CBSE Set Qa1 Maths Sample Test Papers For Class 12th for students online
Maths Class 
XII (CBSE)
You are on Set no 1 Answer 1 to 8
{(1  Cos x)/x^{2}} 
Ans1)
{(1  Cos x)/x^{2}}  =  (2Sin^{2}x/2)/x^{2}  
=  2/4 (Sin(x/2)/(x/2))^{2} =2/4 . 1^{2} = 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'
acc^{n} = dv/dt = 1/2s^{2} ds/dt
= 2/4s^{2} ds/dt
= (2/(1/v^{2})). v = 2v^{3}
acc^{n} is negative and is proportional to cube of velocity.
Cos^{4}xdx 
Ans3) let
I =  Cos^{4}xdx =  (Cos^{2}x)^{2}dx  
=  ((1 + Cos 2x)/2)^{2}dx = 1/4  (1 + Cos^{2}2x + 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 
(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 ] 
^{2 1}  (x  1)/x^{2} . e^{x} dx 
Ans5)
^{2 }I =^{ 1}  (1/x  1/x^{2})e^{x} dx  
= (e^{x}/x)_{1}^{2}_{ }  [as  [f'(x) + f(x)]e^{x}dx = e^{x}f(x) + C  
= (e^{2}/2)  (e/1)  
= (e^{2}  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  R_{X}  R_{Y}  d_{i}=R_{X}  R_{Y}  d_{i}^{2} 
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 
d_{i}^{2} = 8 
R, the rank Correlation
= 1  (6d_{i}^{2}/n(n^{2}
 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 x^{p} y^{q} = (x + y)^{p +q} , prove that
dy/dx = y/x
Ans12) Given x^{p} y^{q} = (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} 
(x^{2} + 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 + (n1)h) 
where nh = b  a
^{2} _{0} 
(x^{2} + 2)dx 
=  h  f(0) + f(h) +....+ f((n1)h)  
=  h  (0^{2} + 2) + (h^{2} + 2) +...+{(n1)^{2}h^{2} +2}  
=  h  2n + h^{2}{ 1^{2} + 2^{2} + 3^{2} +...+ (n1)^{2}}  
=  h  2n + h^{2} {(n1)(n)(2n1)}/6  
=  2nh + {(nhh)(nh)(2nhh)}/6  
=  2 x 2 + 1/6{(2h)(2)(2 x 2 h)}  
= 4 + 1/6 . 2 . 2 . 4 = 4 + 8/3 = 20/3 
Q14)
Solve the differential equation (x + 2y^{2})dy/dx = y , given that when
x = 2, y = 1
Ans14) (x + 2y^{2})dy/dx = y
= ydx/dy = x + 2y^{2}
= ydx/dy  x = 2y^{2}
= dx/dy  x/y = 2y
This is a linear differential equation with x as dependent variable.
Here P = (1/y) , Q = 2y^{
} I.F. = e^{pdy}
= e^{1/ydy}
= e^{log y} = e ^{log y1} = 1/y
required equation is
x . (1/y) =  2y . 1/y dy + C 
= x/y = 2y + C
or x = 2y^{2} + Cy is the required solution of the given equation
Q15) Using properties of determinants prove that:
1 a
a^{3} 1 b b^{3} 1 c c^{3} 
=(a  b)(b  c)(c  a)(a + b + c) 
Ans15) Let
=  1 a
a^{3} 1 b b^{3} 1 c c^{3} 
=(a  b)(b  c)(c  a)(a + b + c) 
=  0 a  b
(a  b)(a^{2} + ab + b^{2}) 0 b  c (b  c)(b^{2} + bc + c^{2}) 1 c c^{3} 
=  0 1
a^{2} + ab + b^{2} 0 1 b^{2} + bc + c^{2} 1 c c^{3} 

=  1
a^{2} + ab + b^{2} 1 b^{2} + bc + c^{2} 

=  0
a^{2} + ab  bc  c^{2} 1 b^{2} + bc + c^{2} 

=  0
(a + c)(a  c) + b(a  c) 1 b^{2 }+ bc + c^{2} 
(applying R1 R1  R2)  
=  (a  b)(b  c)(a  c)  0 a
+ c +b 1 b^{2} + bc + c^{2} 
= (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/x^{2}) (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/c^{2}) = ((3 + (1/3))  (1 + (1/1)))/(3  1)
= 1  (1/c^{2}) = ((10/3)  2)/2 = 4/6 = 2/3
= 1/c^{2} = 1  (2/3) = 1/3
= c^{2} = 3 = c = + 3
but c = 3
(1,3)
= LMVT is applicable.
2x + 1 dx (x^{2} + 4x + 3) 
Ans17) Let
I= 
2x + 1 dx (x^{2} + 4x + 3) 

= 
2x + 4  3 dx (x^{2} + 4x + 3) 

= 
2x + 4 dx
 3 (x^{2} + 4x + 3) 
1 dx (x^{2} + 4x + 3) 
Now
I_{1}= 
2x + 4 dx
= 2(x^{2}
+ 4 +3) + C (x^{2} + 4x + 3) 

I_{2}= 
1 dx
= (x^{2} + 4x + 3) 
1
dx (x^{2} + 2x. 2 + 4  1) 

= 
1 dx ((x+ 2)^{2}  1^{2}) 
_{ } = log  x + 2 + (x^{2} + 4x + 3)  + C_{2}
I = 2(x^{2} + 4x + 3)  3 log  x + 2 + (x^{2} + 4x + 3) + C_{1} + C_{2}
I = 2(x^{2} + 4x + 3)  3 log  x + 2 + (x^{2} + 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 r^{2} = 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 semicircular 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 senicircle. 
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 .x^{2}
or A = x[20  ( + 2)x] +
1/2 . x^{2}
dA/dx = 20  2x (
+ 2) + x
= 20 +
x(  2
 4)
putting dA/dx = 0
x = 20/( + 4)
also d^{2}A/dx^{2} = (
+ 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
C_{11} = 0, C_{12} = 7, C_{13} = 7
C_{21} = 1, C_{22} = 7, C_{23} = 5
C_{31} = 2, C_{32} = 7, C_{33} = 4
A^{1} = 1/7  0 1
2 7 7 7 7 5 4 
X = A^{1}B
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:
2x^{2} + 2y^{2} + 2z^{2}  4x  6y + 2z + 3 = 0.
Ans23) The equation of the sphere is
x^{2} + y^{2} + z^{2}  2x  3y + z + 3/2 = 0
its centre is (1, 3/2, 1/2)
r = (1^{2} + (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 60^{o}.
Ans24) Here P = 10N , Q = 15N ,
= 60^{o}
Let their resultant be R inclined at an angle
with force P
we know
R = (P^{2} + Q^{2}
+ 2PQCos )
= (100 +
225 + 2 x 10 x 15 x 1/2) = 475
= 519 N
and tan = (QSin )
/(P + QCos ) = (15 Sin 60^{o})/(10
+ 15Cos 60^{o})
= (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/s^{2} 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/s^{2}
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
30^{o} 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/s^{2}]
Ans30) (i) The equation of the path of particle is
y = x tan  ((gx^{2})/(2u^{2}Cos^{2}))
= y = x tan 30^{o}  ((10x^{2})/(2(100)^{2}Cos^{2}30^{o})
or y = x/3  x^{2}/1500
(ii) The length of latus rectum
= (2u^{2}Cos^{2})/g
= 1500m
(iii) The greatest height
= (u^{2}Sin^{2})/g
= (100 x 100 x Sin^{2}30^{o})/10 = 250 m
(iv) The time of flight
= (2uSin)/g = (2 x 100 x
Sin^{ }30^{o})/10 = 10 sec
(v) The horizontal range
=^{ }(u^{2}Sin2)/g
= (100 x 100 x Sin 60^{o})/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 ,
M_{d} = 48 and S_{KP} = 0.4
Now S_{KP} = 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)
= ^{6}C_{5}p^{5}q^{1} + ^{6}C_{6}p^{6}
=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 semiannually.
[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 = 4x^{2}  x + 70
find (i) the average cost (ii) the marginal cost , when x = 3.
Ans25) Here C = 4x^{2 } x + 70
= AC = C/x = 1/x(4x^{2 } 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 1^{2}/
= 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
H_{1} : Event of choosing a scooter produced by Plant  I
H_{2} : Event of choosing a scooter produced by Plant  II
E : Event of selecting a scooter of standard quality
Here we have to find
P(H_{2}/E) , Given P(H_{1}) = 70/100, P(H_{2}) = 30/100
By Bay's theorem
P(H_{2}/E) = (P(E/H_{2}) . P(H_{2}))/(P(E/H_{1}).P(H_{1})
+ P(E/H_{2})P(H_{2}))
= (9/10 x 3/10)/ (8/10 x 7/10 + 9/10 x 3/10) = 27/(56 + 27) = 27/83
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