Linear Programming Test

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Linear Programming Test - 28

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Question 1
1 / -0

The corner points of the feasible region determined by the system of linear constraints are (0, 0), (o, 40), (20, 40), (60, 20), (60, 0). The objective function is $$Z = 4x + 3y$$.
Compare the quantity in Column A and Column B

A
The quantity in column A is greater

B
The quantity in column B is greater

C
The two quantities are equal

D
The relationship cannot be determined on the basis of the information Supplied

Solution

Hence, maximum value of $$Z = 300 < 325$$.
So, the quantity in column B is greater.
Question 2
1 / -0

The value of objective function is maximum under linear constraints

A
at the centre of feasible region

B
at (0,0)

C
at the vertex of feasible region

D
the vertex which is of maximum distance from (0,0)

Solution
Question 3
1 / -0

Of all the points of the feasible region, the optimal value of z obtained at the point lies

A
inside the feasible region

B
at the boundary of the feasible region

C
at vertex of feasible region

D
outside the feasible region
Question 4
1 / -0

Object function of LPP is

A
a constraint

B
a function to be maximized or minimized

C
a relation between the decision variables

D
equation of a straight line

Solution
Question 5
1 / -0

The corner points of the feasible solution given by the inequation $$x + y \leq 4, 2x + y \leq 7, x \geq 0, y \geq 0$$ are

A
(0 ,0), (4, 0), (7, 1), (0, 4)

B
(0, 0), $$(\dfrac{7}{2}, 0)$$, (3, 1), (0, 4)

C
(0, 0), $$(\dfrac{7}{2}, 0)$$, (3, 1), (0, 7)

D
(0, 0), (4, 0), (3, 1), (0, &)
Question 6
1 / -0

Solution of LPP to minimize z = 2x + 3y, such that $$x \geq 0, y \geq 0, 1 \leq x + 2y \leq 10 $$ is

A
$$x = 0, y = \dfrac{1}{2}$$

B
$$x = \dfrac{1}{2}, y = 0$$

C
$$x = 1, y = 2$$

D
$$x = \dfrac{1}{2}, y = \dfrac{1}{2}$$
Question 7
1 / -0

Find the output of the program given below if$$ x = 48$$
and $$y = 60$$
10 $$ READ x, y$$
20 $$Let x = x/3$$
30 $$ Let y = x + y + 8$$
40 $$ z = \dfrac y4$$
50 $$PRINT z$$
60 $$End$$

A
$$21$$

B
$$22$$

C
$$23$$

D
$$24$$

Solution

After the step $$ 10 $$ READ $$ x, y $$, the value of $$ x = 48, y = 60 $$

After the step $$ 20 $$ Let $$ x = \frac {x}{3} $$, the value of $$ x = \frac {48}{3} = 16 , y = 60 $$

After the step $$ 30 $$ Let $$ y = x +y + 8 $$, the value of $$ x = 16, y = 16 + 60 + 8 = 84 $$

After the step $$ 40 $$ Let $$ z = \frac {y}{4} $$, the value of $$ x = \frac {84}{4} = 21 $$

Hence $$ z = 21 $$ is the final output printed.
Question 8
1 / -0

10 students of class X took part in a Mathematics quiz If the number of girls is 4 more than the number of boys find the number of boys and girls who took part in the quiz Which graph represents the solution of the problem?

A

B

C

D

Solution

let no.of boys =$$x $$
no of girls = $$y$$

Total number of students took part in quiz -$$ x+y-io$$

$$2+y=10 $$ $$-----(1)$$

if no. of girls is $$ y $$ more than no . of $$\rightarrow =x+y$$

$$x-y=-y$$ $$-----(2)$$

From $$(1) , x+y=10$$

so , $$ y=10-x$$
$$x$$ 0 1 2 3 4 5 7
$$y$$ 10 9 8 7 6 5 3
From $$(2), x-y=-y$$
so $$ y= x+y$$

$$x$$ 0 1 2 3 4
$$y$$ 4 5 6 7 8
hence , the correct option is $$C$$
Question 9
1 / -0

A retired person wants to invest an amount of Rs. $$50, 000.$$ His broker recommends investing in two type of bonds A and B yielding $$10 \%$$ and $$9 \%$$ return respectively on the invested amount. He decides to invest at least $$Rs. 20,000$$ in bond A and at least $$Rs. \ 10,000$$ in bond B. He also wants to invest at least as much in bond A as in bond B. Solve this linear programming problem graphically to maximize his returns.

A
$$4900$$

B
$$2900$$

C
$$5400$$

D
$$4000$$

Solution

$$Points$$ $$Z=\dfrac{10}{100}x+\dfrac{9}{100}y$$
$$ A(20000, 10000)$$ $$Z= Rs. 2900$$
$$B(40000, 10000)$$ $$ Rs. 4900$$ (Maximize)
$$C(25000, 25000)$$ $$Rs. 4750$$
$$D(20000, 20000)$$ $$Rs. 3800$$
Let $$Rs.x$$ invest in bond A and $$Rs. y$$ invest in bond B.
Then A.T.P.
Maximise, $$z =\dfrac{10}{100}x+\dfrac{9}{100}y$$ --- (1)
Subject to constraints
$$x+y\leq 50,000$$ --- (a)
$$x\geq 20,000$$---(b)
$$y\geq 10,000$$--- (c)
and $$x\geq y$$
or $$x-y\geq0$$--- (d)
and $$x\geq 0, y\geq 0$$
Now change inequality into equations
$$x+y=50,000, x = 20,000, y = 10,000, x = y$$
Region: put (0, 0) in (a), (b), (c), (d)
$$0\leq 50000$$ (towards origin)
$$0\geq 20,000$$ (away from origin)
$$0\geq 10,000$$ (away from origin)
As shown in the graph.
So from the tabular column, we conclude that he has to invest $$Rs.40,000$$ in 'A' and $$Rs. 10,000$$ in bond 'B' to get maximum return $$Rs.4,900.$$
Question 10
1 / -0

A manufacturer produces nuts and bolts. It takes $$1$$ hour of work on machine A and $$3$$ hours on machine B to produce a package of nuts. It takes $$3$$ hours on machine A and $$1$$ hour on machine B to produce a package of bolts. He earns a profit of $$Rs. 17.50$$ per package on nuts and $$Rs 7$$ per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operate his machines for at the most $$12$$ hours a day?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let x package nuts and y package bolts are produced
Let z be the profit function, which we have to maximize.
Here, $$z = 17.50x + 7y .... (1)$$ is objective function.
And constraints are
$$x+3y\leq12$$ .... (2)
$$3x+y\leq12$$ .... (3)
$$x\geq 0$$ ....(4)
$$y\geq 0$$ ....(5)
On plotting graph of above constraints or inequalities $$(2), (3), (4)$$ and $$(5)$$ we get shaded region as feasible region having corner points A, O, B and C.
For coordinate of 'c' two equations,
$$x + 3y = 12 .... (6)$$
$$3x + y = 12 .... (7)$$
On solving we get, $$x = 3$$ and $$y = 3$$
Hence coordinate of C are $$(3, 3)$$
Now value of z is evaluated at corner point as shown in the graph.
Therefore, maximum profit is Rs. $$73.5$$ when $$3$$ package nuts and $$3$$ package bolt are produced.

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Re-Attempt Weekly Quiz Competition

$$Points$$	$$Z=\dfrac{10}{100}x+\dfrac{9}{100}y$$
$$ A(20000, 10000)$$	$$Z= Rs. 2900$$
$$B(40000, 10000)$$	$$ Rs. 4900$$ (Maximize)
$$C(25000, 25000)$$	$$Rs. 4750$$
$$D(20000, 20000)$$	$$Rs. 3800$$

$$x$$	0	1	2	3	4	5	7
$$y$$	10	9	8	7	6	5	3

$$x$$	0	1	2	3	4
$$y$$	4	5	6	7	8

Linear Programming Test - 28

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Linear Programming Test - 28

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Question 1

The quantity in column A is greater

The quantity in column B is greater

The two quantities are equal

The relationship cannot be determined on the basis of the information Supplied

Solution

Question 2

at the centre of feasible region

at (0,0)

at the vertex of feasible region

the vertex which is of maximum distance from (0,0)

Solution

Question 3

inside the feasible region

at the boundary of the feasible region

at vertex of feasible region

outside the feasible region

Question 4

a constraint

a function to be maximized or minimized

a relation between the decision variables

equation of a straight line

Solution

Question 5

(0 ,0), (4, 0), (7, 1), (0, 4)

(0, 0), $$(\dfrac{7}{2}, 0)$$, (3, 1), (0, 4)

(0, 0), $$(\dfrac{7}{2}, 0)$$, (3, 1), (0, 7)

(0, 0), (4, 0), (3, 1), (0, &)

Question 6

$$x = 0, y = \dfrac{1}{2}$$

$$x = \dfrac{1}{2}, y = 0$$

$$x = 1, y = 2$$

$$x = \dfrac{1}{2}, y = \dfrac{1}{2}$$

Question 7

$$21$$

$$22$$

$$23$$

$$24$$

Solution

Question 8

Solution

Question 9

$$4900$$

$$2900$$

$$5400$$

$$4000$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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