Linear Programming Test

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Linear Programming Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Minimize: $$z=\sum _{ j=1 }^{ n }{ \sum _{ i=1 }^{ m }{ { c }_{ ij }.{ x }_{ ij } } } $$
subject to : $$\sum _{ j=1 }^{ n }{ { x }_{ ij }={ a }_{ i } } ,i=1,....m;\quad \sum _{ i=1 }^{ m }{ { x }_{ ij }={ b }_{ j } } ,j=1,....n$$ is a LPP with number of constraints

A
$$m+n$$

B
$$m0n$$

C
$$mn$$

D
$$\cfrac{m}{n}$$

Solution
Question 2
1 / -0

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

A
$$x\ge -4,y\le 0$$

B
$$x\le 4,y\ge 0$$

C
$$x\le -4,y=0$$

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

Solution

$$x \le 4$$ and $$x \le -4$$
$$\Rightarrow x \le -4$$
Also,
$$y \ge 0$$ and $$y \le 0$$
$$\Rightarrow y = 0$$
Hence the solutione is $$x \le -4, \; y = 0$$.
Question 3
1 / -0

The ratio of the rate of flow of water in pipes varies inversely as the square of the radius of the pipes. What is the ratio of the rates of flow in two pipes diameters 2 cm and 4 cm?

A
1 : 6

B
1 : 4

C
1 : 2

D
3 : 1

Solution

Given:$${ d }_{ 1 }=2cm\\ { d }_{ 2 }=4cm$$
Solution:Since the diameter are 2cm and 4cm.
The replacement ratio of the two pipes are 1cm and 2cm
$${ r }_{ 1 }=1cm\\ r_{ 2 }=2cm$$
Square of the ratio of the pipes are 1 and 4
$$\therefore $$The ratio of rates of flow in two $$pipes=1:\dfrac { 1 }{ 4 }$$
$$ \Rightarrow$$ $$\dfrac { 1 }{ 4 } $$
Question 4
1 / -0

The diagram shows a scale drawing of a garden which is p metre long and w metre wide.
(a) Write an inequality between p and w.
(b) If a path of 2 m width is added to two sides as shown in the diagram, write a new inequality between the length and the width.

A
(a) $$p < w$$
(b) $$p+2 > w+2$$

B
(a) $$p > w$$
(b) $$p+2 >w+2$$

C
Data insufficient

D
None of these

Solution

Clearly from the figure, length is more than width
$$ => p > w $$

Also
$$ p +2 > w + 2 $$
Question 5
1 / -0

If $$A=$${1,2,3};$$ B=$$ {3,4,5};$$ C=$${4,6}, then $$A\times (B\cap C)=?$$

A
{(2,4)(1,4)}

B
{(2,4)(3,4)(5,6)}

C
{(1,4)(2,4)(3,4)}

D
None of these

Solution

Given,
$$A=\{ 1,2,3\} \\ B=\{ 3,4,5\} \\ C=\{ 4,6\} $$
Now, $$B\cap C=\{ 4\} $$
$$\therefore A\times (B\cap C)=\{ (1,4),(2,4),(3,4)\} $$
Question 6
1 / -0

The value of $$\displaystyle \frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24 }{0.76\times0.76-0.76\times0.24+0.24\times0.24}$$ is

A
0.52

B
1

C
0.01

D
0.1

Solution

Formula used:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
Now,
$$\displaystyle \frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24 }{0.76\times0.76-0.76\times0.24+0.24\times0.24}$$= $$\displaystyle \frac{(0.76)^3+(0.24)^3}{0.76\times0.76-0.76\times0.24+0.24\times0.24}$$

$$=\displaystyle \frac{(0.76+0.24)(0.76\times0.76-0.76\times0.24+0.24\times0.24)}{0.76\times0.76-0.76\times0.24+0.24\times0.24}$$

$$=(0.76+0.24)=1$$

Option B is correct.
Question 7
1 / -0

The incomplete graph shows Dinesh's savings from January to April.
Dinesh saved a total of RS. 730 during the four months. The amount of money saved in February was as much as that saved in March. How much did he save in March?

A
Rs. 245

B
Rs. 275

C
Rs. 250

D
Rs. 295

Solution
Question 8
1 / -0

?

A
$$212$$

B
$$532$$

C
$$244$$

D
$$733$$

Solution

At the start $$ A = 1, C = 2 $$

For the first iteration $$ C = 2 \times 2 + 1 = 5 $$ And $$ A = 1 \times 2 = 2 $$
Since $$ A \, i.e. 2 < 8 $$ the loop starts again

For the second iteration $$ C = 5 \times 5 +2 = 27 $$ And $$ A = 2 \times 2 = 4 $$
Since $$ A \, i.e. 4 < 8 $$ the loop starts again

For the third iteration $$ C = 27 \times 27 + 4 = 733 $$ And $$ A = 4 \times 2 =8 $$
Since $$ A \, i.e. \, 8 \nless \, 8 $$ the loop ends

Printing C gives $$ C = 733 $$
Question 9
1 / -0

Study the graph carefully and answer the question given below it.
The import in $$1976$$ was approximately how many times that of the year $$1971$$?

A
$$0.31$$

B
$$1.68$$

C
$$2.41$$

D
$$3.22$$

Solution

Import in $$1976=5832$$ tonnes

Import in $$1971=1811$$ tonnes

So, import in $$1976/$$import in $$1971=\dfrac{5832}{1811}=3.22$$

It is approximately more than three i.e $$3.22$$
So,

$$Answer (D) \Rightarrow 3.22$$
Question 10
1 / -0

Evaluate $$\displaystyle \frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$$

A
0

B
1

C
2

D
3

Solution

Given, $$\displaystyle \frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$$
$$=\displaystyle \frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(a-b)(b-c)(c-a)}$$
$$=\displaystyle \frac{(a^3-b^3+3b^2a-3a^2b)+(b^3-c^3-3b^2c+3c^2b)+(c^3-a^3-3c^2a+3a^2c)}{(a-b)(b-c)(c-a)}$$
$$=\displaystyle \frac{3b^2a-3a^2b+3c^2b-3b^2c+3a^2c-3c^2a}{(a-b)(b-c)(c-a)}$$
$$=\displaystyle \frac{3(b^2a-a^2b+c^2b-b^2c+a^2c-c^2a)}{(a-b)(b-c)(c-a)}$$
$$=\displaystyle \frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}$$
$$= 3 $$
Option D is correct.