Linear Inequalities Test -4

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Linear Inequalities Test -4

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Question 1
1 / -0

Directions For Questions

Solve the linear inequations graphically:
$$2x+3y > 2$$
$$x-y < 0$$
$$y \geq 0$$
$$x \leq 3$$

...view full instructions

How many solutions does this set include?

A
$$5$$

B
$$10$$

C
$$20$$

D
Infinite

Solution

Given, $$2x+3y>2$$
$$x-y<0 \implies y>x$$
$$y\geq 0$$
$$x\leq 3$$
first, draw the graph for equations $$2x+3y=2$$
$$x-y=0 \implies x=y$$
$$y= 0 \implies x-axis$$
$$x= 3$$

$$x=y$$ is the line which passes through the origin as shown in the above fig
Hence, $$y>x$$ includes the above region of the line.

$$y= 0 \implies x-axis$$. Hence, $$y \geq 0$$ includes the region above the x axis.

$$x= 3 \implies $$line parallel to x-axis. Hence, $$x\leq 3$$ includes the left region of the line.

for $$2x+3y=2$$
substituting y=0, we get $$2x=2\implies x=1$$
substituting x=0, we get $$3y=2\implies y=\dfrac{2}{3}$$
Therefore, line $$2x+3y=2$$ passes through (1,0) and (0,2/3) as shown in the above figure
Hence, $$2x+3y>2$$ includes the above region of the line.

Therefore, the blue shaded region is the feasible region which is not a closed region or it is not bounded. Hence, there are infinite number of solutions.
Question 2
1 / -0

Directions For Questions

Solve the linear inequations graphically:
$$2x+3y > 2$$
$$x-y < 0$$
$$y \geq 0$$
$$x \leq 3$$

...view full instructions

Which of the following points lie in the solution set?

A
$$(1,1)$$

B
$$(1,2)$$

C
$$(2,1)$$

D
$$(3,2)$$

Solution

By option verification,
i.e., substituting the options in given linear inequations and verifying

Given, $$2x+3y>2$$
$$x-y<0$$
$$y\geq 0$$
$$x\leq 3$$

substituting option A (i.e.,) $$(x,y)=(1,1)$$
$$2x+3y>2\implies 2\times 1+3\times 1>2 \implies 5>2$$ True
$$x-y<0 \implies 1-1<0 \implies 0<0$$ False
$$y\geq 0 \implies 1\geq 0$$ True
$$x\leq 3 \implies 1\leq 3$$ True

substituting option B (i.e.,) $$(x,y)=(1,2)$$
$$2x+3y>2\implies 2\times 1+3\times 2>2 \implies 8>2$$ True
$$x-y<0 \implies 1-2<0 \implies -1<0$$ True
$$y\geq 0 \implies 2\geq 0$$ True
$$x\leq 3 \implies 1\leq 3$$ True

substituting option C (i.e.,) $$(x,y)=(2,1)$$
$$2x+3y>2\implies 2\times 2+3\times 1>2 \implies 7>2$$ True
$$x-y<0 \implies 2-1<0 \implies 1<0$$ False
$$y\geq 0 \implies 1\geq 0$$ True
$$x\leq 3 \implies 2\leq 3$$ True

substituting option D (i.e.,) $$(x,y)=(3,2)$$
$$2x+3y>2\implies 2\times 3+3\times 2>2 \implies 12>2$$ True
$$x-y<0 \implies 3-2<0 \implies 1<0$$ False
$$y\geq 0 \implies 2\geq 0$$ True
$$x\leq 3 \implies 3\leq 3$$ True

Therefore option B satisfies the above linear inequalities.
Question 3
1 / -0

Consider the linear inequations and solve them graphically:
$$3x-y-2 > 0;\,\,\, x+y \leq 4;\,\,\, x >0;\,\,\, y \geq 0$$.
The solution region of these inequations is a convex polygon with _____ sides.

A
$$3$$

B
$$4$$

C
$$5$$

D
$$7$$

Solution

Given, $$3x-y-2>0\implies y<3x-2$$
$$x+y\leq 4$$
$$x>0$$ and $$y\geq 0$$
First, draw the graph for equations $$3x-y-2=0$$
$$x+y= 4$$
$$x=0$$ and $$y= 0$$
$$x=0$$ is the Y-axis.
Hence, $$x>0$$ includes the right side region of the line.
$$y=0$$ is the X-axis.
Hence, $$y\geq 0$$ includes the upper side region of the line.

Similarly, for $$y=3x-2$$
Substitute y=0 we get, $$3x=2 \implies x=\dfrac{2}{3}$$
Substitute x=0 we get, $$y=-2$$
Therefore, $$y=3x-2$$ line passes through (2/3,0) and (0,-2).
Hence, $$y<3x-2$$ includes the region below the line.

Similarly, for $$x+y= 4$$
Substitute y=0 we get, $$x=4$$
Substitute x=0 we get, $$y=4$$
Therefore, $$x+y= 4$$ line passes through (2/3,0) and (0,-2).
Hence, $$x+y\leq 4$$ includes the region below the line.

As shown in the above figure, the blue shaded region is the feasible region which is the intersection of all 4 solution sets. Feasible region is the polygon with $$3 \space sides$$ as shown in the figure.
Question 4
1 / -0

Find solution of following inequality also show it graphically.
$$x+3<5,x\in R$$

A

B

C

D

Solution

Given, $$x+3<5\implies x<2$$ and $$x\in R$$
The real numbers includes all the integers(e.g., -4), fractions(e.g., 3/4) and irrational numbers(e.g., $$\sqrt{2}$$).
Option C, represents only integers.
Option B, represents all the numbers $$<2$$
Question 5
1 / -0

Find solution of following inequality, also show it graphically.
$$x-5\geq-7,x\in R$$

A

B

C

D

Solution

Given, $$x-5\geq -7\implies x\geq -2$$ and $$x\in R$$
The real numbers includes all the integers(e.g., -4), fractions(e.g., 3/4) and irrational numbers(e.g., $$\sqrt{2}$$).
option c represents only integers.
option B represents all the numbers $$\geq -2$$
Question 6
1 / -0

Find solution of following inequality, also show it graphically:
$$x+3\leq5,x\in Z$$

A

B

C

D

Solution

Given, $$x+3\leq 5 \implies x\leq 5-3 \implies x\leq 2$$ and
$$x \in Z$$

ntegers are the numbers which include all the positive and negative numbers including zero. (i.e.,) $$Z=\{...,-2,-1,0,1,2,...\}$$

Therefore, $$\{...,-2,-1,1,2\}$$ represents $$x\leq 2$$.
Option C represents the set $$\{...,-2,-1,1,2\}$$ which are $$x\leq 2$$ and $$x\in Z$$.
Question 7
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Directions For Questions

A manufacturer produces two products $$A$$ and $$B$$. Product $$A$$ fetches him a profit of Rs. $$24$$ and product $$B$$ fetches him a profit of Rs. $$14$$. It takes $$15$$ minutes to manufacture one unit of product $$A$$ and $$5$$ minutes to manufacture one unit of product $$B$$. There is a limit of $$30$$ units to the quantity of $$B$$ of manufactured due to a constraint on the availability of materials. The minimum profit that the company decides to make is Rs. $$1000$$. The workers work for a maximum of $$10$$ hours a day.

...view full instructions

Formulate the equations for the above problem.
($$x$$ and $$y$$ are the number of units of $$A$$ and $$B$$ manufactured in a day respectively)

A
$$15x+5y \leq 10;\, 24x + 14y \geq 1000$$

B
$$15x+5y \leq 600;\, 24x + 14y \geq 1000$$

C
$$5x+15y \leq 600;\, 24x + 14y \geq 1000$$

D
$$5x+15y \leq 10;\, 24x + 14y \geq 1000$$

Solution

Given, The profit from the product $$A$$ is $$Rs. 24$$ and
The profit from the product $$B$$ is $$Rs. 14$$
The maximum number of units of $$B$$ manufactured in a day is $$30$$
i.e., $$y\leq 30$$ --------- (1)

Let the number of units of $$A$$ manufactured in a day be $$x$$
Therefore the profit for a day from the product $$A$$ is $$24x$$

Let the number of units of $$B$$ manufactured in a day be $$y$$
Therefore the profit for a day from the product $$B$$ is $$14y$$

The minimum profit for a day is $$Rs.1000$$
Therefore, the total profit from products $$A$$ and $$B$$ should be more than $$Rs.1000$$
i.e., $$24x+14y\geq 1000$$ --------- (2)

Given, time taken to manufacture one product of $$A$$ is $$15\space min$$ and
time taken to manufacture one product of $$B$$ is $$5\space min$$

Therefore, time taken to manufacture $$x$$ products of $$A$$ is $$15x\space min$$ and
time taken to manufacture $$y$$ products of $$B$$ is $$5y\space min$$

In a day, a worker works for a maximum of $$10\space hrs = 600\space min$$
Therefore, the time taken to manufacture products $$A$$ and $$B$$ should be less than $$600\space min$$
I.e., $$15x+5y\leq 600$$ --------- (3)
Question 8
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Find solution of following inequality also show it graphically.
$$x<5,x\in Z$$.

A

B

C

D

Solution

given $$x<5$$ and $$x\in Z$$
Integers are the numbers which include all the positive and negative numbers including zero. (i.e.,) $$Z=\{...,-2,-1,0,1,2,...\}$$

Therefore, $$\{...,-2,-1,1,2,3,4\}$$ represents $$x<5$$
Option A includes all the decimal values. Hence it is false.
Option B represents the set $$\{...,-2,-1,1,2,3\}$$ which are $$x<5$$ and $$x\in Z$$.
Question 9
1 / -0

Find solution of following inequality, also show it graphically:
$$x+3\leq5,x\in N$$

A

B

C

D

Solution

Given, $$x+3\leq 5\implies x\leq 2$$ and $$x \in N$$
Natural numbers are counting numbers whose set is $$N=\{1,2,3,...\}$$
Therefore,$$\{1,2\}$$ represents $$x<5$$
Option D graph has $$\{1,2\}$$ solution set.
Question 10
1 / -0

Consider the linear inequations and solve them graphically:
$$3x-y-2 > 0;\,\,\, x+y \leq 4;\,\,\, x >0;\,\,\, y \geq 0$$.
Which of the following are corner points of the convex polygon region of the solution?

A
$$(0,0)$$

B
$$(2,3)$$

C
$$(0,4)$$

D
$$\left(\dfrac32, \dfrac52\right)$$

Solution

Given, $$3x-y-2>0\implies y<3x-2$$
$$x+y\leq 4$$
$$x>0$$ and $$y\geq 0$$

first, draw the graph for equations $$3x-y-2=0$$
$$x+y= 4$$
$$x=0$$ and $$y= 0$$

$$x=0$$ is the Y-axis.
Hence, $$x>0$$ includes the right side region of the line.

$$y=0$$ is the X-axis.
Hence, $$y\geq 0$$ includes the upper side region of the line.

similarly, for $$y=3x-2$$
substitute y=0 we get, $$3x=2 \implies x=\dfrac{2}{3}$$
substitute x=0 we get, $$y=-2$$
therefore, $$y=3x-2$$ line passes through (2/3,0) and (0,-2).
Hence, $$y<3x-2$$ includes the region below the line.

similarly, for $$x+y= 4$$
substitute y=0 we get, $$x=4$$
substitute x=0 we get, $$y=4$$
therefore, $$x+y= 4$$ line passes through (2/3,0) and (0,-2).
Hence, $$x+y\leq 4$$ includes the region below the line.

solving $$3x-y-2=0$$ and $$x+y= 4$$ we get the intersecting point.
adding the equations
$$\implies 3x+x=2+4\implies 4x=6 \implies x=\dfrac{3}{2}$$

substituting $$x=\dfrac{3}{2} \implies y=4-\dfrac{3}{2} \implies y=\dfrac{5}{2}$$

As shown in the above figure, the blue shaded region is the feasible region with the three corner points $$ (\dfrac{3}{2},\dfrac{5}{2}), (\dfrac{2}{3},0), (4,0)$$