6 Most-Tested GRE Problem Solving Concepts

Noticing that your scores on your GRE practice test isn’t quite as high as you’d like? One quick way to get better GRE Quantitative scores is to increase your content-knowledge in the most-tested Problem Solving areas. Here are the top seven most-tested GRE Quant concepts to review; get these down and you’ll ace the GRE section!

1. Functions and Symbols. A function is a different way of writing an equation. Instead of y = mx + b, we’d have f(x) = mx + b. It’s helpful to think of a function as simply replacing the “y” with a symbol called “f(x).” The GRE may also present made-up symbol functions; pay attention to any definitions you are given, and expand accordingly.

2. Number Properties. The properties of integers, primes, odds and evens, integers, fractions, positives, and negatives will all appear in various questions on your GRE test. The more comfortable you are with them, the more quickly you will arrive at the correct answer. This concept will bleed over into Quantitative Comparisons as well.

3. Plane and Coordinate Geometry. Not only will you need to know the standard equations for a line, parabola, and circle, but also you will need to memorize the distance formula, the midpoint formula, the slope formula, the relationship between slopes and the different quadrants, properties of parallel, perpendicular, vertical, and horizontal lines, as well as the quadratic formula/discriminant. For Plane Geometry, triangles are tested the most often on the GRE. You should know the Pythagorean Theorem, Triangle Inequality Theorem, the special right triangles: 45-45-90 and 30-60-90, as well as the properties of isosceles and equilateral triangles. Other plane geometry concepts to review include angles, circles, and polygons. Make sure you know how to find the perimeter and area of all shapes, and be comfortable dividing irregular shapes into manageable pieces.

4. Linear & Quadratic Equations. y = mx + b is the standard equation for a straight line, or a linear equation, where m is the slope and b is the y-intercept. You’ll need to know how to graph them and how to find the slope given two points. Quadratic equations look like y = ax2 + bx + c, and make a parabola, or curved line. Quadratics have two factors, and two solutions (also called “roots”). You will need to know how to factor quadratic equations to find the roots, how to find the quadratic if given the roots, and how to graph a quadratic on a grid given the equation.

5. Ratios and Proportions. A ratio is a relationship between two things. Given a ratio and one “real world” number, you can always set up a proportion to solve for the other missing “real world” number. Sometimes you will need to do this for similar triangles in Geometry, and sometimes in algebraic word problems.

6. Data Analysis. Data Analysis questions are like an open-book test. Make sure you read every tiny piece of writing on or near the data, including titles, the labels for the x and y-axes, column names, and even footnotes if there are any. Pay attention to the units of measurement, and notice any trends in the data BEFORE reading the questions.

Circular Permutations Questions

We’ve all seen those tricky Permutation/Combination questions involving people around a circular table. How do we solve them? Well, they’re actually pretty ridiculously easy!

Let’s examine one:

There are 7 people and a round table with 5 seats. How many arrangements are possible?

This question might seem complex at first because there are more people than there are seats. It’s kind of a Permutation AND a Combination in one!

So first we’re wondering, how many ways to choose 5 from 7? This is a simple Comb:

7C5 = 7! / 5!2! = 7 x 6 / 2 = 42/2 = 21 ways

So now for each of those ways, we’re wondering, how many ways can we order 5 people around a table?

For any table with “x” seats, the number of possible arrangements is (x-1)!, so here 4! = 4 x 3 x 2 = 24.

21 x 24 = 504

The correct answer is 504.

The key takeaway here is that “choosing” X from Y always allows for the Combination formula (x! / (x-y)! y!), and the number of arrangements around a circular table with “X” seats is always (X-1)! There’s actually not that much to memorize!

Two Types of GRE “Averages”: Mean and Rates

The word “average” on the GRE can refer to two concepts: arithmetic mean, and the average speed (or average rate) formula. It’s important not to confuse the two on the Test Day, as they require different formulas to solve.

Mean is the mathematical average. This is defined as the sum of the terms divided by the number of terms. Mean = Sum / # of terms. For a list of consecutive integers or evenly spaced numbers, the mean is equal to the median, or the middle number. For example, the “average” of 3, 5, and 9 is 5.67.

Average Speed or Average Rate is often found in complex word problems. This type of question is one many students are less familiar with so you may not have seen it before. Let’s review two important equations to remember and look at how this concept appears on the GRE.

The first formula to memorize is: D = R x T. This stands for Distance = Rate x Time (referred to as the “DIRT” formula). It is perfectly acceptable to also think of it as Time = Distance / Rate or as Rate = Distance / Time as well. Usually the “Rate” is speed but it could be anything “per” anything. In a word problem, if you see the word “per” you know this is a question involving rates.

The second formula is: Average Rate = Total Distance / Total Time. This is its own special concept and you will notice that it is NOT a simple Average of the Speeds (which would be something like the Sum of the Speeds / the Number of Different Speeds or what we know as the Arithmetic Mean). Average Rate is a completely different concept, so do not let the common word “average” confuse you. Let’s look at a sample question from Grockit’s GRE question bank:

Question 1: The average (arithmetic mean) of four numbers is 30, after one of the numbers is removed, the average of the remaining three numbers is 10. What number was removed?

We know that the four original numbers sum to 30*4 = 120. The new equation becomes:

4*30 – x/3 = 10
120 – x/3 = 10
120 – x = 30 (add an x to each side and subtract a 30)
90 = x

Strategies and Formulas for Tough GRE Sets

For some advanced Data Analysis and Probability questions, it will help you achieve better scores to know the logic and formulas behind set theory. Set theory hinges on two concepts: union and intersection. The union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets.

Let’s call our sets A, B, and C, and use a Venn diagram to express their relationship.

If n = intersection and u = union, then we can describe the relationship between the sets thusly:
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

To find the number of people in exactly one set: P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)

To find the number of people in exactly two sets: P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)

To find the number of people in exactly three sets: P(A n B n C)

To find the number of people in two or more sets: P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

To find the number of people in at least one set: P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

To find the union of all set: (A + B + C + X + Y + Z + O)

Number of people in exactly one set: (A + B + C)

Number of people in exactly two of the sets: (X + Y + Z)

Number of people in exactly three of the sets: O

Number of people in two or more sets: (X + Y + Z + O)

If you’re like me, and formulas like these sometimes seem complicated and intimidating, let’s look at how making a Venn diagram and applying it to a tough GRE question can provide a little relief!

In 1997, N people graduated from college. If 1/3 of them received a degree in the applied sciences, and, of those, 1/4 graduated from a school in one of six northeastern states, which of the following expressions represents the number of people who graduated from college in 1997 who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?

(A) 11N/12
(B) 7N/12
(C) 5N/12
(D) 6N/7
(E) N/7

The key to understanding this question lies in the last sentence:

…who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?

We have two categories to sum: the people who ONLY received a science degree but NOT from one of the 6 schools, and the people who ONLY went to the 6 schools but did NOT receive a science degree. I made up variables for these categories (x and y).

If N = 12, there are 4 applied science students, 1 of which is both. That means x = 3. If 4 students are applied science, then 12-4 = 8 are from one of the six states but NOT applied science. y = 8.

3 + 8 = 11

So we are looking for an answer choice that gives us 11 when N = 12; the answer is A.

You aren’t likely to see many questions at this difficulty level on the actual GRE, but if you continue to challenge yourself, the medium GRE sets questions will soon look easy!

What’s the Triangle Inequality Theorem?

Triangle Inequality Theorem is fair game on the SAT, ACT, GRE, or GMAT. It’s often forgotten by test-takers, but when it pops up, you’ll be glad you know it! The theorem essentially states that the third side of a triangle must be between the difference and sum of the other two sides.

For example, if we had a triangle in which two sides were 6 and 9, then the third side must be between 3 (9-6) and 15 (9+6). The third side cannot actually equal 3 or 15, it’s important to remember.

Let’s try a practice question utilizing this math rule!

If two sides of a triangle have lengths 2 and 5, which of the following could be the perimeter of the triangle?

I. 9
II. 15
III. 19

A) None
B) I only
C) II only
D) II and III only
E) I,II and III

If two of the sides are 2 and 5. Then the range of possible values for the third side can be expressed as:

3 < x < 7

Perimeter is the sum of the sides. Let’s choose 3 and 7 as values for the 3rd side (even though we know they are the end-limits only) to create a range for the perimeter.

On the low end:

2 + 5 + 3 = 10

On the upper end:

2 + 5 + 7 = 14

So the perimeter range can be expressed as:

10 < x < 14

The perimeter must be BETWEEN 10 and 14. The answer is (A).

Here’s a link to a lot of great Triangle review topics if you want more Geometry practice: http://www.beatthegmat.com/mba/category/tags/gmat-math/geometry/triangles.

GRE Quant Question of the Day: Rates!

Try this “rates” question on your own!

A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?

(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph

First, let’s consider Day 1 and Day 2’s hours.

If x = hours on Day 1, then x + 2 = hours on Day 2. The question said he walked 18 hours total, so we can set up a simple equation:

x + (x + 2) = 18
2x + 2 = 18
2x = 16
x = 8

Therefore he walked 8 hours on Day 1 and 10 hours on Day 2.

We are told he went 1mph FASTER on Day 2. So if Day 1’s mph is y, then Day 2’s mph is y + 1.

Let’s look at the D = R x T formula.

D1 = R1 x T1

D2 = R2 x T2

If we plug in what we know:

D1 = (y) x 8 hrs

D2 = (y + 1) x 10 hrs

We know that D1 + D2 must equal 64, so we can sum the two equations and set them equal to 64.

(y) x 8hrs + (y + 1) x 10hrs = 64

Simplifying…

64 = 8y + 10y + 10

64 = 18y + 10

54 = 18y

3 = y

The answer is (B).