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ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming MCQs
Choose the correct answer from the given four options in questions (1 to 10) :
Question 1.
If the objective function for a L.P.P. is L = 5x + ly and the corner points of the bounded feasible region are (0,0), (7,0), (3,4) and (0,2), then the maximum value
of Z occurs at
(a) (0, 0)
(b) (7, 0)
(c) (3, 4)
(d) (0, 2)
Answer:
(c) (3, 4)
| Corner points | Z = 5x + 7y |
| (0, 0) | Z = 5 × 0 + 7 × 0 = 0 |
| (7, 0) | Z = 5 × 7 + 7 × 0 = 35 |
| (3,4) | Z = 5 × 3 + 7 × 4 = 43 |
| (0, 2) | Z = 5 × 0 + 7 × 2 = 14 |
∴ Zmax = 43 occurs at point (3, 4)
Question 2.
If the objective function for an L.P.P. is Z = 3x – 4y and the corner points for the bounded feasible region are (0,0), (5,0), (6, 5), (6, 8), (4, 10) and (0, 8), then the minimum value of Z occurs at
(a)(0,0)
(b) (0,8)
(c) (5, 0)
(d) (4,10)
Answer:
(b) (0,8)
| Corner points | Z = 3x – 4y |
| (0, 0) | Z = 3 × 0 – 4 × 0 = 0 |
| (5,0) | Z = 3 × 5 – 4 × 0 = 15 |
| (6, 5) | Z = 3 × 6 – 4 × 5 = -2 |
| (6, 8) | Z = 3 × 6 – 4 × 8 = -14 |
| (4,10) | Z = 3 × 4 – 4 × 10 =-28 |
| (0, 8) | Z = 3 × 0 – 4 × 8 = -32 |
∴ Zmin = – 32 attains at (0, 8)
Question 3.
Refer to above, the maximum of Z occurs at
(a) (5,0)
(b) (6, 5)
(c) (6, 8)
(d) (4,10)
Answer:
(a) (5,0)
Zmax = 15 at (5, 0)
Question 4.
If the objective function for an L.P.P. is Z = 3x + 4y and the corner points for unbounded feasible region are (9,0), (4,3), (2,5) and (0,8), then the minimum value of Z occurs at
(a) (0, 8)
(b) (2, 5)
(c) (4, 3)
(d) 9, 0)
Answer:
(c) (4, 3)
| Corner points | Z = 3x + 4y |
| (9,0) | Z = 3 × 9 + 4 × 0 = 27 |
| (4, 3) | Z = 3 × 4 + 4 × 3 = 24 |
| (2, 5) | Z = 3 × 2 + 4 × 5 = 26 |
| (0, 8) | Z = 3 × 0 + 4 × 8 = 32 |
∴ Zmin = 24 at (4, 3)
Question 5.
Corner points of the feasible region for an LPP are (0,2), (3, 0), (6,0), (6,8) and (0,5).
Answer:
Let F = 4x + 6y be the objective function. The Minimum value of F occurs at
(a) (0,2) only
(b) (3, 0) only
(c) the mid point of the line sgment joining the points (0, 2) and (3, 0)
(d) any point on the line segment joining the points (0, 2) and (3, 0).
Answer:
| Corner Points | F = 4x + 6y |
| O (0, 2) | 4 × 0 + 6 × 2 = 12 |
| A (3, 0) | 4 × 3 + 6 × 0 = 12 |
| B (6, 0) | 4 × 6 + 6 × 0 = 24 |
| C (6, 8) | 4 × 6 + 6 × 8 = 72 |
| D (0, 5) | 4 × 0 + 6 × 5 = 30 |
Thus Zmin = 12 at 0(0,2) and A (3,0) i.e. at all points of line joining points 0(0,2) and A(3,0).
Question 6.
Corner points of the feasible region for an LPP are (0,2), (3, 0), (6,0), (6,8) and (0, 5).
Let F = 4x + 6y be the objective function. Then, Maximum of F – Minimum of F =
(a) 60
(b) 48
(c) 42
(d) 18
Answer:
(a) 60
Here Fmax = 72; Fmin = 12
Fmax – Fmin = 72 – 12 = 60
Question 7.
Corner points of the feasible region deter¬mined by the system of constraints are (0, 3), (1, 1) and (3, 0). Let Z=px + qy, where p,q> 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is
(a) p = 2q
(b) p = \(\frac{q}{2}\)
(c) p = 3q
(d) p = q
Answer:
| Corner Points | Z = px + qy |
| (0, 3) | p × 0 + q × 3 = 3q |
| (0, 1) | p × 1 + q × 1 = p + q |
| (3, 0) | p × 3 + q × 0 = 3p |
Since Zmin occurs at (3, 0)and (1, 1).
∴ 3p = p + q
⇒ 2p = q
⇒ p = \(\frac{q}{2}\)
Question 8.
The corner points of the feasible region determined by the system of linear con-straints are (0,10), (5,5), (15,15), (0,20). Let z-px + qy, where p,q> 0. Condition on p and q so that the maximum of z occurs at both the points (15, 15) and (0,20) is
(a)p = q
(b)p = 2q
(c) q = 2p
(d) q = 3p
Answer:
(d) q = 3p
| Corner points | Z = px + qy |
| (0, 10) | p × 0 + q × 10 = 10q |
| (5, 5) | p × 5 + q × 5 = 5p + 5q |
| (15, 15) | p × 15 + q × 15 = 15p + 15q |
| (0, 20) | p × 0 + q × 20 = 20 q |
15p + 15q = 20q
⇒ 15p = 5q
⇒ q = 3p
Question 9.
If an L.P.P. if the objective function Z = ax + by has same maximum value on two corner points of the feasible region, then the number of points at which maximum value of Z occurs is
(a) 0
(b) 2
(c) finite
(d) infinite
Answer:
(d) infinite
Clearly the objective function Z has same maximum value on two comer points of feasible region. Thus Z has same maximum value on whole line joining these two points. Since there are infinite number of points on a line. Therefore, Z has maximum value at infinite no. of points.
Question 10.
The graph of the inequality 2x + 3y > 6 is
(a) half plane that contains the origin
(b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6
(c) whole XOY-plane excluding the points on the line 2x + 3y = 6
(d) entire XOY plane
Answer:
(b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6
Given inequation be 2x + 3y > 6
Clearly origin i.e. (0, 0) is not satisfies the given inequation.
Also, 2x + 3y = 6 meets coordinate axes at (3,0) and (0,2) and both points doesnot lie on 2x + 3y > 6.
its solution set be a half plane that neither contains the origin nor the points on the line 2x + 3y = 6.