This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.20: You Need Both to Reach the Arrow

At long last, we come to a problem in which both charts are required. Specifically, we need to reason our way from Germany’s total tax revenue (mentioned in the pie chart) to Germany’s total retail gasoline sales (mentioned in the bar graph). The bridge between the two is Germany’s gas tax revenue, which appears in both charts.

First, let’s figure out the dollar amount of Germany’s gas tax revenue, using the pie chart. We know that the gas tax accounts for 20.4% of the total tax revenue, and we know that the total tax revenue is $170 billion. Combining those two, we get

Because the answer is expressed in billions, there’s no benefit to writing out the $170 billion longhand.

Sanity check: Before we move on, notice that the gas tax revenue has to be less than the total retail gasoline sales. This means that our answer (Germany’s total retail gasoline sales) has to be greater than $34.68 billion. We can rule out (A), (B), and (C) without calculating the total retail gasoline sales directly.

To finish the calculation, we go to the bar chart, where we find that Germany’s gas tax revenue is 70.7% of its total retail gasoline sales. That fact, combined with our previous calculation of the gas tax revenue in dollars, is enough to give us a figure for total sales:

Of the answers provided, the closest approximation to total sales is $50 billion, making (E) the correct answer choice.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 4.1: Graphical Methods for Describing Data (pp. 62-65)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.19: Percent Comparisons

This problem is best approached in terms of percent change, using the same principles as we’d employ to figure out a percent discount or markup. Following the examples in the Math Review (pp. 11-12), we can write the basic formula for such problems as

We’re instructed to find the “percent less than” the income tax revenue, so the income tax revenue will be our base (the denominator). The numerator will be the difference between the income tax revenue and the gas tax revenue. Putting that all together, we get:

The gasoline tax revenue is 28.7% less than the income tax revenue. Of the answer choices, the closest approximation to this figure is 30% (answer C).

Math Review Reference

For more on this topic, see the following sections of the GRE Math Review:

• 1.7: Percent (pp. 9-12)
• 4.1: Graphical Methods for Describing Data (pp. 64-65)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.18: Gas Tax Facts

Here’s another one of those “triple true-false” questions. Just as in a Reading Comprehension passage, we have to step through each answer choice and check to see if it’s supported by the data. Moreover, just like a Reading Comprehension question, the devil is in the details: we need to read carefully, so as not to be taken in by tricky wording.

First, let’s consider answer (A). At a glance it might seem that this statement is supported by the bar chart, but look again: the bar chart lists gas tax revenue “as a percent of total retail gasoline sales,” but the statement concerns France’s gas tax revenue “as a percent of total tax revenue.” We don’t have any information about the distribution of France’s total tax revenue — only Germany’s. Answer (A) is therefore incorrect.

How about answer (B)? Again, this statement refers to figures that aren’t actually presented in the graphs, which don’t mention the actual “price per gallon” in any of the nine countries. It may be tempting to assume that gas will be more expensive in Norway, since gas taxes are higher there than in Spain. But there are all sorts of other factors — transport costs, for example — that might affect the per-gallon price in each of the two countries. Because we can’t say how those other factors will influence prices, we reject answer (B) for insufficient evidence.

At this point, we know answer (C) is correct, because we’ve ruled out (A) and (B), and at least one answer is correct in every multiple-choice GRE question. To see why (C) is right, look at the pie chart. Germany’s gas tax revenue is given as 20.4% of the total, and the tobacco tax revenue is 5.6%. Because 20.4/5.6 is greater than 3, the gas tax revenue must be more than triple the tobacco tax revenue.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 4.1: Graphical Methods for Describing Data (pp. 62-65)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.17: Median Gas Tax

This is the first problem in the “chart block,” which is like Reading Comprehension for math. The questions in this subsection tend to involve lots of arithmetic and some basic statistics (esp. mean and median). You may also need to employ some simple word-problem algebra, but the emphasis here is on using the charts to find the data requested, then performing simple calculations on that data.

In most cases, the chart block will include two charts or graphs of different kinds. It could be a bar graph and a pie chart (as it is here), or a table and a scatter plot, or some other combination. In any case, an important preliminary step for each question is to figure out which chart(s) are relevant. Easier problems will use data from only one of the charts on the page, but in harder problems, you’ll need to synthesize or compare data from both.

In 5.17, we have it relatively easy: the problem stem itself instructs us to look at the bar graph. There are an odd number of countries represented, so to find the median, all we need to do is arrange the values in order:

{24.1%, 40.0%, 56.9%, 62.0%, 67.6%, 70.0%, 70.7%, 72.7%, 76.8%]

The median, which corresponds to answer choice (A), is underlined above.

In fact, we don’t even need to write down a separate list, because the graph is drawn to scale. Instead, we can read the graph from left to right, stopping when we reach the fifth of the nine values. (On the GRE, data graphics like the ones here are always drawn to scale, as are coordinate planes and number lines. Other geometric figures, however, are not necessarily drawn to scale — a point we’ll revisit in later questions.)

Math Review Reference

For more on this topic, see the following sections of the GRE Math Review:

• 4.1: Graphical Methods for Describing Data (pp. 61-64)
• 4.2: Numerical Methods for Describing Data (p. 69)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.16: Powers of Five

When it comes to exponents, the GRE is a little like an old Dos Equis commercial: “I don’t always test exponents, but when I do, I prefer to test multiple properties in the same question.” To get started, we observe that 25 is a power of 5 and can be rewritten as 5²:

From there, we can use the product rule to combine exponents with the same base:

The final step involves a rule known as the property of equality of exponential functions, which is introduced (but not named) on p. 18 of the Math Review:

This is true for any positive base (other than 1) and for any integers a and b. Our base here is 5, and both 5x + 2 and n are integers, so this equation meets all of the criteria for applying the rule:

The answer, therefore, is (B). But take a look at answer (D) for a moment. This is the result you’d get if you accidentally used the power rule (which doesn’t apply to this problem) instead of the product rule. Sneaky, right? “Near-miss” answers like this are actually quite common in GRE algebra problems. It’s important, therefore, to be familiar with all the exponent rules given in the Math Review, and to know when to apply each of them.

Math Review Reference

For more on this topic, see the following sections of the GRE Math Review:

• 1.3: Exponents and Roots (p. 5)
• 2.2: Rules of Exponents (pp. 18-20)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.15: An Inequality Ready Check

This problem is designed to test your comfort level in manipulating inequalities. When they hand you a problem like this, the testmakers are checking to see if you know one rule in particular:

When you multiply or divide both sides of an inequality by a negative number, the direction of the inequality reverses.

As a little warm-up, consider the inequality

If we multiply both sides by (-1), we have to reverse the sign, or the statement won’t make sense anymore:

This is important in problem 5.15 because it asks us to consider both positive and negative values of x. We can simplify our task by considering the positive values first, but ultimately we’ll need to deal with the negatives as well. Let’s step through the answer choices one at a time.

Answer Choice A

Let x be any positive number. Then we can divide out the x from both sides as follows:

But this is a contradiction, because 5/3 is actually greater than 1. Since assuming that x is positive leads to an incorrect result, we can conclude that inequality (A) has no positive solution. Answer choice (A) is therefore incorrect.

Answer Choice B

Again, we begin by assuming x is positive and dividing out an x to simplify the expression:

Because x is positive, taking the square root of both sides gives us

In other words, inequality (B) is true for any positive x less than 1. So answer choice (B) passes our first test. Now what about negative values of x?

Let’s go back to the original inequality, this time assuming that x < 0. Under this new assumption, if we divide out an x from both sides, we have to flip the sign of the inequality.

This time around, we don’t even have to take the square root: almost any negative value of x is a solution. (Try x = -2, for example, or x = -10.) Because we can find both positive and negative solutions for the inequality, answer choice (B) is a keeper.

Answer Choice C

That leaves us with one final option to evaluate. This one’s actually a little simpler, because we can subtract out the x rather than dividing or multiplying. When you add or subtract a number from both sides of an inequality, it doesn’t change the sign. Thus, we don’t have to decide in advance whether x is negative or positive.

Indeed, it doesn’t end up mattering whether x is negative, positive, or even zero. As shown above, any real value of x leads to a contradiction, meaning that inequality (C) has no real solution at all. We can therefore reject answer (C), leaving (B) as the only correct answer.

A Note on Time Management

As you may have noticed, 5.15 is more time-consuming than most of the other math problems we’ve encountered so far. That’s largely because it requires you to test out a bunch of different cases and see which ones are valid. In other problems, like 5.11 and 5.13, you set up a single equation and solve for a single variable, but in this problem, it’s imperative that you test out each of the answer choices. Even if algebra is your strong suit, “case-by-case” problems like 5.15 may be worth skipping on your first pass through the section. That way, you’ll be sure not to miss any of the faster, easier problems that might appear later on.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 2.5: Solving Linear Inequalities (pp. 23-24)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.14: An Average Problem

The trick to this problem is recognizing that we can’t solve for x and y separately; there just isn’t enough information. Instead, we have to solve for the sum of x and y and use that to figure out what’s going on in list M.

To find an average, we sum all of the values in a data set, then divide by the number of values in that set:

We can use this formula to find any of these three pieces of information, provided that we have the other two:

• If we know the total and the number of values, we can find the average.
• If we know the average and the total, we can find the number of values.
• If we know the average and the number of values, we can find the total.

That last application is what we need here. For list L, we know the average (10/3) and the number of values (3), so we can solve for the total:

Once we know the total of list L, we can determine the value of x + y:

That, in turn gives us enough information to find the total of list M:

We already know the number of values in M (5), so we can now solve for the average:

In this case, no further simplification was necessary. But remember: on an NE problem, you don’t need to convert your answer to lowest terms unless explicitly instructed to do so. If your answer is equal to the right answer, it is the right answer.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 4.2: Numerical Methods for Describing Data (pp. 68-69)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.13: Simple Interest

In word problems like 5.12 (and 6.13 in the next section), we need to come up with our own ad hoc equation. This time, however, there’s a ready-made formula we can apply. Simple annual interest problems are fairly common on the GRE, so the formula for solving them is well worth committing to memory:

• V is the final value of the investment
• P is the principal (the amount initially deposited)
• r is the interest rate
• t is the number of years that interest accrues

Note that we needn’t worry about converting r from percent to decimal — the formula does that for us. So if a question mentions a simple annual interest rate of 5%, the correct value of r is 5, not 0.05.

Reading through the problem stem, we can identify three of the variables we need to complete the interest formula:

• P = $6,000
• V = $6,840
• t = 1

Only r — the value for which we’re supposed to solve — is missing. Plugging in our values of P, V, and t gives the following:

From there, we follow Thoreau’s advice and “simplify, simplify”:

Because r = 14, the correct answer is (D).

Sanity check: the final value V represents both principal and interest, so if we subtract out the principal P, we’re left only with the interest paid:

$840 is more than 10 percent of $6000, so we’d expect our answer to be larger than 10. Thus, even without working the problem in its entirety, we can rule out answers (A) and (B) as too small.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 2.7: Applications (p. 28)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.12: To Market, To Market

In GRE word problems, the underlying algebra is usually fairly simple. The real challenge is finding a mathematical expression that accurately captures all of the relevant details of the situation, without including any extraneous information. Fortunately, this section and the next provide plenty of practice in translating from words to equations.

Here, we can begin by finding an expression for the rate at which space is rented out, as stated in the first sentence of the problem stem:

Sanity check: It’s important to keep track of units, especially in word problems. In this case, we’re looking for a difference in dollars. Keeping track of the dollars, square feet, and days in our calculations will allow us to catch errors caused by multiplying or dividing incorrectly. If our answer comes back in square feet, for example, we know that a mistake has been made somewhere along the way.

Next, each woman’s “rental commitment” can be expressed in terms of the square feet rented times the duration of the rental:

To find out what each woman paid, we multiply her rental commitment by the rate we found in step one:

Cancelling out units gives the cost in dollars:

Finally, subtracting Alice’s cost from Betty’s cost yields our answer:

Betty paid $90 more than Alice, so our answer is (D).

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 2.7: Applications (pp. 25-28)

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.11: The Variable That Wasn’t

This problem illustrates another GRE scare tactic: “variables” that are actually constants. The initial expression looks like an equation in two variables, a and b, which could be a real pain to work with. In the very next equation, however, we’re told that b = 1; in other words, b is a constant. Consequently, we can go right back to that first equation and plug in 1 everywhere we see a b:

At this point, as in many of the QC problems we saw earlier, there are two distinct ways to proceed: solve algebraically or try plugging in answers. Because the algebra here is relatively simple, the first method is the one I’d recommend in most cases.

Solving Algebraically

We begin by multiplying out the denominator:

Next, we combine like terms:

Divide out the coefficient (-1), and we get our solution:

Answer (E) is correct.

Plugging In

But suppose for a moment that you have only 30 seconds on the clock. This is the last problem you’ll have time for, and you won’t be able to finish it in its entirety. In this situation, plugging in may actually be more useful than attempting to solve algebraically, because plugging in lets you eliminate wrong answers along the way. Here’s how the process might work out for this problem:

• If a = 1, the numerator (and thus the fraction) equals 0, because 1 – 1 = 0. So answer (A) can’t be right.
• If a = 0, the fraction evaluates to (-1)/(1) = -1. Cross (B) off the list.
• If a = -1, the denominator is (-1 + 1) = 0, which results in an undefined number. Get rid of answer (C).
• If a = -2, the fraction evaluates to (-3)/(-1) = 3. There goes answer (D).
• Since (A), (B), (C), and (D) don’t work, the answer must be (E).

When I teach problem 5.11 in class, most of my students find the algebraic route to be faster and more straightforward than plugging in. In an end-of-section time crunch, though, it makes sense to focus on eliminating wrong answers as quickly as possible. Although there’s no partial credit on the GRE, it can still be helpful to think in terms of “statistical partial credit,” in the sense that every wrong answer you eliminate increases your expected score by a fraction of a point.

Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

• 2.1: Operations With Algebraic Expressions (pp. 17-18)