Algebra
See if you can notice the difference between most of the math problems
we’ve considered so far, and these:
Eek! There are letters in there! As you no doubt know,
the letters are called variables, and they make possible the
wonderful world of algebra. Whoever thought of adding variables into math
certainly made tens of millions of school kids’ lives harder, because without
algebra, there would be no geometry, no trigonometry, no calculus, and possibly
no misery at all in the world. Thanks a lot, Mr. or Mrs. Math
Genius . . . Of course, there would probably be no microwave ovens or
iPods either, so it’s not all bad.
Since the variables represent unspecified quantities, algebra brings
arithmetic into the world of the unknown. Of course, much of algebra deals with
making that unknown known; that is, solving equations so that variables can be
replaced by good oldfashioned numbers. We’ll get to all that soon enough, but
first a little vocabulary is in order.
Algebra Terms
There are six main terms that describe the world of algebra. The GRE
won’t ask you to define them but will nonetheless give you questions that
require you to work with them.

Constant. A numerical quantity that does not
change.

Variable. An unknown quantity written as a
letter. A variable can be represented by any letter in the English
alphabet; x or y are common on the
GRE, but you’ll see others as well. Variables may be associated with
specific things, like x number of apples or
y dollars. Other times, variables have no
specific association, but you’ll need to manipulate them to show
that you understand basic algebraic principles.

Coefficient. A coefficient is a number that
appears next to a variable and tells how many of that variable there
are. For example, in the term 4x, 4 is the
coefficient and tells us there are four xs. In the
term 3x^{3}, 3 is the
coefficient and tells us there are three
x^{3}s.

Term. The product of a constant and a
variable. Or, a quantity separated from other quantities by addition
or subtraction. For example, in the equation
3x^{3} +
2x^{2} – 7x
+ 4 = x – 1, the side to the left of the
equal sign contains four terms
(3x^{3},
2x^{2},
–7x, 4), while the right side contains two terms
(x, –1). The constants, 4 and –1, are
considered terms because they are coefficients of variables raised
to the zero power: 4 =
4x^{0}. So technically,
every algebraic term is the product of a constant and a variable
raised to some power.

Expression. A funny look on your face. Or in
math, any combination of terms. An expression can be as simple as a
single constant term, like 5, or as complicated as the sum or
difference of many terms, each of which is a combination of
constants and variables, such as . Expressions don’t include an
equal sign—this is what differentiates expressions from equations.
If you’re given the value of every variable in the expression, then
you can calculate the numerical value of the expression. Lacking
those values, expressions can’t be solved, although they can often
be simplified.

Equation. Two expressions linked by an equal
sign. Much of the algebra on the GRE consists of solving
equations.
Inputs and Outputs: Simple Substitutions
We mentioned above that we can calculate the numerical value of an
expression if we’re given the value of every variable in it. In this case,
the expression itself is like a machine that takes an input (the variable
value) and outputs a solution. One of the simplest kinds of algebraic
problems on the GRE operates like this, as in the following example:
We mentioned above that we can calculate the numerical value of an
expression if we’re given the value of every variable in it. In this case,
the expression itself is like a machine that takes an input (the variable
value) and outputs a solution. One of the simplest kinds of algebraic
problems on the GRE operates like this, as in the following example:
If
x = 2, what is the value of
?
There’s nothing to do but simply input 2 into the expression in place
of x and do the math:
So here, an input of 2 yields an output of 6.
You may see something like this in the beginning of the Math section,
but things will most likely get more complicated after that. For example,
the inputs themselves may be a bit more complex than a single number; in
fact, an input may itself contain variables:
If 2y + 8x = 11, what is the
value of 3(2y + 8x)?
You might see this equation bubbling over with variables and panic.
Don’t. Since the expression 2y + 8x
appears in both parts of the question, and we’re told this
expression equals 11, we can simply substitute that figure in place of the
expression on the right to get 3(11) = 33.
A question may also involve multiple substitutions. For instance:
z =
,
y = 3
x, and
x = 2,
then what is the value of
z?
To approach this problem, you just have to input 2 for x
to find y, and then input those values into the
equation for z. Substituting 2 for x into
y = 3x gives y = 3(2)
= 6. Inputting x= 2 and y = 6 into the
equation for z gives:
Simplifying Algebraic Expressions
Before we move on to solving more complicated equations, we need to
cover a few simplification tools that allow us to change algebraic
expressions into simpler but equivalent forms.
Distributing
The rule of distribution states:
The a in this expression can be any kind of term,
meaning it could be a variable, a constant, or a combination of the two.
When you distribute a factor into an expression within parentheses,
multiply each term inside the parentheses by the factor outside the
parentheses. 4(x + 2), for example, would become
4x + (4)(2), or 4x + 8. Let’s try
a harder one:
3y(y^{2} – 6).
Distributing the 3y term across the terms in the
parentheses yields:
Seems logical enough. But the true value of distributing becomes
clear when you see a distributable expression in an equation. We’ll see
an example of this later in the section on linear equations.
Factoring
Factoring an expression is the opposite of distributing.
4x^{3} –
8x^{2} is one meanlooking
expression, right? Or so it seems, until you realize that both terms
share the greatest common factor
4x^{2}, which you can factor
out:
By distributing and factoring, you can group or ungroup quantities
in an equation to make your calculations simpler, depending on what the
other terms in the equation look like. Sometimes distributing will help;
other times, factoring will be the way to go. Here are a few more
examples of both techniques:
Combining Like Terms
After factoring and distributing, you can take additional steps to
simplify expressions or equations. Combining like terms is one of the
simplest techniques you can use. It involves adding or subtracting the
coefficients of variables that are raised to the same power. For
example, by combining like terms, the expression
can be simplified by adding the coefficients of the variable
x^{3} (–1 and 3) together
and the coefficients of x^{2}
(1 and 4) together to get:
Variables that have different exponential values are not like
terms and can’t be combined. Two terms that do not share a variable are
also not like terms and cannot be combined regardless of their
exponential value. For example, you can’t combine:
You can, however, factor the first expression to get
x^{2}(x^{2}
+ 1), which you should do if it helps you answer the
question.
Linear Equations with One Variable
Simplifying is nice, and helpful to boot, but solving is really where
it’s at. To solve an equation, you have to isolate the variable you’re
solving for. That is, you have to “manipulate” the equation until you get
the variable alone on one side of the equal sign. By definition, the
variable is then equal to everything on the other side of the equal sign.
You can’t manipulate an equation the way you used to manipulate your little
brother or sister. When manipulating equations, there are rules. Here’s the
first and most fundamental. In fact, it’s so important we’re going to bold
it:
Whatever you do to one side of an equation, you must do to the other
side.
If you divide one side of an equation by 3, divide the other side by
3. If you take the square root of one side of an equation, take the square
root of the other. If you fall in love with one side of the equation, fall
in love with the other. Neither side will think you’re a twotimer. They’ll
think you’re a highly skilled mathematician.
By treating the two sides of the equation in the same way, you don’t
change what the equation means. You change the form of the
equation into something easier to work with—that’s the point of manipulating
it—but the equation remains true since both sides stay equal.
Take, for instance, the equation 3x + 2 = 5. You can
do anything you want to it as long as you do the same thing to both sides.
Here, since we’re trying to get the variable x alone on the
left, the thing to do is subtract 2 from that side of the equation. But we
can only do that if we subtract 2 from the other side as well:
Ah, that’s better. Now we can just divide both sides by 3 to get
x = 1, and we’re done.
You should use the simplification techniques you learned above
(distributing, factoring, and combining like terms) to help you solve
equations. For example:
That seems fairly nasty, since there aren’t any like terms to combine.
But wait a sec . . . what if you distribute that 3y on the
left side of the equation? That would give:
Shiver our timbers! Now we can subtract
3y^{3} from both sides to get
and then simply divide both sides by 18 to get y = 2.
Reverse PEMDAS
Many equations include a combination of elements you learned about
in our arithmetic discussion. Remember PEMDAS, the acronym you learned
to help you remember the order of operations? Well, what do you get if
you do PEMDAS in reverse? SADMEP, of course. Or, using the old corny
mnemonic device, Sally Aunt Dear My Excuse Please. Wait, scratch
that—mnemonics don’t work in reverse.
Why do we want to reverse our trusty order of operations, anyway?
The idea is to undo everything that has been done to
the variable so that it will be isolated in the end. So you should first
subtract or add any extra terms on the same side as the variable. Then
divide and multiply anything on the same side as the variable. Next,
raise both sides of the equation to a power or take their roots
according to any exponent attached to the variable. Finally, work out
anything inside parentheses. In other words, do the order of operations
backward: SADMEP!
We’ll need to demonstrate with an example. Here’s a little
monstrosity that at first glance might make you reconsider your decision
to go to grad school:
At second glance you might feel the same way. But at third glance
you’d know what to do.
In this equation, poor little x is being square
rooted, multiplied by 2, added to 3, and encased in parentheses—all in
the numerator of a fraction. That’s hardly what we’d call “alone time.”
You’ve got to get him out of there! Undo all of these operations to
liberate x and solve the equation.
Let SADMEP be your guide: First, subtract 2 from both sides of the
equation:
There’s no addition or division possible at this point, but we can
multiply both sides by 2 to get rid of the fraction:
Now divide both sides by 3 (see you later, parentheses!):
Now we’re in position to subtract 3 from each side:
Divide both sides by 2:
Finally, square each side to get rid of the square root:
Success! You’ve freed poor x from all of those
bullying operations.
Variables in the Denominator
Remember, the key to solving equations is to isolate the variable,
but how to do this depends on where the variable is located. A variable
in the numerator of a fraction is usually pretty easy to isolate. But if
the variable is in the denominator, things get more complicated. See
what you can make of this one:
Following SADMEP, start by subtracting the 3:
But now you have to get the x out of the
denominator, and the only way to do that is to multiply both sides of
the equation by that denominator, x + 2:
Divide both sides by 4:
Subtract 2 from each side:
Equations with Absolute Value
To solve an equation in which the variable is within absolute
value bars, you have to follow a twostep process:

Isolate the expression within the absolute value
bars.

Divide the equation in two.
Divide the equation in two? What is this, a magic trick? Kind of.
Watch:
If x + 3 = 5, then x =
Since both 5 and –5 within absolute value bars equal 5, the
expression inside the bars can equal 5 or –5 and the equation will work
out. That’s why we have to work through both scenarios. So we’re
actually dealing with two equations:
x + 3 = 5
x + 3 = –5
For a complete solution, we need to solve both. In the first
equation, x = 2. In the second equation, x
= –8. So the solutions to the equation x + 3
= 5 are x = {–8, 2}. Both work. Substitute them back
into the equation if you have any doubts and to reinforce why we need to
solve two equations to get a full answer to the question.
Equations with Exponents and Radicals
Absolute value equations aren’t the only ones with more than one
possible answer. Exponents and radicals can also have devilish effects
on algebraic equations that are similar to those caused by absolute
value. Consider the equation
x^{2} = 25. Seems pretty
simple, right? Just take the square root of both sides, and you end up
with
x = 5. But remember the rule of multiplying
negative numbers? When two negative numbers are multiplied together the
result is positive. In other words, –5 squared
also
results in 25: –5 ×–5 = 25. This means that whenever you have to take
the square root to simplify a variable brought to the second power, the
result will be two solutions, one positive and one negative:
. (The only exception is if
x = 0.) You’ll see what we mean by working through
this question:
If 2x^{2} = 72,
what is the value of x?
To solve this problem, we first divide both sides by 2 to get
x^{2} = 36. Now we need to
take the square root of both sides:
.
Recognizing when algebraic equations can have more than one
possible answer is especially important in Quantitative Comparison
questions. Sometimes it may appear that the relationship between columns
A and B leans in a definite direction, until you notice that one of the
columns can actually have more than one possible value. Keep your eyes
peeled for this common nuance.
Linear Equations with Two or More Variables
So you’re kicking butt and taking names on those old onevariable
equations, huh? Good. But some GRE questions contain two
variables. Lucky for you, those questions also contain two equations, and
you can use the two equations in conjunction to solve for the variables.
These two equations together are called a system of equations
or simultaneous equations. We said earlier
that manipulating equations isn’t like manipulating your younger brother or
sister. Actually, that’s not entirely true: Solving simultaneous equations
is like manipulating your younger brother and sister. You use one equation
against the other, and in the end you get whatever you want.
There are two ways to solve simultaneous equations. The first method
involves substitution, and the second involves adding or subtracting one
equation from the other. Let’s look at both techniques.
Solving by Substitution
You’ve already seen examples of substitution in the Input/Output
discussion earlier in the chapter, so you should be familiar with the
mechanics of plugging values or variables from one place into another
place to find what you’re looking for. In these next examples, both
pieces of the puzzle are equations. Our method will be to find the value
of one variable in one equation and then plug that into a second
equation to solve for a different variable. Here’s an example:
If x – 4 = y – 3 and
2y = 6, what is x?
You’ve got two equations, and you have to find x.
The first equation contains both x and
y. The second equation contains only
y. To solve for x, you first have to
solve for y in the second equation and substitute that
value into the first equation. If 2y = 6, then dividing
both sides by 2, y = 3. Now, substitute 3 for
y in the equation x – 4 =
y – 3:
That’s all there is to it. But here’s one that’s more likely to
give you trouble:
If 3x = y + 5 and
2y – 2 = 12k, what is x
in terms of k?
Notice anything interesting? There are three
variables in this one. To solve for x in terms of
k, we have to first get x and
k into the same equation. To make this happen, we
can solve for y in terms of k in the
second equation and then substitute that value into the first equation
to solve for x:
Then substitute 6k + 1 for y in
the equation 3x = y + 5:
This is our answer, since x is now expressed in
terms of k. Note that you could also solve this problem
by solving for y in terms of x in the
first equation and substituting that expression in for
y in the second equation. Either way works.
Solving by Adding or Subtracting
The amazing thing about simultaneous equations is that you can
actually add or subtract the entire equations from each other. Here’s an
example:
If 6x + 2y = 11 and 5x
+ y = 10, what is x
+ y?
Look what happens if we subtract the second equation from the
first:
To add or subtract simultaneous equations, you need to know what
variable or expression you want to solve for, and then add or subtract
accordingly. We made the example above purposely easy to show how the
method works. But you won’t always be given two equations that you can
immediately add or subtract from each other to isolate the exact
variable or expression you seek, as evidenced by this next example:
If 2x + 3y = –6 and –4x
+ 16y = 13, what is the value of
y?
We’re asked to solve for y, which means we’ve got
to get rid of x. But one equation has 2x
and the other has –4x, which means the
x terms won’t disappear by simply adding or
subtracting them. Don’t despair; our Golden Rule of Algebra comes to the
rescue: If you do the same thing to both sides of an equation, you don’t
change the meaning of the equation. That means that in this case, we
could multiply both sides of 2x + 3y =
–6 by 2, which would give us 2(2x +3y)
= 2(–6). Using the trusty distributive law, and multiplying out the
second part, gives us 4x +6y = –12.
Now we’re in a position to get rid of those pesky x
terms by adding this new (but equivalent) form of equation 1 to
equation 2:
On the GRE, you will almost always be able to manipulate one of
the two equations in a pair of simultaneous equations so that they can
be added and subtracted to isolate the variable or expression you want.
If you can’t see how to do this, or for questions with easy numbers, go
ahead and solve by using substitution instead. As you practice with
these types of problems, you’ll get a sense for which method works best
for you.
Binomial and Quadratic Equations
A binomial is an expression containing two terms. The terms (
x
+ 5) and (
x – 6) are both binomials. A quadratic
expression takes the form
ax2 +
bx +
c, where
.
Quadratics closely resemble the products formed when binomials are
multiplied. Coincidence? Fat chance. That’s why we treat these topics
together. We’ll start off with binomials, and work our way to quadratics.
Multiplying Binomials
The best acronym ever invented (other than SCUBA: “selfcontained
underwater breathing apparatus”) will help you remember how to multiply
binomials. This acronym is FOIL, and it stands for First + Outer + Inner
+ Last. The acronym describes the order in which we multiply the terms
of two binomials to get the correct product.
For example, let’s say you were kidnapped by wretched forktongued
lizardmen whose only weakness was binomials. Now what if the
lizardking asked you to multiply these binomials:
(x + 1)(x + 3)
What would you do? Follow FOIL, of course. First, multiply the
first (F) terms of each binomial:
x × x =
x^{2}
Next, multiply the outer (O) terms of the binomials:
x × 3 = 3x
Then, multiply the inner (I) terms:
1 × x = x
And then multiply the last (L) terms:
1 × 3 = 3
Add all these terms together:
Finally, combine like terms, and you get:
Here are a few more examples of multiplied binomials to test your
FOILing faculties:
Note that the last one doesn’t form a quadratic equation, and that
none of the terms can be combined. That’s okay. When presented with
binomials, follow FOIL wherever it leads.
Working with Quadratic Equations
A quadratic equation will always have a variable raised to the
power of 2, like this:
x^{2} =
10x – 25
Your job will be to solve for the given variable, as we’ve done
with other algebraic equations throughout this section. The basic
approach, however, is significantly different from what you’ve done so
far. Instead of isolating the variable on the left, you’ll want to get
everything on the left side of the quadratic equation, leaving 0 on the
right. In the example above, that means moving the 10x
and –25 to the left side of the equation:
x^{2} –
10x + 25 = 0
Now it’s looking like most of the products of binomials we saw in
the previous section, except instead of being just a quadratic
expression, it’s a quadratic equation because it’s set equal to 0.
To solve the equation, we need to factor it. Factoring a quadratic
equation means rewriting it as a product of two terms in parentheses,
like this:
x^{2} –
10x + 25 = (x – 5)(x
– 5)
How did we know to factor the equation into these binomials?
Here’s the secret: Factoring quadratic equations on the GRE always fits
the following pattern:
(x ± m)(x
± n)
Essentially, we perform FOIL in reverse. When we approach a
quadratic like x^{2} –
10x + 25, the two numbers we’re looking for as
our m and n terms need to
multiply to give the last number in the equation.
The last number in the equation is 25, so we need to find two numbers
whose product is 25. Some pairs of numbers that work are:
1 and 25
–1 and –25
5 and 5
–5 and –5
Further, the sum of the two numbers needs to give the
middle number in the equation. Be very careful that you
don’t ignore the sign of the middle number:
x^{2} –
10x + 25
Since you’re subtracting 10x, the middle number
is –10. That means the m and n numbers
we seek not only need to multiply to 25 but also need to add to –10.
Going back to our factor list above, –5 and –5 is the only pair that
works. Substituting this into the pattern gives:
(x – 5)(x – 5)
And since we originally set the equation equal to 0, we now have:
(x – 5)(x – 5) = 0
For the product of two terms to equal 0, that means that either
one could be 0. Here both terms are (x – 5), so
x – 5 must equal 0.
x – 5 = 0
x = 5
The final answer is x = 5.
In this example, m and n are
equal, which is why we end up with only one answer. But that’s usually
not the case. Let’s look at another example, using different numbers:
x^{2} =
–10x – 21
To solve for x, first move everything to the left
to set the equation equal to 0:
x^{2} +
10x + 21 = 0
Now we need to figure out what numbers fit our pattern:
(x ± m)(x
± n)
We know from our equation that m × n
needs to equal 21, and m + n
needs to equal 10. So, which numbers work? Let’s look
at m × n = 21 first:
1 and 21
–1 and –21
3 and 7
–3 and –7
We can eliminate (1 and 21) and (–1 and –21), since neither of
these pairs add up to 10. The third pair, 3 and 7, adds to 10, so we can
stop right there and plug these numbers into our pattern:
(x + 3)(x + 7) = 0
If you need to doublecheck your factoring, just FOIL the
resulting binomials, which should bring you right back to the original
quadratic. Since we now have the product of two different binomials sets
equal to 0, one of the two terms needs to be 0. So, either (x
+ 3) = 0 or (x + 7) = 0, which means x
could be equal to –3 or –7. There’s no way to determine for
sure, since both values work.
Quadratic Factoring Patterns
There are three patterns of quadratics that commonly appear on
the GRE. Learn them now, and you’ll work faster on test day.
Pattern 1: x^{2} +
2xy +
y^{2} =
(x + y)(x +
y) = (x +
y)^{2}
Example: x^{2} +
6xy + 9 = (x +
3)^{2}
Pattern 2: x^{2} – 2
xy + y 2 = (x –
y )( x – y) = (x –
y)^{2}
Example: x^{2} –
10xy + 25 = (x –
5)^{2}
Pattern 3: (x +
y)(x – y) =
x^{2} –
y^{2}
Example: (x + 4)(x – 4) =
x^{2} – 16
You may be wondering why we wrote the last pattern with the
factored form first. We wrote it this way because this is the way it
often appears on the GRE: You’ll be given an expression that fits
the pattern on the left of the equal sign [(x
+ y)(x –
y)], and you’ll need to recognize its equivalent form of
(x^{2} –
y^{2}).
Here’s an example of how you can use this pattern on test day:
What is the value of
?
This looks horrific, and, well, it is if you attempt to
perform a full FOIL treatment on it. Who wants to multiply 72 by 85
and deal with all those radical signs? Not us. Not you. Not anyone.
Luckily, quadratic pattern 3 helps us avoid all that work. That
pattern states that whenever we have the sum of two values
multiplied by the difference of those same values, the whole messy
expression is equal to
x^{2} –
y^{2}. Here, if we let
x = and
y =
, then
x^{2} –
y^{2} =
. If this doesn’t look any
better to you, then you’re not realizing that the squared symbol
(the little 2) and the square root symbol (the
) cancel each other out. This
gives the much simpler expression 85 – 72, which equals 13.
This is an excellent example of how changing your math
mindset, something we implored you to do in the previous chapter,
will help you on the GRE. In the difficultlooking problem we just
tackled, your first instinct may have been to hack your way through
the numbers. However, if you instead suspected that the GRE test
makers probably wouldn’t present a problem like this if there wasn’t
a more elegant solution, then you might have searched for an easy
way in. Quadratic factoring pattern 3 does the trick.
Inequalities
Life isn’t always fair. That’s why there are inequalities. An
inequality is like an equation, but instead of relating equal quantities, it
specifies exactly how two quantities are not equal. There
are four types of inequalities:

x > y
x is greater than
y.

x < y
x is less than y.

x ≥ y
x is greater than or equal to y.

x ≤ y
x is less than or equal to y.
So, for example, x + 3 ≤ 2x may be
read as “x + 3 is less than or equal to
2x.” Similarly, y > 0 is another
way of saying “y is greater than 0.” Inequalities may also
be written in compound form, such as 4 < y – 7
< 3y – 10. This is really just two separate
inequalities: 4 < y – 7 and y – 7
< 3y – 10. Another way to think about this compound
inequality is that y – 7, the expression stuck in the
middle, is between 4 and 3y – 10.
Solving Inequalities
Solving inequalities is a lot like solving equations: Get the
variable on one side of the inequality and all the numbers on the other,
using the algebraic rules you’ve already learned. The one exception to
this, and it’s a crucial exception, is that multiplying
or dividing both sides of an inequality by a negative number requires
that you flip the direction of the inequality.
This exception is crucial, so we’ll repeat it:
The Inequality Exception
Multiplying or dividing both sides of an inequality by a
negative number requires you to flip the direction of the
inequality.
Let’s try some examples.
Solve for
x in the inequality
.
First we knock that 3 away from the x by adding 3
to both sides:
That 2 in the denominator is quite annoying, but by now you should
know the fix for that—just multiply both sides by 2 and it will
disappear from the left side of the inequality:
Almost there. We can simplify the right side with our handy
distributive law, multiplying the 2 by both terms in the parentheses to
get a final answer of:
We’re done. The x stands alone, and we know that
it’s less than the expression 4y + 6. Now try this one:
Solve for
x in the inequality
.
Here are the steps, all at once:
Notice that in this example the inequality had to be flipped,
since both sides had to be divided by –2 to isolate the variable in the
end.
To help remember that multiplication or division by a negative
number reverses the direction of the inequality, remember that if
x > y, then –x
< –y. Just as 5 is greater than 4, –5
is less than –4. The larger the number, the smaller it becomes when you
make it negative. That’s why multiplying or dividing inequalities by
negatives requires switching the direction of the inequality
sign.
Inequalities with Two Variables
Another type of inequality problem involves two variables. For
these, you’ll be given a range of values for each of the two variables.
For example:
–8 ≤ a ≤ 0
5 ≤ b ≤ 25
This is just another way of saying that a is
between –8 and 0, inclusive, and that b is between 5
and 25, inclusive. Inclusive means that we include the
values at each end, which is what’s meant by the greater than or
equal to and less than or equal to signs.
If the test makers didn’t want –8 and 0 to be possible values for
a, for example, they would have to use simple
greaterthan and lessthan signs (> and <). That scenario
corresponds to the word exclusive, which means you
should exclude the values at each end.
Once the range of the two variables has been established, the
problem will then ask you to determine the range of values for some
expression involving the two variables. For example, you could be asked
for the range of values of a – b. This is really just
asking for the smallest and largest possible values of a –
b.
One good way to tackle these problems is to whip up a handy
Inequality Table, a table with columns for each of the two variables and
one column for the expression whose range you’re trying to determine.
First write in the largest and smallest values for a
and b from the original inequalities. In this
example, the extreme values for a are –8 and 0, and
for b are 5 and 25. Write these values in the table
so that each combination of a and b is
represented. There will be four combinations total: the smallest value
of a with the smallest value of b; the
smallest value of a with the largest value of
b; the largest value of a with the
smallest value of b; and the largest value of
a with the largest value of b:
a

b

a – b

–8

5


–8

25


0

5


0

25


Now simply evaluate the expression you’re asked about for each of
the four combinations in the table—in this case, a – b:
a

b

a – b

–8

5

–8 – 5 = –13

–8

25

–8 – 25 = –33

0

5

0 – 5 = –5

0

25

0 – 25 = –25

The Inequality Table shows us that the smallest possible value of
a – b is –33 and the largest is –5. Writing this as
a compound inequality gives our final answer:
–33 ≤ a – b ≤ –5
Inequality Ranges
The previous question demonstrated how inequalities can be used to
express the range of values that a variable can take. There are a few
ways that inequality problems may involve ranges. We consider three
scenarios below.
Operations on Ranges
Ranges can be added, subtracted, or multiplied. Consider the
following:
If 4 < x < 7, what is the range
of 2x + 3?
To solve this problem, manipulate the range like an inequality
until you have a solution. Begin with the original range:
4 < x < 7
Since the range we’re ultimately looking for contains
2x, we need to turn the x in
the inequality above into that. We can do this by multiplying the
whole inequality by 2:
8 < 2x < 14
Since we’re doing the same thing to all parts of the
inequality, this manipulation doesn’t change its value or meaning.
In other words, we’re simply invoking the Golden Rule: Do unto one
part what we do unto the other part. But we’re not there yet,
because the range we seek is 2x + 3, not plain old
2x. No problem: Just add 3 to the inequality
across the board, and you have the final answer:
11 < 2x + 3 < 17
Always remember the crucial rule about multiplying
inequalities: If you multiply or divide a range by a negative
number, you must flip the greaterthan or lessthan signs. For
example, if you multiply the range 2 < x
< 8 by –1, the new range will be –2 >
–x > –8.
Absolute Value and Single Ranges
Absolute values do the same thing to inequalities that they do
to equations. You have to split the inequality into two parts, one
reflecting the positive value of the inequality and one reflecting
the negative value. You’ll see an example just below. If the
absolute value is less than a given quantity, then the solution will
be a single range with a lower and an upper bound. An example of a
single range would be the numbers between –5 and 5, as seen in the
following number line:
A single range question will look something like this:
Solve for x in the inequality 2x
– 4 ≤ 6.
First, split the inequality into two. In keeping with the rule
for negative numbers, you’ll have to flip around the inequality sign
when you write out the inequality for the negative scenario:
2x – 4 ≤ 6
2x – 4 ≥ –6
Solve the first:
2x – 4 ≤ 6
2x ≤ 10
x ≤ 5
Then solve the second:
2x – 4 ≥ –6
2x ≥ –2
x ≥ –1
So
x is greater than or equal to –1 and less
than or equal to 5. In other words,
x lies between
those two values. So you can write out the value of
x
in a single range,
.
Absolute Value and Disjointed Ranges
You won’t always find that the value of the variable lies
between two numbers. Instead, you may find that the solution is
actually two separate ranges: one whose lower bound is negative
infinity and whose upper bound is a real number, and one whose lower
bound is a real number and whose upper bound is infinity. Yeah,
words make it sound confusing. A number line will make it clearer.
An example of a disjointed range would be all the numbers smaller
than –5 and larger than 5, as shown below:
On the GRE, disjointed ranges come up in problems in which the
absolute value is greater than a given quantity, such as the
following:
Solve for x in the inequality 3x
+ 4 > 16.
You know the drill. Split ’er up, then solve each inequality:
3x + 4 > 16
3x + 4 < –16
Again, notice that we have to switch the inequality sign in
the second case because of the negativenumber rule.
Solving the first:
3x + 4 > 16
3x >12
x > 4
And the second:
3x + 4 < –16
3x < –20
Notice that
x is greater than the positive
number and smaller than the negative number. In other words, the
possible values of
x don’t lie
between the two numbers; they lie outside the two
numbers. So you need two separate ranges to show the possible values
of
x: –∞ <
x <
–
and 4 <
x < ∞. There are two distinct ranges for
the possible values of
x in this case, which is why
the ranges are called
disjointed. It doesn’t mean
they can bend their fingers back all the way—that’s
doublejointed.
Madeup Symbols
As if there aren’t enough real math symbols in the world, the GRE test
makers occasionally feel the need to make up their own. You may see a
madeup symbol problem on the GRE involving little graphics you’ve never
seen before in a math context. Sure, they look weird, but they often involve
variables and equations, which is why we thought we’d cover them here in the
algebra section. Anyway, there’s a silver lining to this weirdness: Madeup
symbol problems always give you exact instructions of what
to do. Follow the instructions precisely, and you’ll do just fine. Let’s see
how this works with an example:
If x Ω y = 5x +
2y – 10, what is 3 Ω 4?
Some test takers will see a problem like this and think, “Ω is one of
those symbols that only math geniuses learned. I never learned this symbol.
I’m not a math genius. I majored in Spanish. AAAAGGGHHHH!!!!”
Relax: The fact is, no one learned what this symbol means, because the
test makers made it up. Fortunately, the test makers have also made up a
definition for it—that is, they tell us exactly what do to when we see a Ω.
All it requires is some simple calculating.
Look at the first part of the problem again: x Ω
y = 5x + 2y – 10. All
that’s really saying is that whenever we have two numbers separated by a Ω,
we need to take 5 times the first number and 2 times the second number, add
them together, and subtract 10. That’s it—these are the only instructions we
need to follow.
For the expression 3 Ω 4, then, x is 3, which we
multiply by 5 to get 15. Similarly, y is 4, which we’re
told to multiply by 2, which gives us 8. Adding these together gives us 23,
and subtracting 10 brings us to 13 and a quick and easy point. Quick and
easy, that is, if you don’t panic and just follow the directions to a T.
Try one more example to get the hang of this symbol business:
If
j@
k =
, what is 5@–1?
Forget the actual variables for a moment and focus on the instructions
given: Whenever we have two numbers separated by an @, we need to divide the
second number by 2, multiply the first number by 13 and subtract the two
numbers. So, substituting 5 for j and –1 for
k:
The test makers may keep the choices as fractions, or they may decide
to write them in decimal form, in which case –65.5 will be
correct.
GeometryAlgebra Hybrids: Coordinate Geometry
Once upon a time, geometry and algebra hooked up after a drunken
night. The result was coordinate geometry: a geometryalgebra hybrid.
Coordinate geometry combines the graphical figures found in geometry with
the variables used in algebra. So where does it belong, in algebra or
geometry? Geometry wins on the name front, since it appears in the title and
algebra doesn’t. But the GRE Official Guide includes this as an algebra
topic, so we’ll defer to that and cover it here.
The Coordinate Plane
The coordinate plane is where all the magic happens. It’s the
space in which coordinate geometry exists. Pretty snazzy.
Every point on a coordinate plane can be mapped by using two
perpendicular number lines. The horizontal xaxis
defines the space from left to right. The vertical
yaxis defines the space up and down. And the two meet
at a point called the origin.
Every point on the plane has two coordinates. Because it’s the
center of the plane, the origin gets the coordinates (0,0). The
coordinates of all other points indicate how far they are from the
origin. These coordinates are written in the form (x,
y). The xcoordinate is the point’s location
along the xaxis (its distance either to the left or
right of the origin). If a point is to the right of the origin,
its xcoordinate is positive. If a point is to the
left of the origin, its xcoordinate is negative. If a
point is anywhere on the yaxis, its
xcoordinate is 0.
The ycoordinate of a point is its location along
the yaxis (either up or down from the origin). If a
point is above the origin, its ycoordinate is
positive, and if a point is below the origin, its
ycoordinate is negative. If a point is anywhere on
the xaxis, its ycoordinate is 0.
So the point labeled (2, 2) is 2 units to the right and 2 units
above the origin. The point labeled (–7, –5) is 7 units to the left and
5 units below the origin.
Quadrants
Each graph has four quadrants, or sections. The upperright
quadrant is quadrant I, and the numbering continues counterclockwise. At
the exact center of the graph, the origin is not in any of the four
quadrants. Similarly, the xaxis and
yaxis are not in any quadrant, either.
You may be asked where a particular point exists. Although you
could graph the point on a sketched graph, then use the diagram above, a
faster method is to memorize the chart below:
xcoordinate

ycoordinate

quadrant

positive

positive

I

negative

positive

II

negative

negative

III

positive

negative

IV

This chart makes coordinatelocation questions a breeze. Check it
out:
In which quadrant is coordinate pair (–8.4, –2) located?
Sure, you could draw the coordinate plane and plot the pair. Or
you could simply note that both the xcoordinate and
the ycoordinate are negative, which means that (–8.4,
–2) lives in quadrant III.
Distance on the Coordinate Plane
You may come across a GRE math question that asks you to find the
distance between two points on the coordinate plane or to find the
midpoint between two points. This news should make you happy. Why?
Because these are fairly easy as long as you know the necessary
formulas. There are two methods for finding distance and a formula for
finding midpoints. We’ll tackle distance first.
Finding Distance Using the Distance Formula
If you know the coordinates of any two points—we’ll call them
(x1, y1) and
(x2, y2)—you can find their
distance from each other with the aptly named distance formula:
Let’s say you were suddenly overcome by the desire to
calculate the distance between the points (4,–3) and (–3, 8). Just
plug the coordinates into the formula:
Finding Distance Using Right Triangles
You can also solve distance questions using right triangles.
Consider this one:
What is the distance between the points (2, 4) and (5,
0)?
Plot the two points on a graph and connect them with a
straight line:
The distance between these two points is the length of the
dark line connecting them. To calculate this length, let the dark
line be the longest side of a right triangle. Then draw two
perpendicular lines to construct the short sides, as indicated by
the slashed lines in the diagram.
Since the horizontal side went from an
xcoordinate of 2 to an xcoordinate of 5,
its length is 5 – 2 = 3.
Since the vertical side went from a
ycoordinate of 0 to a
ycoordinate of 4, its length is 4 – 0 = 4.
These two sides fit the {3, 4, 5} righttriangle pattern, and
so the long side must be 5. And that’s the distance between the two
points. (No doubt you’ve learned about right triangles, but don’t
worry, we’ll review them extensively in the upcoming geometry
section.)
So which method for calculating distance should you use? If
you’re given the two points, there’s no reason why you shouldn’t be
able to plug them into the distance formula and get the answer
without bothering to sketch out a picture. The reason we also teach
you the righttriangle method is because, hey, you may just like it
better, but more important, a question may present you with a
diagram to start, in which case the triangle method may be
easier.
Finding Midpoints
To find the midpoint between points (x1,
y1) and (x2,
y2) in the coordinate plane, use this formula:
In other words, the x and
ycoordinates of the midpoint are simply the
averages of the x and
ycoordinates of the endpoints. (Again, we know
you’ve heard of and worked with averages before, but we’ll review
the concept in the data analysis section later in the chapter.)
Applying the formula, the midpoint of the line in the coordinate
plane connected by the points (6, 0) and (3, 7) is:
Slope
Coordinate geometry isn’t all the fun and games of calculating
distances and naming quadrants—it’s also about finding slopes and
solving equations. What follows isn’t particularly difficult, but terms
such as slope and yintercept rarely
evoke positive feelings. Stay with us, and you’ll be okay.
A line’s slope is a measurement of how steeply that line climbs or
falls as it moves from left to right. The slopes of some lines are
positive; the slopes of others are negative. Whether a line has a
positive or negative slope is easy to tell just by looking at a graph of
the line. If the line slopes uphill as you trace it from left to right,
the slope is positive. If a line slopes downhill as you trace it from
left to right, the slope is negative. Uphill = positive. Downhill =
negative.
You can get a sense of the magnitude of the slope of a line by
looking at the line’s steepness. The steeper the line, the greater the
slope; the flatter the line, the smaller the slope. Note that an
extremely positive slope is larger than a moderately positive slope,
while an extremely negative slope is smaller than a moderately negative
slope.
Check out the lines below and try to determine whether the slope
of each line is negative or positive and which has the greatest slope:
Lines a and b have positive
slopes, and lines c and d have
negative slopes. In terms of slope magnitude, line a > b
> c > d.
Slopes You Should Know by Sight
There are certain easytorecognize slopes that it pays to
recognize by sight:
 A horizontal line has a slope of zero.
 A vertical line has an undefined slope.
 A line that makes a 45º angle with a horizontal line has a
slope of either 1 or –1, depending on whether it’s going up or
down from left to right.
Of the four lines pictured below, which has a slope of 0,
which has a slope of 1, which has a slope of –1, and which has an
undefined slope?
Line a has slope 0 because it’s horizontal.
Line b has slope –1 because it slopes downward
at 45º as you move from left to right. Line c has
slope 1 because it slopes upward at 45º as you move from left to
right. The slope of line d is undefined because it
is vertical.
Calculating Slope
If you want the technical jargon, slope is a line’s vertical
change divided by its horizontal change. Or, if you prefer the
poetic version, slope is “the rise over run.” Okay, it’s not
Hemingway (The Sun Also Rises), but it’ll do. For
some strange reason, slope is symbolized by the letter
m. We think s would be a better
choice. Hello . . . Maybe s was
taken. Anyway, you have two points on a line—once again
(x1, y1) and (x2,
y2)—the slope of that line can be calculated
using the following formula:
Rise is how far the line goes up between two
points. Run is how far the line goes to the right
between two points. The formula is just shorthand for the difference
between the two ycoordinates divided by the
difference between the two xcoordinates. Let’s
work out an example:
Calculate the slope of the line passing through (–4, 1)
and (3, 4).
The difference in
ycoordinates,
y2 –
y1, is 4 – 1 = 3. The
difference in
xcoordinates,
x2
–
x1, is 3 – (–4) = 7. The slope is simply the
ratio of these two differences,
.
Slopes of Parallel and Perpendicular Lines
The slopes of parallel and perpendicular lines exhibit the
following relationships:
 The slopes of parallel lines are always the
same. If one line has a slope of m, any line parallel to
it will also have a slope of m.
 The slopes of perpendicular lines are always the
opposite reciprocals of each other. A line with
slope m is perpendicular to a line with a
slope of .
In the figure below, lines
q and
r
both have a slope of 2, so they are parallel. Line
s is perpendicular to both lines
q and
r, so it has a slope of
.
Finding the Equation of a Line
Now that you’ve mastered slope, you’re ready to determine the
entire equation for a line that passes through two points. There are two
ways to do this, and we’ll cover both.
The PointSlope Formula
The equation of a line passing through two points in the
coordinate plane can be expressed by the pointslope formula:
(y – y1) =
m(x –
x1)
You already know how to calculate the slope,
m. And x and y
are just variables. You don’t have to do anything special with them
in the formula other than write them down, unchanged.
As for x1 and y1, these are
just the x and ycoordinates of
one of the points given. You can use whichever point you like; just
make sure you use the same point for both the x1
and y1 values.
Let’s use the formula to work through an example, using the
same points we used in the previous question:
What is the equation of the line passing through (–4,
1) and (3, 4)?
We already calculated
. Now we have to decide which of the two points to use in the
equation. Again, either point will do. Maybe (3, 4) in this case
would be a little easier, since it doesn’t have a negative sign.
Letting
x1 = 3 and
y1 = 4 gives:
Believe it or not, we’re almost done! We’ve actually written
the equation of the line passing between the two points; all that’s
left is to simplify it. First, eliminate the fraction by multiplying
both sides by 7:
7(
y – 4) = 7
(
x – 3)
7y – 28 = 3(x – 3)
Then distribute the 3 on the right side:
7y – 28 = 3x – 9
Finally, move the 3x to the left and the –28
to the right to isolate the variables:
–3x + 7y = 19
This is the final equation of the line passing through (–4, 1)
and (3, 4), written in simplest form.
xintercept and yintercept
The x and yintercepts are
the points at which the graph intersects the xaxis
and the yaxis, respectively. A useful way to think
about this is that at the x intercept,
y = 0, and at the yintercept, x
= 0.
Graphically, this is what the x and
yintercepts look like for the equation we
derived above:
To calculate the precise values of these intercepts, recall
the equation of this line, which we figured out earlier: –3x
+ 7y = 19. Since x = 0 at
the yintercept, you can calculate it simply by
setting x = 0 in the equation, like this:
0 + 7y = 19
y =
So
is where the
graph intersects the
yaxis. Since the
xcoordinate at this point is 0, the coordinate pair for
the
yintercept is
. Following the same
mysterious naming conventions that cause slope to be named
m, the point where the graph intersects the
yaxis is called
b.
For the xintercept, let y =
0 in the equation –3x + 7y = 19.
This gives:
–3x + 0 = 19
x =
So
is where the
graph intersects the
xaxis. Since the
ycoordinate at this point is 0, the coordinate
pair for the
xintercept is
.
The SlopeIntercept Equation
The second method for calculating the equation of a line
designated by two points in the coordinate plane is the
slopeintercept equation. It looks like this:
y = mx +
b
It’s likely that you had this one drilled into your head in
high school, and you therefore recall the sound of it while having
absolutely no idea what it means. Relearning it now is like being
reacquainted with an old friend. Aren’t you glad you’re taking the
GRE?
You’re now familiar with each part of this equation:
 x and y are variables,
which you don’t need to change.
 m is the slope, calculated by .
 b is the yintercept,
the point where the graph intersects the
yaxis.
For the line we’ve been working with, the slope,
m, is
.
The
yintercept,
b, is
. Plugging these values into
the slopeintercept equation gives:
This is the equation in slopeintercept format. You could
simplify it by multiplying both sides by 7, then moving the
variables to the left and the numbers to the right. As you may or
may not expect, this gives exactly the same result as the
pointslope format above:
–3x + 7y = 19
This actually makes a lot of sense: The equation for the line
should be the same, regardless of the method you use to calculate
it.
Which Equation Rules?
You might be wondering when you should use each of the two
equations, (y – y1) =
m(x – x1) and
y = mx + b.
Decisions, decisions. Although slopeintercept is more commonly
taught (most of us have at least heard of y =
mx + b), the pointslope
format is actually more useful when it comes to certain problems on
the GRE. Both equations require the slope, so they’re similar in
that respect. The slopeintercept formula requires a very specific
point, the yintercept, but the pointslope formula
can be calculated with any point. Since it’s less restrictive, you
may therefore find more uses for pointslope. But if they flatout
give you the slope and the yintercept, then
y = mx + b
may be the way to go. The bottom line: Know both, and
decide which to use depending on the question’s
parameters.
Word Problems
Perhaps you’ve had nightmares in high school, certainly around SAT
time, involving two trains, traveling in opposite directions, at different
speeds, over different distances . . . That might be the point where you
woke up screaming, or your dream morphed into Aunt Mabel offering you a
piece of cheesecake.
Well, relax. Word problems are nothing more than algebra problems,
complicated by one small inconvenience: You have to come up with your own
equations. Word problems present particular scenarios; the trick is to
translate the information into math. The good news is that once you have
your equations, they usually aren’t hard to solve.
Pretty much any math concept is fair game for a word problem, but
there are certain types that show up with some regularity. We’ll work
through a few examples here, and you’ll see more later in the book.
Using Simultaneous Equations
Try this:
In a sack of 50 marbles, there are 20 more red
marbles than blue marbles. All of the marbles in the sack are either
red or blue. How many blue marbles are in the sack?
Well, these are very nice words and all, but talk is cheap—we want
equations to crunch. And to get equations, we’re going to need to round
ourselves up some variables. We don’t need a variable for total marbles,
since that’s given: 50. However, we do need variables for the red and
blue marbles, since those are unknown quantities. Unlike those nutty
folks who come up with things like m for slope
and b for yintercept, we’ll be
reasonable and just use r for red and b
for blue. Since there’s no yintercept for
miles around in this one, we doubt anyone will mind if we borrow that
b.
Now for some translating: Is there a math expression that means
the same thing as “there are 20 more red marbles than blue marbles”? You
bet:
r = b + 20
Since both r and b are unknown,
we can’t solve the problem from this equation alone, so we’ll need to
squeeze another equation out of this scenario. The only other facts are
that there are 50 marbles total, and only red and blue ones in the sack.
So it must also be true that:
r + b = 50
Eureka! Why so excited? Because we have two different equations,
each containing two variables, which means we have enough information to
solve the problem. And if you remember from earlier in the chapter, we
even have a choice of how to solve such simultaneous equations. We’ll go
with substitution. Since the first equation tells us that r
= b + 20, we can go ahead and
substitute b + 20 for r in the
second equation, giving us:
b + 20 + b
= 50 2b + 20 = 50 2b
= 30 b = 15
Done.
Now perhaps the CAT software would present you with the problem in
the form we just solved. Or, perhaps, if you answered one or two more
questions correctly before this one, that wily CAT would up the ante and
make the question a tad harder by requiring an additional step. For
example, instead of merely asking you to find the number of blue marbles
in the sack, the question could ask you to find the ratio of blue to red
marbles. No problem; it requires a few more steps, but it’s not so
difficult.
First, use either equation to find the number of red marbles.
We’ll just plug 15 for b into r
= b + 20, to get 35 reds. Then set up the
ratio as 15:35 and simplify to 3:7.
Ratios
And speaking of ratios, here’s another kind of word problem based
on this concept.
In a certain class, the ratio of boys to girls
is 4 to 5. If the class has 12 boys, what is the total number of
boys and girls in the class?
Again, the first step is to translate the English into algebra.
Since ratios are parttopart fractions, the ratio of boys to girls in
the class is really just the fraction
. A common careless mistake is to
write the fraction upside down—that is, as
. A great way to avoid making
this mistake is to remember
.
In other words, whatever quantity appears after the word of in the
problem goes in the numerator, and whatever quantity appears after the
word to goes in the denominator.
So, the ratio of boys to girls is 4 to 5, which, in math terms,
can be written like this:
We’re also told that the class has 12 boys. Substituting this into
our equation gives:
We’ll now use the Magic X to crossmultiply diagonally and up:
5 × 12 = girls × 4
60 = girls × 4
= girls
15 = girls
Think 15 is the answer? Think again. The question didn’t ask for
the number of girls, but for the total number of boys
and girls in the class. Easy: Total = 12 + 15 = 27, and we’re
done.
Distance
Since practically everyone who ever attended high school has had
that nightmare about the trains traveling in different directions, the
GRE test makers figure it’s a good problem to torment people with. Of
course, they don’t always use trains. But these problems are really not
that bad if you learn how to handle them; in fact, distance problems
require only one fairly basic formula:
rate × time = distance
The concept is fairly intuitive: If you bike at a rate of 10 miles
per hour, and you bike for 2 hours, you’re going to cover 10 × 2 = 20
miles. What gets confusing is the various ways they state these
problems, but rest assured they always give you enough information to
set up an equation and solve for the variable you seek. Here’s an
example:
Jim rollerskates 6 miles per hour. One
morning, Jim rollerskates continuously for 60 miles. How many hours
did Jim spend rollerskating?
One thing that makes word problems annoying is that they sometimes
include unnecessary facts. Here, for example, we’re told such exciting
things as how Jim is traveling (by roller skates) and
when he started (in the morning). Heck, it doesn’t even matter that his
name is Jim—he could be Egbert the marble collector for
all we care. You have to ignore the nonessentials and focus on the facts
you need to solve the problem. Begin with the trusty distance formula:
rate × time = distance
Then fill in the values you know:
6 miles per hour × time = 60 miles
Divide both sides by 6 miles per hour to calculate the time spent
as 10 hours.
This next distance problem is a bit more difficult because it
requires one extra step. Instead of being given two variables and simply
setting up an equation to solve for the third, you need to first
calculate one of the variables before getting to that point. Try it out:
A traveler begins driving from California and
heads east across the United States. If she drives at a rate of
528,000 feet per hour, and drives 4.8 hours without stopping, how
many miles has she traveled? (1 mile = 5,280 feet)
The reason you need to go the extra mile (so to speak) is because
the rate is given in feet per hour but the distance traveled is given in
miles. The units must jibe for the problem to work, so a conversion is
necessary. Luckily, the numbers are easy to work with: The driver drives
528,000 feet per hour, and 5,280 feet equal one mile. So to change her
rate into miles per hour, simply divide:
= 100 miles per
hour
That’s some fast traveler. Now plug the values from the problem,
including this new one, into the formula:
100 miles per hour × 4.8 hours = distance
The distance traveled is 480 miles.
Work
Work sucks. You’re there from at least 9 to 5, get two weeks off
per year (at best), and you’ve got a boss constantly checking up on you.
Work word problems on the GRE are a breeze in comparison. Work word
problems are very similar to distance word problems, except the final
outcome is units of something produced instead of distance covered. The
other two variables are analogous to those in distance problems: the
rate at which the units are produced, and the time someone spends
producing them. If you knit 2 sweaters per hour, and knit for 8 hours,
then you’ll knit 2 × 8 = 16 sweaters. This generalizes to the same basic
formula we saw earlier:
rate × time = units produced
Since you’ve seen this basic mechanism in action in the previous
section, we’ll jump right to a difficult work word problem:
Four workers can dig a 40foot well in 4 days.
How long would it take for 8 workers working at the same rate to dig
a 60foot well?
Sure, it’s complicated, but the three parameters are the same:
rate, time worked, and amount of work completed. We know the latter: 60
feet dug. The middle amount, time, is what we’re looking for, so we’ll
just leave that as our unknown variable for now. The question then
hinges on the rate, which we need to calculate ourselves. Well, since 4
workers dig 40 feet in 4 days, we can divide by 4 to determine that
those 4 workers can dig 10 feet in one day. The 8 workers that we’re
interested in dig at the same rate, and since they represent twice as
many people, they’ll be able to dig twice as many feet in a day: 20. Now
we have the variables to plug into our formula:
20 feet per day × time = 60 units produced (in this case, feet
dug)
Divide both sides by 20 to get a time of 3 days of work for the 8
workers to dig the 60foot well.
Many other word problems follow the same basic form as the
examples above. For example, in a price word problem, the price of an
item times the number of items bought will equal the total cost. The
particulars are slightly different, but the concept is basically the
same. Use the information given in the problem to get two of the values,
and you’ll always be able to calculate the third. And you’ll never have
a word problem nightmare again.
That wraps it up for algebra. Now we move on to triangles,
circles, and all of the other fun shapes of our next math subject,
geometry.