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Structure of linear programming model - class-X

Description: structure of linear programming model
Number of Questions: 47
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Tags: maths linear programming problems further solving of equations and inequalities business maths linear programming operations research
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" the relation S is defined on $N\times N\left ( a,b \right )S\left ( c,d \right )\Leftrightarrow bc\left ( a+d \right )=ad$ in an equivalance " that statement is ?

  1. True

  2. False


Correct Option: A

If a = b then ax = ...........

  1. $b + x$

  2. bx

  3. b - x

  4. $b\, \div\, x$


Correct Option: B
Explanation:

Given,

a = b
Multiplying both sides by x.
ax = bx

If  x + y = 3 and xy = 2, then the value of $\displaystyle x^{3}-y^{3}$ is equal to 

  1. 6

  2. 7

  3. 8

  4. 0


Correct Option: B
Explanation:

Formula used:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
               $=(\sqrt{(x+y)^2-4xy})[(x+y)^2-xy]$
               $=(\sqrt{(3)^2-4(2)})[(3)^2-2]$
               $=(\sqrt 1)(7)=7$
Option B is correct.

What is the solution of $x\le 4,y\ge 0$ and $x\le -4,y\le 0$ ?

  1. $x\ge -4,y\le 0$

  2. $x\le 4,y\ge 0$

  3. $x\le -4,y=0$

  4. $x\ge -4,y=0$


Correct Option: C
Explanation:

$x \le 4$ and $x \le -4$

$\Rightarrow x \le -4$
Also,
$y \ge 0$ and $y \le 0$
$\Rightarrow y = 0$
Hence the solutione is $x \le -4, \; y = 0$.

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that

  1. The values of decision variables obtained by rounding off are always very close to the optimal values.

  2. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.

  3. The value of the objective function for a minimization problem will likely be less than that for the simplex solution.

  4. All constraints are satisfied exactly.

  5. None of the above.


Correct Option: B
Explanation:

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

If the constraints in linear programming problem are changed

  1. the problem is to be re-evaluated

  2. solution is not defined

  3. the objective function has to be modified

  4. the change in constraints is ignored


Correct Option: A
Explanation:

The above question asks for the impact of change in constraints on the Linear programming problem. In this scenario, when there is a change in constraint, the solution will change definitely. Whether the solution exists or not, we can only find once the problem is re-evaluated. 

In an LPP, the objective function is related to the main objective of any problem, either we have to maximize or minimize the function based on the situation whereas the constraints is related to physical restrictions in achieving the defined objective function. In real life problems, there might be situations when the constraints change, but objective function does not changes to accommodate the change in constraints.
Thus, if constraints in linear programming problem is changed, the problem has to be re-evaluated for the same objective function and after solving we can find whether the solution exists or not.

A wholesale merchant wants to start the business of cereal with Rs. $24000$. Wheat is Rs. $400$ per quintal and rice is Rs. $600$ per quintal. He has capacity to store $200$ quintal cereal. He earns the profit Rs.$25$ per quintal on wheat and Rs. $40$ per quintal on rice. If he stores $x$ quintal rice and $y$ quintal wheat, then for maximum profit the objective function is

  1. $25x+40y$

  2. $40x+25y$

  3. $400x+600y$

  4. $\dfrac{400}{40}x + \dfrac{600}{25}y$


Correct Option: B

The feasible solution of an LP problem, is ________

  1. must satisfies all of the problem's constraints simultaneously

  2. must be a corner point of the feasible region

  3. need not satisfy all of the constraints, only some of them

  4. must optimize the value of the objective function


Correct Option: A
Explanation:

the feasibe solution of a inear programming probem(LP) is a solution that must satisfy all of the problem's constraints simultaniously

The solution of the set of constraints of a linear programming problem is a convex (open or closed) is called ______ region.

  1. feasible

  2. active

  3. linear

  4. none of these


Correct Option: A

Let $P(-1, 0), Q(0, 0)$ and $R(3, 3\sqrt{3})$ be three points. The equation of the bisector of the angle PQR is?

  1. $x+\sqrt{3}y=0$

  2. $\sqrt{3}x+y=0$

  3. $x+\dfrac{\sqrt{3}}{2}y=0$

  4. $\dfrac{\sqrt{3}}{2}x+y=0$


Correct Option: B

Maximum value of $z = 6 x + 11 y$ ,  subject to $2 x + y \leq 104 , x + 2 y \leq 76 , x \geq 0 , y \geq 0$ is

  1. $240$

  2. $540$

  3. $440$

  4. $410$


Correct Option: A

The taxi fare in a city is as follows. For the first km the fare is $Rs.10$ and subsequent distance is $Rs.6 / km.$ Taking the distance covered as $x \ km$ and fare as $Rs\ y$ ,write a linear equation.

  1. $y=4+6x$

  2. $y=4+5x$

  3. $y=3+6x$

  4. $y=3+5x$


Correct Option: A
Explanation:

First $km$ fare =$Rs.10$

Subsequent distance fare=$Rs\ 6/km$

Then fare $x\ km$ of distance 
$y=(x-1) \times 6+10$
$y=6x-6+10$

$y=6x+4$

$A$ is correct 



The problem associated with  $ LPP$  is

  1. single objective function

  2. Double objective function

  3. No any objective function

  4. None


Correct Option: A
Explanation:

The problem associated with LLP is single objective.

A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is

  1. $400 m$

  2. $200 m$

  3. $600 m$

  4. $250 m$


Correct Option: A
Explanation:

Let the length of the first train be $x$ metres


length of the second train$=\dfrac{x}{2}$m


relative speed of the two trains$=48+42$

                                                    $=90kmph$

                                                    $=90\times\dfrac{5}{18}$

                                                    $=25m/s$

$25=\dfrac{x+\dfrac{x}{2}}{12}$

$25=\dfrac{2x+x}{24}$

$3x=25\times24$

$x=25\times8$

$x=200$

So, the length of the train $=200m$

Let the length of the platform be y

Speed of the train $=48 kmph$

                                 $=48\times\dfrac{5}{18}$

                                  $=\dfrac{40}{3}m/s$

$\dfrac{40}{3}=\dfrac{200+y}{45}$

$40\times45=600+3y$

$1800=600+3y$

$3y=1200$

$y=400$

So, the length of the platform$=400m$


Linear programming used to optimize mathematical procedure and is

  1. subset of mathematical programming

  2. dimension of mathematical programming

  3. linear mathematical programming

  4. all of above


Correct Option: A
Explanation:

Linear programming is an extremely powerful tool for addressing a wide range of applied optimization problems. A short list of application areas is resource allocation, production scheduling, warehousing, layout, transportation scheduling, facility location, flight crew scheduling, portfolio optimization, parameter estimation.
So ,linear programming is used to subset mathematical programming.

In linear programming, oil companies used to implement resources available is classified as

  1. implementation modeling

  2. transportation models

  3. oil model

  4. resources modeling


Correct Option: B
Explanation:

In linear programming, $\text{transportation model}$ are applied to problems related to the study of efficient transportation routes.

For oil companies, how effectively the available resources are transported to different destinations with minimum cost.

Linear programming model which involves funds allocation of limited investment is classified as

  1. ordination budgeting model

  2. capital budgeting models

  3. funds investment models

  4. funds origin models


Correct Option: B
Explanation:
In linear programming, $\text{Capital budgeting models}$ to minimize the total capital cost. 
The solutions from the model are used to decide the best combination of capital resources and best times to start and finish projects and to determine the net capital cost.

Which of the following is a property of all linear programming problems?

  1. alternate courses of action to choose from

  2. minimization of some objective

  3. a computer program

  4. usage of graphs in the solution

  5. usage of linear and nonlinear equations and inequalities


Correct Option: A
Explanation:
According to Robbins, the resources(capital, land, labour, materials, ...) are always limited. Every resource have multiple uses.

The problem before any organisation or manager is to choose the best alternatives which can maximize the profit or minimize the cost of production. Linear programming is the method which is used to select the best possible alternatives from the all alternatives.

According to William M. Fox, "Linear programming is a planning technique that permits some objective function to be maximized or minimized within the framework of given situational restrictions"

Therefore, the linear programming is the process of selecting best courses of action to choose from various alternatives.

In transportation models designed in linear programming, points of demand is classified as

  1. ordination

  2. transportation

  3. destinations

  4. origins


Correct Option: C
Explanation:

In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes. i.e., how effectively the available resources are transported to different destinations with minimum cost.

Therefore, the points of demand is classified as destinations.

Consider the following linear programming problem:

Maximize $12X + 10Y$
Subject to: $4X + 3Y ≤ 480$
  $2X + 3Y ≤ 360$
  all variables $ ≥0$

Which of the following points $(X,Y)$ could be a feasible corner point?

  1. $(40,48)$

  2. $(120,0)$

  3. $(180,120)$

  4. $(30,36)$

  5. None of these


Correct Option: B
Explanation:

Given constraints, $4x+3y\leq 480$ and $2x+3y\leq 360$


first, draw the graph for equations $4x+3y= 480$ and $2x+3y= 360$

for $4x+3y= 480$
substitute y=0 we get, $4x=480 \implies x=120$
substitute x=0 we get, $3y=480 \implies y=160$
therefore, $4x+3y= 480$ line passes through (120,0) and (0,160) as shown in fig.
Hence, $4x+3y\leq 480$ includes the region below the line.


for $2x+3y= 360$
substitute y=0 we get, $2x=360 \implies x=180$
substitute x=0 we get, $3y=360 \implies y=120$
therefore, $4x+3y= 480$ line passes through $(180,0)$ and $(0,120)$ as shown in fig.
Hence, $2x+3y\leq 360$ includes the region below the line.

the shaded region as shown in figure is intersection region. From the figure, $(0,120)$ and $(120,0)$ are feasible corner points.

Consider the following linear programming problem:

Maximize $12X + 10Y$
Subject to: $4X + 3Y ≤ 480$
  $2X + 3Y ≤ 360$
all variables $ ≥0$

Which of the following points $(X,Y)$ is feasible?

  1. $(10,120)$

  2. $(120,10)$

  3. $(30,100)$

  4. $(60,90)$

  5. None of the above


Correct Option: C
Explanation:

Consider point $(10,120)$ , It doesnt satisfy the inequality $2X+3Y \le 360$

Consider point $(120,10)$ , it doesnt satisfy the inequality $4X+3Y \le 480$
Consider point $(30,100)$ , it satisfies all the inequalities . So this point is feasible
Consider point $(60,90)$ , it doesnt satisfies the inequalities $2X+3Y \le 360$ and $4X+3Y \le 480$
Therefore the correct option is $C$

Unboundedness is usually a sign that the LP problem.

  1. has finite multiple solutions.

  2. is degenerate.

  3. contains too many redundant constraints.

  4. has been formulated improperly.

  5. none of the above.


Correct Option: D
Explanation:

A linear programming problem is said to have unbounded solution if it has infinite number of solutions. I.e., the problem has been formulated improperly

The first step in formulating an LP problem is

  1. graph the problem.

  2. perform a sensitivity analysis.

  3. identify the objective and the constraints.

  4. define the decision variables.

  5. understand the managerial problem being faced.


Correct Option: E
Explanation:

The first step in formulating an linear programming problem is to understand the managerial problem being faced i.e., determine the quantities that are needed to solve the problem.

Consider the following linear programming problem:

Maximize $5X + 6Y$
Subject to: $4X + 2Y ≤ 420$
  $1X + 2Y ≤ 120$
  all variables  $≥0$

Which of the following points $(X,Y)$ is in the feasible region?

  1. $(30,60)$

  2. $(105,0)$

  3. $(0,210)$

  4. $(100,10)$

  5. None of the above


Correct Option: B,D
Explanation:
Feasible points are the points that satisfy the constraints.
Therefore, substitute the options in the constraint equations and verify. 

$\mathrm A.$ substituting  $(30,60)$ in $4x+2y\leq 420$
we get $4\times 30+2\times 60\leq 420$
$\implies 120+120 \leq 420 \implies 240\leq 420$ True
substituting  $(30,60)$ in $1x+2y\leq 120$
we get $1\times 30+2\times 60\leq 120$
$\implies 30+120 \leq 120 \implies 150\leq 120$ False

$\mathrm B.$ substituting  $(105,0)$ in $4x+2y\leq 420$
we get $4\times 105+2\times 0\leq 420$
$\implies 420+0 \leq 420 \implies 420\leq 420$ True
substituting  $(105,0)$ in $1x+2y\leq 120$
we get $1\times 105+2\times 0\leq 120$
$\implies 105+0 \leq 120 \implies 105\leq 120$ True

$\mathrm C.$ substituting  $(0,210)$ in $4x+2y\leq 420$
we get $4\times 0+2\times 210\leq 420$
$\implies 0+420 \leq 420 \implies 420\leq 420$ True
substituting  $(0,210)$ in $1x+2y\leq 120$
we get $1\times 0+2\times 210\leq 120$
$\implies 0+240 \leq 120 \implies 240\leq 120$ False


$\mathrm D.$ substituting  $(100,10)$ in $4x+2y\leq 420$
we get $4\times 100+2\times 10\leq 420$
$\implies 400+20 \leq 420 \implies 420\leq 420$ True
substituting  $(100,10)$ in $1x+2y\leq 120$
we get $1\times 100+2\times 10\leq 120$
$\implies 100+20 \leq 120 \implies 120\leq 120$ True

Therefore option B and D are the points in the feasible region.

In order for a linear programming problem to have a unique solution, the solution must exist

  1. at the intersection of the nonnegativity constraints.

  2. at the intersection of a nonnegativity constraint and a resource constraint.

  3. at the intersection of the objective function and a constraint.

  4. at the intersection of two or more constraints.

  5. none of the above


Correct Option: D
Explanation:

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints. Then the problem becomes convex and has a single optimum(maximum or minimum) solution. 

Consider the following linear programming problem:

Maximize $5X + 6Y$
Subject to: $4X + 2Y ≤ 420$
  $1X + 2Y ≤ 120$
  all variables $≥ 0$

Which of the following points $(X,Y)$ is feasible?

  1. $(50,40)$

  2. $(30,50)$

  3. $(60,30)$

  4. $(90,20)$

  5. None of these


Correct Option: C
Explanation:
Feasible points are the points that satisfy the constraints.
Therefore, substitute the options in the constraint equations and verify. 

$\mathrm A.$ substituting  $(50,40)$ in $4x+2y\leq 420$
we get $4\times 50+2\times 40\leq 420$
$\implies 200+80 \leq 420 \implies 280\leq 420$ True
substituting  $(50,40)$ in $1x+2y\leq 120$
we get $1\times 50+2\times 40\leq 120$
$\implies 50+80 \leq 120 \implies 130\leq 120$ False

$\mathrm B.$ substituting  $(30,50)$ in $4x+2y\leq 420$
we get $4\times 30+2\times 50\leq 420$
$\implies 120+100 \leq 420 \implies 220\leq 420$ True
substituting  $(30,50)$ in $1x+2y\leq 120$
we get $1\times 30+2\times 50\leq 120$
$\implies 30+100 \leq 120 \implies 130\leq 120$ False

$\mathrm C.$ substituting  $(60,30)$ in $4x+2y\leq 420$
we get $4\times 60+2\times 30\leq 420$
$\implies 240+60 \leq 420 \implies 300\leq 420$ True
substituting  $(60,30)$ in $1x+2y\leq 120$
we get $1\times 60+2\times 30\leq 120$
$\implies 60+60 \leq 120 \implies 120\leq 120$ True

$\mathrm A.$ substituting  $(90,20)$ in $4x+2y\leq 420$
we get $4\times 90+2\times 20\leq 420$
$\implies 360+40 \leq 420 \implies 400\leq 420$ True
substituting  $(90,20)$ in $1x+2y\leq 120$
we get $1\times 90+2\times 20\leq 120$
$\implies 90+40 \leq 120 \implies 130\leq 120$ False

Therefore option C (60,30) is the feasible point

Which of the following statements about an LP problem and its dual is false?

  1. If the primal and the dual both have optimal solutions, the objective function values for both problems are equal at the optimum

  2. If one of the variables in the primal has unrestricted sign, the corresponding constraint in the dual is satisfied with equality

  3. If the primal has an optimal solution, so has the dual

  4. The dual problem might have an optimal solution, even though the primal has no (bounded) optimum


Correct Option: D
Explanation:

if one of the problems(primal, dual) is infeasible then the other problem is infeasible. Hence, the option D is the false statement.

Mark the wrong statement:

  1. The primal and dual have equal number of variables.

  2. The shadow price indicates the change in the value of the objective function, per unit increase in the value of the RHS.

  3. The shadow price of a non-binding constraint is always equal to zero.

  4. The information about shadow price of a constraint is important since it may be possible to purchase or, otherwise, acquire additional units of the concerned resource.


Correct Option: A
Explanation:

The number of variables in dual is equal to the number of constraints in the primal and the  number of variables in primal is equal to the number of constraints in the dual.


Therefore, the primal and dual doesn't have equal number of variables.

In linear programming context, sensitivity analysis is a technique to

  1. Allocate resources optimally.

  2. Minimize cost of operations.

  3. Spell out relation between primal and dual.

  4. Determine how optimal solution to LPP changes in response to problem inputs.


Correct Option: D
Explanation:

A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.
Option D is correct.

Choose the wrong statement:

  1. In order that dual to an LPP may be written, it is necessary that it has at least as many constraints as the number of variables.

  2. The dual represents an alternate formulation of LPP with decision variables being implicit values.

  3. The optimal values of the dual variables can be obtained by inspecting the optimal tableau of the primal problem as well.

  4. Sensitivity analysis is carried out having reference to the optimal tableau alone.


Correct Option: A
Explanation:

In order to write LPP, it is not necessary that it has at least as many constraints as the number of variables.

The number of constraints allowed in a linear program is which of the following?

  1. Less than 5

  2. Less than 72

  3. Less than 512

  4. Less than 1,024

  5. Unlimited


Correct Option: E
Explanation:

there is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.

Which of the following is an essential condition in a situation for linear programming to be useful?

  1. Linear constraints

  2. Bottlenecks in the objective function

  3. Non-homogeneity

  4. Uncertainty

  5. None of the above


Correct Option: A
Explanation:

For linear programming, the constraints must be linear.

Choose the most correct of the following statements relating to primal-dual linear programming problems:

  1. Shadow prices of resources in the primal are optimal values of the dual variables.

  2. The optimal values of the objective functions of primal and dual are the same.

  3. If the primal problem has unbounded solution, the dual problem would have infeasibility.

  4. All of the above.


Correct Option: D
Explanation:

From the primal-dual relationship,

The shadow prices of resources in the primal are optimal values of the dual variables.

If one of the problems has an optimal feasible solution then the other problem also has an optimal feasible solution. The optimal objective function value is same for both primal and dual problems.

If one problem has an unbounded solution then the other problem is infeasible.

Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is $100 and the profit in the manufacture of a unit of product E is $87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?

  1. $5 D + 7 E≤ 5,000$

  2. $9 D + 3 E ≥4,000$

  3. $5 D + 7 E = 4,000$

  4. $5 D + 9 E ≤5,000$

  5. $9 D + 3 E ≤5,000$


Correct Option: E
Explanation:

Given, product D takes 5 hours per unit of labour, and
product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes $5D$ hours and
to produce E units of product E takes $7E$ hours 
Given, total labour hours per week are $4000$ hours.
Hence, $5D+7E\leq 4000$

Given, product D takes 9 hours per unit of machine time, and
product E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes $9D$ hours and
to produce E units of product E takes $3E$ hours 
Given, total machine hours per week are $5000$ hours.
Hence, $9D+3E\leq 5000$

To write the dual; it should be ensured that  
I. All the primal variables are non-negative.
II. All the bi values are non-negative.
III. All the constraints are $≤$ type if it is maximization problem and $≥$ type if it is a minimization problem.

  1. I and II

  2. II and III

  3. I and III

  4. I, II and III


Correct Option: C
Explanation:

To write the dual, then all the primal variables must be non-negative.

All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.

If $x=\log _{2^2}2+\log _{2^3}2^2+\log _{2^4}2^3......+\log _{2^{n+1}}2^n+$, then the minimum value of $x$ will be-

  1. $\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  2. $n\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  3. $\left(\dfrac n{n+1}\right)^{\tfrac 1n}$

  4. None of the above.


Correct Option: B
Explanation:
given $x={log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}$

we know that $A.M. \geq G.M.$

$\implies \dfrac{log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}{n}\geq \sqrt[n]{log _{2^2}2*log _{2^3}{2^2}*log _{2^4}{2^3}*...*log _{2^{n+1}}{2^n}}$

$\implies \dfrac{x}{n}\geq \sqrt[n]{\dfrac{log2}{log2^2}*\dfrac{log2^2}{log2^3}*\dfrac{log2^3}{log2^4}*...*\dfrac{log2^n}{log2^{n+1}}}$

$\implies x\geq n\sqrt[n]{\dfrac{log2}{log2^{n+1}}}$

$\implies x\geq n({\dfrac{log2}{(n+1)log2}})^{\dfrac 1n}$

$\implies x\geq n({\dfrac{1}{n+1}})^{\dfrac 1n}$

therefore the minimum value of $x$ is $ n({\dfrac{1}{n+1}})^{\dfrac 1n}$

If $a,b >0$, $a+b=1$, then the least value of $(1+\dfrac 1a)(1+\dfrac 1b)$, is

  1. $3$

  2. $6$

  3. $9$

  4. $12$


Correct Option: C
Explanation:
Given, $a+b=1$
we know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

$\implies \dfrac{1}{2}\geq \sqrt{ab}$

$\implies \sqrt{ab}\leq \dfrac{1}{2}$

squaring on both sides

$\implies ab \leq \dfrac{1}{4}$  --------------(1)

Similarly

$\implies \dfrac{1+a+1+b}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+(a+b)}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+1}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{3}{2}\geq \sqrt{(1+a)(1+b)}$

squaring on both sides

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{4}{9}$  ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{4}*\dfrac{4}{9}$

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{9}$

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{1}{9ab}$

$\implies \dfrac{(1+a)(1+b)}{ab}\geq 9$

$\implies \dfrac{(1+a)}{a}*\dfrac{(1+b)}{b}\geq 9$

$\implies (1+\dfrac{1}{a})(1+\dfrac{1}{b})\geq 9$

Therefore, the minimum value of $ (1+\dfrac{1}{a})(1+\dfrac{1}{b})$ is $ 9$

If $l,m,n$ be three positive roots of the equation $x^3-ax^2+bx+48=0$, then the minimum value of $\dfrac 1l +\dfrac 2m+\dfrac 3n$ is

  1. $1$

  2. $2$

  3. $\dfrac {-3}{2}$

  4. $\dfrac 52$


Correct Option: C
Explanation:

we Know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let $a=\dfrac{1}{l}, b=\dfrac{2}{m}, c=\dfrac{3}{n}$

Therefore,

$\dfrac{1}{3}(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq \sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$


$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$

Given, the roots of the polynomial $x^3-ax^2+bx+48=0$ are $l,m,n$
Therefore, the product of the roots $lmn=-(\dfrac{48}{1})=-48$

Substituting $lmn=-48$ in the above equation

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{6}{-48})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1}{-8})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(-\dfrac{1}{2})^3}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times(-\dfrac{1}{2})$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq (-\dfrac{3}{2})$

therefore, the minimum value is $-\dfrac{3}{2}$

Let $a _1,a _2....,a _n$ be a non negative real number such that $a _1+a _2....+a _n=m$ and let $S=\underset{i<j}\sum a _ia _j$, then

  1. $S\leq \dfrac {m^2}2$

  2. $S> \dfrac {m^2}4$

  3. $S< \dfrac {m}2$

  4. $S> \dfrac {m^2}2$


Correct Option: A

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.

Rawmaterial 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.

  1. $3{x} _{1}-4{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  2. $4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  3. $4{x} _{1}-3{x} _{2}=0$ and $4{x} _{2}-5{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  4. $4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\le0$


Correct Option: B
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 $P, 6{x} _{1}+5{x} _{2}+2{x} _{3}\le5,000$
For raw material $Q, 4{x} _{1}+7{x} _{2}+3{x} _{3}\le6,000$
Product $B$ requires $\frac{1}{2}$ and product $C$ requires ${\left(\frac{1}{3}\right)}^{rd}$ the time required for product $A.$
Then $\frac{t}{2}$ and $\frac{t}{3}$ are the times in hours to produce $B$ and $C$ and since $1,600$ units of $A$ will need time $1,600t$ hours, we get the constraint,
$t{x} _{1}+\frac{t}{2}{x} _{2}+\frac{t}{3}{x} _{3}\le 1,600t$ or 
${x} _{1}+\frac{{x} _{2}}{2}+\frac{{x} _{3}}{3}\le1,600$ or
$6{x} _{1}+3{x} _{2}+2{x} _{3}\le9,600$
Market demand requires
${x} _{1}\ge300, {x} _{2}\ge250,$ and ${x} _{3}\ge200$ 
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 $\frac{{x} _{1}}{3}=\frac{{x} _{2}}{4},$
and $\frac{{x} _{2}}{4}=\frac{{x} _{3}}{5}.$
Thus, there are two additional constraints
$4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$ 

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  $End$

  1. $21$

  2. $22$

  3. $23$

  4. $24$


Correct Option: A
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.

Conclude from the following:
$n^2 > 10$, and n is a positive integer.
A: $n^3$
B: $50$
  1. The quantity A is may be greater or smaller than B.

  2. The quantity B is greater than A.

  3. The two quantities are equal.

  4. The relationship cannot be determined from the information given.


Correct Option: A
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

For any positive real number $a$ and for any $n \in N$, the greatest value of 
$\dfrac {a^n}{1+a+a^2....a^{2n}}$ is

  1. $\dfrac 1{2n}$

  2. $\dfrac 1{2n+1}$

  3. $\dfrac 1{2n-1}$

  4. None of the above.


Correct Option: B
Explanation:

We know that $A.M.\geq G.M.$


Therefore, $\dfrac{1+a+a^2+...+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{1*a*a^2*...*a^{2n}}$

$\implies \dfrac{1+a+a^2+....+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{a^{(1+2+...+2n)}}$

We know that sum of first $n$ numbers is $1+2+...+n=\dfrac{n(n+1)}{2}$

Therefore $1+2+...+2n=\dfrac{2n(2n+1)}{2}=n(2n+1)$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq (a^{n(2n+1)})^{\dfrac{1}{2n+1}}$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq a^n$

$\implies \dfrac{a^n}{1+a+...+a^{2n}}\leq \dfrac{1}{2n+1}$

Therefore the greatest value of $\dfrac{a^n}{1+a+...+a^{2n}}$ is $\dfrac{1}{2n+1}$

The rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a,0)$ and $B(0,b)$, then the minimum value of $\left(a+\dfrac 1a\right) ^2+\left(b+\dfrac 1b\right) ^2$ is

  1. $0$

  2. $8$

  3. $k^2-4+\dfrac 4{k^2}$

  4. $k^2+4+\dfrac 4{k^2}$


Correct Option: B
Explanation:

since, $A.M. \geq G.M. \implies \dfrac{a+b}{2}\geq \sqrt{ab}$


let $a=1 $ and $b=a^2$

Therefore, $\dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq a$

$\implies \dfrac{1+a^2}{a}\geq 2$

$\implies \dfrac{1}{a}+\dfrac{a^2}{a}\geq 2$

$\implies a+\dfrac{1}{a}\geq 2$
 squaring on both sides
$\implies (a+\dfrac{1}{a})^2\geq 4$ ------- (1)

similarly, $ (b+\dfrac{1}{b})^2\geq 4$ --------(2)

adding (1) and (2) we get

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 4+4$

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 8$

Therefore the minimum value is $8$

If $a>0$, then least value of $(a^3+a^2+a+1) ^2$ is

  1. $64a^2$

  2. $16a^4$

  3. $16a^3$

  4. None of the above.


Correct Option: C
Explanation:

we know that $A.M.\geq G.M.$


therefore, $\dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^3*a^2*a*1}$

$\implies \dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^6}$

squaring on both sides 

$\implies (\dfrac{a^3+a^2+a+1}{4})^2\geq ({a^{\dfrac{6}{4}}})^2$

$\implies ({a^3+a^2+a+1})^2\geq 16a^3$

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