{"id":"cb_question_bank:73857a20-b09c-4ca2-a92e-587e4a2e0732","source_id":"cb_question_bank","source_item_id":"73857a20-b09c-4ca2-a92e-587e4a2e0732","external_id":"3e613a2d-72f8-4e08-8fce-8abf51849968","ibn":null,"question_id":"ee7b1de1","program":"SAT","module":"math","domain_code":"H","domain":"Algebra","skill_code":"H.E.","skill":"Linear inequalities in one or two variables","difficulty":"H","score_band":7,"answer_type":"spr","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"A small business owner budgets dollar sign 2,200 to purchase candles. The owner must purchase a minimum of 200 candles to maintain the discounted pricing. If the owner pays dollar sign 4.90 per candle to purchase small candles and dollar sign 11.60 per candle to purchase large candles, what is the maximum number of large candles the owner can purchase to stay within the budget and maintain the discounted pricing?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: left;\">A small business owner budgets&nbsp;<math alttext=\"dollar sign 2,200\"><mo>$</mo><mrow><mn>2,200</mn></mrow></math> to purchase candles. The owner must purchase a minimum of <math alttext=\"200\"><mn>200</mn>\n</math> candles to maintain the discounted pricing. If the owner pays <math alttext=\"dollar sign 4.90\"><mo>$</mo><mn>4.90</mn></math> per candle to purchase small candles and <math alttext=\"dollar sign 11.60\"><mo>$</mo><mn>11.60</mn></math> per candle to purchase large candles, what is the maximum number of large candles the owner can purchase to stay within the budget and maintain the discounted pricing?</p>","rationale_html":"<p style=\"text-align: left;\">The correct answer is <math alttext=\"182\"><mn>182</mn>\n</math>. Let <math alttext=\"s\"><mi>s</mi>\n</math> represent the number of small candles the owner can purchase, and let <math alttext=\"script l\"><mi>l</mi></math> represent the number of large candles the owner can purchase. It’s given that the owner pays <math alttext=\"dollar sign 4.90\"><mo>$</mo><mn>4.90</mn></math> per candle to purchase small candles and&nbsp;<math alttext=\"dollar sign 11.60\"><mo>$</mo><mn>11.60</mn></math> per candle to purchase large candles. Therefore, the owner pays <math alttext=\"4.90 s\"><mn>4.90</mn><mi>s</mi></math> dollars for <math alttext=\"s\"><mi>s</mi>\n</math> small candles and&nbsp;<math alttext=\"11.60 script l\"><mn>11.60</mn><mi>l</mi></math> dollars for&nbsp;<math alttext=\"script l\"><mi>l</mi></math> large candles, which means the owner pays a total of <math alttext=\"4.90 s plus 11.60 script l\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi></math> dollars to purchase candles. It’s given that the owner budgets <math alttext=\"dollar sign 2,200\"><mo>$</mo><mn>2,200</mn></math> to purchase candles. Therefore,&nbsp;<math alttext=\"4.90 s plus 11.60 script l less than or equals 2,200\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math>. It’s also given that the owner must purchase a minimum of <math alttext=\"200\"><mn>200</mn>\n</math> candles. Therefore, <math alttext=\"s plus script l greater than or equals 200\"><mi>s</mi><mo>+</mo><mi>l</mi><mo>≥</mo><mn>200</mn></math>. The inequalities <math alttext=\"4.90 s plus 11.60 script l less than or equals 2,200\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math> and <math alttext=\"s plus script l greater than or equals 200\"><mi>s</mi><mo>+</mo><mi>l</mi><mo>≥</mo><mn>200</mn></math> can be combined into one compound inequality by rewriting the second inequality so that its left-hand side is equivalent to the left-hand side of the first inequality. Subtracting <math alttext=\"script l\"><mi>l</mi></math> from both sides of the inequality&nbsp;<math alttext=\"s plus script l greater than or equals 200\"><mi>s</mi><mo>+</mo><mi>l</mi><mo>≥</mo><mn>200</mn></math> yields&nbsp;<math alttext=\"s greater than or equals 200 minus script l\"><mi>s</mi><mo>≥</mo><mn>200</mn><mo>-</mo><mi>l</mi></math>. Multiplying both sides of this inequality by <math alttext=\"4.90\"><mn>4.90</mn></math> yields&nbsp;<math alttext=\"4.90 s greater than or equals 4.90 left parenthesis 200 minus script l right parenthesis\"><mn>4.90</mn><mi>s</mi><mo>≥</mo><mn>4.90</mn><mfenced><mrow><mn>200</mn><mo>-</mo><mi>l</mi></mrow></mfenced></math>, or&nbsp;<math alttext=\"4.90 s greater than or equals 980 minus 4.90 script l\"><mn>4.90</mn><mi>s</mi><mo>≥</mo><mn>980</mn><mo>-</mo><mn>4.90</mn><mi>l</mi></math>. Adding&nbsp;<math alttext=\"11.60 script l\"><mn>11.60</mn><mi>l</mi></math> to both sides of this inequality yields <math alttext=\"4.90 s plus 11.60 script l greater than or equals 980 minus 4 period 90 script l plus 11 period 60 script l\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≥</mo><mn>980</mn><mo>-</mo><mn>4</mn><mo>.</mo><mn>90</mn><mi>l</mi><mo>+</mo><mn>11</mn><mo>.</mo><mn>60</mn><mi>l</mi></math>, or&nbsp;<math alttext=\"4.90 s plus 11.60 script l greater than or equals 980 plus 6.70 script l\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≥</mo><mn>980</mn><mo>+</mo><mn>6.70</mn><mi>l</mi></math>. This inequality can be combined with the inequality&nbsp;<math alttext=\"4.90 s plus 11.60 script l less than or equals 2,200\"><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math>, which yields the compound inequality <math alttext=\"980 plus 6.70 script l less than or equals 4.90 s plus 11.60 script l less than or equals 2,200\"><mn>980</mn><mo>+</mo><mn>6.70</mn><mi>l</mi><mo>≤</mo><mn>4.90</mn><mi>s</mi><mo>+</mo><mn>11.60</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math>. It follows that&nbsp;<math alttext=\"980 plus 6.70 script l less than or equals 2,200\"><mn>980</mn><mo>+</mo><mn>6.70</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math>. Subtracting <math alttext=\"980\"><mn>980</mn>\n</math> from both sides of this inequality yields&nbsp;<math alttext=\"6.70 script l less than or equals 2,200\"><mn>6.70</mn><mi>l</mi><mo>≤</mo><mn>2,200</mn></math>. Dividing both sides of this inequality by&nbsp;<math alttext=\"6.70\"><mn>6.70</mn></math> yields approximately <math alttext=\"script l less than or equals 182.09\"><mi>l</mi><mo>≤</mo><mn>182.09</mn></math>. Since the number of large candles the owner purchases must be a whole number, the maximum number of large candles the owner can purchase is the largest whole number less than <math alttext=\"182.09\"><mn>182.09</mn>\n</math>, which is <math alttext=\"182\"><mn>182</mn>\n</math>.</p>","correct_answer":["182"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.820000+00:00","source_updated_at":"2023-08-02T20:25:59.820000+00:00","options":[],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}