Lesson Business-related problems to solve them using quadratic equations
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<H2>Business-related problems to solve them using quadratic equations</H2> <H3>Problem 1</H3>A speculator sold some shares of stock worth $2880. Several days later, the stock having dropped in price $2 per share, he repurchased, for the same amount of money, 6 more shares than he sold. How many shares did he sell? <B>Solution</B> <pre> Let x be the number of shares he sold. Then the number of shares he purchased was (x+6). The selling price was {{{2880/x}}} dollars per share. The buying price was {{{2880/(x+6))}}}. According to the problem, {{{2880/x}}} - {{{2880/(x+6)}}} = 2 dollars per share. At this point, the setup is done. To solve the equation, multiply both sides by x*(x+6). You will get then 2880(x+6) - 2880x = 2x*(x+6) 1440(x+6) - 1440x = x(x+6) 1440*6 = x^2 + 6x x^2 + 6x - 8640 = 0 (x-90)*(x+96) = 0. The only positive root, which makes sense is x= 90. So, the number of shares he sold was 90. <U>ANSWER</U> <U>CHECK</U>. The selling price was {{{2880/90)}}} = 32 dollars per share. The buying price was {{{1280/(90+6)}}} = 30 dollars per share. The difference in prices was 32 - 30 = 2 dollars per share. ! Correct ! </pre> <H3>Problem 2</H3>Paul bought a number of shares of stock for a total of $3000. Three months later the stock had increased in value by $5 per share and he sold all but 50 shares and regained his original investment of $3000. How many shares did he sell? <B>Solution</B> <pre> Let n be the number of shares Paul sold. Then the original number of shares he bought was (n+50). The buying price per share was {{{3000/(n+50)}}} dollars. The selling price per share was {{{3000/n}}} dollars. The difference of these prices was 5 dollars, according to the problem, which gives this "price" equation {{{3000/n}}} - {{{3000/(n+50)}}} = 5 dollars. (1) To solve this equation, multiply both sides by n*(n+50), then simplify step by step 3000*(n+50) - 3000*n = 5n*(n+50) 3000n + 150000 - 3000n = 5n^2 + 250n 5n^2 + 250n - 150000 = 0 n^2 + 50n - 30000 = 0 (n+200)*(n-150) = 0 The equation has two roots, -200 and 150. Of these two roots, we discard the negative one and accept the positive value n= 150. <U>CHECK</U>. To check, substitute n= 150 into equation (1). You will get {{{3000/150}}} - {{{3000/200}}} = 20 - 15 = 5, which is precisely correct. <U>ANSWER</U>. Paul sold 150 shares. </pre> <H3>Problem 3</H3>The margin of profit of a company is the net income divided by the total sales. A company's margin of profit increased by 0.02 from last year. Last year the company sold its product at $3.00 each and had a net income of $4500. This year it increased the price of its product by $0.50 each, sold 2000 more, and had a net income of $7140. The company never has had a margin of profit greater than 0.15. How many of its product were sold last year and how many were sold this year? <B>Solution</B> The margin of profit of a company is the net income divided by the total sales. [1] A company's margin of profit increased by 0.02 from last year. [2] Last year the company sold its product at $3.00 each and had a net income of $4500. [3] This year it increased the price of its product by $0.50 each, sold 2000 more, and had a net income of $7140. [4] The company never has had a margin of profit greater than 0.15. How many of its product were sold last year and how many were sold this year? I numbered the sentences for easy referring. <pre> Let x be the number of their product that were sold last year. From [2], we have this expression for margin profit last year {{{4500/(3x)}}}, or {{{1500/x}}}. (1) From [3], we have this expression for margin profit this year {{{7140/(3.50*(x+2000))}}}, or {{{2040/(x+2000)}}} (2) From [1], quantity (2) is 0.02 greater than quantity (1). It gives this equation {{{2040/(x+2000)}}} - {{{1500/x}}} = 0.02. +-------------------------------------------------------------------+ | In this problem, the MAJOR STEP is to get a setup equation, | | and we successfully overcome it (!) | +-------------------------------------------------------------------+ +---------------------------------------+ | To solve it is just a brute force. | +---------------------------------------+ Multiply both sides by x*(x+2000) 2040x - 1500(x + 2000) = 0.02x*(x+2000) Simplify step by step 2040x - 1500x - 3000000 = 0.02x^2 + 40x 0.02x^2 - 500x + 3000000 = 0 The roots of this quadratic equation are (use the quadratic formula) {{{x[1]}}} = 10000; {{{x[2]}}} = 15000. Value {{{x[1]}}} = 10000 gives the profit margin {{{4500/(3*10000)}}} = 0.15 = 15%, which contradicts to condition [4]; so we reject this root. Value {{{x[2]}}} = 15000 gives the profit margin {{{4500/(3*15000)}}} = 0.10 = 10%, which agrees with condition [4]; so we accept this solution. Thus last year the company sold 15000 units of the product. This year they sold 2000 units more, i.e. 15000 + 2000 = 17000 units of the product. </pre> My other lessons on quadratic equations in this website are - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/proof-of-quadratic-by-completing-the-square.lesson>PROOF of quadratic formula by completing the square</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Learning-by-examples-on-HOW-TO-complete-the-square.lesson>HOW TO complete the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-solve-quadratic-equation-by-completing-the-square-Learning-by-examples.lesson>HOW TO solve quadratic equation by completing the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A> - 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