Lesson Joint-work problems for 3 participants
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<H2>Joint-work problems for 3 participants</H2> <H3>Problem 1</H3>A lion can eat a sheep in 2 hours, a wolf can eat a sheep in 3 hours, and a dog can eat the sheep in 6 hours. How long would it take a lion, a wolf, and a dog all together to eat the poor sheep? <B>Solution</B> <pre> Their individual eating rates are {{{1/2}}}, {{{1/3}}} and {{{1/6}}} of a sheep per hour. Their combined eating rate is {{{1/2}}} + {{{1/3}}} + {{{1/6}}} of a sheep per hours, which is {{{1/2}}} + {{{1/3}}} + {{{1/6}}} = {{{3/6}}} + {{{2/6}}} + {{{1/6}}} = {{{6/6}}} = 1 (one sheep per hour). It means that they will complete eating in one hour. <U>ANSWER</U> </pre> <H3>Problem 2</H3> The pool has 3 pipes, A, B and C. Pipes A and B can fill the pool in 10 hours, pipes A and C can fill it in 12 hours, and pipes B and C can fill it in 15 hours. How long will it take to fill the pool if all 3 pipes are turned on? <B>Solution</B> <pre> Let "a" be the pipe's A rate-of-work; "b" be the pipe's B rate-of-work; and "c" be the pipe's C rate-of-work. In other words, "a" is the part of the tank volume filled by the pipe A when it works alone, measured as the ratio {{{the_tank_volume/hour}}}. And similar for "b" and for "c". The condition says that the pipes A and B fill the pool in 14 hours. It means that they fill {{{1/14}}} of the pool volume per hour. It gives an equation a + b = {{{1/10}}}. (1) Similarly, a + c = {{{1/12}}}, (2) b + c = {{{1/15}}}. (3) Add all three equations (1), (2) and (3). Yu will get 2(a + b + c) = {{{1/10 + 1/12 + 1/15}}} = {{{6/60 + 5/60 + 4/60}}} = {{{15/60}}} = {{{1/4}}}, or a + b + c = {{{1/8}}}. Thus the three pipes working together fill {{{1/8}}} of the tank volume per hour. Hence, it will take 8 hour for three pipes to fill the tank. <U>Answer</U>. The three pipes will fill the tank in 8 hours. </pre> <H3>Problem 3</H3> The pool has 3 pipes, A, B and C. Pipes A and B can fill the pool in 10 hours, pipes A and C can fill it in 12 hours, and pipes B and C can fill it in 15 hours. How long will it take for the pipe A to fill the pool working alone? For pipe B? For pipe C? <B>Solution</B> <pre> Notice that the input data is the same as in <B>Problem 1</B> above, only the question is different. Let again, as in the solution to the <B>Problem 1</B>, "a" be the pipe's A rate-of-work; "b" be the pipe's B rate-of-work; and "c" be the pipe's C rate-of-work. Then similarly to the <B>Problem 1</B>, you have equations a + b = {{{1/10}}}, (1) a + c = {{{1/12}}}, (2) b + c = {{{1/15}}}. (3) Add all three equations (1), (2) and (3). Yu will get 2(a + b + c) = {{{1/10 + 1/12 + 1/15}}} = {{{6/60 + 5/60 + 4/60}}} = {{{15/60}}} = {{{1/4}}}, or a + b + c = {{{1/8}}}. (4) Now subtract equation (1) from (4). You will get c = {{{1/8 - 1/10}}} = {{{10/80 - 8/80}}} = {{{2/80}}} = {{{1/40}}}. Thus the rate of work of the pipe C is {{{1/40}}}. In other words, pipe C fills {{{1/40}}} volume of the tank per hour. So, pipe C can fill the tank in 40 hours working alone. Next, subtract equation (2) from (4). You will get b = {{{1/8 - 1/12}}} = {{{3/24 - 2/24}}} = {{{1/24}}}. Thus the rate of work of the pipe B is {{{1/24}}}. It means that pipe B fills {{{1/24}}} volume of the tank per hour. So, pipe B can fill the tank in 24 hours working alone. Finally, subtract equation (3) from (4). You will get a = {{{1/8 - 1/15}}} = {{{15/120 - 8/120}}} = {{{7/120}}}. Thus the rate of work of the pipe A is {{{7/120}}}. It means that pipe A fills {{{7/120}}} volume of the tank per hour. So, pipe A can fill the tank in {{{120/7}}} = {{{17}}}{{{1/7}}} hours working alone. <U>Answer</U>. Pipe A can fill the tank in {{{120/7}}} = {{{17}}}{{{1/7}}} hours working alone. Pipe B can fill the tank in 24 hours. Pipe C can do it in 40 hours. </pre> <H3>Problem 4</H3>Three machines are filling water bottles. The machines can fill the daily quota of water bottles in 90 h, 99 h, and 110 h, respectively. How long would it take to fill the daily quota of water bottles with all three machines working? <B>Solution</B> <pre> The first machine can fill {{{1/90}}} of the daily quota of bottles in one hour (in each hour). It is its rate of work. The second machine can fill {{{1/99}}} of the daily quota of bottles in one hour (in each hour). It is its rate of work. The third machine can fill {{{1/110}}} of the daily quota of bottles in one hour (in each hour). It is its rate of work. If all three machines works simultaneously, they fill {{{1/90 + 1/99 + 1/110}}} = {{{11/990 + 10/990 + 9/990}}} = {{{(11+10+9)/990}}} = {{{30/990}}} = {{{1/33}}} of the daily quota of bottles in one hour (in each hour). Hence, it will take 33 hours for the three machine to fill the daily quota of bottles working simultaneously. <U>Answewr</U>. It will take 33 hours for the three machine to fill the daily quota of bottles working simultaneously. </pre> <H3>Problem 5</H3>Three boys, Alex, Bill and Crist work to complete the same job. Alex and Bill's combined rate-of-work is {{{3/20}}} job per hour; Alex and Crist's combined rate-of-work is {{{2/15}}} of the job per hour; Bill and Crist's combined rate-of-work is {{{7/60}}} of job per hour. Find the rate-of-work of each boy. Find the combined rate-of-work of the three boys working togrther. How long will it take for three boys to complete the job working together? <B>Solution</B> <pre> Let "a", "b" and "c" be the rate-of-work of each of the boys Alex, Bill, and Crist respectively. We are given that a + b = {{{3/20}}}, (1) a + c = {{{2/15}}}, (2) b + c = {{{7/60}}}. (3) To solve the system (1), (2), (3), let us start adding the equations (1), (2) and (3). You will get 2a + 2b + 2c = {{{3/20 + 2/15 + 7/60}}} = {{{9/60 + 8/60 + 7/60}}} = {{{(9+8+7)/60}}} = {{{24/60}}} = {{{2/5}}}. Hence, a + b + c = {{{1/5}}}. Thus we just found the combined rate-of-work of the three boys working together. It is {{{1/15}}} job per hour. It implies that it will take 15 hours for three boys to complete the job working together. Now we have to find individual rate-of-work for each boy. For it, let us first subtract the equation (1) from (4). You will get c = {{{1/5 - 3/20}}} = {{{12/60 - 9/60}}} = {{{3/60}}} = {{{1/20}}}. Next, subtract the equation (2) from (4). You will get b = {{{1/5 - 2/15}}} = {{{3/15 - 2/15}}} = {{{1/15}}}. Finally, subtract the equation (3) from (4). You will get a = {{{1/5 - 7/60}}} = {{{12/60 - 7/60}}} = {{{5/60}}} = {{{1/12}}}. <U>Answer</U>. The individual rates of work are {{{1/12}}} for Alex, {{{1/15}}} for Bill and {{{1/20}}} for Crist (in job-per-hour units). The combined rate-of-work of the tree boys working together is {{{1/5}}} of the job-per-hour. If the three boys work together, it will take 5 hours for them to complete the job. </pre> <H3>Problem 6</H3>Jack and Jill can mow the park together in 10 hours. Jack and Joe can mow the same park together in 15 hours. Jill and Joe can mow the same park together in 18 hours. Determine the number of hours it would take Jill alone to mow the park. <B>Solution</B> <pre> Let "a", "b" and "c" be the rate-of-work of each of the persons Jack, Jill, and Joe, respectively. We are given that a + b = {{{1/10}}}, (1) a + c = {{{1/15}}}, (2) b + c = {{{1/18}}}. (3) To solve the system (1), (2), (3), let us start adding the equations (1), (2) and (3). You will get 2a + 2b + 2c = {{{1/10 + 1/15 + 1/18}}} = {{{9/90 + 6/90 + 5/90}}} = {{{(9+6+5)/90}}} = {{{20/90}}} = {{{2/9}}}. Hence, a + b + c = {{{1/9}}}. Thus we just found the combined rate-of-work of the three persons working together. It is {{{1/9}}} job per hour. Now, whose productivity we must to estimate? Jill's alone? Distract the equation (2) from (4). You will get b = {{{1/9 - 1/15}}} = {{{10/90 - 6/90}}} = {{{4/90}}} = {{{2/45}}}. It means that Jill alone can mow {{{2/45}}} of the area per hour. Hence, it will take {{{45/2}}} = 22.5 hours for Jill to mow the park working alone. <U>Answer</U>. It will take 22.5 hours for Jill to mow the park working alone. </pre> <H3>Problem 7</H3>Each of valves A, B and C when open release water into a tank at it's own constant rate. With all three valves open, the tank fills in 1 hr. With only valves A and C open, it takes 1.5 hr to fill the tank and with only valves B and C open, it takes 2 hr. How long does it take to fill the tank with valves A and B open? <B>Solution</B> <pre> Let "a" be the rate of the valve A in the "{{{tank_volume/hour}}}" unit. Let "b" be the rate of the valve B in the same unit, and Let "c" be the rate of the valve C in the same unit. Then you are given a + b + c = {{{1/1}}} = 1. (1) a + c = {{{1/1.5}}} = {{{2/3}}}, (2) b + c = {{{1/2}}}. (3) We can rewrite this system in an equivalent form 2a + 2b + 2c = 2, (1) a + c = {{{2/3}}}, (2) b + c = {{{1/2}}}. (3) Now distract both equations (2) and (3) from (1) (both sides). You will get a + b = 2 - {{{2/3}}} - {{{1/2}}} = {{{12/6 - 4/6 - 3/6}}} = {{{5/6}}}. The last equality says that the combined rate of the valves A and B is {{{5/6}}} of the tank volume per hour. Hence, the valves A and B will fill the tank in {{{6/5}}} on an hour, or in 1 hour and 12 minutes working together. </pre> <H3>Problem 8</H3>Jane wants to buy a photocopier. The sales person has the following information on three models. If all three are used, a specific job can be completed in 50 minutes. If copier A operates for 20 minutes and copier B operates for 50 minutes, then one-half the job can be is completed. If copier B operates for 30 minutes and copier C operates for 80 minutes, then three-fifths of the job can be completed. Which is the fastest copier, and how long does it take for this copier to complete the entire job? <B>Solution</B> <pre> Let x = rate of work of the copier A in units {{{job/minute}}}; y = rate of work of the copier B -------- " -----------; z = rate of work of the copier C -------- " -----------. As the condition says, if all three copiers work, the job is done in 50 minutes: 50x + 50y + 50z = 1. If copier A operates for 20 minutes and copier B operates for 50 minutes, then one-half the job can be is completed. It means 20x + 50y = {{{1/2}}}. If copier B operates for 30 minutes and copier C operates for 80 minutes, then three-fifths of the job can be completed. It means 30y + 80z = {{{3/5}}}. So you have three equations for three unknowns: 50x + 50y + 50z = 1, (1) 20x + 50y = {{{1/2}}}, (2) 30y + 80z = {{{3/5}}}. (3) Notice, this system is <U>LINEAR</U>. Now, to solve it, I will do this <U>TRICK</U>. 1. Add equations (2) and (3) (both sides). You will get 20x + 80y + 80z = {{{11/10}}}. (4) ( <--- {{{1/2 + 3/5}}} = {{{11/10}}}). 2. Multiply equation (1) by 8, multiply equation (4) by 5 and write them one under the other, like this: 400x + 400y + 400z = 8, (5) 100x + 400y + 400z = {{{11/2}}}. (6) 3. Distract (6) from (5). You will get 300x = 8 - {{{11/2}}} = {{{(16-11)/2}}} = {{{5/2}}}. Hence, x = {{{5/(300*2)}}} = {{{1/120}}}. It is the rate of work of the copier A. Almost all is done. Now from equation (2) you can easily find y: 20x + 50y = {{{1/2}}} ---> 20x = {{{1/2 - 50/120}}} = {{{60/120 - 50/120}}} = {{{10/120}}} = {{{1/12}}} ---> y = {{{1/240}}} and from equation (3) you can find "z": 30y + 80z = {{{3/5}}} ---> 80z = {{{3/5 - 30/240}}} = {{{3/5 - 1/8}}} = {{{24/40 - 5/40}}} = {{{19/40}}} ---> z = {{{19/(30*80)}}} = {{{19/2400}}}. Thus the rate of work of the three copiers is: {{{1/120}}} for A, {{{1/240}}} for B and {{{19/2400}}} for C. The fastest copier is A. It will take 120 minutes for the copier A to complete the job. The problem is solved. </pre> My other lessons on <B>rate-of-work</B> problems in this site are - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Rate-of-work-problem.lesson>Rate of work problems</A> - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Word-problems-WORKING-TOGETHER-by-Fractions.lesson>Using Fractions to solve word problems on joint work</A> - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Solving-more-complicated-word-problems-on-joint-work.lesson>Solving more complicated word problems on joint work</A> - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Using-quadr-eqns-to-solve-word-problems-on-joint-work.lesson>Using quadratic equations to solve word problems on joint work</A> - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Solving-rate-of-work-problem-by-reducing-to-a-system-of-linear-equations.lesson>Solving rate of work problem by reducing to a system of linear equations</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Solving-joint-work-problems-by-reasoning.lesson>Solving joint work problems by reasoning</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Selected-problems-from-the-archive-on-joint-work-word-problems.lesson>Selected joint-work word problems from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/HOW-TO-algebreze-and-solve-these-joint-work-problems.lesson>HOW TO algebreze and solve these joint work problems ?</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Had-the-number-of-workers-be-more-the-job-would-be-completed-sooner.lesson>Had there were more workers, the job would be completed sooner</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/One-unusual-joint-work-problem.lesson>One unusual joint work problem</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Special-joint-wok-problems-that-admit-and-require-an-alternative-solution-method.lesson>Special joint work problems that admit and require an alternative solution method</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Snow-removal-problem.lesson>Snow removal problem</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Entertainment-problems-on-joint-work.lesson>Entertainment problems on joint work</A> - <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Joint-work-word-problem-for-the-day-of-April-first.lesson>Joint work word problems for the day of April, 1</A> - <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/OVERVIEW-of-lessons-on-rate-of-work-problems.lesson>OVERVIEW of lessons on rate-of-work problems</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
