Lesson Typical problems on buying and selling items
Algebra
->
Equations
-> Lesson Typical problems on buying and selling items
Log On
Algebra: Equations
Section
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Source code of 'Typical problems on buying and selling items'
This Lesson (Typical problems on buying and selling items)
was created by by
ikleyn(52776)
:
View Source
,
Show
About ikleyn
:
<H2>Typical problems on buying and selling items</H2> <H3>Problem 1</H3>White chocolate cost $20.00 per bar, and dark chocolate cost $25.00 per bar. If Anne bought 15 bars of chocolate for $340.00, how many bars of dark chocolate did she buy? <B>Solution</B> <pre> Let x be the number of dark chocolate bars Anne bought. Then the number of the white chocolate bars is 15-x. Write the total money equation 25x + 20*(15-x) = 340 dollars. Simplify and find x 25x + 300 - 20x = 340 25x - 20x = 340 - 300 5x = 40 x = 40/5 = 8 <U>ANSWER</U>. Anne bought 8 dark chocolate bars. <U>CHECK</U>. 25*8 + 20*(15-8) = 200 + 140= 340 dollars, total. ! Correct ! </pre> <H3>Problem 2</H3>At the city museum, child admission is 6.30 and adult admission is $9.60. On Monday, 158 tickets were sold for a total sales of $1193.40. How many child tickets were sold that day? <B>Solution</B> <pre> Let x be the number of child tickets sold that day. Then the number of adult tickets sold is (158-x). Write the total money equation 6.30*x + 9.60*(158-x) = 1193.40 dollars. Simplify step by step and find x 6.30x + 9.60*158 - 9.60x = 1193.40 6.30x - 9.60x = 1193.40 - 9.60*158 -3.30x = -323.40 x = {{{(-323.40)/(-3.30)}}} = 98. <U>ANSWER</U>. 98 child tickets. <U>CHECK</U>. 6.30*98 + 9.60*(158-98) = 1193.40 dollars, the total revenue. ! correct ! </pre> <H3>Problem 3</H3>30 books and magazines were on a shelf. Each book costs $5 while each magazine costs $3. The total cost of the books was $6 more than the magazines. How many books were on the shelf. <B>Solution</B> <pre> x books, 30-x magazines. Money equation is 5x -3*(30-x) = 6 dollars. Simplify and find x 5x - 90 + 3x = 6 8x = 90+6 = 96 x = 96/8 = 12. <U>ANSWER</U>. 12 books. </pre> <H3>Problem 4</H3>A machine requires 6 hours to make a unit of Product A and 5 hours to make a unit of Product B. Last month the machine operated for 481 hours, producing a total of 89 units. How many units of Product A were produced? <B>Solution</B> <pre> Let x be the number of units of A-product. Then the number of units of B-product is 89-x. Write the full elapsed time equation 6x + 5*(89-x) = 481. Simplify and find x 6x + 445 - 5x = 481 6x - 5x = 481 - 445 x = 36. <U>ANSWER</U>. 36 units of A-product were produced. </pre> My other 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-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/Using-price-equation-to-solve-some-business-related-problems.lesson>Using price equation to solve some business related 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.
