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)/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.
 |
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 | ||||
|
(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)/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
P
Q = {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) |
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.
|
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) |
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) |
||
((x + 1)2
- 12) | + C2= log | x + 2 +
(x2
+ 4x + 3) | + C2I = 2
(x2 + 4x +
3) - 3 log | x + 2 + (x2
+ 4x + 3)| + C1 + C2I = 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 . 2
x
= 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 = 4x2 - x + 70
= AC = C/x = 1/x(4x2 - 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
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