Jane and Zane need 36 weeks to make 15 drums. Translating the word problems in to algebraic expressions. If they both work together, how many weeks will it take for them to produce 15 handcrafted drums? Zane can make a similar handcrafted drum in 6 weeks. OK, if these problems frustrate the bejeebers out of you, this is the post for you! Since the sum of the individual rates of each student equals their rate when working together, you can also use that information to solve for x. 2) Jane can make a handcrafted drum in 4 weeks. If initially the tank was empty and someone started the two pumps together but left the drainage hole open, how long does it take for the tank to be filled? We know, (P’s rate alone) + (Q’s rate alone) = (P and Q’s combined rate), (Q’s rate alone) = (P and Q’s combined rate) – (P’s rate alone), (Q’s rate alone) = 1/n – 1/m = m/ (nm) – n/ (nm) = (m – n)/(nm). Problem 2:It takes 6 hours for pump A, used alone, to fill a tank of water. Basically, it means that the amount of goods produced is equal to the rate at which the goods are produced multiplied by the time spent producing the goods. In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars? The fact that rates are ratios means: we can solve these problems by setting up proportions and using proportional thinking! Notice the difference with problems 1 and 2: I'm setting the variables to be rates of work instead of "time to finish a job". The variable will represent the amount of time that it takes the faster worker to complete the homework: Let x = the time needed for the faster worker to complete the homework. “Machines P and Q are two different machines that cover jars in a factory. For a job that can be completed in a certain amount of time t, the rate of work done per unit of time is . At what time will the swimming pool be filled if pump B is started at 10 am? The extension of this idea is that if you have N identical machines, and each one works at a rate of R, then the combined rate is N*R. With just these three ideas, you can unlock any work problem on the GMAT. GMAT® (plus a listing of any other GMAC® trademarks used on this web site) is a registered trademark of the Graduate Management Admission Council®. When the tank is full and a drainage hole is open, the water is drained in 20 hours. Here A and B can be two people, two machines, etc. Solution to Problem 1: We first calculate the rate of work of John and Linda John: 1 / 1.5 and Linda 1 / 2 Let t be the time for John and Linda to mow the Lawn. First of all, consider these three GMAT like questions. The word “rate” and the word “ratio” have the same Latin root: in fact, they also share a Latin root with the “rationality” of our minds, but that’s a discussion that would bring up to our noses into Pythagorean and Platonic philosophies. I will show solutions to these practice questions at the end of this post. Pump B used alone takes 8 hours to fill the same tank. If they work by themselves, how long will it take each child to complete the homework? Answer = C. “Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks.”  Jane’s rate is (1 drum)/(4 weeks) = 1/4. Since the number “1500 jars” appears over and over, let’s arbitrarily say 1500 jars = 1 lot, and we’ll use units of lots per hour to simplify our calculations. As you will see in the solutions below, that’s an extremely powerful strategy for solution. We will learn how to solve math work problems that involve two persons. ), and the equation becomes A = RT. It turns out, that equation is just a specific instance of a much more general equation. Solution to Problem 4:the rates of the two pumps arepump A: 1 / 3 , pump B: 1 / 6Working together, If pump A works for t hours then pump B works t - 1 hours since it started 1 hour late. We now know Q’s rate, and we want the amount of 1 lot, so we use the “art” equation. Follow carefully how they are applied in the solutions below. Zane’s rate is (1 drum)/(6 weeks) = 1/6. Word problems on unit rate ... L.C.M method to solve time and work problems. In that equation, R, the rate, is distance per time, but in non-distance problems, rate can be anything over time — wrenches produced per hour, houses painted per day, books written per decade, etc.