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CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Linear Programming
Quiz 4
The maximum profit that can be made in a day is:
Report Question
0%
Rs.
450
0%
Rs.
800
0%
Rs.
650
0%
Rs.
525
Explanation
let the number of normal calculators produced in a day be
x
and
the number of scientific calculators produced in a day be
y
the minimum of total calculators to be produced per day is
200
⟹
x
+
y
≥
200
Given, the minimum number of
normal calculators to be produced per day is
100
⟹
x
≥
100
and
the minimum number of scientific
calculators to be produced per day is
80
⟹
y
≥
80
Also g
iven, the maximum number of
normal calculators can be produced per day is
200
⟹
x
≤
200
and
the maximum number of scientific
calculators can be produced per day is
170
⟹
y
≤
170
A normal calculator incurred a loss of
R
s
.2
For
x
normal calculators, the loss is
R
s
.2
x
A scientific calculator gained a profit of
R
s
.5
For
x
y
scientific calculators, the gain is
R
s
.5
y
Therefore, profit of the manufacturer
P
=
5
y
−
2
x
In the above figure, the blue shaded region is the feasible region with five corner points.
(
100
,
170
)
,
(
200
,
170
)
,
(
200
,
80
)
,
(
120
,
80
)
,
(
100
,
100
)
(
100
,
170
)
is the point where
x
=
100
intersects
y
=
170
(
200
,
170
)
is the point where
x
=
200
intersects
y
=
170
(
200
,
80
)
is the point where
x
=
200
intersects
y
=
80
(
120
,
80
)
is the point where
x
+
y
=
200
intersects
y
=
80
I.e., substituting
y
=
80
⟹
x
+
80
=
200
⟹
x
=
200
−
80
⟹
x
=
120
(
100
,
100
)
is the point where
x
+
y
=
200
intersects
x
=
100
I.e., substituting
x
=
100
⟹
100
+
y
=
200
⟹
x
=
200
−
100
⟹
x
=
100
Now substituting the corner points the profit equation,
substituting
(
100
,
170
)
⟹
P
=
5
×
170
−
2
×
100
=
650
substituting
(
200
,
170
)
⟹
P
=
5
×
170
−
2
×
200
=
450
substituting
(
200
,
80
)
⟹
P
=
5
×
80
−
2
×
200
=
0
substituting
(
120
,
80
)
⟹
P
=
5
×
80
−
2
×
120
=
160
substituting
(
100
,
100
)
⟹
P
=
5
×
100
−
2
×
100
=
300
650
is the maximum profit
The Convex Polygon Theorem states that the optimum (maximum or minimum) solution of a LPP is attained at atleast one of the ______ of the convex set over which the solution is feasible.
Report Question
0%
origin
0%
corner points
0%
centre
0%
edge
Explanation
The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.
In order to obtain maximum profit, the quantity of normal and scientific calculators to be manufactured daily is:
Report Question
0%
(
100
,
170
)
0%
(
200
,
170
)
0%
(
100
,
80
)
0%
(
200
,
80
)
Explanation
let the number of normal calculators produced in a day be
x
and
the number of scientific calculators produced in a day be
y
the minimum of total calculators to be produced per day is
200
⟹
x
+
y
≥
200
Given, the minimum number of
normal calculators to be produced per day is
100
⟹
x
≥
100
and
the minimum number of scientific
calculators to be produced per day is
80
⟹
y
≥
80
Also g
iven, the maximum number of
normal calculators can be produced per day is
200
⟹
x
≤
200
and
the maximum number of scientific
calculators can be produced per day is
170
⟹
y
≤
170
A normal calculator incurred a loss of
R
s
.2
For
x
normal calculators, the loss is
R
s
.2
x
A scientific calculator gained a profit of
R
s
.5
For
x
y
scientific calculators, the gain is
R
s
.5
y
Therefore, profit of the manufacturer
P
=
5
y
−
2
x
substituting the options in the equation
P
=
5
y
−
2
x
and finding the maximum value and verifying whether the option satisfies the given constraints.
substituting option A i.e.,
(
x
,
y
)
=
(
100
,
170
)
x
+
y
≥
200
⟹
100
+
170
=
270
≥
200
True
x
≥
100
and
x
≤
200
True
y
≥
80
and
x
≤
170
True
Hence,
P
=
5
y
−
2
x
=
5
∗
170
−
2
∗
100
=
650
substituting option B i.e.,
(
x
,
y
)
=
(
200
,
170
)
x
+
y
≥
200
⟹
200
+
170
=
370
≥
200
True
x
≥
100
and
x
≤
200
True
y
≥
80
and
x
≤
170
True
Hence,
P
=
5
y
−
2
x
=
5
∗
170
−
2
∗
200
=
450
substituting option C i.e.,
(
x
,
y
)
=
(
100
,
80
)
x
+
y
≥
200
⟹
100
+
80
=
180
≥
200
False
substituting option D i.e.,
(
x
,
y
)
=
(
200
,
80
)
x
+
y
≥
200
⟹
200
+
80
=
280
≥
200
True
x
≥
100
and
x
≤
200
True
y
≥
80
and
x
≤
170
True
Hence,
P
=
5
y
−
2
x
=
5
∗
80
−
2
∗
200
=
0
Therefore,
(
100
,
170
)
satisfies all constraints and produces the maximum value than the other options.
The region on the graph sheet with satisfies the constraints including the non- negativity restrictions is called the _______ space.
Report Question
0%
solution
0%
interval
0%
concave
0%
convex
Explanation
In linear programming, the feasible region is defined as the region in which all the set of points which satisfy the given constraints and the non-negativity restrictions.
In the above figure, the blue shaded region is the feasible region which contains the solution set.
In order to maximize the profit of the company, the optimal solution of which of the following equations is required?
Report Question
0%
P
=
x
+
y
−
200
0%
P
=
5
y
−
2
x
0%
P
=
y
−
80
0%
P
=
200
−
x
Explanation
let the number of normal calculators produced in a day be
x
and
the number of scientific calculators produced in a day be
y
the minimum of total calculators to be produced per day is
200
⟹
x
+
y
≥
200
Given, the minimum number of
normal calculators to be produced per day is
100
⟹
x
≥
100
and
the minimum number of scientific
calculators to be produced per day is
80
⟹
y
≥
80
Also g
iven, the maximum number of
normal calculators can be produced per day is
200
⟹
x
≤
200
and
the maximum number of scientific
calculators can be produced per day is
170
⟹
y
≤
170
A normal calculator incurred a loss of
R
s
.2
For
x
normal calculators, the loss is
R
s
.2
x
A scientific calculator gained a profit of
R
s
.5
For
x
y
scientific calculators, the gain is
R
s
.5
y
Therefore, profit of the manufacturer
P
=
5
y
−
2
x
How many acres of each (wheat and rye) should the farmer plant in order to get maximum profit?
Report Question
0%
(
5
,
5
)
0%
(
4
,
4
)
0%
(
4
,
5
)
0%
(
4
,
3
)
Explanation
let
x
be the acres of wheat planted and
y
be the acres of rye planted
Given that there are a total of
10
acres of land to plant.
Atleast
7
acres is to be planted i.e.,
x
+
y
≥
7
Given that the cost to plant one acre of wheat is
$
200
Therefore, the cost for
x
acres of wheat is
200
x
Given that the cost to plant one acre of rye is
$
100
Therefore, the cost for
y
acres of rye is
100
y
Given that, amount for planting wheat and rye is
$
1200
Therefore the total cost to plant wheat and rye is
200
x
+
100
y
≤
1200
⟹
2
x
+
y
≤
12
Given that, the time taken to plant one acre of wheat is
1
hr
Therefore, the time taken to plant
x
acres of wheat is
x
hrs
Given that, the time taken to plant one acre of rye is
2
hrs
Therefore, the time taken to plant
y
acres of rye is
2
y
hrs
Given that, the total time for planting is
12
hrs
Therefore, the total time to plant wheat and rye is
x
+
2
y
≤
12
Given that, one acre of wheat yields a profit of
$
500
Therefore, the profit from
x
acres of wheat is
500
x
Given that, one acre of rye yields a profit of
$
300
Therefore, the profit from
y
acres of wheat is
300
y
therefore the total profit from the wheat and rye is
P
=
500
x
+
300
y
Now substituting the options in the profit expression and verifying
Substituting option A
(
x
,
y
)
=
(
5
,
5
)
x
+
y
≥
0
⟹
5
+
5
≥
7
⟹
10
≥
7
True
2
x
+
y
≤
12
⟹
2
(
5
)
+
5
≤
12
⟹
15
≤
12
False
Substituting option B
(
x
,
y
)
=
(
4
,
4
)
x
+
y
≥
0
⟹
4
+
4
≥
7
⟹
8
≥
7
True
2
x
+
y
≤
12
⟹
2
(
4
)
+
4
≤
12
⟹
12
≤
12
True
x
+
2
y
≤
12
⟹
4
+
2
(
4
)
≤
12
⟹
12
≤
12
True
Substituting option C
(
x
,
y
)
=
(
4
,
5
)
x
+
y
≥
0
⟹
4
+
5
≥
7
⟹
9
≥
7
True
2
x
+
y
≤
12
⟹
2
(
4
)
+
5
≤
12
⟹
13
≤
12
False
Substituting option D
(
x
,
y
)
=
(
4
,
3
)
x
+
y
≥
0
⟹
4
+
3
≥
7
⟹
7
≥
7
True
2
x
+
y
≤
12
⟹
2
(
4
)
+
3
≤
12
⟹
11
≤
12
True
x
+
2
y
≤
12
⟹
4
+
2
(
3
)
≤
12
⟹
10
≤
12
True
(
4
,
4
)
,
(
4
,
3
)
satisfies the constraints. Therefore finding the profit
for
(
4
,
4
)
,
P
=
500
x
+
300
y
=
500
(
4
)
+
300
(
4
)
=
3200
for
(
4
,
3
)
,
P
=
500
x
+
300
y
=
500
(
4
)
+
300
(
3
)
=
2900
Therefore the maximum profit is attained at
(
4
,
4
)
How many acres of land will be left unplanted if the farmer aims for a maximum profit?
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
let
x
be the acres of wheat planted and
y
be the acres of rye planted
Given that there are a total of
10
acres of land to plant.
Atleast
7
acres is to be planted i.e.,
x
+
y
≥
7
Given that the cost to plant one acre of wheat is
$
200
Therefore, the cost for
x
acres of wheat is
200
x
Given that the cost to plant one acre of rye is
$
100
Therefore, the cost for
y
acres of rye is
100
y
Given that, an amount for planting wheat and rye is
$
1200
Therefore the total cost to plant wheat and rye is
200
x
+
100
y
≤
1200
⟹
2
x
+
y
≤
12
Given that, the time taken to plant one acre of wheat is
1
hr
Therefore, the time taken to plant
x
acres of wheat is
x
hrs
Given that, the time taken to plant one acre of rye is
2
hrs
Therefore, the time taken to plant
y
acres of rye is
2
y
hrs
Given that, the total time for planting is
12
hrs
Therefore, the total time to plant wheat and rye is
x
+
2
y
≤
12
Given that, one acre of wheat yields a profit of
$
500
Therefore, the profit from
x
acres of wheat is
500
x
Given that, one acre of rye yields a profit of
$
300
Therefore, the profit from
y
acres of wheat is
300
y
therefore the total profit from the wheat and rye is
P
=
500
x
+
300
y
In the above figure, the blue shaded region is the feasible region with three corner points.
(
4
,
4
)
,
(
2
,
5
)
,
(
5
,
2
)
Now substituting the corner points the profit equation,
substituting
(
4
,
4
)
⟹
P
=
500
x
+
300
y
=
500
(
4
)
+
300
(
4
)
=
3200
substituting
(
2
,
5
)
⟹
P
=
500
x
+
300
y
=
500
(
2
)
+
300
(
5
)
=
2500
substituting
(
5
,
2
)
⟹
P
=
500
x
+
300
y
=
500
(
5
)
+
300
(
2
)
=
3100
$
3200
is the maximum profit is attained by planting
4
acres of wheat and
4
acres of rye.
i.e., a total of
4
+
4
=
8
acres of land is planted. But, the total land is
10
acres. therefore, the unplanted land is
10
−
8
=
2
acres
If
a
,
b
,
c
∈
+
R
such that
λ
a
b
c
is the minimum value of
a
(
b
2
+
c
2
)
+
b
(
c
2
+
a
2
)
+
c
(
a
2
+
b
2
)
, then
λ
=
Report Question
0%
1
0%
3
0%
4
0%
None of the above.
Explanation
We know that
A
.
M
.
≥
G
.
M
.
Therefore,
b
2
+
c
2
2
≥
√
b
2
c
2
⟹
b
2
+
c
2
≥
2
b
c
Multiplying
a
on both sides doesn’t change the inequality. Since, given that
a
is positive.
⟹
a
(
b
2
+
c
2
)
≥
2
a
b
c
————(1)
Similarly,
b
(
a
2
+
c
2
)
≥
2
a
b
c
————(2)
and
c
(
a
2
+
b
2
)
≥
2
a
b
c
—————(3)
adding (1), (2) and (3) we get
a
(
b
2
+
c
2
)
+
b
(
a
2
+
c
2
)
+
c
(
a
2
+
b
2
)
≥
2
a
b
c
+
2
a
b
c
+
2
a
b
c
⟹
a
(
b
2
+
c
2
)
+
b
(
a
2
+
b
2
)
+
c
(
a
2
+
b
2
)
≥
6
a
b
c
Therefore
λ
is
6
Consider the objective function
Z
=
40
x
+
50
y
The minimum number of constraints that are required to maximize
Z
are
Report Question
0%
4
0%
2
0%
3
0%
1
Explanation
Since in the given function
Z
=
40
x
+
50
y
, two variables are used.
So, the two constraints will be
x
≥
0
,
y
≥
0
and the third one will be of the type
a
x
+
b
y
≥
c
.
Hence, at least
3
constraints are required.
Maximum value of
z
=
3
x
+
4
y
subject to
x
−
y
≤
−
1
,
−
x
+
y
≤
0
,
x
,
y
≥
0
is given by?
Report Question
0%
1
0%
4
0%
6
0%
None of the these
Explanation
No common feasible area
No commonpoint,
So,
Z
m
a
x
cant find for these curves.
Hence none of these.
The objective function
z
=
x
1
+
x
2
, subject to
x
1
+
x
2
≤
10
,
−
2
x
1
+
3
x
2
≤
15
,
x
1
≤
6
,
x
1
,
x
2
≥
0
has maximum value .................. of the feasible region.
Report Question
0%
at only one point
0%
at only two points
0%
at every point of the segment joining two points
0%
at every point of the line joining two points
Explanation
For the equation,
x
1
+
x
2
=
10
,
x
0
10
5
y
10
0
5
For the equation,
−
2
x
1
+
3
x
2
=
15
,
x
0
−
3
3
y
5
3
7
The feasible region for the given lines is as indicated in the figure.
The values of
z
at the corner points are:
At
(
0
,
0
)
,
z
=
0
At
(
0
,
5
)
,
z
=
5
At
(
3
,
7
)
,
z
=
3
+
7
=
10
At
(
6
,
4
)
,
z
=
6
+
4
=
10
At
(
6
,
0
)
,
z
=
6
Hence, the objective function
z
, has maximum value at every point of the segment joining two points of the feasible region.
The objective function
z
=
4
x
1
+
5
x
2
, subject to
2
x
1
+
x
2
≥
7
,
2
x
1
+
3
x
2
≤
15
,
x
2
≤
3
,
x
1
,
x
2
≥
0
has minimum value at the point.
Report Question
0%
On x-axis
0%
On y-axis
0%
At the origin
0%
On the parallel to x-axis
Explanation
Value of
z
=
4
x
1
+
5
x
2
Convert the given inequalities into equalities to get the corner points
2
x
1
+
x
2
=
7
.......
(
i
)
At
x
1
=
0
,
x
2
=
7
and
x
2
=
0
,
x
1
=
3.5
So, the corner points of
(
i
)
are
(
0
,
7
)
and
(
3.5
,
0
)
2
x
1
+
3
x
2
=
15
......
(
i
i
)
At
x
1
=
0
,
x
2
=
5
and
x
2
=
0
,
x
1
=
7.5
So, the corner points of
(
i
)
are
(
0
,
5
)
and
(
7.5
,
0
)
x
2
=
3
......
(
i
i
i
)
Plot these corner points on the graph paper and the line given in
(
i
i
i
)
The shaded part shows the feasible region.
At
x
2
=
3
,
x
1
=
2
in
(
i
)
and
x
1
=
3
in
(
i
i
)
The corner points of the feasible region are
(
3.5
,
0
)
,
(
7.5
,
0
)
,
(
3
,
3
)
and
(
2
,
3
)
Corner points
Z
=
4
x
2
+
5
x
2
(
3.5
,
0
)
14
(
7.5
,
0
)
30
(
3
,
3
)
27
(
2
,
3
)
23
Minimum value of
Z
=
14
and it lies on
x
-axis.
The objective function of LPP defined over the convex set attains its optimum value at
Report Question
0%
atleast two of the corner points
0%
all the corner points
0%
atleast one of the corner points
0%
none of the corner points
Explanation
Let
Z
=
a
x
+
b
y
be the objective function
When
Z
has optimum value(maximum or minimum), where the variables
x
and
y
are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.
Thus, the function attains its optimum value at one of the corner points.
Hence, C is correct.
The constraints
−
x
1
+
x
2
≤
1
−
x
1
+
3
x
2
≤
9
x
1
,
x
2
≥
0
defines on
Report Question
0%
Bounded feasible space
0%
Unbounded feasible space
0%
Both bounded and unbounded feasible
0%
None of the above
Explanation
Draw the following curves and find common area, which will give you feasible solutions.
So, our constraints are:
−
x
1
+
x
2
≤
1
-------------(i)
−
x
1
+
3
x
2
≤
9
----------(ii)
x
1
x
2
≥
0
-------------(iii)
We can see from the graph clearly that the constraints have unbounded feasible space
An article manufactured by a company consists of two parts
X
and
Y
. In the process of manufacture of the part
X
.
9
out of
100
parts may be defective. Similarly
5
out of
100
are likely to be defective in part
Y
. Calculate the probability that the assembled product will not be defective.
Report Question
0%
0.86
0%
0.864
0%
0.8456
0%
0.8645
Explanation
Let
A
=
Part
X
is not defective
Probability of
A
is
P
(
A
)
=
91
100
B
=
Part
Y
is not defective.
Probability of
B
is
P
(
B
)
=
95
100
Required probability
=
P
(
A
∩
B
)
=
P
(
A
)
P
(
B
)
=
91
100
×
95
100
=
8645
10000
The corner points of the feasible region determined by the system of linear constraints are
(
0
,
10
)
,
(
5
,
5
)
,
(
15
,
15
)
,
(
0
,
20
)
. Let
z
=
p
x
+
q
y
where
p
,
q
>
0
. Condition on p and q so that the maximum of z occurs at both the points
(
15
,
15
)
and
(
0
,
20
)
is __________.
Report Question
0%
q
=
2
p
0%
p
=
2
p
0%
p
=
q
0%
q
=
3
p
Explanation
Let
z
0
be the maximum value of
z
in the feasible region.
Since maximum occurs at both
(
15
,
15
)
and
(
0
,
20
)
,
t
h
e
v
a
l
u
e
z
0
i
s
a
t
t
a
i
n
e
d
a
t
b
o
t
h
(
15
,
15
)
a
n
d
(
0
,
20
)
$.
⟹
z
0
=
p
(
15
)
+
q
(
15
)
and
z
0
=
p
(
0
)
+
q
(
20
)
⟹
p
(
15
)
+
q
(
15
)
=
p
(
0
)
+
q
(
20
)
⟹
15
p
=
5
q
⟹
3
p
=
q
Hence, the answer is option (D).
Corner points of the bounded feasible region for an LP problem are
A
(
0
,
5
)
B
(
0
,
3
)
C
(
1
,
0
)
D
(
6
,
0
)
. Let
z
=
−
50
x
+
20
y
be the objective function. Minimum value of z occurs at ______ center point.
Report Question
0%
(
0
,
5
)
0%
(
1
,
0
)
0%
(
6
,
0
)
0%
(
0
,
3
)
Explanation
We check the value of the
z
at each of the corner points.
At
A
(
0
,
5
)
−
z
=
−
50
x
+
20
y
=
−
50
(
0
)
+
20
(
5
)
=
100
At
B
(
0
,
3
)
−
z
=
−
50
x
+
20
y
=
−
50
(
0
)
+
20
(
3
)
=
60
At
C
(
1
,
0
)
−
z
=
−
50
x
+
20
y
=
−
50
(
1
)
+
20
(
0
)
=
−
50
At
D
(
6
,
0
)
−
z
=
−
50
x
+
20
y
=
−
50
(
6
)
+
20
(
0
)
=
−
300
Hence, we see that
z
is minimum at
D
(
6
,
0
)
and minimum value is
−
300
.
The corner points of the feasible region are
A
(
0
,
0
)
,
B
(
16
,
0
)
,
C
(
8
,
16
)
and
D
(
0
,
24
)
. The minimum value of the objective function
z
=
300
x
+
190
y
is _______
Report Question
0%
5440
0%
4800
0%
4560
0%
0
Explanation
We know that, for a cartesian polygon , the maximum value occurs at the corner points or vertices of the polygon.
Given
z
=
300
x
+
190
y
By substituting
A
(
0
,
0
)
in the equation we get
z
=
0
By substituting
B
(
16
,
0
)
in the equation we get
z
=
4800
By substituting
C
(
8
,
16
)
in the equation we get
z
=
5440
By substituting
D
(
0
,
24
)
in the equation we get
z
=
4560
Hence the minimum value of Z occured at
C
(
0
,
0
)
with
z
=
0
The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.
Colour
Number of cars manufactured
Vento
Creta
WagonR
Red
65
88
93
White
54
42
80
Black
66
52
88
Sliver
37
49
74
Which car was twice the number of silver Vento?
Report Question
0%
Silver WagonR
0%
Red WagonR
0%
Red Vento
0%
White Creta
Explanation
The number of silver Vento car
=
37
(from the table)
Twice the number of silver Vento cars
=
2
×
37
=
74
Now from table we can see that silver WagonR is only car type having
74
cars
A firm manufactures three products
A
,
B
and
C
. Time to manufacture product
A
is twice that for
B
and thrice that for
C
and if the entire labour is engaged in making product
A
,
1600
units of this product can be produced.These products are to be produced in the ratio
3
:
4
:
5.
There is demand for at least
300
,
250
and
200
units of products
A
,
B
and
C
and the profit earned per unit is Rs.
90
,
Rs
40
and Rs.
30
respectively.
Raw
material
Requirement per unit product(Kg)
A
Requirement per unit product(Kg)
B
Requirement per unit product(Kg)
C
Total availability (kg)
P
6
5
2
5
,
000
Q
4
7
3
6
,
000
Formulate the problem as a linear programming problem and find all the constraints for the above product mix problem.
Report Question
0%
3
x
1
−
4
x
2
=
0
and
5
x
2
−
4
x
3
=
0
where
x
1
,
x
2
,
x
3
≥
0
0%
4
x
1
−
3
x
2
=
0
and
5
x
2
−
4
x
3
=
0
where
x
1
,
x
2
,
x
3
≥
0
0%
4
x
1
−
3
x
2
=
0
and
4
x
2
−
5
x
3
=
0
where
x
1
,
x
2
,
x
3
≥
0
0%
4
x
1
−
3
x
2
=
0
and
5
x
2
−
4
x
3
=
0
where
x
1
,
x
2
,
x
3
≤
0
Explanation
Formulation of L.P Model
Let
x
1
,
x
2
and
x
3
denote the number of units of products
A
,
B
and
C
to be manufactured .
Objective is to maximize the profit.
i.e., maximize
Z
=
90
x
1
+
40
x
2
+
30
x
3
Constraints can be formulated as follows:
For raw material
P
,
6
x
1
+
5
x
2
+
2
x
3
≤
5
,
000
For raw material
Q
,
4
x
1
+
7
x
2
+
3
x
3
≤
6
,
000
Product
B
requires
1
2
and product
C
requires
(
1
3
)
r
d
the time required for product
A
.
Then
t
2
and
t
3
are the times in hours to produce
B
and
C
and since
1
,
600
units of
A
will need time
1
,
600
t
hours, we get the constraint,
t
x
1
+
t
2
x
2
+
t
3
x
3
≤
1
,
600
t
or
x
1
+
x
2
2
+
x
3
3
≤
1
,
600
or
6
x
1
+
3
x
2
+
2
x
3
≤
9
,
600
Market demand requires
x
1
≥
300
,
x
2
≥
250
,
and
x
3
≥
200
Finally, since products
A
,
B
and
C
are to be produced in the ratio
3
:
4
:
5
,
x
1
:
x
2
:
x
3
::
3
:
4
:
5
or
x
1
3
=
x
2
4
,
and
x
2
4
=
x
3
5
.
Thus, there are two additional constraints
4
x
1
−
3
x
2
=
0
and
5
x
2
−
4
x
3
=
0
where
x
1
,
x
2
,
x
3
≥
0
Equation of normal drawn to the graph of the function defined as
f
(
x
)
=
sin
x
2
x
,
x
≠
0
and
f
(
0
)
=
0
at the origin is?
Report Question
0%
x
+
y
=
0
0%
x
−
y
=
0
0%
y
=
0
0%
x
=
0
Explanation
let slope of tangent=T, slope of normal=N
T
=
d
y
d
x
=
x
.
c
o
s
x
2
×
2
x
−
s
i
n
x
2
x
2
T
a
t
(
x
=
0
)
=
c
o
s
x
2
×
2
x
2
−
s
i
n
x
2
x
2
x
=
0
⟹
2
c
o
s
x
2
−
s
i
n
x
2
x
2
(
s
i
n
x
x
a
t
x
=
1
)
⟹
2
−
1
=
1
N
×
T
=
−
1
N
=
−
1
a
t
x
=
0
,
y
=
0
y
=
m
x
y
=
−
x
x
+
y
=
0
The taxi fare in a city is as follows. For the first km the fare is
R
s
.10
and subsequent distance is
R
s
.6
/
k
m
.
Taking the distance covered as
x
k
m
and fare as
R
s
y
,write a linear equation.
Report Question
0%
y
=
4
+
6
x
0%
y
=
4
+
5
x
0%
y
=
3
+
6
x
0%
y
=
3
+
5
x
Explanation
First
k
m
fare =
R
s
.10
Subsequent distance fare=
R
s
6
/
k
m
Then fare
x
k
m
of distance
y
=
(
x
−
1
)
×
6
+
10
y
=
6
x
−
6
+
10
y
=
6
x
+
4
A
is correct
The feasible region for anLPP is shown shaded in the figure. Find the maximum value of the objective function
z
=
11
x
+
7
y
.
Report Question
0%
35
0%
47
0%
21
0%
14
Explanation
the cover point of feasible
(
0
,
3
)
,
(
0
,
5
)
,
(
3
,
2
)
→
Z
(
0
,
3
)
=
21
Z
(
0
,
5
)
=
35
Z
(
3
,
2
)
=
47
= max value of
Z
" the relation S is defined on
N
×
N
(
a
,
b
)
S
(
c
,
d
)
⇔
b
c
(
a
+
d
)
=
a
d
in an equivalance " that statement is ?
Report Question
0%
True
0%
False
The inequalities
f
(
−
1
)
≤
−
4
,
f
(
1
)
≤
0
&
f
(
3
)
≤
5
are known to hold for
f
(
x
)
=
a
x
2
+
b
x
+
c
then the least value of
′
a
′
is:
Report Question
0%
−
1
/
4
0%
−
1
/
3
0%
1
/
4
0%
1
/
8
The problem associated with
L
P
P
is
Report Question
0%
single objective function
0%
Double objective function
0%
No any objective function
0%
None
Explanation
The problem associated with LLP is single objective.
Shade region is represented by
Report Question
0%
2
x
+
5
y
≥
80
,
x
+
y
≤
20
,
x
≥
0
,
y
≤
0
0%
2
x
+
5
y
≥
80
,
x
+
y
≥
20
,
x
≥
0
,
y
≥
0
0%
2
x
+
5
y
≤
80
,
x
+
y
≤
20
,
x
≥
0
,
y
≥
0
0%
2
x
+
5
y
≤
80
,
x
+
y
≤
20
,
x
≥
0
,
y
≤
0
The maximum value of
P
=
3
x
+
4
y
subject to the constraints
x
+
y
≤
40
,
x
+
2
y
≤
60
,
x
≥
0
and
y
≥
0
is
Report Question
0%
120
0%
140
0%
100
0%
160
The maximum value of
4
x
+
5
y
subject to the constraints
x
+
y
≤
20
,
x
+
2
y
≤
35
,
x
−
3
y
≤
12
is
Report Question
0%
84
0%
95
0%
100
0%
96
If
x
is any real number, then which of the following is correct?
Report Question
0%
x
2
1
+
x
4
≥
1
2
0%
x
2
1
+
x
4
≥
1
4
0%
x
2
1
+
x
4
≤
1
2
0%
none of these
Explanation
We know that
(
x
−
1
x
)
2
≥
0
x
2
+
1
x
2
−
2
≥
0
x
2
+
1
x
2
≥
2
0
<
1
x
2
+
1
x
2
≤
1
2
0
≤
x
2
x
4
+
1
≤
1
2
(
∵
x
2
and
x
4
are positive or zero)
x
2
x
4
+
1
∈
[
0
,
1
2
]
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
1
Not Answered
29
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Correct : 0
Incorrect : 0
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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