# Systems of Equations in Data Sufficiency:

If the GMAT Quantitative section were 3 hours, all of our lives would be easier. We could take our time, work through each solution, check our choice, and sharpen our pencil before each question. Since we are not afforded this luxury, we must take back every second we can. Data Sufficiency questions are a great place to start, since most students take too much time *solving* the problem instead of assessing whether it’s *possible* to solve the problem. Here we’ll look at common DS time-saving techniques through the lens of systems of linear equations.

**2 Linear Equations, 2 Variables, 1 Solution**

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We learned back in 9^{th} Grade that if you have 2 linear equations with 2 variables, there will always be one unique solution. For Data Sufficiency questions, we can seek out these equations in the stimulus and statements and (without calculating) know if we have enough information. For example,

If the mileage on car X is currently exactly 3 times the mileage on car Y, what is the mileage on car X?

(1) When each car had 6,000 miles less than they have now, car X had 4.5 times the mileage that car Y had.

(2) When each car has 2,000 miles more than they have now, car X will have 2.75 times the mileage that car Y will have.

Stimulus:

x = 3y

Statement 1:

(x – 6000) = 4.5(y – 6000)

Statement 2:

(x + 2000) = 2.75(y + 2000)

Visually, this is a lot easier than dealing with cars and mileage. Again, we do not need to calculate. Since an equation is GIVEN in the stimulus, we just need one more to find the unique solution for x. Each statement is sufficient. Choice (D).

**Caveat #1: Same Equation Disguised**

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A dry cleaning store charges a certain price to clean either shirts or pants and a certain price to clean jackets. Robert, Scott, and Ted each had some clothing items cleaned at this store. How much did Scott pay to have 1 shirt and 1 jacket dry cleaned?

(1) Robert paid $15 to have 3 shirts, 1 pair of pants, and 2 jackets dry cleaned.

(2) Ted paid $30 to have 5 shirts, 3 pairs of pants, and 4 jackets dry cleaned.

The stimulus has no equation, but does show that shirts and pants costs the same, and in turn should be considered the same variable when extracting our equations.

Statement 1:

15 = 4s + 2j (s is the combined number of shirts and pants)

Statement 2:

30 = 8s + 4j

Here, we have two equations and two unknowns, so the answer should be (C), right? What’s interesting about these two equations is that they are *exactly the same!* Divide (2) by 2 and you get (1). Graphically, this is a line on top of itself, with an infinite number of solutions. Beware of caveat #1—one equation disguised as many. Choice (E) is correct.

**Caveat #2: Non-Linear Equations**

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What is the value of x?

(1) 3y – 8 = x

(2) x(x+1) = 2y

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Two equations and two unknowns. No problem, right? Well, first, let’s think of this graphically. Statement 1 is a simple line with a slope of 1/3 and a y-intercept of +8. Statement 2, however, is a parabola, not a line. Because the coefficient of the term is positive, the parabola will open upwards, and intersect the line from Statement 1 in two places. There is *not* a unique solution, because x is squared in Statement 2. However, depending on the restrictions in the stimulus (say, if x>0) can still assist in finding a unique situation. So beware of exponential equations, but also note any restrictions that may limit your options. (In fact, if this stimulus included x>0, then there *is* a unique solution for this question.)

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**Set up Equation Quickly and Move On, If At All**

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In many instances, the actual numbers do not matter for Quantitative DS questions. In fact, the more “wacky” the number looks to you, the more likely you won’t need to “use” it. For the question below, you may note $5,750 and think to yourself, “I’ve never done any calculation with that number, what the hell!” And you’re right. The more important piece of information is that there *is *a number, and for DS, sometimes that’s all you need.

For a certain company X, the average daily payroll for each 30-day payroll cycle is the average (arithmetic mean) of the daily payroll totals for each of the 30 days. During the first part of a recent 30-day payroll cycle, the daily payroll was a constant $5,750. When a new employee was hired during this 30-day cycle, the total payroll for each day rose by $280. If the new daily payroll total remained constant for the remainder of the cycle, what was the average daily payroll for the 30-day cycle?

(1) The new employee was hired on the 11th day of the payroll cycle.

(2) The average daily payroll was $5,890 through the first 20 days of the cycle.

To know any weighted average, we simply need the number of observations at each rate. Statement 1 tells us that we will be at one rate ($5,750) for 10 days, and another rate for the remaining 20 days ($5,750 + $280). Statement 2 tells us that for x days we are at a $5,750 rate, and then for (20 – x) days at that rate rose by $280, and provides the weighted average. We can solve for x in each, since each has one linear variable in an equation, and we have all the other information we need.

Spend time to set up equations only if there is confusion as to what information is missing, or as to what is needed. Otherwise, don’t waste your time on DS questions finding a concrete solution. Save that for the Problem Solving questions.

Any other time-saving tips? That’s what the comments field is for!

## 7 comments

Kartik Sharma on October 19th, 2009 at 6:55 am

Hi Jake,

In Caveat 2, i couldnt figure out a unique solution when x>0 as you mentioned. Can you please explain that?

Jake Becker on October 19th, 2009 at 9:55 am

Kartik,

If we were to take the time to solve for the exact solutions ( + or -), we'd plug in y in terms of x from one equation to the other. You would then be left with a quadratic equation that can be reorganized to read the following:

3x^2 + 3x - 16 = 0

While not easily factored, you will notice that there must be one positive factor and one negative factor.

Hesham on October 20th, 2009 at 11:36 pm

In Caveat 2, shouldn't y-intercept be 8/3 ?

Jake Becker on October 21st, 2009 at 4:18 pm

Yes, my apologies for the typo. Y-intercept should be 8/3. Point remains the same, though.

Hesham on October 21st, 2009 at 12:25 am

could you write the equations for:

(2) The average daily payroll was $5,890 through the first 20 days of the cycle.

How do i get the mean for thirty days?

Jeff on October 21st, 2009 at 4:03 pm

In your last example with the payroll situation, I do not understand your explanation for statement 2.

Can you please either clarify this statement

"Statement 2 tells us that for x days we are at a $5,750 rate, and then for (20 – x) days at that rate rose by $280, and provides the weighted average. We can solve for x in each, since each has one linear variable in an equation, and we have all the other information we need."

or just show the calculations from that statement (even though I understand the point of your article was to show that calculations are unnecessary in this situation) so I can try to understand what you were saying.

Thanks.

Jake Becker on October 21st, 2009 at 4:28 pm

Jeff and Hesham,

Here's the equation we'd set up for Statement 2, in an easier way that how it was described in the article:

Average daily payroll for the month =

[(20*5890) + (10*6170)] / 30

This is the standard weighted average formula.

20 days @ 5890 + 10 days @ (5890 + 280) = 30 days @ 179500.

The total payroll / total days = total average.