Welcome back to our series on GMAT quant rates problems. This article will build upon the last and prepare you to easily solve combined work problems. To begin, let’s review the standard formula we learned last time:

(1 / *c*) = (1 / *a*) + (1 / *b*)

**where c is the combined time for the machines, working together, to complete the job, and a and b are the independent times for each machine, working alone, to complete the job.**

We saw in the last article that this formula typically requires finding a least common denominator in order to add the fractions (1 / *a*) and (1 / *b*), once the *a* and *b* values are plugged in. We can improve the formula by applying this “least common denominator” function to the equation’s general form, like this:

(1 / *c*) = (1 / *a*)(*b* / *b*) + (1 / *b*)(*a */* a*)

The product *ab* is not necessarily the least common denominator for the fractions. If the values *a* and *b* share any prime factors, then their least common denominator is actually their product *ab* divided by the product of their shared primes (for more on this, see the article on least common multiples in our number properties series). This is okay, because we don’t have to have a *least* common denominator in order to add fractions. We just need *a* common denominator, and taking the product of the denominators guarantees this. In order to achieve this common denominator while preserving the values of the fractions, we will multiply (1 / *a*) by (*b* / *b*) and (1 / *b*) by (*a* / *a*). This works because (*b* / *b*) and (*a* / *a*) are just clever versions of the number 1.

(1 / *c*) = (1 / *a*)(*b* / *b*) + (1 / *b*)(*a */* a*)

(1 / *c*) = (*b* / *ab*) + (*a* / *ab*)

Now that we have the common denominator of *ab*, we can add the fractions:

(1 / *c*) = (*a* + *b*) / *ab*

And we can “flip it” to solve for *c*:

*c* = (*a* * *b*) / (*a* + *b*)

This formula represents an equivalent but improved version of the one we started with: (1 / *c*) = (1 / *a*) + (1 / *b*). For any combined work problem, you can choose which version of the formula to work with. Generally the “new and improved” version is easier, because it has the common denominator step “baked in.” To verbalize this improved version: the combined time *c* for two machines to complete a job is the product of their independent times *a* and *b* divided by the sum of these times. Sometimes I call this the “*product-sum quotient*” of the independent times *a* and *b*.

Let’s try this out on an official combined work problem:

**If it would take one machine 10 minutes to fill a large production order and another machine 12 minutes to fill the same order, how many minutes would it take both machines working together, at their respective rates, to fill the order?**

**(A) 4 1/60**

**(B) 5**

**(C) 5 5/11**

**(D) 5 ½**

**(E) 11**

For review, and for the sake of comparison, let’s do this first with the “original” version of the formula.

(1 / *c*) = (1 / *a*) + (1 / *b*)

(1 / *c*) = (1 / 10) + (1 / 12)

(1 / *c*) = (6 / 60) + (5 / 60)

(1 / *c*) = (11 / 60)

*c* = 60 / 11 = 5 5/11 **(answer C)**

You may also have used the common denominator of 120 (by simply multiplying 10 and 20) instead of finding the least common denominator of 60.

Here’s how this problem works with the new version of the formula:

*c* = (*a* * *b*) / (*a* + *b*)

*c* = (10 * 12) / (10 + 12)

*c* = 120 / 22 = 60 / 11 = 5 5/11 **(answer C)**

Either way, you have to do a bit of division at the end. But I usually prefer this new version of the formula to the version with all those reciprocals.

The good news is that there is an even easier way to solve these problems!** It involves distinguishing between the “fast machine” and the “slow machine” in a combined work problem.** So let’s talk about lawyers and lumberjacks.

Imagine that three teams of two people are each given the job of chopping down a large spruce tree. Team 1 is two lumberjacks. Team 2 is two lawyers. And team 3 is one lumberjack and one lawyer. Each team is assigned a tree with a similar trunk circumference and density. What will be the order in which the teams finish the job of chopping down their assigned tree?

No one ever has trouble answering this question. Team 1, the two lumberjacks, will finish first. When their tree falls to the ground, the two lawyers on Team 2 will still be arguing about whether or not it’s legal to chop down their tree. What about Team 3, the lumberjack and the lawyer? The lumberjack on the team will do most of the work, and the team will finish after Team 1 but still before Team 2.

Here’s the point of my silly analogy: *one fast machine and one slow machine, working together, are slower than two fast machines and faster than two slow machines.*

Since combined work problems involve one fast machine and one slow machine working together, we can apply this principle to solve these problems *very quickly*. Let’s look at the same problem again:

**If it would take one machine 10 minutes to fill a large production order and another machine 12 minutes to fill the same order, how many minutes would it take both machines working together, at their respective rates, to fill the order?**

**(A) 4 1/60**

**(B) 5**

**(C) 5 5/11**

**(D) 5 ½**

**(E) 11**

In this problem, the fast machine takes 10 minutes, and the slow machine takes 12 minutes. It follows that two fast machines, working together, take 5 minutes, while two slow machines, working together, take 6 minutes. Therefore the time for one fast machine and one slow machine (the time we were asked for) is somewhere between 5 and 6 minutes. It cannot be exactly 5 or exactly 6. The lumberjack/lawyer team will not tie with the lumberjack/lumberjack team or with the lawyer/lawyer team.

On many combined work problems, this is enough to determine the correct answer. But this problem has the answers C and D of 5 5/11 and 5 ½, respectively. What to do? The key now is to remember that *the time for one fast machine and one slow machine is always closer to the faster time*. In this case, the answer must be closer to 5 than to 6, and **the correct answer is C.** A full explanation of this fact is beyond the scope of this article.

**Here’s what this means for combined work problems:** you can solve them by dividing each of the given independent times by 2 and selecting the answer choice nearer to the lower quotient (closer to the time for two fast machines than to the time for two slow machines). If this fails because there are multiple answer choices in the range, you can always use either version of the standard formula. But this method is so easy and works often enough that it’s always worth trying first.

Let’s try out another official combined work problem:

**Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to ½ of capacity in 3 hours and a second inlet pipe fills the same empty tank to ⅔ of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?**

**(A) 3.25**

**(B) 3.6**

**(C) 4.2**

**(D) 4.4**

**(E) 5.5**

The first step on this problem is to get rid of the “½ of capacity” and “⅔ of capacity” values, which are not very helpful. If the first pump fills the tank to ½ of capacity in 3 hours, then it fills the tank completely in 6 hours. If the second pump fills the tank to ⅔ of capacity in 6 hours, then it fills the tank completely in 9 hours. So 6 hours and 9 hours are the fast and slow times we need for this problem.

How fast can two fast pumps fill the tank? They could do it in 6/2 = 3 hours. How long would it take two slow pumps? It would take 9/2 = 4.5 hours. So our answer must be between 3 and 4.5 hours, but closer to 3 hours. **So somewhere between 3 hours and 3.75 hours.**

**This leaves answer choices A and B.** You might suspect that answer choice A, 3.25 hours, is a little *too* close to the “two fast machines” time of 3 hours and a little *too* far from the “two slow machines” time of 4.5 hours. You’d be correct. But we can always use the formula to make sure.

*c* = (*a* * *b*) / (*a* + *b*)

*a* = 6

*b* = 9

*c* = (6 * 9) / (6 + 9)

*c* = 54 / 15 = 3 9/15 = 3 ⅗ = 3.6

**Yes, B is the correct answer.** The final resolution of the fraction 54 / 15 isn’t even necessary. If you recognize that 15 is not divisible by 4, you can rule out A. The remainder of *x* / 15 can never be 0.25.

Since 6 and 9 have an easy least common multiple, the “original” version of the formula wouldn’t be a bad option here either:

(1 / *c*) = (1 / *a*) + (1 / *b*)

(1 / *c*) = (1 / 6) + (1 / 9)

(1 / *c*) = (3/18) + (2 / 18)

(1 / *c*) = (5 / 18)

*c* = 18 / 5 = 3 ⅗ = 3.6

This brings us full circle in our study of combined work problems. Start by trying the “2 fast machines, 2 slow machines” approach we learned in this article. Then if you have to, use whichever version of the standard formula seems more workable with the time values given in the problem.

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**Contributor: ***Elijah Mize (Apex GMAT Instructor)*