Lesson Solving more complicated problems by the backward method
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<H2>Solving more complicated problems by the backward method</H2> <H3>Problem 1</H3>The Knave of Hearts stole some tarts. He ate half of them, and half a tart more. The Knave of Diamonds ate half of what was left, and half a tart more. Then the Knave of Clubs ate half of what remained, and half a tart more. This left just one tart for the Knave of Spades. How many tarts did the Knave of Hearts steal? <B>Solution</B> <pre> I will solve the problem from the ending state to the starting step. I will organize my logic, my solution and my calculations using this table below. Name starting consumed ending (step) amount amount amount H (Heart) x 0.5x+0.5 h = x - (0.5x+0.5) = 0.5x-0.5 (1) D (diamond) h 0.5h+0.5 d = h - (0.5h+0.5) = 0.5h-0.5 (2) C (club) d 0.5d+0.5 c = d - (0.5d+0.5) = 0.5d-0.5 (3) S (spade) c = 1 (4) So, I have 4 equations (1), (2), (3) and (4) for 4 unknowns x, h, d and c. Now I will move backward from (4) to (1) to determine the unknowns one after another. The last equation c = 1 (4) in the table gives me the opportunity to find "d" from (3) 1 = 0.5d-0.5 ---> 1 + 0.5 = 0.5d ---> 1.5 = 0.5d ---> d = 1.5/0.5 = 3; d = 3. (5) Now, knowing d= 3, I substitute it into equation (2) and find h 3 = 0.5h-0.5 ---> 3 + 0.5 = 0.5h ---> 3.5 = 0.5h ---> h = 3.5/0.5 = 7; h = 7. (6) Finally, knowing h= 7, I substitute it into equation (1) and find x 7 = 0.5x-0.5 ---> 7 + 0.5 = 0.5x ---> 7.5 = 0.5x ---> x = 7.5/0.5 = 15; x = 15. (7) Thus the problem is just solved, and the <U>ANSWER</U> is +-----------------------------------------+ | The Knave of Hearts stole 15 tarts. | +-----------------------------------------+ </pre> The first solution method is complete. Now I will show you the second method, which is wrapping the entire problem around. <pre> Let the original number of tarts be T. 1st time around: Ate {{{1/2}}} of the tarts, and {{{1/2}}} more gives us a remainder of {{{(1/2)*T - (1/2)}}} = {{{T/2 - 1/2)}}}. 2nd time around: Ate {{{1/2}}} of REMAINING tarts, plus {{{1/2}}} more gives us a remainder of {{{(1/2)*(T/2 - 1/2) - 1/2}}} = {{{T/4 - 1/4 - 1/2}}} = {{{T/4 - 3/4)}}}. 3rd time around: Ate {{{1/2}}} of REMAINING tarts, plus {{{1/2}}} more gives us a remainder of {{{(1/2) * (T/4 - 3/4) - 1/2}}} = {{{T/8 - 3/8 - 1/2}}} = {{{T/8 - 7/8}}} After 3rd time around, 1 tart remained; so we get {{{T/8 - 7/8}}} = 1. Multiplying by 8, we obtain T - 7 = 8. It gives the final answer for the original number of tarts T = 8 + 7 = 15. </pre> <H3>Problem 2</H3>Henry spent 1/4 of his money and an additional $3 on a number of DVDs. He then spent 3/5 of the remaining money and an additional $6 on a number of batteries. Given that he was left with $24, how much money did Henry have at first? <B>Solution</B> It is a nice problem to be solved backward and explained accordingly. <pre> First, let x be the remainder money amount after buying a number of DVDs. Then for this amount of money you have this equation x - {{{((3/5)x + 6)}}} = 24. (1) To solve it, multiply everything in equation (1) by 5 to get 5x - (3x + 30) = 24*5 5x - 3x = 24*5 + 30 2x = 120 + 30 = 150 x = 150/2 = 75. Thus you found that after buying a number of DVDs, the remainder was 75 dollars. OK. It gives you NEXT EQUATION from the first part of the problem M - {{{((1/4)M + 3)}}} = 75 (2) where M is the unknown money amount Henry had at first. From this equation, you get {{{(3/4)M}}} = 75 + 3 = 78, M = {{{(78*4/3)}}} = 26*4 = 104. <U>ANSWER</U>. At first, Henry had $104. </pre> <H3>Problem 3</H3>Gillian and Ben had a total of 4850 toys. Ben gave 3/7 of his toys to Gillian. She then gave 1/3 of her toys to Ben. The ratio of the number of toys Gillian left to the number of toys Ben finally had was 2:3. a) How many toys did Gillian have at the end? b) How many toys did Ben have at first? <B>Solution</B> I will solve the problem using backward solution method. To make the solution totally clear, I will present / (will re-tell) the problem's description in steps. <pre> 1. At the start, the total number of toys was 4850. 2. Ben gave 3/7 of his toys to Gilian. 3. Gilian then gave 1/3 of her toys to Ben. 4. After that, the ratio of Gilian's toys to Ben toys was 2:3. </pre> At this point, the real solution is starting. <pre> Notice that, althought Gilian and Ben exchanged toys, the total number of toys remained 4850, with no change. So, at the final step 4, there were totaly 4850 toys in the ratio Gilian to Ben as 2:3. It means, there were 5 equal parts of toys, each part counted 4850/5 = 970 toys, and Gilian had 2 such parts, or 2*970 = 1940 toys, while Ben had 3 such parts, or 3*970 = 2910 toys. In particular, we just know the <U>ANSWER</U> to question (a): Gilian had 1940 toys at the end. Next, the last action that Gilian did was giving 1/3 of her toys to Ben. HENCE, her final number of toys, 1940, is 2/3 of what she had before this giving. So, we conclude that before giving to Ben, Gilian had (3/2)*1940 = 2910 toys. Repeating it again, immediately before step 3, Gilian had 2910 toys. Hence, immediately before step 3, Ben had the rest 4850 - 2910 = 1940 toys. Thus we analysed the final stage and then restored from the condition, that immediately before step 3, Gilian had 2910 toys, while Ben had 1940 toys. Next, moving back, we know that after step 2, Ben had 1940 toys; Gilian had 2910 toys. In other words, after giving Gilian 3/7 of his toys at the step 2, Ben left with 1940 toys; it is 4/7 what he had before his giving. HENCE, before this giving, Ben had 7/4*1940 = 3395 toys and Gilian had the rest 4850-3395 = 1455 toys. Again, immediately before step 2, Ben had 3395 toys anad Gilian had the rest 1455 toys. It gives the <U>ANSWER</U> to question (a): at first, Ben had 3395 toys. </pre> <H3>Problem 4</H3>Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book C. First, he moved 2/9 of the stamps from Book B to Book C. Then, he moved 3/5 of the remaining stamps from Book B to Book A. Next, he moved 1/3 of his stamps from Book A to Book C. In the end, there were 240 fewer stamps in Book B than in Book A. Book C had 460 more stamps than Book B, and there were 600 stamps left in Book C. How many more stamps were there in Book B than Book A at first? <B>Solution</B> Again, it is a good problem to solve it using the backward method. <pre> So, I will accurately divide the entire process by steps. (1) First, he moved 2/9 of the stamps from Book B to Book C. (2) Then, he moved 3/5 of the remaining stamps from Book B to Book A. (3) Next, he moved 1/3 of his stamps from Book A to Book C. (4) In the end, there were 240 fewer stamps in Book B than in Book A, and Book C had 460 more stamps than Book B and there were 600 stamps left in Book C. First of all, based on description of final state (4), we can easily find the number of stamps in each book A, B and C: {{{c[4]}}} = 600; {{{b[4]}}} = 600 - 460 = 140; {{{a[4]}}} = 140 + 240 = 380. Next, I will introduce arrays of numbers {{{a[1]}}}, {{{b[1]}}}, {{{c[1]}}} ( distribution of stamps immediately before step 1; same as the initial distribution of stamps ) {{{a[2]}}}, {{{b[2]}}}, {{{c[2]}}} ( distribution of stamps immediately before step 2) {{{a[3]}}}, {{{b[3]}}}, {{{c[3]}}} ( distribution of stamps immediately before step 3) {{{a[4]}}}, {{{b[4]}}}, {{{c[4]}}} ( distribution of stamps immediately before step 4) We just found out the values {{{a[4]}}}, {{{b[4]}}}, {{{c[4]}}}: they are {{{a[4]}}} = 380; {{{b[4]}}} = 140; {{{c[4]}}} = 600. We will now move from step (4) to step (3). Marvin's action from step (3) to step (4) was "he moved 1/3 of the stamps from Book A to Book C". So, we can write {{{a[4]}}} = {{{(2/3)*a[3]}}}, {{{b[4]}}} = {{{b[3]}}}, {{{c[4]}}} = {{{c[3]}}} + {{{(1/3)*a[3]}}}. First of these equations gives {{{a[3]}}} = {{{(3/2)*a[4]}}} = {{{(3/2)*380}}} = 570; second gives {{{b[3]}}} = 140; third gives {{{c[3]}}} = 600 - {{{(1/3)*570}}} = 410. Thus we can fill the row 3 : {{{a[3]}}} = 570; {{{b[3]}}} = 140; {{{c[3]}}} = 410. We will now move from step (3) to step (2). Marvin's action from step (2) to step (3) was "he moved 3/5 of the stamps from Book B to Book A." So, we can write {{{a[3]}}} = {{{a[2] + (3/5)*b[2]}}}, {{{b[3]}}} = {{{(2/5)*b[2]}}}, {{{c[3]}}} = {{{c[2]}}}. First of these equations gives 570 = {{{a[2] + (3/5)*b[2]}}}; second gives 140 = {{{(2/5)*b[2]}}}; third gives {{{c[2]}}} = 410. It implies {{{b[2]}}} = {{{(5/2)*140}}} = 350; 570 = {{{a[2] + (3/5)*350}}} = {{{a[2]}}} + 210; hence, {{{a[2]}}} = 570-210 = 360. Thus we can fill the row 2 : {{{a[2]}}} = 360; {{{b[2]}}} = 350; {{{c[2]}}} = 410. We will now move from step (2) to step (1). Marvin's action from step (1) to step (2) was "he moved 2/9 of the stamps from Book B to Book C." So, we can write {{{a[2]}}} = {{{a[1]}}}, {{{b[2]}}} = {{{(7/9)*b[1]}}}, {{{c[2]}}} = {{{c[1]+(2/9)*b[1]}}}. First of these equations gives {{{a[1]}}} = 360; second gives {{{b[1]}}} = {{{(9/7)*350}}} = 450; third gives {{{c[1]}}} = {{{410-(2/9)*450}}} = 310. Thus we can fill the row 1 : {{{a[1]}}} = 360; {{{b[1]}}} = 450; {{{c[1]}}} = 310. So the answer to the problem question is {{{b[1]}}} - {{{a[1]}}} = 450-360 = 90 stamps. <U>ANSWER</U> </pre> <H3>Problem 5</H3>Cindy, Derrick and Edmund had $410 altogether. Cindy gave 1/3 of her money to Edmund, Derrick then gave 2/5 of his money to Cindy. As a result, Derrick had $20 less than Cindy and $30 less than Edmund. How much money did Cindy have at first? <B>Solution</B> Let solve the problem in a backward manner. <pre> Let x be what Derrick had finally. Then Cindy had (x+20) dollars at the end; Edmund had (x+30) dollars at the end. The total is $410; it did not change after these interior changes. So x + (x+20) + (x+30) = 410 3x + 50 = 410 3x = 410 - 50 = 360 x = 360/3 = 120 dollars. Thus at final, Derrick had 120 dollars; Cindy had 120+20 = 140 dollars and Edmund had 120+30 = 150 dollars. Now we will make one step back, from the end to the beginning. As the problem says, Derrick gave 2/5 of his money to Cindy. Hence, Derrick left with only 3/5 of his money, and this 3/5 of his original money is 120 dollars. It means that originally Derrick had {{{(5/3)*120}}} = 5*40 = 200 dollars. HENCE, Derrick gave 2/5 of $200, or 80 dollars to Cindy. The problem says: "Cindy gave 1/3 of her money to Edmund". So, Cindy left with 2/3 of her original money, PLUS 80 dollars she got from Derrick. So, we write this equation for the Cindy original amount "C" {{{(2/3)*C + 80}}} = 140 dollars. We solve this equation easily and get {{{(2/3)*C}}} = 140-80 = 60; hence C = {{{(3/2)*60}}} = 90. <U>ANSWER</U>. Cindy had 90 dollars, at first. </pre> <H3>Problem 6</H3>Stephen and Siew Mei had $540 altogether. Stephen gave 1/7 of his money Siew Mei. In return Siew Mei gave 1/4 of the total amount she had to Stephen. Then they both had the same amount of money. How much did Stephen had at first ? <B>Solution</B> <pre> In course of mutual exchanging, the total amount of dollars does not change: at the end, the total amount is the same $540, as at the beginning. Therefore, at the end, Stephen and Siew Mei have 540/2 = 270 dollars each. Let x be the amount which Stephan had at first. Then Siew Mei had (540-x) dollars at first. When Stephen gave 1/7 of his money to Siew Mei, Stephen left with {{{(6/7)x}}} dollars; Siew Mei had {{{(540-x) + (1/7)x}}}. When Siew Mei gave 1/4 of the total amount she had to Stephen, Stephen had finally {{{(6/7)x}}} + {{{(1/4)*((540-x)+(1/7)x)}}} dollars. This final amounts of Stephen's money is 270 dollars, so we write this equation {{{(3/4)*((540-x)+(1/7)x)}}} = 270. Simplify this equation and find x. First multiply both sides by 4*7 = 28 3*7*((540-x) + (1/7)*x) = 270*28 21*(540-x) + 3x = 270*28 21*540 - 21x + 3x = 270*28 21*540 - 270*28 = 18x 3780 = 18x x = {{{3780/18}}} = 210. <U>ANSWER</U>. Stephen had initially 210 dollars; hence, Siew Mei had initially 540-210 = 330 dollars. <U>CHECK</U>. After first giving, Stephen had 210-30 = 180 dollars; Siew Mei had 330+30 = 360 dollars. After second giving, Stephen had 180 + 90 = 270 dollars; Siew Mei had 360-90 = 270 dollars. ! Equal amounts, correct ! </pre> Thus this solution is completed. The problem can be solved MENTALLY, without using equations. The mental solution uses the backward methodology. <pre> In course of mutual exchanging, the total amount of dollars does not change: at the end, the total amount is the same $540, as at the beginning. Therefore, at the end, Stephen and Siew Mei have 540/2 = 270 dollars each. According to the problem, at the last exchange Siew Mei gave 1/4 of her money to Stephen. Since Siew Mei left after that with 270 dollars, it means that 270 dollars is 3/4 of the amount she had immediately before this transaction; so, she had {{{(4/3)*270}}} = 360 dollars before this transaction. Hence, Siew Mei gave 360-270 = 90 dollars to Stephen at this last transaction. From it, we conclude, that Stephen had 270-90 = 180 dollars before the last transaction. These 180 dollars are {{{6/7}}} of what he had at the beginning; hence, at the beginning Stephen had {{{(7/6)*180)}}} = 210 dollars, and the problem is just SOLVED and completed. </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/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-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.
