CBSE Set Qa1 Maths Sample Test Papers For Class 12th for students online

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Maths Class - XII (CBSE)
You are on Set no 1 Answer 1 to 8

Q1) Evaluate

  {(1 - Cos x)/x2}

Ans1)

   {(1 - Cos x)/x2}   = (2Sin2x/2)/x2
=  2/4 (Sin(x/2)/(x/2))2   =2/4 . 12 = 1/2 (as  Sinx/x=1)
 

Q2) A particle moves along a straight line so that S = t . Show that the acceleration is negative and is proportional to the cube of velocity.
Ans2)
Here the equation of motion is 
S = t
differentiating both sides w.r.t. 't'
v = dS/dt  = 1/2t = 1/2S  ( S = t )
or v = 1/2S
differentiating again w.r.t. 't'
accn = dv/dt = -1/2s2 ds/dt
                   = -2/4s2 ds/dt
                   = -(2/(1/v2)). v = -2v3
accn is negative and is proportional to cube of velocity.              

Q3) Evaluate

Cos4xdx

Ans3) let          

I =  Cos4xdx =  (Cos2x)2dx
 =  ((1 + Cos 2x)/2)2dx = 1/4  (1 + Cos22x + 2Cos 2x)dx
 =1/4  [1 + (1 + Cos 4x)/2 + 2Cos 2x]dx
 = 1/4[x + x/2 + (Sin 4x)/8 + (2Sin 2x)/2] + C 
 = 3x/8 + (Sin 4x)/32 + (1/4)Sin 2x + C

Q4) Evaluate           

(1 + tan x)/(x + log sec x) dx

Ans4) Let

I = (1 + tan x)/(x + log sec x) dx
 =  log | x + log sec x | + C
[ as  f'(x)/f(x) dx = log | f(x) | + C ]

Q5)  

2
1
(x - 1)/x2 . ex dx

Ans5)

2
I =
1
(1/x - 1/x2)ex dx
= (ex/x)12  [as  [f'(x) + f(x)]exdx = exf(x) + C
= (e2/2) - (e/1)
= (e2 - 2e)/2

Q6) A die is rolled. If the outcome is an even number, what is the probability that it is a prime number ?
Ans6) Let P be the event of getting an even no.
Q be the event of getting an prime no.
Then P = { 2,4,6} , Q = {2,3,5} and
PQ = {2}
we have to find
p (Q/P) = n(QP)/nP = 1/3

Q7) Calculate Spearman's rank correlation from the following Data:

x 1 2 3 4 5 6
y 1 3 2 6 4 5

Ans7) Here :-

X Y RX RY di=RX - RY di2
1 1 6 6 0 0
2 3 5 4 1 1
3 2 4 5 -1 1
4 6 3 1 2 4
5 4 2 3 -1 1
6 5 1 2 -1 1
di2 = 8

R, the rank Correlation
 = 1 - (6di2/n(n2 - 1))
 = 1 - (6 x 8)/6(35) 
 = 1 - 8/35 = 27/35 = 0.77

Q8) Solve the differential equation: dy/dx = 1 + x + y + xy 
Ans8)
Given 
dy/dx = 1 + x + y + xy
dy/dx = (1 + x) + y(1 + x)
dy/dx = (1 + x) + (1 + y)
dy/(1 + y) = (1 + x)dx
Integrating both sides:          

dy/(1 + y) = (1 + x)dx

log |1 + y| = (1 +x)2/2 + C  is the required solution.

Q9) Find a matrix X such that 2A + B + X = 0,
where           

A= -1  2
 3   4
B= 3 -2
1   5 

Ans9) Here

A= -1  2
 3   4
B= 3 -2
1   5 

2A + B + X = 0
X = -2A - B

= -2 -1  2
 3   4
 - 3 -2
1   5 
=  2   4
-6 -8
 - 3  -2
1   5
= -1  -2
-7  -13

Q10) Using differentials, find the approximate value of 26.
Ans10) Let y + y = 26 = (25 + 1)   - (i)
here function is
y =x
y + y = (x + x)
comparing with (i)
x = 25 , x = 1
y = 25 = 5
hence 5 + y = 26  - (ii)
Also y = (dy/dx) .x
y = (1/(2x))at x = 25 .x
y = 1/(225) .(1) = 1/10 = 0.1
from (ii)
26 = 5 + y = 5 + 0.1 = 5.1

Q11) Three bags contain 7 white, 8 red and 
Ans11) Bag I           Bag II             Bag III
           W   R          W   R             W   R
            7    8           9    6              5    7
 The following cases arises:
 White ball from each bag
Or
Red ball from each bag
required probability
 = 7/15 x 9/15 x 5/12 + 8/15 x 6/15 x 7/12
 = 651/2700 = 217/900

Q12) If xp yq = (x + y)p +q , prove that
 dy/dx = y/x
Ans12) Given xp yq = (x + y)p +q 
taking log on both sides
= plog x +qlog y = (p +q)log (x +y)
Differentiating both sides w.r.t. 'x'    
= p/x + q/y dy/dx = ((p +q)/(x + y))(1 + dy/dx) 
= (q/y - (p +q)/(x + y))dy/dx = (p + q)/(x + y) - p/x
= (qx - py)/(y(x + y)) dy/dx = (qx - py)/(x(x + y))
= dy/dx = y/x  Hence proved.

Q13) Evaluate as limit of a sum.

2
0
(x2 + 2)dx

Ans13) Here  f(x) = x2 + 2 , a = 0 , b = 2
nh = b - a = 2 - 0 = 2
By differentiation of integral as limit of a sum:-

b
a
f(x)dx = h f(a) + f(a + h) +....+ f(a + (n-1)h)

where nh = b - a

2
0
(x2 + 2)dx 
= h f(0) + f(h) +....+ f((n-1)h)
h (02 + 2) + (h2 + 2) +...+{(n-1)2h2 +2}
= h 2n + h2{ 12 + 22 + 32 +...+ (n-1)2}     
= h 2n + h2 {(n-1)(n)(2n-1)}/6
= 2nh + {(nh-h)(nh)(2nh-h)}/6
= 2 x 2 + 1/6{(2-h)(2)(2 x 2 -h)}
= 4 + 1/6 . 2 . 2 . 4 = 4 + 8/3 = 20/3

Q14) Solve the differential equation (x + 2y2)dy/dx = y , given that when x = 2, y = 1
Ans14)
(x + 2y2)dy/dx = y
= ydx/dy = x + 2y2
= ydx/dy - x = 2y2
= dx/dy - x/y = 2y
This is a linear differential equation with x as dependent variable.
Here P = -(1/y) , Q = 2y
I.F. = epdy = e-1/ydy = e-log y = e log y-1 = 1/y
required equation is

x . (1/y) = 2y . 1/y dy + C

= x/y = 2y + C
or x = 2y2 + Cy is the required solution of the given equation

Q15) Using properties of determinants prove that:

1   a   a3
1   b   b3
1   c   c3
=(a - b)(b - c)(c - a)(a + b + c)

Ans15) Let

= 1   a   a3
1   b   b3
1   c   c3
=(a - b)(b - c)(c - a)(a + b + c)
          applying  R1 R1 - R2 ; R2 R2 - R3
= 0   a - b     (a - b)(a2 + ab + b2)
0   b - c     (b - c)(b2 + bc + c2)
1   c           c3
= 0   1     a2 + ab + b2
0   1     b2 + bc + c2
1   c      c3
= 1     a2 + ab + b2
1     b2 + bc + c2
0     a2 + ab - bc - c2
1     b2 + bc + c2
= 0     (a + c)(a - c) + b(a - c)
1     b2 + bc + c2
(applying R1 R1 - R2)
= (a - b)(b - c)(a - c) 0     a + c +b
1     b2 + bc + c2
                 = (a - b)(b - c)(a - c) [ 0 - (a + c + b)]
                  = (a - b)(b - c)(a - c) [-(a + c + b)]
                 = (a - b)(b - c)(a - c)(a + b + c)    Hence Proved.

Q16) Find if lagranges mean value theorem is applicable to the function f(x) = x + (1/x) on [1, 3]
Ans16) Here f(x) = x + (1/x) on [1, 3]
(a) f(x) is defined finitely at all points on [1, 3] = f(x) is continous at all points on [1, 3]
(b) f'(x) = 1 - (1/x2)  (which is not defined only at x = 0)
which , exists finitely at all points (1,3)
f(x) is differentiable at all points (1,3).
Conditions of LMVT are satisfied,
at least one point C (1,3)
s.t
f'(c) = (f(b) - f(a))/(b - a) = (f(3) - f(1))/(3 - 1)
= 1 - (1/c2) = ((3 + (1/3)) - (1 + (1/1)))/(3 - 1)
= 1 - (1/c2) = ((10/3) - 2)/2 = 4/6 = 2/3
= 1/c2 = 1 - (2/3) = 1/3
= c2 = 3 = c = + 3
but c = 3 (1,3)
= LMVT is applicable.

 

Q17) Evaluate

      2x + 1       dx
(x2 + 4x + 3)

Ans17) Let 

I=       2x + 1       dx
(x2 + 4x + 3)
    2x + 4 - 3     dx
(x2 + 4x + 3)
     2x + 4         dx    - 3
(x2 + 4x + 3)
          1         dx
(x2 + 4x + 3)
or I = I1 - 3 I2
Now
I1=       2x + 4       dx      =  2(x2 + 4 +3) + C
(x2 + 4x + 3)
I2           1          dx =
(x2 + 4x + 3)
                1             dx 
(x2 + 2x. 2 + 4 - 1)
=          1           dx 
((x+ 2)2 - 12)
                = log | x + 2 + ((x + 1)2 - 12) | + C2
                 = log | x + 2 + (x2 + 4x + 3) | + C2
              I = 2(x2 + 4x + 3) - 3 log | x + 2 + (x2 + 4x + 3)| + C1 + C2
              I = 2(x2 + 4x + 3) - 3 log | x + 2 + (x2 + 4x + 3) | + C

Q18) Find the regression coefficients and hence the coefficient of correlation from the following regression lines:
 3x + 2y - 3 = 0 ; 2x + y - 4 = 0
Ans18) Let 3x + 2y - 3 = 0 be regression line of y on x
= y = -(3/2)x + 3/2
= byx = -3/2
hence 2x + y - 4 = 0 is regression line of x on y
= 2x = - y + 4
= x = -(1/2)y + 2
= bxy = -1/2
Now r2 = byx . bxy = 3/4 < 1
our supposition is correct
Hence r = -(3/4) = -(1.732)/2
            = - 0.866   ( byx, bxy both are negative)

Q19) A window is in the form of a rectangle surmounted by a semi-circular opening. If the perimeter of the window is 20m, find the dimensions of the window so that the maximum possible light is admitted through the whole opening.
Ans19)
 

Let 2x be the length, y be breadth of the rectangular portion, thus x is the radius of the seni-circle.

The perimeter of window = 20 = 2x + y + y + 1/2 . 2x
= y = 1/2 [20 - ( + 2)x]
Now A the area of window
= 2xy + 1/2 .x2
or A = x[20 - ( + 2)x] + 1/2 . x2
dA/dx = 20 - 2x ( + 2) + x
             = 20 + x( - 2 - 4)
putting dA/dx = 0
x = 20/( + 4)
also d2A/dx2 = -( + 4) < 0
= x = 20/( + 4) is pt of maxima
for maximum possible, the dimensions of the window
2x = 40/( + 4)  and y = 20/( + 4) m

Q20) Using matrices solve the following system of equation for x, y and z:
            x + 2y - 3z = 6
            3x + 2y - 2z = 3
            2x - y + z = 2
Ans20)
The given eqns. becomes AX =B

where A = 1  2  -3
3  2  -2
2  -1  1
; X = x
y
z
; B = 6
3
2
Now |A| = 1  2  -3
3  2  -2
2  -1  1

= 1(2 - 2) - 2(3 + 4) - 3(-3 - 4)
= 7 0
= A-1 exists and it is given by
 A-1 = adj A / |A|
C11 = 0, C12 = -7, C13 = -7
C21 = 1, C22 = 7, C23 = 5
C31 = 2, C32 = -7, C33 = -4

A-1 = 1/7  0    1   2
-7   7  -7
-7   5  -4

X = A-1B

x
y
z
 = 1/7  0   1    2
-7   7   -7
-7   5   -4
6
3
2
 = 1/7  7
-35
-35
=  1
-5
-5

x = 1, y = -5, z = -5 Ans.

Q21) Using the properties of integral, Evaluate:

/2
 
-/2
f(x)dx, where f(x) = Sin |x| + Cos |x|

Ans21) Let

I =  /2
 
-/2
f(x)dx,

Now f(-x) = Sin |-x| + Cos |-x|
              = Sin |x| + Cos |x|
              = f(x)

I = 2 /2
 
0
[Sin |x| + Cos |x|]dx
= 2 /2
 
0
[Sin x + Cos x]dx    [ |x| = x if x 0 ]

= 2 [ -Cos x + Sin x]o/2 
= 2 [(-Cos /2 + Sin /2) - (-Cos 0 + Sin 0)]
= 2 [1 + 1] = 4 Ans.

(Students are expected to solve either part B or part C )
Part B

Q22) Find the projection of = 2 - + on = - 2 + .
Ans22)
Projection of on
 = ./ || = ((2 - + ) . ( - 2 + ))/6
 = 5/6.

Q23) Find the centre and radius of the following sphere:
2x2 + 2y2 + 2z2 - 4x - 6y + 2z + 3 = 0.
Ans23)
The equation of the sphere is
x2 + y2 + z2 - 2x - 3y + z + 3/2 = 0
its centre is (1, 3/2, -1/2)
r = (12 + (3/2)2 + (1/2)2 - (3/2)) = 2

Q24) Find the magnitude and direction of resultant of two forces of magnitude 10N and 15N, which are inclined to each other at an angle of 60o.
Ans24)
Here P = 10N , Q = 15N , = 60o
Let their resultant be R inclined at an angle with force P
we know 
R = (P2 + Q2 + 2PQCos )
   = (100 + 225 + 2 x 10 x 15 x 1/2) = 475 = 519 N
and tan = (QSin ) /(P + QCos ) = (15 Sin 60o)/(10 + 15Cos 60o)
 = (153)/35
 = = tan-1 ((33)/7)

 

Q25) Find the two like parallel forces, acting at a distance of 3m apart which is equivalent to a force 12N acting at a distance of 0.5m from one of the forces.
Ans25)
Let P and Q be the like parallel forces acting at A and B and their resultant 12N acting at C

 

Now P + Q = 12  - (i)
Also P x AC = Q x CB
= P x 0.5 = Q x 2.5
= P = 5Q

from (i) 6Q = 12 = Q = 2N P = 10N
P= 10N , Q = 2N
(The resultant will act nearer to the larger force)

Q26) A particle moving with uniform acceleration passes over 720cm in the 11th second and 960cm in the 15th second of its motion .Find the initial velocity of the particle.
Ans26)
Let u m/sec be the initial velocity and a m/s2 be the acceleration of the particle. Then the distance travelled in 11th second
St = u + (1/2)a(2t - 1)
= 720 = u + (1/2)a(22 -1) = u + (21/2)a  - (i)
also 960 = u + (1/2)a(30 -1) = u + (29/2)a  - (ii)
(ii) - (i) =
240 =8a/2 = a = 60 m/s2
u = 720 - (21/2)60 = 90 m/sec

Q27) Find so that the four points with positive vector -6 + 3 + 2 ,  3 + + 4, 5 + 7 + 3 and -13 + 17 - are coplanar.
Ans27)
Let A, B, C, D be the points whose position vectors are given.
= - = (3 + + 4) - (-6 + 3 + 2)
 = 9 + ( - 3) + 2
= - = 11 + 4 +
= -7 + 14 - 3
A, B, C, D are coplanar
, , are also coplanar
= [ , ,   ] = 0

=  9   - 3   2
11     4      1
-7     14    -3
= 0

= 9(-12 - 14) - ( - 3)(-33 + 7) + 2(154 + 28) = 0
= + 2 = 0 = = -2

Q28) Solve the differential equation:
dx/dy = (x + 2y - 1)/(x + 2y + 1)
Ans28) Here dx/dy = (x + 2y - 1)/(x + 2y + 1)   - (i)
putting x + 2y = t
= 1 + 2(dy/dx) = dt/dx
from (i)
(1/2)[(dt/dx) - 1] = (t - 1)/(t + 1)
or dt/dx = (3t - 1)/(t + 1)
= ((t + 1)/(3t - 1))dt = dx
Integrating both sides

1/3 (3t - 1) dt + 4/3 
(3t - 1)
    1      dt =
(3t - 1)
dx

= (t/3) + (4/3)((log (3t - 1))/3) = x + c
or (x + 2y) + (4/3)log (3x + 6y - 1) = 3x + c' is the req. sol.

Q29) Find the vector equation of plane passing through the intersection of planes
. (2 - 7 + 4) = 3 and  . (3 - 5 + 4) + 11 = 0
and passing through the point (-2, 1, 3)
Ans29)
The two planes are 
2x - 7y + 4z - 3 + k(3x - 5y + 4z +11) = 0
or (3k + 2)x - (5k + 7)y + (4k +4)z + (11k - 3) = 0 - (i)
(i) passes through the point (-2, 1, 3)
= -2(3k + 2) - 1(5k + 7) + 3(4k +4) + (11k - 3) = 0
= 12k - 2 =0
= k = 1/6
hence the required equation of plane is
((3/6) + 2)x - ((5/6) + 7)y + ((4/6) + 4)z + ((11/6) - 3) = 0
= (15/6)x - (47/6)y + (28/6)z +(-7/6) = 0
= 15x - 47y + 28z - 7 = 0 is the rquired equation of plane in cartesian form
in vector form the equation of plane is
. (15 - 47 + 28) = 7

Q30) A particle is projected with a velocity of 100m/sec, at an angle of elevation of 30o find:
(i) the equation of its path.
(ii) the length of the latus rectum of its path.
(iii) the greatest height attained.
(iv) the time of flight and
(v) the horizontal range.
[take g = 10m/s2]
Ans30) (i) The equation of the path of particle is
y = x tan - ((gx2)/(2u2Cos2))
= y = x tan 30o - ((10x2)/(2(100)2Cos230o)
or y = x/3 - x2/1500

(ii) The length of latus rectum
= (2u2Cos2)/g = 1500m

(iii) The greatest height
= (u2Sin2)/g = (100 x 100 x Sin230o)/10 = 250 m

(iv) The time of flight 
= (2uSin)/g = (2 x 100 x Sin 30o)/10 = 10 sec

(v) The horizontal range
= (u2Sin2)/g = (100 x 100 x Sin 60o)/10 = 5003

 

Part C

Q22) For a certain distribution the arithmetic mean is 45, median is 48 and Karl Pearson's coefficient of Skewness is -0.4. Calculate the S.D. of the distribution.
Ans22)
Here = 45 , Md = 48 and SKP = -0.4
Now SKP = 3(Mean - Median)/S.D.
= -0.4 = 3(45 - 48)/
= 9/0.4 = 22.5

Q23) An unbiased coin is tossed 6 times. Find using Binomial distribution, the probability of getting at least 5 heads.
Ans23)
Here probability of success (getting head) in one trial = p = 1/2
q = 1 - 1/2 = 1/2 , n = 6
p(at least 5 heads)
= p(r 4)
= p(5 or 6)
= 6C5p5q1 + 6C6p6
=6(1/2)5(1/2) + 1(1/2)6
(1/2)6(6 + 1) = 7/64.

Q24) Find the present value of an annuunity of Rs. 1,200 payable at the end of each of 6 month for 3 years when the interest is earned at 8% per year compounded semi-annually. [Take (1.04)6 = 1.2653]
Ans24) Here Annuity A = Rs. 1200
No. of years = 3 (m)
No. of compounding in a year = 2 (n)
rate of interest (r) = 8% per annum
i = r/(2 x 100) = 0.04
hence the present value,
P = A/i [1 - (1 + i)-m.n]
P = 1200/0.04 [ 1 - (1.04)-6]
   = 30000 [ 1 - 1/1.2653]
or P = 30000 [ 1 - 0.7903]
    P = 30000 x 0.2.97 = Rs. 6291

Q25) The cost function of a firm is given by
C = 4x2 - x + 70
find (i) the average cost (ii) the marginal cost , when x = 3.
Ans25) Here C = 4x- x + 70
= AC = C/x = 1/x(4x- x + 70) =  4x - 1 + 70/x
(AC) at x = 3 = 1/3 (4 x 9 - 3 + 70) = 1/3(103) = 103/3
Also MC = dC/dx = 2 x 4(x) - 1 = 8x - 1
(MC) at x = 3 = 24 - 1 = 23

Q26) X draws a bill on Y for Rs. 3,000 on 1st March, 1999 payable after 3 months. The bill is discounted by X, as he is in need of money, Compute the discount in each of the following cases:
(i) The bill is discounted for Rs 2,840
(ii) The bill is discounted at 8% per annum.
Ans26)(i) The value of bill = Rs. 3,000
Discounted value = Rs. 2,840
Discount = Rs.(3,000 - 2,840) = Rs. 160.

(ii) Discount period = 3 months
rate of interest = 8%
Amount of discount = 8% of 3000 for 3 months
= (8/100) x (3/12) x 3000
= Rs. 60.

Q27) Ten percent of bulbs produced in a certain factory turn out to be defective. Find the probability using poisson distribution, that in a sample of 10 bulbs chosen at random, (i) Exactly 2 (ii) more than 2 bulbs, will be defective (Take e-1 = 0.368).
Ans27)
Here n = 10, p = 10/100 = 0.10
= np = 10 x 0.10 = 1
According to poisson distribution,
P(r) = e- r/

(i) P (exactly 2) = P(r = 2)
 = (e- 2)/ = (e-1 2)/= 0.368 x 12/ = 0.184

(ii) P (more than 2) = P(2 < r 10)
 = 1 - P(0 or 1)
 = 1 - [(e- 0)/ + (e- 1)/ ]
 = 1 - [e-1 + (e-1.1)/ ]
 = 1 - 2e-1 = 1 - 2 x 0.368 = 0.264

Q28) A began a business unit with Rs. 85,000. After 2 month B joined with a certain capital. At the end of the year, the profit were divided in the ratio of 5 : 2. Find how much money was invested by B.
Ans28)
Investment of A for 12 months = Rs. 85,000
Let investment of B for 10 months be Rs. X.
Then the ratio of investment per month (ratio of equivalent capital)
= 85,000 x 12 : X x 10
= 102000 : X
This is given to be 5 : 2
102000/X  = 5/2
= X = 102000 x 2/5 = 20400 x 2 = Rs. 40,800 Ans.

Q29) A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods X and Y. To produce one unit of X , 2 units of labour and 3 unit of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of Y. If X and Y are priced as Rs 100 and Rs 120 per unit respectively, how should the producer use his resources to maximise the total revenue? Solve the problem, graphically.
Ans29)
Let x and y be the number of units produced of X and Y respectively.
Let us compile the given information in a tabular form:

For producing X For producing Y Total Available units
Labour 2 3 30
Capital 3 1 17
Revenue per unit Rs.100 Rs.120

This information can be formulated as under:
we have to maximize
R = 100x + 120y
Subject to the constraints:
x 0, y 0, 2x + 3y 30, 3x + y 17 } (A)
Let us plot the lines
2x + 3y = 30 (meet X axis at (15,0) Y axis at (0,10))
3x + y = 17 (meet X axis at (17/3,0) Y axis at (0,17))
 
These lines meet at (3,8). The feasible region of the system of inequations (A) is the region OABC (the shaded region). It is a bounded region whose vertices (extreme points) are O, A, B, C
Let us calculate the value of R at these points
R at 0 = 0 + 0 = 0
R at A(17/3,0) = 100 x 17/3 + 0 = 1700/3 = 566.66 (approx)
R at B(3,8) = 100 x 3 + 120 x 8 = 300 + 960 = 1260
R at C(0,10) = 0 + 120 x 10 = 1200
Hence R is maximum at B (3,8)
He should produce 3 units of X and 8 units of Y to maximize his revenues and the maximum revenue is Rs 1260.

Q30) A company has two plants to manufacture scooters. Plant - I manufactures 70% of the scooters and Plant - II manufactures 30%. At Plant - I, 80% of the scooters are rated of standard quality and at Plant - II 90% of the scooters are found to be of standard quality. A scooter is chosen at random and is found to be of standard quality. Find the probability that it has come from Plant - II.
Ans30)
Let
H1 : Event of choosing a scooter produced by Plant - I
H2 : Event of choosing a scooter produced by Plant - II
E : Event of selecting a scooter of standard quality
Here we have to find
P(H2/E) , Given P(H1) = 70/100, P(H2) = 30/100
By Bay's theorem
P(H2/E) = (P(E/H2) . P(H2))/(P(E/H1).P(H1) + P(E/H2)P(H2))
= (9/10 x 3/10)/ (8/10 x 7/10 + 9/10 x 3/10) = 27/(56 + 27) = 27/83