FORMAT OF THE GRE EXAM – GRE TEST MATH SECTION
The math section consists of three types of questions: Quantitative Comparisons, Standard Multiple Choice, and Graphs. They are designed to test your ability to solve problems, not to test your mathematical knowledge. The section is 45 minutes long and contains 28 questions. The questions can appear in any order.
About 14 Quantitative Comparisons
About 9 Standard Multiple Choice
About 5 Graphs
LEVEL OF DIFFICULTY of the GRE Math Section
Except for the graphs, GRE math is just like SAT math, though surprisingly slightly easier. The mathematical skills tested are very basic: only first year high school algebra and geometry (no proofs). However, this does not mean that the math section is easy. The medium of basic mathematics is chosen so that everyone taking the test will be on a fairly even playing field. This way students who majored in math, engineering, or science don’t have an undue advantage over students who majored in humanities. Although the questions require only basic mathematics and all have simple solutions, it can require considerable ingenuity to find the simple solution. If you have taken a course in calculus or another advanced math topic, don’t assume that you will find the math section easy. Other than increasing your mathematical maturity, little you learned in calculus will help on the GRE.
As mentioned above, every GRE math problem has a simple solution, but finding that simple solution may not be easy. The intent of the math section is to test how skilled you are at finding the simple solutions. The premise is that if you spend a lot of time working out long solutions you will not finish as much of the test as students who spot the short, simple solutions. So if you find yourself performing long calculations or applying advanced mathematics–stop. You’re heading in the wrong direction.
Substitution is a very useful technique for solving GRE math problems. It often reduces hard problems to routine ones. In the substitution method, we choose numbers that have the properties given in the problem and plug them into the answer-choices.
If n is an odd integer, which one of the following is an even integer?
(A) 3n + 2 (B) n/4 (C) 2n + 3 (D) n(n + 3) (E) nn
We are told that n is an odd integer. So choose an odd integer for n, say, 1 and substitute it into each answer-choice. In Choice (A), 3(1) + 2 = 5, which is not an even integer. So eliminate (A). Next, n/4 = 1/4 is not an even integer–eliminate (B). Next, 2n + 3 = 2(1) + 3 = 5 is not an even integer–eliminate (C). Next, n(n + 3) = 1(1 + 3) = 4 is even and hence the answer is possibly (D). Finally, in Choice (E), the nn = 1(1) = 1, which is not even–eliminate (E). The answer is (D).
When using the substitution method, be sure to check every answer-choice because the number you choose may work for more than one answer-choice. If this does occur, then choose another number and plug it in, and so on, until you have eliminated all but the answer. This may sound like a lot of computing, but the calculations can usually be done in a few seconds.
When substituting in quantitative comparison problems, don’t rely on only positive whole numbers. You must also check negative numbers, fractions, 0, and 1 because they often give results different from those of positive whole numbers. Plug in the numbers 0, 1, 2, -2, and 1/2, in that order.
Determine which of the two expressions below is larger, whether they are equal, or whether there is not enough information to decide. The answer is (A) if Column A is larger, (B) if Column B is larger, (C) if the columns are equal, and (D) if there is not enough information to decide.
Column A x does not equal 0 Column B x xx If x = 2, then Column B is 4. In this case, Column B is larger. However, if x equals 1, then Column B is 1. In this case, the two columns are equal. Hence, the answer is (D)–not enough information to decide.
Note: If, as above, you get a certain answer when a particular number is substituted and a different answer when another number is substituted (Double Case), then the answer is (D)–not enough information to decide.
Sometimes instead of making up numbers to substitute into the problem, we can use the actual answer-choices. This is called Plugging In. It is a very effective technique but not as common as Substitution.
Example: The digits of a three-digit number add up to 18. If the ten’s digit is twice the hundred’s digit and the hundred’s digit is 1/3 the unit’s digit, what is the number?
(A) 246 (B) 369 (C) 531 (D) 855 (E) 893
First, check to see which of the answer-choices has a sum of digits equal to 18. For choice (A), 2 + 4 + 6 = 12. Eliminate. For choice (B), 3 + 6 + 9 = 18. This may be the answer. For choice (C), 5 + 3 + 1 = 9. Eliminate. For choice (D), 8 + 5 + 5 = 18. This too may be the answer. For choice (E), 8 + 9 + 3 = 20. Eliminate. Now, in choice (D), the ten’s digit is not twice the hundred’s digit, 5 does not equal 2(8). Eliminate. Hence, by process of elimination, the answer is (B). Note that we did not need the fact that the hundred’s digit is 1/3 the unit’s digit.
Defined functions are very common on the GRE, and most students struggle with them. Yet once you get used to them, defined functions can be some of the easiest problems on the test. In this type of problem, you will be given a symbol and a property that defines the symbol.
Example: Define x # y by the equation x # y = xy – y. Then 2 # 3 =
(A) 1 (B) 3 (C) 12 (D) 15 (E) 18
From the above definition, we know that x # y = xy – y. So all we have to do is replace x with 2 and y with 3 in the definition: 2 # 3 = 2(3) – 3 = 3. Hence, the answer is (B).
Define x @ y to be xx.
Column A Column B z @ 2 z @ 3 Most students who are unfamiliar with defined functions are unable to solve this problem. Yet it is actually quite easy. By the definition given above, @ merely squares the first term.
So z @ 2 = zz, and z @ 2 = zz. In each case, the result is zz. Hence the two expressions are equal. The answer is (C).
This broad category is a popular source for GRE questions. At first, students often struggle with these problems since they have forgotten many of the basic properties of arithmetic. So before we begin solving these problems, let’s review some of these basic properties.
• “The remainder is r when p is divided by q” means p = qz + r; the integer z is called the quotient. For instance, “The remainder is 1 when 7 is divided by 3” means 7 = 3(2) + 1.
Example: When the integer n is divided by 2, the quotient is u and the remainder is 1. When the integer n is divided by 5, the quotient is v and the remainder is 3. Which one of the following must be true?
(A) 2u + 5v = 4
(B) 2u – 5v = 2
(C) 4u + 5v = 2
(D) 4u – 5v = 2
(E) 3u – 5v = 2
Translating “When the integer n is divided by 2, the quotient is u and the remainder is 1” into an equation gives n = 2 u + 1. Translating “When the integer n is divided by 5, the quotient is v and the remainder is 3” into an equation gives n = 5v + 3. Since both expressions equal n, we can set them equal to each other: 2u + 1 = 5v + 3. Rearranging and then combining like terms yields 2u – 5v = 2. The answer is (B).
• A number n is even if the remainder is zero when n is divided by 2: n = 2z + 0, or n = 2z.
• A number n is odd if the remainder is one when n is divided by 2: n = 2z + 1.
• The following properties for odd and even numbers are very useful–you should memorize them:
even x even = even
odd x odd = odd
even x odd = even
even + even = even
odd + odd = even
even + odd = odd
• Consecutive integers are written as x, x + 1, x + 2, . . .
• Consecutive even or odd integers are written as , x + 2, x + 4, . . .
• The integer zero is neither positive nor negative, but it is even: 0 = 2(0).
• A prime number is an integer that is divisible only by itself and 1.
The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, . . .
• A number is divisible by 3 if the sum of its digits is divisible by 3.
For example, 135 is divisible by 3 because the sum of its digits (1 + 3 + 5 = 9) is divisible by 3.
• The absolute value of a number, | |, is always positive. In other words, the absolute value symbol eliminates negative signs.
For example, | -7 | = 7. Caution, the absolute value symbol acts only on what is inside the symbol, | |. For example, -| -7 | = -(+7) = -7. Here, only the negative sign inside the absolute value symbol is eliminated.
Example: If a, b, and c are consecutive integers and a < b < c, which of the following must be true?
I. b – c = 1
II. abc/3 is an integer.
III. a + b + c is even.
(A) I only (B) II only (C) III only (D) I and II only (E) II and III only
Let x, x + 1, x + 2 stand for the consecutive integers a, b, and c, in that order. Plugging this into Statement I yields b – c = (x + 1) – (x + 2) = -1. Hence, Statement I is false.
As to Statement II, since a, b, and c are three consecutive integers, one of them must be divisible by 3. Hence, abc/3 is an integer, and Statement II is true.
As to Statement III, suppose a is even, b is odd, and c is even. Then a + b is odd since even + odd = odd. Hence, a + b + c = (a + b) + c = (odd) + even = odd. Thus, Statement III is not necessarily true. The answer is (B).
Generally, quantitative comparison questions require much less calculating than do multiple-choice questions. But they are trickier.
Substitution is very effective with quantitative comparison problems. But you must plug in all five major types of numbers: positives, negatives, fractions, 0, and 1. Test 0, 1, 2, -2, and 1/2, in that order.
General Principles For Solving Quantitative Comparisons
• You Can Add or Subtract the Same Term (Number) from Both Sides of a Quantitative Comparison Problem.
• You Can Multiply or Divide Both Sides of a Quantitative Comparison Problem by the Same Positive Term (Number). (Caution: This cannot be done if the term can ever be negative or zero.)
You can think of a quantitative comparison problem as an inequality/equation. Your job is to determine whether the correct symbol with which to compare the columns is <, =, >, or that it cannot be determined. Therefore, all the rules that apply to solving inequalities apply to quantitative comparisons. That is, you can always add or subtract the same term to both columns of the problem. If the term is always positive, then you can multiply or divide both columns by it. (The term cannot be negative because it would then invert the inequality. And, of course, it cannot be zero if you are dividing.)
Column A Column B 1/5 + 1/3 + 1/8 1/8 + 1/5 + 1/4 Don’t solve this problem by adding the fractions in each column; that would be too time consuming–the LCD is 120! Instead, merely subtract 1/5 and 1/8 from both columns:
Column A Column B 1/3 1/4 Now 1/3 is larger than 1/4, so Column A is larger than Column B.
Note: If there are only numbers (i.e., no variables) in a quantitative comparison problem, then “not-enough-information” cannot be the answer. Hence (D), not-enough-information, cannot be the answer to Example 1 above.
• Caution: You Must Be Certain That the Quantity You Are Multiplying or Dividing by Can Never Be Zero or Negative. (There are no restrictions on adding or subtracting.)
The following example illustrates the false results that can occur if you don’t guarantee that the number you are multiplying or dividing by is positive.
Column A 0 < x < 1 Column B xx x Solution (Invalid): Dividing both columns by x yields
Column A Column B x 1 We are given that x < 1, so Column B is larger. But this is a false result because when x = 0, the two original columns are equal:
Column A Column B (0)(0) = 0 0 Hence, the answer is actually (D), not-enough-information to decide.
• Caution: Don’t Cancel Willy-Nilly.
Some people are tempted to cancel the 4x from both columns of the following problem:
Column A Column B 4x – 6 5y – 6 – 4x You cannot cancel the 4x’s from both columns because they do not have the same sign. In Column A, 4x is positive. Whereas in Column B, it is negative.
Substitution (Special Cases)
We already studied this method in the section Substitution. Here, we will practice more and learn a couple of special cases.
A. In A Problem with Two Variables, Say, x and y, You Must Check The Case in Which x = y. (This often gives a double case.)
Column A x and y are integers greater than 1. Column B 2(x + y) 2xy If x = y = 2, then 2(x + y) = 2(2 + 2) = 8 and 2xy = 2(2)(2) = 8. In this case, the columns are equal. For all other choices of x and y, Column B is greater. (You should check a few cases.) Hence, we have a double case, and therefore the answer is (D).
B. When You Are Given That x < 0, You Must Plug in Negative Whole Numbers, Negative Fractions, and -1. (Choose the numbers -1, -2, and -1/2, in that order.)
C. Sometimes You Have to Plug in The First Three Numbers (But Never More Than Three) From a Class of Numbers.
Column A x is a positive integer. Column B The number of distinct prime factors of x The number of distinct prime factors of x cubed We need only look at x = 1, 2, and 3. If x = 1, then x has no prime factors, likewise for x cubed. Next, if x = 2, then x has one prime factor, 2, and x cubed equals 8 also has one prime factor, 2. Finally, if x = 3, then x has one prime factor, 3, and x cubed equals 27 also has one prime factor, 3. In all three cases, the columns are equal. Hence, the answer is (C). Note, there is no need to check x = 4. The writers of the GRE do not change the results after the third number.
One-fourth of the math problems on the GRE involve geometry. (There are no proofs.) Unfortunately, the figures on the GRE are usually not drawn to scale. Hence, in most cases you cannot solve problems or check your work by “eyeballing” the drawing.
Following are some of the basic properties of geometry. You probably know many of them. Memorize any that you do not know.
1. There are 180 degrees in a straight angle.
2. Two angles are supplementary if their angle sum is 180 degrees.
3. Two angles are complementary if their angle sum is 90 degrees.
4. Perpendicular lines meet at right angles.
5. A triangle with two equal sides is called isosceles. The angles opposite the equal sides are called the base angles.
6. The altitude to the base of an isosceles or equilateral triangle bisects the base and bisects the vertex angle.
7. The angle sum of a triangle is 180 degrees.
8. In an equilateral triangle all three sides are equal, and each angle is 60 degrees.
9. The area of a triangle is bh/2, where b is the base and h is the height.
10. In a triangle, the longer side is opposite the larger angle, and vice versa.
11. Two triangles are similar (same shape and usually different size) if their corresponding angles are equal. If two triangles are similar, their corresponding sides are proportional.
12. Two triangles are congruent (identical) if they have the same size and shape.
13. In a triangle, an exterior angle is equal to the sum of its remote interior angles and is therefore greater than either of them.
14. Opposite sides of a parallelogram are both parallel and congruent.
15. The diagonals of a parallelogram bisect each other.
16. If w is the width and l is the length of a rectangle, then its area is A = lw and its perimeter is P=2w + 2l.
17. The volume of a rectangular solid (a box) is the product of the length, width, and height. The surface area is the sum of the area of the six faces.
18. If the length, width, and height of a rectangular solid (a box) are the same, it is a cube. Its volume is the cube of one of its sides, and its surface area is the sum of the areas of the six faces.
19. A tangent line to a circle intersects the circle at only one point. The radius of the circle is perpendicular to the tangent line at the point of tangency.
20. An angle inscribed in a semicircle is a right angle.
Example: In the figure to the right, what is the value of x?
(E) 47 Since 2x + 60 is an exterior angle, it is equal to the sum of the remote interior angles. That is, 2x + 60 = x + 90. Solving for x gives x = 30. The answer is (A).
Most geometry problems on the GRE require straightforward calculations. However, some problems measure your insight into the basic rules of geometry. For this type of problem, you should step back and take a “birds-eye” view of the problem. The following example will illustrate.
Example: In the figure to the right, O is both the center of the circle with radius 2 and a vertex of the square OPRS. What is the length of diagonal PS?
(E) 2/3 The diagonals of a square are equal. Hence, line segment OR (not shown) is equal to SP. Now, OR is a radius of the circle and therefore OR = 2. Hence, SP = 2 as well, and the answer is (D).
The distance between points (x, y) and (a, b) is given by the following formula:
Example: In the figure to the right, the circle is centered at the origin and passes through point P. Which of the following points does it also pass through?
(A) (3, 3)
(C) (2, 6)
(D) (1.5, 1.3)
(E) (-3, 4) Since the circle is centered at the origin and passes through the point (0,-3), the radius of the circle is 3. Now, if any other point is on the circle, the distance from that point to the center of the circle (the radius) must also be 3. Look at choice (B). Using the distance formula to calculate the distance between and (0, 0) (the origin) yields
Hence, is on the circle, and the answer is (B).
The midpoint M between points (x, y) and (a, b) is given by
M = (x + a/2, y + b/2)
In other words, to find the midpoint, simply average the corresponding coordinates of the two points.
Example: In the figure to the right, polygon PQRO is a square and T is the midpoint of side QR. What are the coordinates of T ?
(A) (1, 1)
(B) (1, 2)
(C) (1.5, 1.5)
(D) (2, 1)
(E) (2, 3)
Since point R is on the x-axis, its y-coordinate is 0. Further, since PQRO is a square and the x-coordinate of Q is 2, the x-coordinate of R is also 2. Since T is the midpoint of side QR, the midpoint formula yields T = (2 + 2/2, 2 + 0/2) = (4/2, 2/2) = (2, 1). The answer is (D).
The slope of a line measures the inclination of the line. By definition, it is the ratio of the vertical change to the horizontal change. The vertical change is called the rise, and the horizontal change is called the run. Thus, the slope is the rise over the run. Given the two points (x, y) and (a, b) the slope is
M = (y – b)/(x – a)
Example: In the figure to the right, what is the slope of line passing through the two points?
(E) 2 The slope formula yields m = (4 – 2)/(5 – 1) = 2/4 = 1/2. The answer is (C).
Multiplying both sides of the equation m = (y -b)/(x – a) by x-a yields
y – b = m(x – a)
Now, if the line passes through the y-axis at (0, b), then the equation becomes
y – b = m(x – 0)
y – b = mx
y = mx + b
This is called the slope-intercept form of the equation of a line, where m is the slope and b is the y-intercept. This form is convenient because it displays the two most important bits of information about a line: its slope and its y-intercept.
Example: Column A The equation of the line above is y = 9x/10 + k Column B AO BO Since y = 9x/10 + k is in slope-intercept form, we know the slope of the line is 9/10. Now, the ratio of BO to AO is the slope of the line (rise over run). Hence, BO/AO = 9/10. Multiplying both sides of this equation by AO yields BO = 9AC/10. In other words, BO is 9/10 the length of AO. Hence, AO is longer. The answer is (A).
Inequalities are manipulated algebraically the same way as equations with one exception:
Multiplying or dividing both sides of an inequality by a negative number reverses the inequality. That is, if x > y and c < 0, then cx < cy.
Example: For which values of x is 4x + 3 > 6x – 8?
As with equations, our goal is to isolate x on one side:
Subtracting 6x from both sides yields -2x + 3 > -8
Subtracting 3 from both sides yields -2x > -11
Dividing both sides by -2 and reversing the inequality yields x < 11/2
Positive & Negative Numbers
A number greater than 0 is positive. On the number line, positive numbers are to the right of 0. A number less than 0 is negative. On the number line, negative numbers are to the left of 0. Zero is the only number that is neither positive nor negative; it divides the two sets of numbers. On the number line, numbers increase to the right and decrease to the left.
The expression x > y means that x is greater than y. In other words, x is to the right of y on the number line.
We usually have no trouble determining which of two numbers is larger when both are positive or one is positive and the other negative (e.g., 5 > 2 and 3.1 > -2). However, we sometimes hesitate when both numbers are negative (e.g., -2 > -4.5). When in doubt, think of the number line: if one number is to the right of the number, then it is larger.
Miscellaneous Properties of Positive and Negative Numbers
1. The product (quotient) of positive numbers is positive.
2. The product (quotient) of a positive number and a negative number is negative.
3. The product (quotient) of an even number of negative numbers is positive.
4. The product (quotient) of an odd number of negative numbers is negative.
5. The sum of negative numbers is negative.
6. A number raised to an even exponent is greater than or equal to zero.
The absolute value of a number is its distance on the number line from 0. Since distance is a positive number, absolute value of a number is positive. Two vertical bars denote the absolute value of a number: | x |. For example, | 3 | = 3 and | -3 | = 3.
Students rarely struggle with the absolute value of numbers: if the number is negative, simply make it positive; and if it is already positive, leave it as is. For example, since -2.4 is negative, | -2.4 | = 2.4 and since 5.01 is positive | 5.01 | = 5.01.
Further, students rarely struggle with the absolute value of positive variables: if the variable is positive, simply drop the absolute value symbol. For example, if x > 0, then | x | = x.
However, negative variables can cause students much consternation. If x is negative, then | x | = -x. This often confuses students because the absolute value is positive but the -x appears to be negative. It is actually positive–it is the negative of a negative number, which is positive. To see this more clearly let x = -k, where k is a positive number. Then x is a negative number. So | x | = -x = -(-k) = k. Since k is positive so is -x. Another way to view this is | x | = -x = (-1)x = (-1)(a negative number) = a positive number.
If x < y and y < z, then x < z.
Column A 1/Q > 1 Column B QQ 1 Since 1/Q > 1 and 1 > 0, we know from the transitive property that 1/Q is positive. Hence, Q is positive. Therefore, we can multiply both sides of 1/Q > 1 by Q without reversing the inequality:
Q(1/Q) > 1(Q)
Reducing yields 1 > Q
Multiplying both sides again by Q yields Q > QQ
Using the transitive property to combine the last two inequalities yields 1 > QQ
The answer is (B).
I. To compare two fractions, cross-multiply. The larger number will be on the same side as the larger fraction.
Column A Column B 9/10 10/11 Cross-multiplying gives (9)(11) versus (10)(10), which reduces to 99 versus 100. Now, 100 is greater than 99. Hence, 10/11 is greater than 9/10.
III. To solve a fractional equation, multiply both sides by the LCD (lowest common denominator) to clear fractions.
Example: If (x + 3)/(x – 3) = y, what is the value of x in terms of y?
(A) 3 – y (B) 3/y (C) (2 + y)/(y – 2) (D) (-3y -3)/(1 – y) (E) 3y/2
First, multiply both sides of the equation by x – 3: (x – 3)(x + 3)/(x – 3) = (x – 3)y
Cancel the (x – 3’s) on the left side of the equation: x + 3 = (x – 3)y
Distribute the y: x + 3 = xy – 3y
Subtract xy and 3 from both sides: x – xy = -3y – 3
Factor out the x on the left side of the equation: x(1 – y) = -3y – 3
Finally, divide both sides of the equation by 1 – y: x = (-3y -3)/(1 – y)
Hence, the answer is (D).
IV. When dividing a fraction by a whole number (or vice versa), you must keep track of the main division bar.
Example: a/(b/c) = a(c/b) = ac/b. But (a/b)/c = (a/b)(1/c) = a/(bc).
V. Two fractions can be added quickly by cross-multiplying: a/b + c/d = (ad + bc)/bd
Example: 1/2 – 3/4 =
(A) -5/4 (B) -2/3 (C) -1/4 (D) 1/2 (E) 2/3
Cross multiplying the expression 1/2 – 3/4 yields 1(4) – 2(3)/2(4) = (4 – 6)/8 = -2/8 = -1/4. Hence, the answer is (C).
VI. To find a common denominator of a set of fractions, simply double the largest denominator until all the other denominators divide into it evenly.
VII. Fractions often behave in unusual ways: Squaring a fraction makes it smaller, and taking the square root of a fraction makes it larger. (Caution: This is true only for proper fractions, that is, fractions between 0 and 1.)
Example: 1/3 squared equals 1/9 and 1/9 is less than 1/3. Also the square root of 1/4 is 1/2 and 1/2 is greater than 1/4.
Problems involving averages are very common on the GRE. They can be classified into four major categories as follows.
I. The average of N numbers is their sum divided by N, that is, average = sum/N.
Column A x > 0 Column B The average of x, 2x, and 6 The average of x and 2x By the definition of an average, Column A equals (x + 2x + 6)/3 = (3x + 6)/3 = 3(x + 2)/3 = x + 2, and Column B equals (x + 2x)/2 = 3x/2. Now, if x is small, then x + 2 is larger than 3x/2. But if x is large, then 3x/2 is larger. (Verify this by plugging in x = 1 and x = 100.)
II. Weighted average: The average between two sets of numbers is closer to the set with more numbers.
Example: If on a test three people answered 90% of the questions correctly and two people answered 80% correctly, then the average for the group is not 85% but rather 3(90) + 2(80)/5 = 430/5 = 86. Here, 90 has a weight of 3–it occurs 3 times. Whereas 80 has a weight of 2–it occurs 2 times. So the average is closer to 90 than to 80 as we have just calculated.
III. Using an average to find a number.
Sometimes you will be asked to find a number by using a given average. An example will illustrate.
Example: If the average of five numbers is -10, and the sum of three of the numbers is 16, then what is the average of the other two numbers?
(A) -33 (B) -1 (C) 5 (D) 20 (E) 25
Let the five numbers be a, b, c, d, e. Then their average is (a + b + c + d + e)/5 = -10. Now three of the numbers have a sum of 16, say, a + b + c = 16. So substitute 16 for a + b + c in the average above: (16 + d + e)/5 = -10. Solving this equation for d + e gives d + e = -66. Finally, dividing by 2 (to form the average) gives (d + e)/2 = -33. Hence, the answer is (A).
IV. Average Speed = Total Distance/Total Time
Although the formula for average speed is simple, few people solve these problems correctly because most fail to find both the total distance and the total time.
Example: In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hours at 60 mph. What was his average speed for the whole trip?
(A) 50 (B) 53 1/2 (C) 55 (D) 56 (E) 57 1/2
The total distance is 1(50) + 3(60) = 230. And the total time is 4 hours. Hence, Average Speed = Total Distance/Total Time = 230/4 = 57 1/2. The answer is (E). Note, the answer is not the mere average of 50 and 60. Rather the average is closer to 60 because he traveled longer at 60 mph (3 hrs) than at 50 mph (1 hr).
RATIO & PROPORTION
A ratio is simply a fraction. Both of the following notations express the ratio of x to y: x:y, x/y. A ratio compares two numbers. Just as you cannot compare apples and oranges, so too must the numbers you are comparing have the same units. For example, you cannot form the ratio of 2 feet to 4 yards because the two numbers are expressed in different units–feet vs. yards. It is quite common for the GRE to ask for the ratio of two numbers with different units. Before you form any ratio, make sure the two numbers are expressed in the same units.
Column A Column B The ratio of 2 miles to 4 miles The ratio of 2 feet to 4 yards Forming the ratio in Column A yields 2 miles/4 miles = 1/2 or 1:2. The ratio in Column B cannot be formed until the numbers are expressed in the same units. Let’s turn the yards into feet. Since there are 3 feet in a yard, 4 yards = 4 x 3 feet = 12 feet. Forming the ratio yields 2 feet/12 feet = 1/6 or 1:6. Hence, Column A is larger.
A proportion is simply an equality between two ratios (fractions). For example, the ratio of x to y is equal to the ratio of 3 to 2 is translated as x/y = 3/2. Two variables are directly proportional if one is a constant multiple of the other:
y = kx, where k is a constant.
The above equation shows that as x increases (or decreases) so does y. This simple concept has numerous applications in mathematics. For example, in constant velocity problems, distance is directly proportional to time: d = vt, where v is a constant. Note, sometimes the word directly is suppressed.
Example: If the ratio of y to x is equal to 3 and the sum of y and x is 80, what is the value of y?
(A) -10 (B) -2 (C) 5 (D) 20 (E) 60
Translating “the ratio of y to x is equal to 3” into an equation yields: y/x = 3
Translating “the sum of y and x is 80” into an equation yields: y + x = 80
Solving the first equation for y gives: y = 3x.
Substituting this into the second equation yields
3x + x = 80
4x = 80
x = 20
Hence, y = 3x = 3(20) = 60. The answer is (E).
In many word problems, as one quantity increases (decreases), another quantity also increases (decreases). This type of problem can be solved by setting up a direct proportion.
Example: If Biff can shape 3 surfboards in 50 minutes, how many surfboards can he shape in 5 hours?
(A) 16 (B) 17 (C) 18 (D) 19 (E) 20
As time increases so does the number of shaped surfboards. Hence, we set up a direct proportion. First, convert 5 hours into minutes: 5 hours = 5 x 60 minutes = 300 minutes. Next, let x be the number of surfboards shaped in 5 hours. Finally, forming the proportion yields
3/50 = x/300
3(300)/50 = x
The answer is (C).
If one quantity increases (or decreases) while another quantity decreases (or increases), the quantities are said to be inversely proportional. The statement “y is inversely proportional to x” is written as
y = k/x, where k is a constant.
Multiplying both sides of y = k/x by x yields
yx = k
Hence, in an inverse proportion, the product of the two quantities is constant. Therefore, instead of setting ratios equal, we set products equal.
In many word problems, as one quantity increases (decreases), another quantity decreases (increases). This type of problem can be solved by setting up a product of terms.
Example: If 7 workers can assemble a car in 8 hours, how long would it take 12 workers to assemble the same car?
(A) 3hrs (B) 3 1/2hrs (C) 4 2/3hrs (D) 5hrs (E) 6 1/3hrs
As the number of workers increases, the amount time required to assemble the car decreases. Hence, we set the products of the terms equal. Let x be the time it takes the 12 workers to assemble the car. Forming the equation yields
7(8) = 12x
56/12 = x
4 2/3 = x
The answer is (C).
To summarize: if one quantity increases (decreases) as another quantity also increases (decreases), set ratios equal. If one quantity increases (decreases) as another quantity decreases (increases), set products equal.
EXPONENTS & ROOTS
There are five rules that govern the behavior of exponents:
Problems involving these five rules are common on the GRE, and they are often listed as hard problems. However, the process of solving these problems is quite mechanical: simply apply the five rules until they can no longer be applied.
There are only two rules for roots that you need to know for the GRE:
To factor an algebraic expression is to rewrite it as a product of two or more expressions, called factors. In general, any expression on the GRE that can be factored should be factored, and any expression that can be unfactored (multiplied out) should be unfactored.
The most basic type of factoring involves the distributive rule:
ax + ay = a(x + y)
For example, 3h + 3k = 3(h + k), and 5xy + 45x = 5xy + 9(5x) = 5x(y + 9). The distributive rule can be generalized to any number of terms. For three terms, it looks like ax + ay + az = a(x + y + z). For example, 2x + 4y + 8 = 2x + 2(2y) + 2(4) = 2(x + 2y + 4).
Example: If x – y = 9, then (x – y/3) – (y – x/3) =
(A) -4 (B) -3 (C) 0 (D) 12 (E) 27
(x – y/3) – (y – x/3) =
x – y/3 – y + x/3 =
4x/3 – 4y/3 =
4(x – y)/3 =
The answer is (D).
Difference of Squares
One of the most important formulas on the GRE is the difference of squares:
Example: If x does not equal -2, then
(A) 2(x – 2) (B) 2(x – 4) (C) 8(x + 2) (D) x – 2 (E) x + 4
In most algebraic expressions involving multiplication or division, you won’t actually multiply or divide, rather you will factor and cancel, as in this problem.
2(x – 2)
The answer is (A).
Perfect Square Trinomials
Like the difference of squares formula, perfect square trinomial formulas are very common on the GRE.
For example, .
A mathematical expression that contains a variable is called an algebraic expression. Some examples of algebraic expressions are 3x – 2y, 2z/y. Two algebraic expressions are called like terms if both the variable parts and the exponents are identical. That is, the only parts of the expressions that can differ are the coefficients. For example, x + y and -7(x + y) are like terms. However, x – y and 2 – y are not like terms.
Adding & Subtracting Algebraic Expressions
Only like terms may be added or subtracted. To add or subtract like terms, merely add or subtract their coefficients:
You may add or multiply algebraic expressions in any order. This is called the commutative property:
x + y = y + x xy = yx
For example, -2x + 5x = 5x + (-2x) = (5 – 2)x = -3x and (x – y)(-3) = (-3)(x – y) = (-3)x – (-3)y = -3x + 3y.
Caution: the commutative property does not apply to division or subtraction.
When adding or multiplying algebraic expressions, you may regroup the terms. This is called the associative property:
x + (y + z) = (x + y) + z x(yz) = (xy)z
Notice in these formulas that the variables have not been moved, only the way they are grouped has changed: on the left side of the formulas the last two variables are grouped together, and on the right side of the formulas the first two variables are grouped together.
For example, (x -2x) + 5x = (x + -2x) + 5x = x + (-2x + 5x) = x + 3x = 4x and 24x = 2x(12x) = 2x(3x4x) = (2x3x)4x = 6x4x = 24x
Caution: the associative property doesn’t apply to division or subtraction.
Notice in the first example that we changed the subtraction into negative addition: (x – 2x) = (x + – 2x). This allowed us to apply the associative property over addition.
When simplifying expressions with nested parentheses, work from the inner most parentheses out:
5x + (y – (2x – 3x)) = 5x + (y – (-x)) = 5x + (y + x) = 6x + y
Sometimes when an expression involves several pairs of parentheses, one or more pairs are written as brackets. This makes the expression easier to read:
2x(1 -y + 2(3 – y)) =
2x(1 -y + 6 – 2y) =
2x(1 –y + 6) =
2x(1 + y – 6) =
2x(y – 5) =
2xy – 10x
Order of Operations: (PEMDAS)
When simplifying algebraic expressions, perform operations within parentheses first and then exponents and then multiplication and then division and then addition and then subtraction. This can be remembered by the mnemonic:
PEMDAS Please Excuse My Dear Aunt Sally
Questions involving graphs rarely involve any significant calculating. Usually, the solution is merely a matter of interpreting the graph.
1. During which year was the company’s earnings 10 percent of its sales?
(A) 85 (B) 86 (C) 87 (D) 88 (E) 90
Reading from the graph, we see that in 1985 the company’s earnings were $8 million and its sales were $80 million. This gives 8/80 = 1/10 = 10/100 = 10%. The answer is (A).
2. During what two-year period did the company’s earnings increase the greatest?
(A) 85-87 (B) 86-87 (C) 86-88 (D) 87-89 (E) 88-90
Reading from the graph, we see that the company’s earnings increased from $5 million in 1986 to $10 million in 1987, and then to $12 million in 1988. The two-year increase from ’86 to ’88 was $7 million–clearly the largest on the graph. The answer is (C).
3. During the years 1986 through 1988, what were the average earnings per year?
(A) 6 million (B) 7.5 million (C) 9 million (D) 10 million (E) 27 million
The graph yields the following information:
Year Earnings 1986 $5 million 1987 $10 million 1988 $12 million Forming the average yields (5 + 10 + 12)/3 = 27/3 = 9. The answer is (C).
4. If Consolidated Conglomerate’s earnings are less than or equal to 10 percent of sales during a year, then the stockholders must take a dividend cut at the end of the year. In how many years did the stockholders of Consolidated Conglomerate suffer a dividend cut?
(A) None (B) One (C) Two (D) Three (E) Four
Calculating 10 percent of the sales for each year yields
Year 10% of Sales (millions) Earnings (millions) 85 .10 x 80 = 8 8 86 .10 x 70 = 7 5 87 .10 x 50 = 5 10 88 10 x 80 = 8 12 89 .10 x 90 = 9 11 90 .10 x 100 = 10 8 Comparing the right columns shows that earnings were 10 percent or less of sales in 1985, 1986, and 1990. The answer is (D).
Although exact steps for solving word problems cannot be given, the following guidelines will help:
(1) First, choose a variable to stand for the least unknown quantity, and then try to write the other unknown quantities in terms of that variable.
For example, suppose we are given that Sue’s age is 5 years less than twice Jane’s and the sum of their ages is 16. Then Jane’s age would be the least unknown, and we let x = Jane’s age. Expressing Sue’s age in terms of x gives Sue’s age = 2x – 5.
(2) Second, write an equation that involves the expressions in Step 1. Most (though not all) word problems pivot on the fact that two quantities in the problem are equal. Deciding which two quantities should be set equal is usually the hardest part in solving a word problem since it can require considerable ingenuity to discover which expressions are equal.
For the example above, we would get (2x – 5) + x = 16.
(3) Third, solve the equation in Step 2 and interpret the result.
For the example above, we would get by adding the x’s: 3x – 5 = 16. Then adding 5 to both sides gives 3x = 21. Finally, dividing by 3 gives x = 7. Hence, Jane is 7 years old and Sue is 2x – 5 = 2(7) – 5 = 9 years old.
Virtually, all motion problems involve the formula Distance = Rate x Time, or
D = R x T
Example : Scott starts jogging from point X to point Y. A half-hour later his friend Garrett who jogs 1 mile per hour slower than twice Scott’s rate starts from the same point and follows the same path. If Garrett overtakes Scott in 2 hours, how many miles will Garrett have covered?
(A) 2 1/5 (B) 3 1/3 (C) 4 (D) 6 (E) 6 2/3
Following Guideline 1, we let r = Scott’s rate. Then 2r – 1 = Garrett’s rate. Turning to Guideline 2, we look for two quantities that are equal to each other. When Garrett overtakes Scott, they will have traveled the same distance. Now, from the formula D = R x T, Scott’s distance is D = r x 2 1/2 and Garrett’s distance is D = (2r – 1)2 = 4r – 2. Setting these expressions equal to each other gives 4r – 2 = r x 2 1/2. Solving this equation for r gives r = 4/3. Hence, Garrett will have traveled D = 4r – 2 = 4(4/3) – 2 = 3 1/3 miles. The answer is (B).
The formula for work problems is Work = Rate x Time, or W = R x T. The amount of work done is usually 1 unit. Hence, the formula becomes 1 = R x T. Solving this for R gives R = 1/T.
Example : If Johnny can mow the lawn in 30 minutes and with the help of his brother, Bobby, they can mow the lawn 20 minutes, how long would take Bobby working alone to mow the lawn?
(A) 1/2 hour (B) 3/4 hour (C) 1 hour (D) 3/2 hours (E) 2 hours
Let r = 1/t be Bobby’s rate. Now, the rate at which they work together is merely the sum of their rates:
Total Rate = Johnny’s Rate + Bobby’s Rate
1/20 = 1/30 + 1/t
1/20 – 1/30 = 1/t
(30 – 20)/(30)(20) = 1/t
1/60 = 1/t
t = 60
Hence, working alone, Bobby can do the job in 1 hour. The answer is (C).
The key to these problems is that the combined total of the concentrations in the two parts must be the same as the whole mixture.
Example : How many ounces of a solution that is 30 percent salt must be added to a 50-ounce solution that is 10 percent salt so that the resulting solution is 20 percent salt?
(A) 20 (B) 30 (C) 40 (D) 50 (E) 60
Let x be the ounces of the 30 percent solution. Then 30%x is the amount of salt in that solution. The final solution will be 50 + x ounces, and its concentration of salt will be 20%(50 + x). The original amount of salt in the solution is 10%(50). Now, the concentration of salt in the original solution plus the concentration of salt in the added solution must equal the concentration of salt in the resulting solution: 10%(50) + 30%x = 20%(50 + x). Multiply this equation by 100 to clear the percent symbol and then solving for x yields x = 50. The answer is (D).
The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.
Example : Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?
(A) 3 (B) 7 (C) 10 (D) 13 (E) 16
Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 – D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 – D). Summarizing this information in a table yields
Dimes Quarters Total Number D 20 – D 20 Value 10D 25(20 – D) 305 Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:
10D + 25(20 – D) = 305
10D + 500 – 25D = 305
-15D = -195
D = 13
Hence, there are 13 dimes, and the answer is (D).
Typically, in these problems, we start by letting x be a person’s current age and then the person’s age a years ago will be x – a and the person’s age a years in future will be x + a. An example will illustrate.
Example : John is 20 years older than Steve. In 10 years, Steve’s age will be half that of John’s. What is Steve’s age?
(A) 2 (B) 8 (C) 10 (D) 20 (E) 25
Steve’s age is the most unknown quantity. So we let x = Steve’s age and then x + 20 is John’s age. Ten years from now, Steve and John’s ages will be x + 10 and x + 30, respectively. Summarizing this information in a table yields
Age now Age in 10 years Steve x x + 10 John x + 20 x + 30 Since “in 10 years, Steve’s age will be half that of John’s,” we get
(x + 30)/2 = x + 10
x + 30 = 2(x + 10)
x + 30 = 2x + 20
x = 10
Hence, Steve is 10 years old, and the answer is (C).