Lesson Advanced word problems to solve using a single linear equation
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<H2>Advanced word problems to solve using a single linear equation</H2> You often may meet word problems that seems to be for several (two or three) unknowns from the first glance, but actually are for one single unknown. The goal of this lesson is to teach you to recognize such problems as one single unknown problems and to show you how to solve them. <H3>Problem 1</H3>A child's piggy bank contains 35 coins and a total of $6.25. The bank contains quarters, dimes and nickels. There are twice as many dimes as there are nickles. How many of each coin is in the bank? <B>Solution</B> <pre> Let N be the number of nickels. Then the number of dimes is 2N, according to the condition. Hence, the number of quarters is (35-N-2N) = (35-3N). The nickels contribute 5N cents towards the total. The dimes contribute 10*(2N) = 20N cents towards the total. The quarters contribute 25*(35-3N) cents towards the total. So, your balance equation is 5N + 20N + 25*(35-3N) = 625 cents. (<<<---=== you counted all the cents) Simplify and solve for N step by step. 25N + 875 - 75N = 625 ====> -50N = 625 - 875 = -250 ====> N = {{{(-250)/(-50)}}} = 5. <U>Answer</U>. 5 nickels, 10 dimes and 35-5-10 = 20 quarters. <U>Check</U>. Can you check it on your own ?? </pre> <U>The lesson to learn from this solution</U> : <pre> This problem is for one unknown, not for three. Your major task, when you start solve it, is to select the key unknown, and then to express other quantities via that unknown. Then write the basic governing equation and solve it accurately. </pre> <H3>Problem 2</H3>Alice has saved $5.00 in dimes, quarters, and half dollars. She finds that she has 30 coins altogether and there are four times as many quarters as there are half dollars. How many dimes does Alice have ? <B>Solution</B> <pre> Let H be the number of half-dollars. Then the number of quarters is 4H, according to the condition. Hence, the dimes are the rest (30-H-4H) = (30-5H) coins. The dimes contribute 10*(30-5H) cents towards to the total. The quarters contribute 25*((4H) = 100H cents towards to the total. Finally, the half-dollars contribute 50*H cents towards to the total. Thus the total is 10*(30-5H) + 100H + 50H, and your money equation is 10*(30-5H) + 100H + 50H = 500 cents. (Should I explain that $5 is 500 cents ?) This is the simple equation in one unknown. Let's solve it together. 300 - 50H + 100H + 50H = 500 ====> 100H = 500 - 300 = 200 ====> H = {{{200/100}}} = 2. So, there are 2 half-dollar coins, 4*2 = 8 quarters, and the rest 30-2-8 = 20 coins are dimes. Let us CHECK the solution: 20*10 + 8*25 + 50*2 = 500 cents. ! Correct ! </pre> <U>The lesson to learn from this solution</U> : <pre> This problem is FOR ONE UNKNOWN. Therefore, I made all efforts from my side to solve it as the problem in ONE UNKNOWN. </pre> <H3>Problem 3</H3>Erwin has 40 pieces of coins consisting of nickels, dimes and quarters. The number of nickels is equal to the number of dimes. If he has a total of $8.25, how many coins of each kind does he have? <B>Solution</B> <pre> Let n be the number of the 5¢ coins. Then the number of the 10¢ coins is also n, according to the condition, and the number of the 25¢ coins is 40-n-n = (40-2n): they are the rest coins. The total money equation is then 5n + 10n + 25*(40-2n) = 825 cents, or 5n + 10n + 1000 - 50n = 825 -35n = 825 - 1000 = -175 n = -175/(-35) = 5. <U>ANSWER</U>. There are five nickels; five dimes and the rest 40-5-5 = 30 are the quartes. <U>CHECK</U>. The total is 5*5 + 5*10 + 30*25 = 825 cents = $8.25. ! Correct ! </pre> <H3>Problem 4</H3>A coffee manufacturer sells a 10-pound package of coffee that consists of three flavors of coffee. Vanilla-flavored coffee costs $12 per pound, hazelnut-flavored coffee costs $13.50 per pound, and mocha-flavored coffee costs $15 per pound. The cost of the 10 pound package is $138. The package contains the same amount of hazelnut coffee as mocha coffee. How many pounds of each type of coffee are in the package? <B>Solution</B> Again, typical problem to solve using SINGLE unknown. <pre> Let x = amount of hazelnut (in pounds). Then the amount of mocha coffee is the same x pounds. Then the amount of vanilla-flavored coffee is (10-2x) pounds. The money equation is 12*(10-2x) + 13.50x + 15x = 138, or 120 - 24x + 28.5x = 138, 4.5x = 138 - 120 = 18 ====> x = {{{18/4.5}}} = 4 pounds. <U>Answer</U>. 4 pounds of hazelnut; 4 pounds of mocha coffee; and 2 pounds of vanilla-flavored coffee. </pre> <H3>Problem 5</H3>A farmer has 1080 acres of land on which he grows corn, wheat, and soybeans. It costs $45 per acre to grow corn, $60 to grow wheat, and $50 to grow soybeans. Because of market demand, the farmer will grow twice as many acres of wheat as of corn. He has allocated $57,300 for the cost of growing his crops. How many acres of each crop should he plant? <B>Solution</B> <pre> Let x be the area for corn, in acres. Then the area for wheat is 2x acres. Hence, the area for soybeans is (1080 - x - 2x) = 1080-3x acres. Now it is easy to write the "money" equation 45*x + 60*(2x) + 50*(1080-3x) = 57300. Simplify and solve for x 45x + 120x + 50*1080 - 150x = 57300 15x = 57300 - 50*1080 = 3300 x = 3300/15 = 220. <U>ANSWER</U>. 220 acres for corn; 2*220 = 440 acres for wheat and the rest, 1080-220-440 = 420 acres for soybeans. </pre> <H3>Problem 6</H3>A local grocery store makes a 9-pound mixture of trail mix. The trail mix contains raisins, sunflower seeds, and chocolate covered peanuts. The raisins cost $2 per pound, the sunflower seeds cost $1 per pound, and the chocolate covered peanuts cost $1.50 per pound. The mixture calls for twice as many raisins as sunflower seeds. The total cost of the mixture is $14.50. How much of each ingredient did the store use ? <B>Solution</B> <pre> Let x = weight of sunflower seeds (in pounds). Then the weight of the raisins is 2x pounds and the weight of the chocolate covered peanuts is 9 - x - 2x = 9-3x pounds. The "cost equation" is 2*(2x) + 1*x + 1.50*(9-3x) = 14.50 dollars. =============> 4x + x + 13.5 - 4.5x = 14.5 ====> 0.5x = 14.5 - 13.5 = 1 ===> x = 2. <U>Answer</U>. 2 pounds of sunflower seeds, 4 pound of raisins and 9-3*2 = 3 pounds of the chocolate covered peanuts. </pre> <H3>Problem 7</H3>Hotel ordered 200 flowers for annual celebration. They ordered carnations at $1.50 each, roses at $5.75 each, and daisies at $2.60 each. They ordered mostly carnations and 20 fewer roses than daisies. The total order came to $589.50. How many of each type of flowers was ordered? <B>Solution</B> <pre> Let D be the number of daisies. Then the number of roses was (D-20). The carnations were the rest (200 - D - (D-20)) = 180-2D flowers. The cost for daisies was 2.60*D dollars. The cost for roses was 5.75*(D-20) dollars. The cost for carnations was 1.50*(180-2D) dollars. The "money" equation is 2.60*D + 5.75*(D-20) + 1.50*(180-2D) = 589.50 dollars (<<<---=== here I counted all the money) It is your basic (balance) equation. As soon as you got it, you completed the setup part of the solution. You got a single equation for only one unknown D. Simplify it and solve for D, the number of daisies. Then calculate the number of other flowers. </pre> <H3>Problem 8</H3>A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3235 tickets overall. It has sold 182 more $20 tickets than $10 tickets. The total sales are $62170. How many tickets of each kind have been sold? How many $10 dollar tickets were sold? <B>Solution</B> <pre> Let T be the number of $10-tickets sold. Then the number of $20-tickets is (T+182), according to the condition, and the number of $30-tickets is (3235 - T - (T+182)) = 3053-2T. Hence, your "money equation" is 10T + 20*(T+182) + 30*(3053-2T) = 62170. Simplify and solve for T: (10T + 20T - 60T) = 62170 - 20*182 - 30*3053, -30T = -33060 =====> T = {{{(-33060)/(-30)}}} = 1102. Thus, 1102 of the $10-tickets, 1102 + 182 = 1284 of the $20-tickets and 3235-1102- 1284 = 849 of the $30-tickets were sold. <U>Check</U>. $10*1102 + $20*1284 + $30*849 = 62170. ! Correct ! </pre> <H3>Problem 9</H3>In a triathlon, Jenny swim for one hour, bike for 1.75 hours, and ran for one hour. Her average biking speed was two times her average running speed, and her average running speed was eight times her average swimming speed. The total distance of the triathlon was 55.5 km. Calculate Jenny's average swimming speed in kilometers per hour. <B>Solution</B> <pre> Let "r" be Jenny's swimming speed, in {{{km/h}}}. Then her running speed was 8r, and her biking speed was 2*(8r) = 16r. Swimming for 1 hour, Jenny covered the distance of 1*r = r kilometers. Biking for 1.75 hour, Jenny covered the distance of 1.75*16r = 28r kilometers. Running for 1 hour, Jenny covered the distance of 1*2r = 8r kilometers. The total distance Jenny covered was r + 28r + 8r = 37r. It gives you an equation 37r = 55.5 kilometers. Then r = {{{55.5/37}}} = 1.5 km/h. It was Jemmy's swimming speed. <U>Answer</U>. Jemmy's swimming speed was 1.5 km/h. </pre> <U>Lessons to learn from the solutions to the above problems</U>: <pre> 1. Think twice or even thrice on how to choose the major/leading unknown. 2. Express other data via this unknown. 3. Make an equation. 4. Simplify and then solve the equation correctly. </pre> <H3>Problem 10</H3>The local theater has three types of seats for Broadway plays: main floor, balcony, and mezzanine. Main floor tickets are $59, balcony tickets are $49, and mezzanine tickets are $34. One particular night, sales totaled $106,306. There were 254 more main floor tickets sold than balcony and mezzanine tickets combined. The number of balcony tickets sold is 444 more than 3 times the number of mezzanine tickets sold. How many of each type of ticket were sold? <B>Solution</B> <pre> Let x be the number of the mezzanine tickets. Then, according to the condition, the number of the balcony tickets is (3x + 444), and the number of the main floor tickets is the sum (x + (3x + 444)) + 254 = 4x + 698. Next, you write the total money equation 59*(4x + 698) + 49*(3x + 444) + 34*x = 106306 dollars. Now you simply this equation and find x (59*4*x + 49*3*3x + 34x) + (59*698 + 49*444) = 106306 417x + 62938 = 106306 417x = 106306 - 62938 = 43368 x = 43368/142 = 104. So, 104 mezzanine tickets were sold; 3*104 + 444 = 756 balcony tickets and 4*104 + 698 = 114 main floor tickets. It is the <U>ANSWER</U>, so the problem is </pre> <H3>Problem 11</H3>Andrea sells flower pot of different sizes. The price of a small flower pot is $10, medium flower pot $15 and large flower pot $40. Every months, the number of small flower pots sold is equal to the number of the medium and large flower pots sold. The number of medium flower pots sold is twice the number of the large flower pots sold. Andrea needs to pay a rent of $300 per month for her business premise. What are the minimum numbers of pots of each size which Andrea has to sell in a month so that she can pay the monthly rent? <B>Solution</B> <pre> Let x be the number of the LARGE flower pots. Then the number of the medium pots is 2x, according to the condition, and the number of the small spots is (2x + x) = 3x, again, in accordance with the condition. The total money equation is THIS 10*(3x) + 15*(2x) + 40x = 300 monetary units. Simplify and find x 30x + 30x + 40x = 300, 100x = 300 x = 300/100 = 3. <U>ANSWER</U>. 3 large pots; 2*3 = 6 medium pots and 3*3 = 9 small pots. </pre> On solving single linear equations and relevant word problems see the lessons - <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/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.
