A radio station is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away at least 20 tickets. The total cost of all the tickets they give away can be no more than $280.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To solve the problem, we need to set up a system of inequalities based on the given conditions.<\/p>\n\n\n\n Let:<\/p>\n\n\n\n We can simplify the second inequality: We now have the following system of inequalities:<\/p>\n\n\n\n To find the feasible solutions, we can graph these inequalities. Here\u2019s how we approach it:<\/p>\n\n\n\n To find the intersection of the lines ( x + y = 20 ) and ( x + 2y = 28 ), we can solve the equations simultaneously.<\/p>\n\n\n\n From ( x + y = 20 ): Substituting into ( x + 2y = 28 ): Substituting ( x = 12 ) back into ( y = 20 – x ): The intersection point ( (12, 8) ) satisfies both inequalities.<\/p>\n\n\n\n The radio station can give away 12 tickets priced at $10 each and 8 tickets priced at $20 each, fulfilling both conditions of giving away at least 20 tickets while keeping the total cost under $280. This demonstrates the effective use of inequalities to model real-world scenarios, providing a systematic approach to decision-making in such situations.<\/p>\n","protected":false},"excerpt":{"rendered":" A radio station is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away at least 20 tickets. The total cost of all the tickets they give away can be no more than $280. The Correct Answer and Explanation is: To […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-152931","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152931","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=152931"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152931\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=152931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=152931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=152931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Given Conditions:<\/h3>\n\n\n\n
\n
[
x + y \\geq 20
]<\/li>\n\n\n\n
[
10x + 20y \\leq 280
]<\/li>\n\n\n\n
[
x \\geq 0, \\quad y \\geq 0
]<\/li>\n<\/ol>\n\n\n\nSimplifying the Cost Inequality<\/h3>\n\n\n\n
[
10x + 20y \\leq 280
]
Dividing the entire inequality by 10 gives:
[
x + 2y \\leq 28
]<\/p>\n\n\n\nSummary of Inequalities<\/h3>\n\n\n\n
\n
Finding the Feasible Region<\/h3>\n\n\n\n
\n
\n
\n
\n
Finding Intersection Points<\/h3>\n\n\n\n
[
y = 20 – x
]<\/p>\n\n\n\n
[
x + 2(20 – x) = 28 \\
x + 40 – 2x = 28 \\
-x + 40 = 28 \\
-x = -12 \\
x = 12
]<\/p>\n\n\n\n
[
y = 20 – 12 = 8
]<\/p>\n\n\n\nFeasible Solution<\/h3>\n\n\n\n
Conclusion<\/h3>\n\n\n\n