MCQ Questions for Class 12 Commerce Maths Linear Programming Quiz 5

The shape of the region determined by

$2x+6y \geq 12, 3x +2y \geq 6, x+y \leq 8, x \geq 0, y \geq 0$

0%
a triangle
0%
a quadrilateral
0%
a pentagon
0%
a hexagon

A_____ in a table represents a relationship among a set of values.

0%
Column
0%
Key
0%
Row
0%
Entry

The feasible region of an LPP is shown in the figure. If

$z=3x+9y$ , then the minimum value of z occurs at?

0%
$(5, 5)$
0%
$(0, 10)$
0%
$(0, 20)$
0%
$(15, 15)$

Explanation

z+3x+9y.

$z+3x+9y$

$z_{(5.5)}=15+45=60$

$z_{(0, 0)}=0+90=90$

Minimum value at

$(5, 5)$ .

For the LPP; maximise

$z=x+4y$ subject to the constraints

$x+2y\leq 2$ ,

$x+2y\geq 8$ ,

$x, y\geq 0$ .

0%
$z_{max}=4$
0%
$z_{max}=8$
0%
$z_{max}=16$
0%
Has no feasible solution

Explanation

$x+2y\leq 2$

$x+2y\geq 8$

$x, y\geq 0$ .

$2x - y +1 = 0$ is tangent to the hyperbola

$\frac {x^2}{a^2} - \frac {y^2}{16} = 1$ then which of the following CANNOT be sides of a right angled triangle?

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a,4,1
0%
2a,4,1
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a,4,2
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2a,8,1

Convex set: A set of point in a plane is said to be convex set it the line segment joining any two points in the set, completely lies in the set.
Which of the following sets represent the convex sets ? jistify your answer.

Explanation

Image A-Convex set,

$\because$ line joining any two points in the set lies inside the set

Image B-convex set

Image C-Not a convex set

Image D-Not a convex set

Image E-Convex set

Image F-Not a convex set

1237645_1525413_ans_1a1d2a76d6ea4fbc918f3c680e95f9e0.png

The minimum value of

$z=10x+25y$ subject to

$0\leq x\leq 3, 0\leq y\leq 3, x+y\geq 5$ is?

0%
$80$
0%
$95$
0%
$105$
0%
$30$

Explanation

From graph it is evident that (x,y) satisfying all required conditions for which z is minimum must lie on x+y=5,

So, for finding minimum we can rewrite z as

$z= 10(5-y)+25y = 15y +50$

Lower the value of y, lower is value of z . The minimum value of y which satisfy all required condition is 2 which is also evident from graph.

So, minimum value of z is

$15(2)+50=80$

1408568_1675860_ans_621e754e3e8f4e3680055544261bdd23.png

The maximum value of

$z=9x+11y$ subject to

$3x+2y\leq 12, 2x+3y\leq 12$ ,

$x\geq 0, y\geq 0$ is _________.

0%
$44$
0%
$54$
0%
$36$
0%
$48$

Explanation

$3x + 2y = 12$ ....

$(1)$

$2x + 3y = 12$ ....

$(2)$

Now, Solving eq.

$(1)$ and

$(2)$ we get,

$x = \dfrac{12}{5}, y = \dfrac{12}{5}$

$\therefore\ B \left (\dfrac{12}{5}, \dfrac{12}{5} \right )$

Now,

$Z = 9x + 11 y$

(a) at

$(0,0) \Rightarrow\ Z = 9(0) + 11 (0) = 0$

(b) at

$(0,4) \Rightarrow\ Z = 9(0) + 11(4) = 44$

$(4,0) \Rightarrow\ Z = 9(4) + 11 (0) = 36$

(d) at

$\left (\dfrac{12}{5}, \dfrac{12}{5} \right ) \Rightarrow\ Z = 9 \left (\dfrac{12}{5} \right) + 11 \left (\dfrac{12}{5} \right )$

$= \dfrac{12}{5} (9 + 11 ) - \dfrac{12}{5} \times 20$

$= 12 \times 4$

$= 48$

$\therefore\ Z (max) = 48$

2133734_1675834_ans_fec5d3caa728423cb629fbcbb3f14cee.png

Two towns A and B are 60 km apart. A school is to built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school be built at

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town B
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45 km from town A
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town A
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45 km from town B

The feasible solution for a LPP is shown in Fig. Let

$Z = 3x - 4y$ be the
Objective function, Minimum of Z occurs at

0%
(0, 0)
0%
(0, 8)
0%
(0, 5)
0%
(4, 10)

The feasible solution for a LPP is shown in Fig. Let Z=3x−4y be the

The objective function, Maximum value of Z + Minimum value of Z is equal to

0%
13
0%
1
0%
-13
0%
-17

The feasible solution for a LPP is shown in Fig. Let Z=3x−4y be the

Objective function, Maximum of Z occurs at

0%
(5, 0)
0%
(6, 5)
0%
(6, 8)
0%
(4, 10)

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

0%
The quantity in column A is greater
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The quantity in column B is greater
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The two quantities are equal
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The relationship cannot be determined on the basis of the information Supplied

Explanation

Hence, maximum value of

$Z = 300 < 325$ .

So, the quantity in column B is greater.

1905579_1799565_ans_ae9b8afb7b9d407db145b9bfae696a5b.png

State true or false.

In a LPP, the minimum value of the objective function Z = ax + by is always 0 if origin is one of the corner point of the feasible region.

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True
0%
False

State true or false.

If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.

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True
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False

State true or false.

In a LPP, the maximum value of the objective function Z = ax + by is always finite.

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True
0%
False

State true or false.

Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.

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True
0%
False

The value of objective function is maximum under linear constraints

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at the centre of feasible region
0%
at (0,0)
0%
at the vertex of feasible region
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the vertex which is of maximum distance from (0,0)

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

0%
inside the feasible region
0%
at the boundary of the feasible region
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at vertex of feasible region
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outside the feasible region

Object function of LPP is

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a constraint
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a function to be maximized or minimized
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a relation between the decision variables
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equation of a straight line

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

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

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

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

$End$

0%
$21$
0%
$22$
0%
$23$
0%
$24$

Explanation

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.

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?

Explanation

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

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

$y= x+y$

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

hence , the correct option is

$C$

2117297_315014_ans_855d3da4f44543df9abc579352e1c528.png

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.

0%
$4900$
0%
$2900$
0%
$5400$
0%
$4000$

Explanation

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

$x\geq y$

$x-y\geq0$ --- (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

$Rs.4,900.$

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?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

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.

Conclude from the following:

$n^2 > 10$ , and n is a positive integer.

$n^3$

$50$

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The quantity A is may be greater or smaller than B.
0%
The quantity B is greater than A.
0%
The two quantities are equal.
0%
The relationship cannot be determined from the information given.

Explanation

given,

$n^2 > 10$ and

$n >0$

multiplying both equations we get

$n^3>0$

so, it may be greater than or less than 50

Hence, quantity A is may be greater or smaller than B

$\displaystyle x\ge 0$

$\displaystyle 3y-2x\ge -12$

$\displaystyle 2x+5y\le 20$
The area of the triangle formed in the xy plane by the system of inequalities above is:

0%
$60$
0%
$30$
0%
$40$
0%
$50$

Explanation

given,

$x\geq 0$

$3y-2x \geq -12$ and

$2x+5y\leq 20$

first, draw the graph for equations

$x= 0$

$3y-2x = -12$ and

$2x+5y = 20$

$x=0$ is Y-axis. Hence

$x\geq 0$ includes the above region of the line.

for

$3y-2x = -12$

substitute y=0 we get,

$-2x=-12 \implies x=6$

substitute x=0 we get,

$3y=-12 \implies y=-4$

therefore,

$3y-2x = -12$ line passes through (6,0) and (0,-4) as shown in fig.

Hence,

$3y-2x \geq -12$ includes the region above the line.

similarly for

$2x+5y = 20$

substitute y=0 we get,

$2x=20 \implies x=10$

substitute x=0 we get,

$5y=20 \implies y=4$

therefore,

$2x+5y = 20$ line passes through (10,0) and (0,4) as shown in fig.

Hence,

$2x+5y\leq 20$ includes the region below the line

the shaded region as shown in figure is intersection region

distance between (0,4) and (0,-4) is 8 units

adding

$3y-2x = -12$ and

$2x+5y = 20 \implies y=1$

substituting y=1 in any one of equation gives

$x=7.5$

therefore, shaded triangle base =8 units and

height = 7.5 unit

Area =

$\dfrac{1}{2}\times base\times height$

$= \dfrac{1}{2}\times 8\times 7.5 = 30$

Minimize :

$z=3x+y$ , subject to

$2x+3y\le 6, x+y \ge 1, x\ge 0, y\ge 0$

0%
$x=1, y=1$
0%
$x=0, y=1$
0%
$x=1, y=0$
0%
$x=-1, y=-1$

A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has

$30$ workers (male and female) and

$17$ units capital; which he uses to produce two types of goods

$A$ and

$B$ . To produce one unit of

$A$ ,

$2$ workers and

$3$ units of capital are required while

$3$ workers and

$1$ unit of capital is required to produce one unit of

$B$ . If

$A$ and

$B$ are priced at

$Rs.\ 100$ and

$Rs.\ 120$ per unit respectively, how should he use his resources to maximise the total revenue? Form the LPP and solve graphically.
Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?

0%
$1260$
0%
$1130$
0%
$1290$
0%
$3421$

Explanation

Let the number of goods of type A and B produced be respectively x and y.

To maximize, $Z=(100x+120y)$

Subject to the constraints:

$2x+3y\leq30$ ---- (1)

$3x+y\leq17$ ---- (2) where $x, y\leq 0.$

Take the testing points as $(0, 0)$ for (1) we have: $2(0) + 3(0) \leq 30\Rightarrow0\leq 30$ , which is true.

Take the testing points as $(0, 0)$ for (2) we have: $3(0) + (0) \leq 17\Rightarrow0\leq 17$ , which is true.

The shaded region $OACBO$ as shown in the given figure is the feasible region, which is bounded.

The coordinates of the corner points of the feasible region are $A(\frac{17}{3},0), E(3,8), C(0, 10) and O(0, 0).$

So, Value of Z at $A\left(\dfrac{17}{3}, 0\right)=\dfrac{1700}{3}$

Value of Z at $B(0, 10)=1200$

Value of Z at $C(3, 8)=1260$

Value of Z at $O(0, 0)=0$

Pts.	Coordinate	$Z^{max} = 100x + 120y$
$O$	$(0,0)$	$Z=0$
$A$	$\left(\dfrac{17}3,0\right)$	$Z = \dfrac{1700}3$
$E$	$(3,8)$	$Z = 300+960 = 1260$
$C$	$(0,10)$	$Z = 1200$

The maximum value of Z is $Rs.1260$ which occurs at $x = 3$ and $y = 8.$

Thus the factory must produce $3$ units and $8$ units of the goods of type A and B respectively. The maximum obtained profit earned by the factory by producing these items is $Rs. 1260.$

Yes, we agree with the view of manufacturer that men and women workers are equally efficient and so should be paid at the same rate.

CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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