{"id":162267,"date":"2023-10-20T09:56:57","date_gmt":"2023-10-20T04:26:57","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=162267"},"modified":"2023-10-20T09:56:57","modified_gmt":"2023-10-20T04:26:57","slug":"ml-aggarwal-class-12-maths-solutions-section-c-chapter-3-mcqs","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-c-chapter-3-mcqs\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming MCQs"},"content":{"rendered":"

Accessing ML Aggarwal Class 12 Solutions ISC<\/a> Chapter 3 Linear Programming MCQs can be a valuable tool for students seeking extra practice.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming MCQs<\/h2>\n

Choose the correct answer from the given four options in questions (1 to 10) :<\/p>\n

Question 1.
\nIf 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
\nof Z occurs at
\n(a) (0, 0)
\n(b) (7, 0)
\n(c) (3, 4)
\n(d) (0, 2)
\nAnswer:
\n(c) (3, 4)<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 5x + 7y<\/td>\n<\/tr>\n
(0, 0)<\/td>\nZ = 5 \u00d7 0 + 7 \u00d7 0 = 0<\/td>\n<\/tr>\n
(7, 0)<\/td>\nZ = 5 \u00d7 7 + 7 \u00d7 0 = 35<\/td>\n<\/tr>\n
(3,4)<\/td>\nZ = 5 \u00d7 3 + 7 \u00d7 4 = 43<\/td>\n<\/tr>\n
(0, 2)<\/td>\nZ = 5 \u00d7 0 + 7 \u00d7 2 = 14<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmax<\/sub> = 43 occurs at point (3, 4)<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf 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
\n(a)(0,0)
\n(b) (0,8)
\n(c) (5, 0)
\n(d) (4,10)
\nAnswer:
\n(b) (0,8)<\/p>\n\n\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 3x – 4y<\/td>\n<\/tr>\n
(0, 0)<\/td>\nZ = 3 \u00d7 0 – 4 \u00d7 0 = 0<\/td>\n<\/tr>\n
(5,0)<\/td>\nZ = 3 \u00d7 5 – 4 \u00d7 0 = 15<\/td>\n<\/tr>\n
(6, 5)<\/td>\nZ = 3 \u00d7 6 – 4 \u00d7 5 = -2<\/td>\n<\/tr>\n
(6, 8)<\/td>\nZ = 3 \u00d7 6 – 4 \u00d7 8 = -14<\/td>\n<\/tr>\n
(4,10)<\/td>\nZ = 3 \u00d7 4 – 4 \u00d7 10 =-28<\/td>\n<\/tr>\n
(0, 8)<\/td>\nZ = 3 \u00d7 0 – 4 \u00d7 8 = -32<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmin<\/sub> = – 32 attains at (0, 8)<\/p>\n

Question 3.
\nRefer to above, the maximum of Z occurs at
\n(a) (5,0)
\n(b) (6, 5)
\n(c) (6, 8)
\n(d) (4,10)
\nAnswer:
\n(a) (5,0)<\/p>\n

Zmax<\/sub> = 15 at (5, 0)<\/p>\n

Question 4.
\nIf 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
\n(a) (0, 8)
\n(b) (2, 5)
\n(c) (4, 3)
\n(d) 9, 0)
\nAnswer:
\n(c) (4, 3)<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 3x + 4y<\/td>\n<\/tr>\n
(9,0)<\/td>\nZ = 3 \u00d7 9 + 4 \u00d7 0 = 27<\/td>\n<\/tr>\n
(4, 3)<\/td>\nZ = 3 \u00d7 4 + 4 \u00d7 3 = 24<\/td>\n<\/tr>\n
(2, 5)<\/td>\nZ = 3 \u00d7 2 + 4 \u00d7 5 = 26<\/td>\n<\/tr>\n
(0, 8)<\/td>\nZ = 3 \u00d7 0 + 4 \u00d7 8 = 32<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmin<\/sub> = 24 at (4, 3)<\/p>\n

Question 5.
\nCorner points of the feasible region for an LPP are (0,2), (3, 0), (6,0), (6,8) and (0,5).
\nAnswer:
\nLet F = 4x + 6y be the objective function. The Minimum value of F occurs at
\n(a) (0,2) only
\n(b) (3, 0) only
\n(c) the mid point of the line sgment joining the points (0, 2) and (3, 0)
\n(d) any point on the line segment joining the points (0, 2) and (3, 0).
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
Corner Points<\/td>\nF = 4x + 6y<\/td>\n<\/tr>\n
O (0, 2)<\/td>\n4 \u00d7 0 + 6 \u00d7 2 = 12<\/td>\n<\/tr>\n
A (3, 0)<\/td>\n4 \u00d7 3 + 6 \u00d7 0 = 12<\/td>\n<\/tr>\n
B (6, 0)<\/td>\n4 \u00d7 6 + 6 \u00d7 0 = 24<\/td>\n<\/tr>\n
C (6, 8)<\/td>\n4 \u00d7 6 + 6 \u00d7 8 = 72<\/td>\n<\/tr>\n
D (0, 5)<\/td>\n4 \u00d7 0 + 6 \u00d7 5 = 30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Thus Zmin<\/sub> = 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).<\/p>\n

Question 6.
\nCorner points of the feasible region for an LPP are (0,2), (3, 0), (6,0), (6,8) and (0, 5).
\nLet F = 4x + 6y be the objective function. Then, Maximum of F – Minimum of F =
\n(a) 60
\n(b) 48
\n(c) 42
\n(d) 18
\nAnswer:
\n(a) 60<\/p>\n

Here Fmax<\/sub> = 72; Fmin<\/sub> = 12
\nFmax<\/sub> – Fmin<\/sub> = 72 – 12 = 60<\/p>\n

Question 7.
\nCorner points of the feasible region deter\u00acmined 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
\n(a) p = 2q
\n(b) p = \\(\\frac{q}{2}\\)
\n(c) p = 3q
\n(d) p = q
\nAnswer:<\/p>\n\n\n\n\n\n\n
Corner Points<\/td>\nZ = px + qy<\/td>\n<\/tr>\n
(0, 3)<\/td>\np \u00d7 0 + q \u00d7 3 = 3q<\/td>\n<\/tr>\n
(0, 1)<\/td>\np \u00d7 1 + q \u00d7 1 = p + q<\/td>\n<\/tr>\n
(3, 0)<\/td>\np \u00d7 3 + q \u00d7 0 = 3p<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Since Zmin<\/sub> occurs at (3, 0)and (1, 1).
\n\u2234 3p = p + q
\n\u21d2 2p = q
\n\u21d2 p = \\(\\frac{q}{2}\\)<\/p>\n

Question 8.
\nThe 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
\n(a)p = q
\n(b)p = 2q
\n(c) q = 2p
\n(d) q = 3p
\nAnswer:
\n(d) q = 3p<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = px + qy<\/td>\n<\/tr>\n
(0, 10)<\/td>\np \u00d7 0 + q \u00d7 10 = 10q<\/td>\n<\/tr>\n
(5, 5)<\/td>\np \u00d7 5 + q \u00d7 5 = 5p + 5q<\/td>\n<\/tr>\n
(15, 15)<\/td>\np \u00d7 15 + q \u00d7 15 = 15p + 15q<\/td>\n<\/tr>\n
(0, 20)<\/td>\np \u00d7 0 + q \u00d7 20 = 20 q<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

15p + 15q = 20q
\n\u21d2 15p = 5q
\n\u21d2 q = 3p<\/p>\n

\"ML<\/p>\n

Question 9.
\nIf 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
\n(a) 0
\n(b) 2
\n(c) finite
\n(d) infinite
\nAnswer:
\n(d) infinite<\/p>\n

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.<\/p>\n

Question 10.
\nThe graph of the inequality 2x + 3y > 6 is
\n(a) half plane that contains the origin
\n(b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6
\n(c) whole XOY-plane excluding the points on the line 2x + 3y = 6
\n(d) entire XOY plane
\nAnswer:
\n(b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6<\/p>\n

Given inequation be 2x + 3y > 6
\nClearly origin i.e. (0, 0) is not satisfies the given inequation.
\nAlso, 2x + 3y = 6 meets coordinate axes at (3,0) and (0,2) and both points doesnot lie on 2x + 3y > 6.
\nits solution set be a half plane that neither contains the origin nor the points on the line 2x + 3y = 6.<\/p>\n","protected":false},"excerpt":{"rendered":"

Accessing ML Aggarwal Class 12 Solutions ISC Chapter 3 Linear Programming MCQs can be a valuable tool for students seeking extra practice. 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 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162267"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=162267"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162267\/revisions"}],"predecessor-version":[{"id":162271,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162267\/revisions\/162271"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=162267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=162267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=162267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}