First of all, consider these three GMAT like questions. Try them on your own before reading the post. I will show solutions to these practice questions at the end of this post.
1) Running at the same rate, 8 identical machines can produce 560 paperclips a minute. At this rate, how many paperclips could 20 machines produce in 6 minutes?

(A) 1344
(B) 3360
(C) 8400
(D) 50400
(E) 67200
2) Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks. If they both work together, how many weeks will it take for them to produce 15 handcrafted drums?

(A) 30
(B) 36
(C) 70
(D) 80
(E) 150
3) Machines P and Q are two different machines that cover jars in a factory. When Machine P works alone, it covers 1500 jars in m hours. When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours. In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars?

(A) m/(m + n)
(B) n/(m + n)
(C) mn/(m + n)
(D) mn/(m – n)
(E) mn/(n – m)
OK, if these problems frustrate the bejeebers out of you, this is the post for you!
Big Idea #1: The “ART” Equation
You may be familiar with the distance equation, D = RT (“distance equals rate times time”), sometimes remembered as the “dirt” equation. It turns out, that equation is just a specific instance of a much more general equation. In that equation, R, the rate, is distance per time, but in nondistance problems, rate can be anything over time — wrenches produced per hour, houses painted per day, books written per decade, etc. In these cases, typical of work problems, we are no longer concerned with “distance” per time, but with the amount of something produced per time. We use A to represent this amount (the number of wrenches, the number of houses, etc.), and the equation becomes A = RT. Sometimes folks remember this as the “art” equation.
Here’s a simple mnemonic. When you travel, you are moving on the Earth, which is made of dirt, so for traveling & distance you use D = RT. Work problems involve machines, and machines make things —– making is creation, and creation is the essence of art, so use the A = RT equation. (I know, I know, what comes out of most machines is hardly worthy of aesthetic elevation, but it works for a mnemonic!)
Big Idea #2: Rates are Ratios
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. The point is: a rate is a ratio, that is to say, a fraction. Technically, any fraction, any ratio, in which the numerator and the denominator have different units is a rate. Fuel efficiency (mpg) and price per unit and most baseball fractions (ERA, BA, OBP, SLG, etc.) are rates. Currency rates and exchanges rates are common financial market rates that, ironically, almost never appear on the GMAT — go figure! Most GMAT rates have time in the denominator, and it’s a rate of how fast work is being done or how fast something is being produced or accomplished.
The fact that rates are ratios means: we can solve these problems by setting up proportions and using proportional thinking! As you will see in the solutions below, that’s an extremely powerful strategy for solution.
Big Idea #3: Add Rates
The vast majority of work problems on the GMAT involve two people or two machines and comparisons of their individual production to their combined production. Questions #2 and #3 above are of this form. The questions will often give you information about times and about amounts, and what you need to know is: you can’t add or subtract times to complete a job and you can’t add or subtract amounts of work; instead, you add and subtract rates.
(rate of A alone) + (rate of B alone) = (combined rate of A & B)
Here A and B can be two people, two machines, etc. 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. At this point, you may want to go back and give another attempt at those three practice questions. Follow carefully how they are applied in the solutions below.
Practice Problem Solutions
1) “Running at the same rate, 8 identical machines can produce 560 paperclips a minute.” That 560 is a combined rate of 8 machines — 560 = 8*R, so the rate of one machine is R = 560/8 = 70 paperclips per minute.
“At this rate, how many paperclips could 20 machines produce in 6 minutes?” Well, the combined rate of 20 machines would be Rtotal = 20*70 = 1400 pc/min. Now, plug that into the “art” equation: A = RT = (1400)*(6) = 8400 pc. Answer = C.
2) Method I: the rates solution
“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. Zane’s rate is (1 drum)/(6 weeks) = 1/6. The combined rate of Jane + Zane is
R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12
That’s the combined rate. We need to make 15 drums — we have a rate and we have an amount, so use the “art” equation to solve for time:
T = A/R = 15/(5/12) = 15*(12/5) = (15/5)*12 = 3*12 = 36
BTW, notice in the penultimate step, the universal fraction strategy: cancel before you multiply (Tip #3: http://magoosh.com/gmat/2012/caniuseacalculatoronthegmat/. Jane and Zane need 36 weeks to make 15 drums. Answer = B.
Method II: the proportion solution
“Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks.” Let’s find the LCM of 4 and 6 — that’s 12 weeks. In a 12 week period, Jane, making a drum every 4 weeks, makes three drums. In a 12 week period, Zane, making a drum every 6 weeks, makes two drums. Therefore, in a 12 weeks period, they produce 5 drums between the two of them. If they make 5 drums in 12 weeks, they need triple that time, 36 weeks, to make 15 drums. Therefore, Jane and Zane need 36 weeks to make 15 drums. Answer = B.
3) This is a particularly challenging, one because we have variables in the answer choices. I will show an algebraic solution, although a numerical solution (http://magoosh.com/gmat/2012/variablesingmatanswerchoicesalgebraicapproachvsnumericalapproach/) is always possible.
“Machines P and Q are two different machines that cover jars in a factory. When Machine P works alone, it covers 1500 jars in m hours. When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours. In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars? ”
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.
P’s individual rate is (1 lot)/(m hours) = 1/m. The combined rate of P & Q is (1 lot)/(n hours) = 1/n. 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)
We now know Q’s rate, and we want the amount of 1 lot, so we use the “art” equation.
1 = [(m – n)/ (nm)]*T
T = (mn)/(m – n)
Answer = D
Here’s another practice question:
http://gmat.magoosh.com/questions/47
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