24. a + b =30
2b =602 2a
Column A
Column B
ab
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (D). An unwary test taker might waste time trying to
find the values of a and b, because the centered data appears at first glance to
provide a system of two linear equations with two unknowns. But you can easily
manipulate the second equation so that it is identical to the first:
2b =602 2a
2b =2~30 2 a!
b =302 a
a + b =30
As you can see, you’re really dealing with only one equation. Since you cannot
solve one equation in two unknowns, you cannot make the comparison.
Whenever you encounter a Quantitative Comparison question that calls for solving
one or more linear equations, stop in your tracks before taking pencil to paper. Size up
the equation to see whether it’s one of the two unsolvable kinds you learned about
here. If so, unless you’re given more information, the correct answer will be choice (D).
FACTORABLE QUADRATIC EXPRESSIONS WITH ONE
VARIABLE
A quadratic expression includes a “squared” variable, such as x
2
. An equation is
quadratic if you can express it in this general form:
ax
2
1 bx 1 c 5 0
Where:
x is the variable
a, b, and c are constants (numbers)
a Þ 0
b can equal 0
c can equal 0
Here are four examples (notice that the b-term and c-term are not essential; in other
words, either b or c, or both, can equal zero (0)):
Chapter 10: Math Review: Number Theory and Algebra 243
ALERT!
If the centered information in
a Quantitative Comparison
consists of one or more linear
equations, never assume you
can solve for the variable(s).
www.petersons.com
Quadratic Equation
Same Equation, but in the Form:
ax
2
1 bx 1 c 5 0
2w
2
5 16
x
2
5 3x
3y 5 4 2 y
2
7z 5 2z
2
2 15
2w
2
2 16 5 0(nob-term)
x
2
2 3x 5 0(noc-term)
y
2
1 3y 2 4 5 0
2z
2
2 7z 2 15 5 0
Every quadratic equation has exactly two solutions, called roots. (But the two roots
might be the same.) On the GRE, you can often find the two roots by factoring.To
solve any factorable quadratic equation, follow these three steps:
Put the equation into the standard form: ax
2
1 bx 1 c 5 0.
Factor the terms on the left side of the equation into two linear expressions (with
no exponents).
Set each linear expression (root) equal to zero (0) and solve for the variable in each
one.
Factoring Simple Quadratic Expressions
Some quadratic expressions are easier to factor than others. If either of the two
constants b or c is zero (0), factoring requires no sweat. In fact, in some cases, no
factoring is needed at all:
A Quadratic with No c-term A Quadratic with No b-term
2x
2
5 x
2x
2
2 x 5 0
x~2x 2 1!50
x 5 0, 2x 2 1 5 0
x 5 0,
1
2
2x
2
2 4 5 0
2~x
2
2 2!50
x
2
2 2 5 0
x
2
5 2
x 5
=
2, 2
=
2
Factoring Quadratic Trinomials
A trinomial is simply an algebraic expression that contains three terms. If a quadratic
expression contains all three terms of the standard form ax
2
1 bx 1 c, then factoring
becomes a bit trickier. You need to apply the FOIL method, in which you add together
these terms:
(F) the product of the first terms of the two binomials
(O) the product of the outer terms of the two binomials
(I) the product of the inner terms of the two binomials
(L) the product of the last (second) terms of the two binomials
PART IV: Quantitative Reasoning244
ALERT!
When dealing with a
quadratic equation, your first
step is usually to put it into the
general form ax
2
1 bx 1 c 5
0. But keep in mind: The only
essential term is ax
2
.
www.petersons.com
Note the following relationships:
(F) is the first term (ax
2
) of the quadratic expression
(O 1 I) is the second term (bx) of the quadratic expression
(L) is the third term (c) of the quadratic expression
You’ll find that the two factors will be two binomials. TheGRE might ask you to
recognize one or both of these binomial factors.
25. Which of the following is a factor of x
2
2 x 2 6?
(A) (x 1 6)
(B) (x 2 3)
(C) (x 1 1)
(D) (x 2 2)
(E) (x 1 3)
The correct answer is (B). Notice that x
2
has no coefficient. This makes the
process of factoring into two binomials easier. Set up two binomial shells:
(x )(x )
The product of the two missing second terms (the “L” term under the FOIL
method) is 26. The possible integral pairs that result in this product are (1,26),
(21,6), (2,23,), and (22,3). Notice that the second term in the trinomial is 2x.
This means that the sum of the two integers whose product is 26 must be 21.
The pair (2,23) fits the bill. Thus, the trinomial is equivalent to the product of the
two binomials (x 1 2) and (x 2 3).
To check your work, multiply the two binomials using the FOIL method:
~x 1 2!~x 2 3!5x
2
2 3x 1 2x 2 6
5 x
2
2 x 1 6
If the preceding question had asked you to determine the roots of the equation
x
2
2 x 2 6 5 0, you’d simply set each of the binomial factors equal to zero (0), then solve
for x in each one. The solution set (the two possible values of x) includes the roots 22
and 3.
26. How many different values of x does the solution set for the equation
4x
2
5 4x 2 1 contain?
(A) none
(B) one
(C) two
(D) four
(E) infinitely many
The correct answer is (B). First, express the equation in standard form: 4x
2
2
4x 1 1 5 0. Notice that the c-term is 1. The only two integral pairs that result in
Chapter 10: Math Review: Number Theory and Algebra 245
www.petersons.com
this product are (1,1) and (21,21). Since the b-term (24x) is negative, the
integral pair whose product is 1 must be (21,21). Set up a binomial shell:
(? 2 1)(? 2 1)
Notice that the a-term contains the coefficient 4. The possible integral pairs that
result in this product are (1,4), (2,2), (21,24), and (22,22). A bit of trial-and-
error reveals that only the pair (2,2) works. Thus, in factored form, the equation
becomes (2x 2 1)(2x 2 1) 5 0. To check your work, multiply the two binomials
using the FOIL method:
~2x 2 1!~2x 2 1!54x
2
2 2x 2 2x 1 1
5 4x
2
2 4x 1 1
Since the two binomial factors are the same, the two roots of the equation are the
same. In other words, x has only one possible value. (Although you don’t need to
find the value of x in order to answer the question, solving for x in the equation
2x21 5 0 yields x 5
1
2
.
Stealth Quadratic Equations
Some equations that appear linear (variables include no exponents) may actually be
quadratic. On the GRE, be on the lookout for either of two situations:
The same variable inside a radical also appears outside:
=
x 5 5x
~
=
x!
2
5~5x!
2
x 5 25x
2
25x
2
2 x 5 0
The same variable that appears in the denominator of a fraction also appears
elsewhere in the equation:
2
x
5 3 2 x
2 5 x~3 2 x!
2 5 3x 2 x
2
x
2
2 3x 1 2 5 0
In both scenarios, you’re dealing with a quadratic (nonlinear) equation with one
variable. So, in either equation, there are two roots. (Both equations are factorable, so
go ahead and find their roots.)
The test makers often use the Quantitative Comparison format to cover this concept.
PART IV: Quantitative Reasoning246
www.petersons.com
27.
63xx=
Column A Column B
x
1
12
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (D). An unwary test taker might assume that the
equation is linear, because x is not squared. Substituting
1
12
for x satisfies the
centered equation. But the two quantities are not necessarily equal. If you clear
the radical by squaring both sides of the equation, then isolate the x-terms on one
side of the equation, you’ll see that the equation is quadratic:
36 3
2
xx=
36 3 0
2
xx−=
To find the two roots, factor out 3x, then solve for each root:
3x~12x 2 1! =0
x =0,12x 2 1=0
x =0,
1
12
Since
1
12
is only one of two possible values for x, you cannot make a definitive
comparison.
NONLINEAR EQUATIONS WITH TWO VARIABLES
In the world of math, solving nonlinear equations with two or more variables can be
very complicated, even for bona-fide mathematicians. But on the GRE, all you need to
remember are these three general forms:
Sum of two variables, squared: (x 1 y)
2
5 x
2
1 2xy 1 y
2
Difference of two variables, squared: (x 2 y)
2
5 x
2
2 2xy 1 y
2
Difference of two squares: x
2
2 y
2
5 (x 1 y)(x 2 y)
Chapter 10: Math Review: Number Theory and Algebra 247
www.petersons.com
You can verify these equations using the FOIL method:
(x 1 y)
2
5 (x 1 y)(x 1 y)
5 x
2
1 xy 1 xy 1 y
2
5 x
2
1 2xy 1 y
2
(x 2 y)
2
5 (x 2 y)(x 2 y)
5 x
2
2 xy 2 xy 1 y
2
5 x
2
2 2xy 1 y
2
(x 1 y)(x 2 y)
5 x
2
1 xy 2 xy 2 y
2
5 x
2
2 y
2
For the GRE, memorize the three equations listed here. When you see one form on the
exam, it’s a sure bet that your task is to rewrite it as the other form.
28. If x
2
2 y
2
5 100, and if x 1 y 5 2, then x 2 y 5
(A) 22
(B) 10
(C) 20
(D) 50
(E) 200
The correct answer is (D). If you’re on the lookout for the difference of two
squares, you can handle this question with no sweat. Use the third equation you
just learned, substituting 2 for (x 1 y), then solving for (x 2 y):
x
2
2 y
2
5~x 1 y!~x 2 y!
100 5~x 1 y!~x 2 y!
100 5~2!~x 2 y!
50 5~x 2 y!
SOLVING ALGEBRAIC INEQUALITIES
You can solve algebraic inequalities in the same manner as equations. Isolate the
variable on one side of the inequality symbol, factoring and eliminating terms
wherever possible. However, one important rule distinguishes inequalities from equa-
tions: Whenever you multiply or divide both sides of an inequality by a negative
number, you must reverse the inequality symbol. Simply put: If a . b, then 2a ,2b.
12 2 4x , 8 Original inequality
24x ,24 12 subtracted from each side; inequality unchanged
x . 1 Both sides divided by 24; inequality reversed
Here are five general rules for dealing with algebraic inequalities. Study them until
they’re second nature to you because you’ll put them to good use on the GRE.
Adding or subtracting unequal quantities to (or from) equal quantities:
If a . b, then c 1 a . c 1 b
If a . b, then c 2 a , c 2 b
Adding unequal quantities to unequal quantities:
If a . b, and if c . d, then a 1 c . b 1 d
PART IV: Quantitative Reasoning248
TIP
You usually can’t solve
quadratics using a shortcut.
Always look for one of the
three common quadratic
forms. If you see it, rewrite it as
its equivalent form to answer
the question as quickly and
easily as possible.
ALERT!
Be careful when handling
inequality problems: The wrong
answers might look right,
depending on the values you
use for the different variables.
www.petersons.com
Comparing three unequal quantities:
If a . b, and if b . c, then a . c
Combining the same positive quantity with unequal quantities by multiplication
or division:
If a . b, and if x . 0, then xa . xb
Combining the same negative quantity with unequal quantities by multiplication
or division:
If a . b, and if x , 0, then xa , xb
29. If a . b, and if c . d, then which of the following must be true?
(A) a 2 b . c 2 d
(B) a 2 c . b 2 d
(C) c 1 d , a 2 b
(D) b 1 d , a 1 c
(E) a 2 c , b 1 d
The correct answer is (D). Inequality questions can be a bit confusing, can’t
they? In this problem, you need to remember that if unequal quantities (c and d)
are added to unequal quantities of the same order (a and b), the result is an
inequality in the same order. This rule is essentially what answer choice (D) says.
SOLVING ALGEBRA “STORY” PROBLEMS
The remainder of this chapter is devoted exclusively to algebra “story” problems, in
which you translate real-world scenarios into algebraic expressions and equations.
Here are the types of story problems you’ll review in the pages ahead (some call for the
application of specific formulas):
• weighted-average problems
• currency problems
• mixture problems
• investment problems
• problems involving rate of production and rate of travel
• problems involving overlapping sets
To illustrate each type, we’ll show you one or two GRE-style questions. Read the
analysis of each question to find out what formulas to apply and what shortcuts (if
any) and strategies you can use to solve the problem.
Weighted-Average Problems
You solve weighted-average problems using the arithmetic mean (simple average)
formula, except you give the set’s terms different weights. For example, if a final exam
score of 90 receives twice the weight of each of two midterm exam scores 75 and 85,
Chapter 10: Math Review: Number Theory and Algebra 249
NOTE
For algebra story problems, the
GRE test makers generally use
the Problem Solving format
(the same format we use
here). Don’t be surprised,
though, if your exam includes
one or two Quantitative
Comparisons that are actually
story problems in disguise.
www.petersons.com
think of the final exam score as two scores of 90—and the total number of scores as 4
rather than 3:
WA 5
75 1 85 1~2!~90!
4
5
340
4
5 85
Similarly, when some numbers among terms appear more often than others, you must
give them the appropriate “weight” before computing an average.
30. During an 8-hour trip, Brigitte drove 3 hours at 55 miles per hour and 5
hours at 65 miles per hour. What was her average rate, in miles per hour,
for the entire trip?
(A) 58.5
(B) 60
(C) 61.25
(D) 62.5
(E) 66.25
The correct answer is (C). Determine the total miles driven: (3)(55) 1 (5)(65) 5
490. To determine the average over the entire trip, divide this total by 8, which is
the number of total hours: 490 4 8 5 61.25.
A tougher weighted-average problem might provide the weighted average and ask for
one of the terms, or require conversions from one unit of measurement to another—or
both.
31. A certain olive orchard produces 315 gallons of oil annually, on average,
during four consecutive years. How many gallons of oil must the orchard
produce annually, on average, during the next six years, if oil production
for the entire 10-year period is to meet a goal of 378 gallons per year?
(A) 240
(B) 285
(C) 396
(D) 420
(E) 468
The correct answer is (D). In the weighted-average formula, 315 annual
gallons receives a weight of 4, while the average annual number of gallons for the
next six years (x) receives a weight of 6:
378 5
1,260 1 6x
10
3,780 5 1,260 1 6x
3,780 2 1,260 5 6x
2,520 5 6x
420 5 x
This solution (420) is the average number of gallons needed per year, on
average, during the next 6 years.
PART IV: Quantitative Reasoning250
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To guard against calculation errors, check your answer by sizing up the
question. Generally, how great a number are you looking for? Notice that the
stated goal is a bit greater than the annual average production over the first
four years. So you’re looking for an answer that is greater than the goal—a
number somewhat greater than 378 gallons per year. You can eliminate
choices (A) and (B) out of hand. The number 420 fits the bill.
Currency Problems
Currency problems are similar to weighted-average problems in that each item (bill or
coin) is weighted according to its monetary value. Unlike weighted-average problems,
however, the “average” value of all the bills or coins is not at issue. In solving currency
problems, remember the following:
• You must formulate algebraic expressions involving both number of items (bills or
coins) and value of items.
• You should convert the value of all moneys to a common currency unit before
formulating an equation. If converting to cents, for example, you must multiply
the number of nickels by 5, dimes by 10, and so forth.
32. Jim has $2.05 in dimes and quarters. If he has four fewer dimes than
quarters, how much money does he have in dimes?
(A) 20 cents
(B) 30 cents
(C) 40 cents
(D) 50 cents
(E) 60 cents
The correct answer is (B). Letting x equal the number of dimes, x 1 4
represents the number of quarters. The total value of the dimes (in cents) is 10x,
and the total value of the quarters (in cents) is 25(x 1 4) or 25x 1 100. Given that
Jim has $2.05, the following equation emerges:
10x 1 25x 1 100 5 205
35x 5 105
x 5 3
Jim has three dimes, so he has 30 cents in dimes.
You could also solve this problem without formal algebra, by plugging in
each answer choice in turn. Let’s try this strategy for choices (A) and (B):
(A) 20 cents is 2 dimes, so Jim has 6 quarters. 20 cents plus $1.50 adds up
to $1.70. Wrong answer!
(B) 30 cents is 3 dimes, so Jim has 7 quarters. 30 cents plus $1.75 adds up
to $2.05. Correct answer!
Chapter 10: Math Review: Number Theory and Algebra 251
TIP
You can solve most GRE
currency problems by working
backward from the answer
choices.
www.petersons.com
Mixture Problems
In GRE mixture problems, you combine substances with different characteristics,
resulting in a particular mixture or proportion, usually expressed as percentages.
Substances are measured and mixed by either volume or weight—rather than by
number (quantity).
33. How many quarts of pure alcohol must you add to 15 quarts of a solution
that is 40% alcohol to strengthen it to a solution that is 50% alcohol?
(A) 4.0
(B) 3.5
(C) 3.25
(D) 3.0
(E) 2.5
The correct answer is (D). You can solve this problem by working backward
from the answer choices—trying out each one in turn. Or, you can solve the
problem algebraically. The original amount of alcohol is 40% of 15. Letting x
equal the number of quarts of alcohol that you must add to achieve a 50% alcohol
solution, .4(15) 1 x equals the amount of alcohol in the solution after adding more
alcohol. You can express this amount as 50% of (15 1 x). Thus, you can express
the mixture algebraically as follows:
~0.4!~15!1x 5~0.5!~15 1 x!
6 1 x 5 7.5 1 0.5x
0.5x 5 1.5
x 5 3
You must add 3 quarts of alcohol to obtain a 50% alcohol solution.
Investment Problems
GRE investment problems involve interest earned (at a certain percentage rate) on
money over a certain time period (usually a year). To calculate interest earned,
multiply the original amount of money by the interest rate:
amount of money 3 interest rate 5 amount of interest on money
For example, if you deposit $1,000 in a savings account that earns 5% interest
annually, the total amount in the account after one year will be $1,000 1 .05($1,000)
5 $1,000 1 $50 5 $1,050.
GRE investment questions usually involve more than simply calculating interest
earned on a given principal amount at a given rate. They usually call for you to set up
and solve an algebraic equation. When handling these problems, it’s best to eliminate
percent signs.
PART IV: Quantitative Reasoning252
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. product of the first terms of the two binomials
(O) the product of the outer terms of the two binomials
(I) the product of the inner terms of the two binomials
(L). equal. If you clear
the radical by squaring both sides of the equation, then isolate the x-terms on one
side of the equation, you’ll see that the equation is