Lesson Rate of work problems
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<H2>Rate of work problems</H2> In this lesson you will find typical introductory rate of work problems and their solutions. <H3>Problem 1</H3>16 workers are able to paint 12 cupboards in 6 days. How many days would it take for 12 workers to paint 45 cupboards? <B>Solution</B> From the given data, the rate of work for 1 worker per 1 day is {{{12/(16*6)}}} = {{{12/96}}} = {{{1/8}}} {{{cupboards/(worker*day)}}}. In other words, 1 worker paints {{{1/8}}} of a cupboard per day. (We assume that all workers have the same rate of work and that all cupboards are identical). If 12 workers paint 45 cupboards in <B>x</B> days, then their rate of work is {{{45/(12*x)}}} {{{cupboards/(worker*day)}}}. It the same rate of work as calculated above, i.e. {{{1/8}}} {{{cupboards/(worker*day)}}}. Thus you have an equation {{{45/(12*x)}}} = {{{1/8}}}. Hence, x = {{{(45*8)/12}}} = {{{30}}} days. <B>Answer</B>. {{{30}}} days. <H3>Problem 2</H3>It takes 40 minutes for 4 hoses to fill a 6000 gallon tank. How long would it take for 5 hoses to fill a 3000 gallon tank? <B>Solution</B> In the first case the rate of the hose is {{{6000/(4*40)}}}. In the second case, when 5 hoses are used, the rate is {{{3000/(5*t)}}}, where t is the unknown time. The hose rate is the same in both cases. It gives an equation {{{6000/(4*40)}}} = {{{3000/(5*t)}}}. Hence, t = {{{3000/6000}}}.{{{(4*40)/5}}} = {{{(1/2)*4*8}}} = 16 minutes. <B>Answer</B>. 16 minutes. <H3>Problem 3</H3>Two men dig a trench 30 m long in 3 days. How long would it take 3 men to dig a trench 15 m long working at the same rate? <B>Solution</B> {{{30/(2*3)}}} = {{{15/(3*x)}}}. This proportion equalize the rate of work in these two cases. x is unknown number of days for 3 men to dig a trench 15 m long. Solve the proportion for x. The answer is: one day. You may get the same answer by noticing that the rate of 1 man is 5 m of the trench per day. It follows the first part of the condition. The way of equalizing the rates is the universal and most direct method to solve problems like this one. <B>Answer</B>. One day. <H3>Problem 4</H3>Suppose that the amount of time it takes to build a highway varies directly with the length of the highway and inversely with the number of workers. Suppose also that it takes 200 workers 3 weeks to build 2 miles of highway. How many workers would be needed to build 5 miles of highway in 6 weeks? <B>Solution</B> {{{2/(200*3)}}} = {{{5/(x*6)}}}. This proportion equalize the rate of work in these two cases. x is unknown number of workers to build 5 miles of highway in 6 weeks. Solve the proportion for x. The answer is: x = {{{(200*3*5)/(2*6)}}} = 250 workers. The way of equalizing the rates is the universal and most direct method to solve problems like this one. <B>Answer</B>. 250 workers. <H3>Problem 5</H3>Eight men can excavate 15 cubic meters of drainage canal in 7 hours. Three men can back-fill 10 cubic meters in 4 hours. How long will it take for the 10 men to excavate and back-fill 20 cubic meters in the project? <B>Solution</B> <pre> Rate of work excavating is e = {{{15/(8*7)}}} cubic meters per hour for 1 man. Rate of work back-filling is b = {{{10/(3*4)}}} cubic meters per hour for 1 man. It will take E = {{{20/(10*e)}}} hours for 10 men to excavate 20 m^3. It will take B = {{{20/(10*b)}}} hours for 10 men to back-fill 20 m^3. In all, it will take E + B = {{{20/(10*e) + 20/(10*b)}}} = {{{2/((15/(8*7))) + 2/((10/(3*4)))}}} = {{{(2*8*7)/15 + (2*3*4)/10}}} = {{{112/10 + 24/15}}} hours for 10 men excavate and back-fill 20 cubic meters in the project. To complete the task elegantly, notice that {{{1/10}}} of an hour is 6 minutes, while {{{1/15}}} of an hour is 4 minutes. Therefore, your answer is 112*6 + 24*4 minutes = 768 minutes = 12 hours and 48 minutes. </pre> <H3>Problem 6</H3>A fort has enough provisions to fed everyone in it for 90 days. After 20 days, 600 more soldiers arrive as reinforcements, and the food only lasts 50 days longer. How many people were in the fort originally? <B>Solution</B> <pre> Initially, there was amount food enough to feed n people for 90 days, i.e. 90*n portions. After 20 days, this amound decreased to (90-20)*n = 70n portion. With 600 more soldiers, there was amount of food enough for 50 days, i.e. (n+600)*50. It gives an equation 70*n = (n+600)*50. 70n = 50n + 3000 20n = 3000 n = {{{3000/20}}} = 1500. <U>Answer</U>. There were 1500 people in the fort initially. </pre> <H3>Problem 7</H3>A contractor agrees to lay a road 3000 m long in 30 days. 50 men are employed and they work for 8 hours per day. After 20 working days, he finds that only 1200 m of the road is completed. How many more men does he need to employ in order to finish the project on time if each man now works 10 hours a day? <B>Solution</B> <pre> 50 men working 8 hours per day 20 working days, spent 50*8*20 = 8000 men-hours. Let n be the number of men to employ additionally. Then the team will count (50+n) workers, that are to complete the remaining 1800 m working 10 hours per day during remaining 10 days. The equation for the constant rate of work is {{{1200/8000}}} = {{{1800/((50+n)*10*10)}}}. From this proportion, 50+n = {{{(1800*8000)/(1200*10*10)}}} = 120. Hence, 120 - 50 = 70 workers should be employed additionally to complete the work under given conditions. </pre> <H3>Problem 8</H3>6 wolves can eat 6 sheep in 6 days. How many wolves does it take to eat 60 sheep in 60 days? <B>Solution</B> <pre> Let x be the number of wolves under the question. The rate of eating, in the first scenario, is {{{6_sheep/(6_Wolves*6_days)}}} = {{{1/6}}} of a sheep per single wolf per day. The rate of eating, in the second scenario, is {{{60_sheep/(x_wolves*60_days)}}} = {{{1/x}}} of a sheep per single wolf per day. We assume that the rate is the same in both case, which gives us an equation {{{1/x}}} = {{{1/6}}}. From this equation, you get the <U>ANSWER</U>. 6 wolves. </pre> My other lessons on joint work problems in this site are - <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/Joint-work-word-problems-for-3-4-5-participants.lesson>Joint-work problems for 3 participants</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.
