- B To avoid working with fractions, cross-multiply and arrive at: 3a = bc. To
make the columns look similar, substitute 3a for bc in Quantity B: 4(3a) = 12a.
The comparison is 9a versus 12a. Since you are told that a is positive, you can
divide both quantities by a. The comparison is thus:
QUANTITY A QUANTITY B
9 12 A B C D
Quantity B is greater. - C
Step 1: To simplify the comparison, add 8 to both quantities:
Quantity A = 6x + 9y and Quantity B = 36.
Step 2: To arrive at a value for Quantity A, manipulate the equation in
the prompt to look similar to the expression in Quantity A:
3(2x + 3y = 12)
6 x + 9y = 36
Step 3: Substitute 36 for 6x + 9y in Quantity A
The two quantities are equal.
- A
Step 1: To make the two quantities look similar, factor the expression in
Quantity A: ax + ay = a(x + y).
Step 2: Divide both quantities by (x + y). Quantity A = a and Quantity B = b.
Since you are told that a > b, the value in Quantity A is greater.
Exponents and Roots
Exponent Basics
In the term 5^3 , five is the base and 3 is the exponent. The exponent represents the
number of times you multiply the base by itself. So 5^3 comes out to 5 × 5 × 5. Both
the exponent and base can be any real number (not just positive integers).
Most GRE questions dealing with exponents will require you to use a few simple
rules and manipulate them in unorthodox situations. But before getting to these
rules, let’s look at some other properties of exponents:
An even exponent always yields a positive result.
The base of an exponential expression can be positive or negative, but when the
base is raised to an even exponent, the result will always be positive. Consider:
x^2 = 16
The most obvious solution to the preceding equation is 4. If you substitute 4 for x,
you arrive at 4^2 = 16, which is a true statement. But notice that –4 is also a solution
for x! If you substitute −4 back into the equation, you arrive at (−4)^2 = (−4)( −4) =
16, which is also true.
CHAPTER 11 ■ ALGEBRA 269
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