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RD Chapter 15- Linear Inequations Ex-15.6 Interview Questions Answers

Question 1 :
Solve the following systems of linear inequations graphically.
(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
(iii) x – y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0
(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0
(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

Answer 1 :

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
We shall plot the graph of the equation and shade the side containing solutions of the inequality,
You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,
2x + 3y ≤ 6
So when,

x

0

1

3

y

2

1.33

0

3x + 2y ≤ 6
So when,

x

0

1

2

y

3

1.5

0

x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

x

0

1

3

y

2

1.33

0

x + 4y ≤ 4

So when,

x

0

2

4

y

1

0.5

0

x ≥ 0, y ≥ 0

(iii) x – y ≤ 1, x + 2y ≤ 8,2x + y ≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x – y ≤ 1

So when,

x

0

2

1

y

-1

1

0

x + 2y≤ 8

So when,

x

0

4

8

y

4

2

0

2x + y ≥ 2

So when,

x

0

2

1

y

2

-2

0

x ≥ 0, y ≥ 0

(iv) x + y ≥ 1, 7x + 9y ≤63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x + y ≥ 1

So when,

x

0

2

1

y

1

-1

0

7x + 9y ≤ 63

So when,

x

0

5

9

y

7

3.11

0

x ≤ 6, y ≤ 5 and x ≥0, y ≥ 0

(v) 2x + 3y ≤ 35, y ≥ 3, x≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

2x + 3y ≤ 35

So when,

x

0

5

17.5

y

11.667

8.33

0

y ≥ 3, x ≥ 2, x ≥ 0, y≥ 0

Question 2 :
Show that the solution set of the following linear inequations is empty set:
(i) x – 2y ≥ 0, 2x – y ≤ –2, x ≥ 0, y ≥ 0
(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

Answer 2 :

(i) x – 2y ≥ 0, 2x – y ≤–2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of the inequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x – 2y ≥ 0

So when,

x

0

2

4

y

0

1

2

2x – y ≤ –2

So when,

x

0

1

-1

y

2

4

0

x ≥ 0, y ≥ 0

The lines do notintersect each other for x ≥ 0, y ≥ 0

Hence, there is nosolution for the given inequations.

(ii) x + 2y ≤ 3, 3x +4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x + 2y ≤ 3

So when,

x

0

1

3

y

1.5

1

0

3x + 4y ≥ 12

So when,

x

0

2

4

y

3

1.5

0

y ≥ 1, x ≥ 0, y ≥ 0

Question 3 :

Find the linear inequations for which the shaded area in Fig. 15.41 isthe solution set. Draw the diagram of the solution set of the linearinequations.

Answer 3 :

Here, we shall applythe concept of a common solution area to find the signs of inequality by usingtheir given equations and the given common solution area (shaded part).

We know that,

If a line is in theform ax + by = c and c is positive constant. (In case of negative c, therule becomes opposite), so there are two cases which are,

If a line is above theorigin:

(i) If the shaded areais below the line then ax + by < c

(ii) If the shadedarea is above the line then ax + by > c

If a line is below theorigin then the rule becomes opposite.

So, according to therules

Question 4 : Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42.

Answer 4 :

Here, we shall applythe concept of a common solution area to find the signs of inequality by usingtheir given equations and the given common solution area (shaded part).

We know that,

If a line is in theform ax + by = c and c is positive constant.

So, according to therules