Lesson Using price equation to solve some business related problems
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<H2>Using price equation to solve some business related problems</H2> <H3>Problem 1</H3>Three chums wanted to buy a complete set of Pelota rackets with 3 balls. However they figure out that each of them would need to pay P100 less if they can find two more chums, to share equally the cost of the sporting equipment they wish to buy. How much is the Pelota Set? <B>Solution</B> <pre> Let x be the cost (the price) the Pelota Set. In the 1st scenario, 3 (three) persons share this cost, so each of them pays {{{x/3}}}. In the 2nd scenario, 3 + 2 = 5 (five) persons share this cost, so each of them pays {{{x/5}}}. We are given that the difference is equal to P100, which gives this "price" equation {{{x/3}}} - {{{x/5}}} = 100. Thus the setup is just done, and now our task is to solve this basic equation. For it, multiply both sides by 3*5 = 15. You will get then 5x - 3x = 1500 2x = 1500 x = 1500/2 = 750. Thus the problem is just solved and the <U>ANSWER</U> is The Pelota Set costs P750. <U>CHECK</U>. {{{750/3}}} - {{{750/5}}} = 250 - 150 = 100 P. ! Correct ! </pre> <H3>Problem 2</H3>James spent $6780 on some watches and clocks. The amount spent on watches was $2820 more than the amount spent on the clocks. He bought 3/5 as many clocks as watches. Each clock cost $25 less than each watch. Find the number of watches and the number of clocks bought by James. <B>Solution</B> The solution consists of two parts (two steps) and uses two clear simple ideas, accompanying with short calculations. <U>First step</U> <pre> Let X be the amount of money spent on watches, Y be the amount of money spent on clocks. From the condition, we have two equations X + Y = 6780 (total dollars) X - Y = 2820 (X is $2820 more than Y) To find X, add the equations. You will get 2X = 6780+2820 = 9600, X = 9600/2 = 4800 dollars. To find Y, subtract second equation from the first one. You will get 2Y = 6780-2820 = 3960, Y = 3960/2 = 1980 dollars. </pre> First step is complete. We just found that the amount spent on watches was $4800; the amount spent on clocks was $1980. <U>Second step</U> <pre> Let W be the number of watches bought by James. Then the number of clocks was {{{(3/5)W}}} = 0.6W. Each watch price was {{{4800/W}}} dollars; each clock price was {{{1980/(0.6*W)}}} = {{{3300/W}}}. Next, the difference of prices is 25 dollars. It gives this "price" equation {{{4800/W}}} - {{{3300/W}}} = 25. Simplify and find W {{{1500/W}}} = 25 W = {{{1500/25}}} = 60. <U>ANSWER</U>. 60 watches and {{{(3/5)*60}}} = 36 clocks. </pre> My lessons on solving single linear equations and word problems in one unknown are - <A HREF=https://www.algebra.com/algebra/homework/equations/How-to-solve-a-linear-equation.lesson>HOW TO solve a linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Using-linear-equations-to-solve-word-problems.lesson>Simple word problems to solve using a single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/More-complicated-word-problems-to-solve-using-single-linear-equation.lesson>More complicated word problems to solve using a single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/06-Typical-word-problems-from-the-archive-to-solve-by-reduction-to-single-linear-equation.lesson>Typical word problems to solve using a single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Typical-problems-on-selling-and-buying-items.lesson>Typical problems on buying and selling items</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Typical-investment-problems.lesson>Typical investment problems</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Advanced-word-problems-to-solve-by-reduction-to-single-linear-equation.lesson>Advanced word problems to solve using a single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/HOW-TO-algebreze-and-solve-these-problems-using-one-eqn-in-one-unknown.lesson>HOW TO algebraize and solve these problems using one equation in one unknown</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Challenging-word-problems-to-solve-using-a-single-linear-equation.lesson>Challenging word problems to solve using a single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Selected-word-problems-to-solve-by-reducing-to-single-linear-equation.lesson>Selected word problems to solve by reducing to single linear equation</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Solving-some-business-related-problems.lesson>Solving some business-related problems</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/HOW-TO-solve-these-simple-word-problems-MENTALLY-without-using-equations.lesson>HOW TO solve these simple word problems MENTALLY without using equations</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Time-equation-to-solve-some-Travel-and-Distance-problems.lesson>Using time equation to solve some Travel and Distance problems</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Solving-simple-and-simplest-problems-by-the-backward-method.lesson>Solving problems by the backward method</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Solving-problems-on-the-remaining-amount.lesson>Solving more complicated problems by the backward method</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/Solving-entertainment-problems-on-shortage-money.lesson>Solving entertainment problems on shortage of money</A> - <A HREF=https://www.algebra.com/algebra/homework/equations/OVERVIEW-of-lessons-on-solving-linear-equations-and-word-problems-in-one-unknown.lesson>OVERVIEW of lessons on solving linear equations and word problems in one unknown</A> in this site.
