , 4, , + C (5, 1) (0.05) (0.95), 5, , ), , + 5(0.05)(, 4, , ), , (, 4, , ), , (, , 19, 20, 6, 5, , +, , 5, 20, , 19, 20, , 4, , ), , ), 4, , ) = 1.2(0.95), , Winning Star Institute, , winningstar.in, , 7200037940
