Lesson HOW TO algebraize and solve these problems using one equation in one unknown
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<H2>HOW TO algebraize and solve these problems using one equation in one unknown</H2> <H3>Problem 1</H3>85 tickets were sold for $2105. The number of reserved tickets sold was one more than twice as many as the number of premium tickets sold. If premium seats cost $39, reserved seats cost $29 and general admission seats cost $19, how many tickets of each kind were sold? <B>Solution</B> <pre> Let P be the number of premium tickets. Then the number of reserved tickets is (2P+1). Then the number of general admission seats is (85 - P - (2P+1)) = (84-3P). The "money" equation is 39*P + 29*(2P+1) + 19*(84-3P) = 2105 39*P + 58*P - 57*P + 29 + 19*84 = 2105 40*P = 2105 - 29 - 19*84 = 480 ====> P = {{{480/40}}} = 12 premium tickets. ====> # of reserved tickets = 2P+1 = 25. # of the general admission seats = 84 - 3*12 = 24. <U>Answer</U>. 12 premium, 25 reserved and 48 general admission. <U>Check</U>. 12 + 25 + 48 = 85; 12*39 + 25*29 + 48*19 = 2105 dollars. ! Correct ! </pre> ============== <U>The lesson to learn from this solution</U>: <pre> This problem is for ONE unknown. It is not for using the systems of equations. Your task is to select the basic unknown by reasonable/rational way, then express other unknowns via that basic value and to built/to construct the equation to solve. </pre> <H3>Problem 2</H3>At a college play production, 360 tickets were sold. The ticket prices were $8, $10, and $12 and the total income from ticket sales was 3436. How many tickets of each type were wold if the number of $8 tickets sold was twice the number of $12 tickets sold? <B>Solution</B> It is a typical problem to solve using a single unknown. <pre> Let x = # of $12 tickets. Then the number of $8 tickets is 2x, according to the condition. Then the number of $10 tickets is (360 - x - 2x) = (360-3x). Now your "money" equation is 8*(2x) + 10*(360-3x) + 12*x = 3436, or 16x + 3600 - 30x + 12x = 3436 -2x = 3436 - 3600 = - 164 ====> x = 82. <U>Answer</U>. 82 $12 tickets; 2*82 = 164 $8 tickets; and (360-3*82) = 114 $10 tickets. <U>Check</U>. 12*82 + 8*164 + 10*114 = 3436 dollars. ! Correct ! </pre> <H3>Problem 3</H3>A farmer buys 100 animals for $100 total. The animals include at least 1 cow, 1 pig and 1 chicken. A cow costs $10, a pig costs $3 and a chicken costs $0.50. If the farmer buys 5 times as many cows as pigs, how many of each animal does he buy? <B>Solution</B> <pre> Let x be the number of pigs. Then the number of cows is 5x and the number of chicken is 100-x-5x = 100-6x. Then the money equation is 10*(5x) + 3*x + 0.50*(100-6x) = 100 dollars. Simplify and solve 50x + 3x + 50 - 3x = 100 50x = 50 x = 50/50 = 1 is the number of pigs. <U>ANSWER</U>. 1 pig, 5 cows and 100-1-5 = 94 chicken. </pre> <H3>Problem 4</H3>You're standing in a line-up at the movie theatre. If 40% of the people in line are in front of you and 58% of the people in the line are behind you, find the total number of people in the queue. <B>Solution</B> <pre> Let x be the number of people in the line. From the given part, we have this equation 0.4x + 1 + 0.58x = x. Simplify and find x 1 = x - 0.4x - 0.58x 1 = 0.02x x = 1/0.02 = 50 persons in the line. <U>ANSWER</U> </pre> <H3>Problem 5</H3>Ben set off his journey from Grand City to Central City at an average speed of 67 km/h. Jerry left Central City for Grand City at the same time. Jerry traveled along the same road as Ben at a speed which is 20 km/h faster than Ben. If the two cities were 770 km apart, how far away from Grand City was Ben when he passed Jerry along the way ? <B>Solution</B> <pre> Let x = "how far away from Grand City was Ben when he passed Jerry along the way". Then you have this "time equation" {{{x/67}}} = {{{(770-x)/(67+20)}}}, or {{{x/67}}} = {{{(770-x)/87}}}. The equation (1) says that the time which Ben spent driving before the meeting was the same as the time that Jerry spent. Solve it for x: 87x = 67*(770-x) 87x = 67*770 - 67x ====> 87x + 67x = 51590 ====> 154x = 51590 ====> x = {{{51590/154}}} = 335. <U>Answer</U>. Ben passed Jerrry at 335 miles from Grand City. </pre> <H3>Problem 6</H3>Your teacher gives a test with 2- and 4-point questions for a total of 100 points. The number of 4 points questions is one-third the number of 2-point questions. How many 2- and 4-point questions are on the test? <B>Solution</B> <pre> Let x be the number of the 4-point questions. then the number of the 2-point questions is 3x. Total point equation is 100 = 2*(3x) + 4x Simplify and solve 100 = 6x + 4x 100 = 10x x = 100/10 = 10. <U>ANSWER</U>. 10 4-point questions and 3*10 = 30 2-point questions. </pre> <H3>Problem 7</H3>In a class of 40 the average score was 70.25. The average scores for boys and girls were 68 and 73 respectively. How many boys were in the class ? <B>Solution</B>. <pre> Let x be the number of boys; then the number of girls is (40-x). Total scores of boys is 68x; total scores of girls is 73*(40-x). Total scores of the class is 40*70.25 = 2810. The equation of total scores is 68x + 73*(40-x) = 2810. From the equation x = {{{(2810-73*40)/(68-73)}}} = 22. <U>ANSWER</U>. There are 22 boys in the class. </pre> <H3>Problem 8</H3>Bob prepared a variety of cookies. 2/3 of the cookies were peanut butter. 2/3 of the remaining cookies were sugar. There were 5 triple fudge cookies. How many cookies did Bob prepare altogether? <B>Solution</B> <pre> Let x be the total number of cookies. Then the number of the peanut butter cookies is {{{(2/3)x}}}; the number of the peanut butter cookies is {{{(2/3)*(1/3)x}}}, or {{{(2/9)x}}} and the number of the triple fudge cookies is 5. An equation for the total number of cookies is {{{(2/3)x}}} + {{{(2/9)x}}} + 5 = x. To solve, multiply all the terms by 9 to ride off the denominators 6x + 2x + 45 = 9x 45 = 9x - 6x - 2x 45 = x x = 45. <U>ANSWER</U>. 45 cookies, in total. </pre> <H3>Problem 9</H3>A man left one-fourth of his estate to his wife, one-fifth to each of his two sons, one-eight to his daughter and the remainder, an amount of $3,600 to charity. Find the amount of estate. <B>Solution</B> <pre> Let x be the amount of estate. Then you write equation as you read the text {{{x/4}}} + {{{2*(x/5)}}} + {{{x/8}}} + 3600 = x. To solve it, multiply all the terms by 40 to ride off the denominators. 10x + 16x + 5x + 3600*40 = 40x 144000 = 40x - 10x - 16x - 5x 144000 = 9x x = 144000/9 = 16000. <U>ANSWER</U>. The amount of estate was 16000 dollars. </pre> <H3>Problem 10</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 "money" 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> 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/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/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.
