Lesson Challenging word problems to solve using a single linear equation
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<H2>Challenging word problems to solve using a single linear equation</H2> <H3>Problem 1. A boat moving upstream and downstream on a river</H3>A motorboat moves on a river that has a current of 3 miles per hour. The trip upstream takes 8 hours, while the return trip takes 5 hours. What is the speed of the motorboat relative to the water? <B>Solution</B> <pre> Let us denote the unknown speed of the motorboat relative to the water as <B>x</B> miles/hour. When motorboat moves upstream, its speed relative to the bank of the river is <B>x-3</B> miles/hour. So, moving 8 hours upstream, the motorboat passes <B>8*(x-3)</B> miles along the river bank. When moving downstream, the motorboat speed relative to the bank of the river is <B>x+3</B> miles/hour. So, moving 5 hours downstream, the motorboat passes <B>5*(x+3)</B> miles along the river bank. This is the same distance, so we have an equation 8*(x-3) = 5(x+3). Note that this is a linear equation. Performing multiplication and collecting common terms and making other steps of the general procedure, we have 8x - 24 = 5x + 15, 8x - 5x = 24 + 15, 3x = 39, x = 13 miles/hour. <U>Check</U> Substitute 13 to both sides of the original equation. You have: left side 8*(13-3) = 8*10 = 80, right side 5*(13+3) = 5*16 = 80. <U>Answer</U>. The motorboat speed relative to the water is equal to 13 miles/hour. </pre> <H3>Problem 2. Mixing water and antifreeze</H3>How much pure antifreeze liquid should be added to 1 gallon of 40% antifreeze to get 60% antifreeze? (Concentrations here are volume concentrations, measured in [vol/vol] units). <B>Solution</B> <pre> First, there is 0.4 gal of pure antifreeze in 1 gallon of 40% antifreeze. Let us denote <B>x</B> a volume of pure antifreeze, which should be added to 1 gallon of 40% antifreeze to get 60% antifreeze. So, the volume of antifreeze after adding is <B>0.4 + x</B> gallons, while the total volume of liquid (water plus antifreeze) after adding is <B>1.0+x</B>. The condition of 60% volume concentration gives an equation {{{(0.4+x)/(1.0+x) = 0.6}}}, or <B>0.4+x = 0.6*(1.0+x)</B>. This is a linear equation. Simplifying the equation step by step gives 0.4+x = 0.6 + 0.6*x, x-0.6*x = 0.6-0.4, 0.4*x = 0.2, x = 0.2/0.4 = 0.5 gallons. <U>Check</U>: {{{(0.4+0.5)/(1.0+0.5) = 0.9/1.5 = 0.6}}}. <U>Answer</U>. 0.5 gallons of pure antifreeze should be added. </pre> <H3>Problem 3. Alloy</H3>A piece of alloy composed of copper and zinc weights 81400 N (Newtons). and is 1000 {{{cm^3}}} in volume. How much copper and how much zinc is there in the alloy? The density of copper is 8.92 {{{g/cm^3}}}, the density of zinc is 7.14 {{{g/cm^3}}}. <B>Solution</B> <pre> Let us denote the unknown volume of copper in the alloy as <B>x</B> {{{cm^3}}}. The weight of copper in the alloy is 981 {{{cm/s^2}}} * 8.92 {{{g/cm^3}}} * x {{{cm^3}}} = 8750.52 {{{g/(cm^3*s^2)}}} * x {{{cm^3}}} = 8750.52 {{{N/cm^3}}} * x {{{cm^3}}}, where 981 {{{cm/s^2}}} is the gravity acceleration. The volume of zinc in alloy is (1000 - x) {{{cm^3}}}, and the weight of zinc in alloy is 981 {{{cm/s^2}}} * 7.14 {{{g/cm^3}}} * (1000-x) {{{cm^3}}} = 7004.34 {{{N/cm^3}}} * (1000-x) {{{cm^3}}}. The total weight of alloy is 81400 Newtons, which gives us an equation 8750.52 * x + 7004.34 * (1000-x) = 81400 N = 81400 {{{g*m/s^2}}} = 8140000 {{{g*cm/s^2}}}. (Note that in the previous line we converted Newtons (that are {{{g*m/s^2)}}}) to {{{g*cm/s^2}}}, because we use grams, centimeters and seconds as units uniformly in the solution of this problem). The last equation is a linear one. Performing multiplication and collecting common terms and making other steps of the general procedure, we have 8750.52 * x + 7004.34*1000 - 7004.34*x = 8140000, 8750.52 * x + 7004340 - 7004.34*x = 8140000, 8750.52 * x - 7004.34*x = 8140000 - 7004340, 1746.18 x = 1135600, x = 1135600/1746.18 = 650.37 So, the volume of copper is 650.37 {{{cm^3}}}, and mass of copper in alloy is 8.92 {{{g/cm^3}}} * 650.37 {{{cm^3}}} = 5801 {{{g}}}. Volume of zinc in alloy is 1000-650.37 = 349 {{{cm^3}}}, and mass of zinc in alloy is 7.14 {{{g/cm^3}}} * 349 {{{cm^3}}} = 2496 {{{g}}}. <U>Answer</U>. Mass of copper in alloy is 5801 g; mass of zinc in alloy is 2496 g. </pre> <H3>Problem 4. Web-site</H3>A website advertises properties for sale. If the property is sold within the month, the website charges $140 for the advertisement, but if the property is not sold within the month, the website pays the advertiser $30. In a particular month, 49 advertisements were placed, and a total of $3120 was made across all advertising. Find the number of advertisements that resulted in a sale of the property. <B>Solution</B> <pre> Let x be the number of advertisements that resulted in a sale of the property. Then 49-x is the number of advertising that did not result in a sale of the property. Your equation to find x is this revenue equation 140*x - 30*(49-x) = 3120. From the equation, x = {{{(3120 + 30*49)/(140+30)}}} = 27. <U>ANSWER</U>. The number of advertisements that resulted in a sale of the property is 27. </pre> <H3>Problem 5</H3>A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 45 pounds each, and the small boxes weigh 35 pounds each. There are 120 boxes in all. If the truck is carrying a total of 4850 pounds in boxes, how many of each type of box is it carrying? <B>Solution</B> <pre> Let x = # of large boxes; then the number of small boxes is (120-x). The total weight equation is 45x + 35*(120-x) = 4850 pounds. From the equation x = {{{(4850 - 35*120)/(45-35)}}} = 65. <U>ANSWER</U>. 65 large boxes and 120-65 = 55 small boxes. <U>CHECK</U>. 45*65 + 35*55 = 4850 (total weight). ! Precisely correct ! </pre> Other lessons on solving single linear equations and relevant word problems 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/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.
