Linear Inequalities Test -3

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Linear Inequalities Test -3

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Question 1
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Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.

A
( 11 , 13 ) , ( 13 , 15 ) , ( 17 , 19 ) , ( 19 , 21 )

B
( 11 , 13 ) , ( 13 , 15 ) , ( 15 , 17 ) , ( 17 , 21 )

C
( 9 , 11 ) , ( 13 , 15 ) , ( 15 , 17 ) , ( 17 , 19 )

D
( 11 , 13 ) , ( 13 , 15 ) , ( 15 , 17 ) , ( 17 , 19 )

Solution

Let the consecutive odd natural numbers be x and x+2.

According to the question , x>10 and x + (x + 2) < 40

Now x +(x + 2) < 40 ⇒ 2x < 38 ⇒ x < 19

Hence the maximum value of x is 17 and minimum value is 11.

So the possible pairs of odd natural numbers are ( 11 , 13 ) , ( 13 , 15 ) , ( 15 , 17 ) , ( 17 , 19 )
Question 2
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Which of the following is correct ?

A
If -4 < 7 , then 4 < - 7

B
If -2 < 5 , then 2 > 5

C
If 0 > -7 , then 0 < 7

D
If 8 > 1 , then -8 > -1

Solution

Given 0 > -7

Multiplying throughout by -1,we get0 < 7 [ When both sides of an inequality are multiplied by a negative number ,then the sign of inequality is reversed]
Question 3
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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 points belong to the feasible solution region?

A
$$\left(\dfrac12,0\right)$$

B
$$\left(\dfrac12,\dfrac52\right)$$

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

D
None of the above

Solution

By option verification,
i.e., substituting the options in the given inequations and verifying
Given, $$3x-y-2>0$$
$$x+y\leq 4$$
$$x>0$$ and $$y\geq 0$$

Substituting option A i.e., $$(x,y)=(\dfrac{1}{2},0)$$
$$3*\dfrac{1}{2}-0-2>0\implies -\dfrac{1}{2}>0$$ False

Substituting option B i.e., $$(x,y)=(\dfrac{1}{2},\dfrac{5}{2})$$
$$3*\dfrac{1}{2}-\dfrac{5}{2}-2>0\implies -3>0$$ False

Substituting option C i.e., $$(x,y)=(\dfrac{3}{2},\dfrac{5}{2})$$
$$3*\dfrac{3}{2}-\dfrac{5}{2}-2>0\implies 0>0$$ False

No option satisfies the given linear inequations, hence option D is correct.
Question 4
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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

The quantity of $$A$$ and $$B$$ in one day for which profit will be maximum is:

A
$$25,30$$

B
$$30,25$$

C
$$25,25$$

D
$$30,30$$

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)

Now, substituting the options and verifying which will be the maximum and satisfies the 3 constraints.

substituting option A i.e., $$(x,y)=(25,30)$$
$$15x+5y\leq 600\implies 15*25+5*30 = 525\leq 600$$ True
$$24x+14y = 24*25+14*30 = 1020$$

substituting option B i.e., $$(x,y)=(30,25)$$
$$24x+14y = 24*30+14*25 = 1070$$
$$15x+5y\leq 600\implies 15*30+5*25 = 575\leq 600$$ True

substituting option C i.e., $$(x,y)=(25,25)$$
$$24x+14y = 24*25+14*25 = 950$$
$$15x+5y\leq 600\implies 15*25+5*25 = 500\leq 600$$ True

substituting option D i.e., $$(x,y)=(30,30)$$
$$24x+14y = 24*30+14*30 = 1140$$
$$15x+5y\leq 600\implies 15*30+5*30 = 600\leq 600$$ True

Therefore, $$(30,30)$$ will produce the maximum profit.
Question 5
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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. Find the maximum daily profit.

A
Rs. $$1020$$

B
Rs. $$1040$$

C
Rs. $$1140$$

D
Rs. $$1240$$

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)

The total profit from products $$A$$ and $$B$$ is $$P=24x+14y$$
In the above figure, the blue shaded region is the feasible region with three corner points.$$(\dfrac{290}{12},30), (30,30), (\dfrac{340}{9},\dfrac{20}{3})$$

$$(\dfrac{290}{12},30)$$ is the point where  $$24x+14y= 1000$$ intersects $$y=30$$
I.e., substituting $$y=30 \implies 24x+14*30 =1000 \implies x=\dfrac{1000-420}{24}\implies x=\dfrac{290}{12}$$

$$(30,30)$$ is the point where $$15x+5y= 600$$ intersects $$y=30$$
I.e., substituting $$y=30 \implies 15x+5*30 =600 \implies x=\dfrac{600-150}{15}\implies x=30$$

$$(\dfrac{340}{9},\dfrac{20}{3})$$ is the point where  $$24x+14y= 1000$$ intersects $$15x+5y= 600$$
I.e., solving the two equations, we get $$x=\dfrac{340}{9}$$and $$y=\dfrac{20}{3}$$

Now substituting the corner points the profit equation,
substituting $$(\dfrac{290}{12},30) \implies P=24*\dfrac{290}{12}+14*30=1000$$

substituting $$(30,30) \implies P=24*30+14*30=1140$$

substituting $$(\dfrac{340}{9},\dfrac{20}{3}) \implies P=24*\dfrac{340}{9}+14*\dfrac{20}{3}=1000$$

$$1140 $$ is the maximum profit
Question 6
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Solve the following inequality and show it graphically:
$$\dfrac{x+4}{x-3}>0,x\in W$$

A

B

C

D

Solution

Given, $$\dfrac{x+4}{x-3}>0$$ and $$x\in W$$
whole numbers are positive numbers including zero whose set is $$W=\{0,1,2,3,...\}$$
$$\implies x\geq 0$$
$$(x+4)*\dfrac{1}{x-3}>0$$
$$\implies x+4>0$$ and $$\dfrac{1}{x-3}>0$$
$$\implies x>-4$$ and $${x-3}>0$$
$$\implies x>-4$$ and $$x>3$$ and $$x\geq 0$$
$$x>-4$$ and $$x\geq 0 \implies x\geq 0$$
Therefore, $$\{4,5,6,...\}$$ represents the $$x>3$$ and $$x\geq 0$$
Question 7
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Solve the following inequality and show it graphically:
$$|x+3|<4 ,x\in R$$

A

B

C

D

Solution

Given that $$|x+3|<4$$
Which implies $$x+3 < 4$$ and $$x+3>-4$$
$$\Rightarrow x<1$$ and $$x>-7$$
$$\Rightarrow -7<x<1$$
Since, $$x\in R$$
$$x$$ can take values $$(-7,1)$$, not including boundary points.
Hence, D is correct.
Question 8
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If the product of $$n$$ positive numbers is $$1$$, then their sum is

A
A positive integer

B
Divisible by $$n$$

C
Equal to $$n+\dfrac 1n$$

D
Greater than or equal to $$n$$

Solution

Since, $$A.M. \geq G.M.$$
let n positive numbers be $$x_1, x_2,...,x_n$$

$$A.M. =\dfrac{x_1+x_2+...+x_n}{n}$$

$$G.M. = x_1*x_2*...*x_n$$

Therefore, $$\dfrac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1*x_2*...*x_n}$$

$$\implies x_1+x_2+...+x_n \geq n*\sqrt[n]{x_1*x_2*...*x_n}$$

$$\implies x_1+x_2+...+x_n \geq n$$
Question 9
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Solve the following inequality and show it graphically:
$$|x+3|<4 ,x\in Z$$

A

B

C

D

Solution

Given that $$|x+3|<4$$
Which implies $$x+3 < 4$$ and $$x+3>-4$$
$$\Rightarrow x<1$$ and $$x>-7$$
$$\Rightarrow -7<x<1$$
Since $$x$$ is natural number
$$x$$ can take values from $$-6$$ to $$0$$.
Hence, option $$C$$ is correct.
Question 10
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Find solution of following inequality, also show it graphically:
$$x-5\geq-7,x\in Z$$

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 A represents all the numbers $$\geq -2$$
Question 11
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Solve the following inequality and show it graphically:
$$-2<x+3<5,x\in N$$

A

B

C

D

Solution

Given, $$-2<x+3<5$$ and $$x \in N$$
Natural numbers are counting numbers whose set is $$N=\{1,2,3,...\}$$
$$\implies x>0$$
$$-2<x+3<5 \implies x+3>-2$$ and $$x+3<5$$
$$\implies x>-2-3$$ and $$x<5-3$$
$$\implies x>-5$$ and $$x<2$$
$$x>0 $$ and $$x>-5\implies x>0$$
Therefore,$$\{1\}$$ represents $$x>0 $$ and $$x<2$$
Question 12
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The shaded region in the figure is the solution set of the inequations.

A
$$5x + 4y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 2$$

B
$$5x + 4y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 2$$

C
$$5x + 4y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$$

D
$$5x + 4y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 2$$

Solution

1. Since, region(shaded) lies above x-axis
$$\therefore y\ge 0$$
2. Shaded region is on left of $$x=6$$
$$\therefore x\le 6$$
3. Shaded region is below $$y=3$$
$$\therefore y\le 3$$
4. Equation of line passing through (4,0) & (0,5) is
$$y-0=\cfrac { 5 }{ -4 } \left( x-4 \right) $$
$$5x+4y-20=0$$
$$5x+4y=20$$
If $$5x+4y\ge 20$$, Put (0,0) we get $$0\ge 20$$ FALSE.
$$\therefore $$ Region lies away from origin which is true
$$\therefore \quad 5x+4y\ge 20$$
5. Shaded region lies on right side of y-axis
$$\therefore x\ge 0$$
$$\therefore 5x+4y\ge 20,x\le 6,y\le 3,x\ge 0,y\ge 0$$