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Mathematics
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Math for 14 year olds
Most of the math on the GMAT is set to what would be called junior high school level in the
United States. It involves simple algebra and geometry. The irony is that most students could
have done better on the GMAT Math section ten years earlier when the material was fresher in
their minds. To help you, we have math chapters dedicated to Algebra, Percentages/Ratios,
Geometry, and Arithmetic.
Test writers must only use pre-high school level math because if they used higher level math, it
would create an advantage for math majors. The result is that there is no calculus or trigonometry
on the exam. For test writers this creates a challenge, "How do you create questions using math
for 14 year olds that Harvard grads will get wrong?" The answer is that they primarily use:
1) trick questions
2) extremely difficult Word Problems
3) questions that require extensive deductive reasoning (Geometry and Data Sufficiency)
Word Problems are the focus of the test, and for that reason, the Math section is more of a
reading test than strictly a "math" test (for those British/Commonwealth students "maths" is
"math" in the U.S.). Test writers devise extremely complex and contorted Word Problems that can
fool 95% of GMAT students using only basic math principles. That is why we have a chapter
dedicated exclusively to Word Problems strategy. Foreign students should pay particular attention
to this section because Word Problems will give them the most trouble. We focus on Word
Problems in chapter 7.
Note: these chapters are particularly dense with material. Take them slowly and digest the
information thoroughly before progressing to the next chapter.
Scrap Paper
We discussed scrap paper earlier in the guide. Doing calculations in your head may increase
errors, particularly under the pressure of test day. This is no-win situation because copying
questions also invites errors. Because you cannot write on the screen, every question has to be
partially recopied onto scrap paper. Students should always practice with unlimited scrap paper
(just like on test day) in order to get used to this awkward process of recopying questions.
This chapter is broken into two parts:
I. General Math Strategies
II. Basics
I. General Math Strategies
A. Backsolving
B. Plug-In
C. Possible Range Strategy
A. Backsolving (otherwise known as "reverse-solving")
This is the most important strategy for the math section! Backsolving involves inserting
answer choices to solve problems. This is often preferable to trying to translate a complicated
question into an algebraic equation. Inserting answer choices into a Word Problem can also make
a complicated question more understandable. Backsolving helps you eliminate or choose a
sample answer. Remember, math questions on the GMAT use very simple principles, so the test
writers have to make the questions as complicated and intimidating as possible. Some questions
are actually written so that backsolving is the only effective way to solve the problem.
How to Backsolve:
1. Decide if the problem is too complicated to solve algebraically (this should take only a
few seconds).
2. Insert the middle answer--the one that would be in the middle of potential answers if it
were on a number line.
3. If a smaller number would work, choose answer choices 1 or 2; if a larger number would
work, choose 4 or 5. (The answer choices are usually arranged from lowest to highest
value--answer choices 1 through 5).
4. Eliminate down to one answer and choose.
Try backsolving on this question:
When the positive integer Z is divided by 24, the remainder is 10. What is the remainder when Z
is divided by 8?
a) 1
b) 2
c) 3
d) 4
e) 5
Solution
Notice that this question seems to defy a quick algebraic solution. The best way to address this
problem is to backsolve, to take potential answers and feed them into the question until one
answer works. Pick a number for Z such that Z/24 has a remainder of 10; that is, a number 10
greater than a multiple of 24. 24 is a multiple of 24, so let Z = 24 + 10 = 34. When 34 is divided by
8 you get 34/8 = 4 with a remainder of 2. Not satisfied? Try 58 (which is 2 × 24 with a remainder
of 10). 58 divided by 8 gives a remainder of 2. It appears that (B) is the correct answer.
Strategy: Plug-in is an even more effective strategy for double checking your answers. When you
arrive at an answer, plug in the answer to test that it works or plug in more numbers. Remember
that you have nearly two minutes to do each question on the GMAT CAT. This gives you plenty of
time to double check yourself; however, pacing is still extremely important, as you read in
Chapter One.
B. Plug-In
Sometimes the best approach to backsolving is not to put in sample answer choices, but to
insert numbers that prove/disprove the question. The numbers you choose for backsolving should
fit the question's parameters. For example, if the question asks for an integer, you should insert
integers. Usually try plugging in a few different numbers (positive, negative, etc.).
1. Decide if the problem is too complicated to solve algebraically (this should take a only
few seconds).
2. Insert sample numbers into the equation. Depending on the question, try negative
numbers, positive numbers, odd/even and fractions.
3. Eliminate down to one answer and choose.
Plug-in Example
If n is an even integer, which of the following must be an odd integer?
a) 3n - 2
b) 3(n + 1)
c) n - 2
d) n/3
e) n/2
Solution
(B) Every time you have variables in the answer choices, you should plug in. Make n equal to 2. If
n is 2, then 3(n + 1) = 9. Since our target is an odd integer, this answer choice works. Try a few
more numbers to double check. For example, 2 may work with choice (e) to make an odd number
(1), but it will not work with any other even numbers.
C. Ballpark Strategy (otherwise known as the possible range strategy)
This strategy allows you to answer questions quicker than doing the calculations, and it is an
effective tool to double-check your answers. Using the Ballpark Strategy, you find an answer by
what could reasonably be in the range of the answer. This is particularly useful when the possible
answers are scattered over a large range.
Try the Ballpark Strategy here:
If 0.303z = 2,727, then z =
a)9,000
b)900
c)90
d)9
e)0.9
Solution
(A) Because the answer choices are so far apart, you can ballpark this problem. Think about it:
.303 is close to 1/3. 1/3 of z = 2,727, then what answer could possibly be correct? You don't even
have to do the math. 2,727 is about 1/3 of 9,000; therefore, the answer must be 9,000, according
to the Ballpark Strategy (note that there are no other answers even in the 9,000 range. Or, you
could multiply both sides by 1000 to eliminate the decimal points, then divide 2,727,000 by 303
and get the same answer.
Strategy: Ballparking is an effective strategy for double checking your answers. When you arrive
at an answer, make sure it is in the ballpark of what the answer could be.
II. The Basics
This chapter contains a basic review of math concepts. Most students should skim through
this section. It is vital that you know all the terminology in this section since the questions on the
test will assume you know it.
A. Integers
B. Positive / Negative Rules
C. Fractions
D. Equivalent Fractions
E. Multiplying and Dividing Fractions
F. Adding and Subtracting Fractions
G. Decimals
H. Adding and Subtracting Decimals
I. Multiplying and Dividing Decimals
J. Averages and Medians
A. Integers
A positive number is a number greater than zero, such as + 5 (usually written as 5).
A negative number is a number less than zero, such as -5.
A whole number is a number that does not include a decimal part or a fractional part. It is also
called an integer.
The absolute value of a number, written as | -5 |, is the magnitude of the number; the absolute
value of + 5 is equal to the absolute value of -5, written as |+5 | = |-5 | = 5.
Zero is an integer
B. Positive / Negative Rules
What happens when positive and negative numbers are added/subtracted/multiplied?
Multiplication/Division
Positive × Positive = Positive, for example 2 × 2 = 4
Positive × Negative = Negative, for example 2 × -2 = -4
Negative × Negative = Positive, for example -2 × -2 = 4
Addition/Subtraction
Adding a negative number is the same as subtraction.
4 + (-5) = 4 - 5 = -1
Subtracting a negative number is the same as addition.
4 - (-5) = 4 + 5 = 9
An even number is an integer that is divisible by 2 (4, 6, 8, 10).
An odd number is an integer not divisible by 2 (3, 5, 7, 9).
A prime number is a positive integer that has exactly two different positive divisors, 1 and itself.
For example, 2, 3, 5, 7. The number 1 is not a prime number since it has only one positive divisor.
Consecutive numbers are a set of numbers in which each member of the set is the successor of
its predecessor.
Examples:
even consecutive numbers: 4, 6, 8, 10
odd consecutive numbers: 3, 5, 7, 9
prime consecutive numbers: 3, 5, 7, 11, 13.
Adding, Subtracting and Multiplying odd/even: The following is true of even and odd whole
numbers (use example numbers in your mind to illustrate (note that zero is an even number):
Even + Even = Even
exg. 4 + 4 = 8 (even)
Odd + Even = Odd,
exg. 3 + 4 = 7(odd)
Odd + Odd = Even
exg. 3 + 3 = 6 (even
Even × Even = Even
exg. 2 × 2 = 4 (even)
Even × Odd = Even
exg. 2 × 3 = 6 (even)
Odd × Odd = Odd
exg. 3 × 3 = 9 (odd)
Even - Even = Even
exg 16 - 8 = 8 (even)
Even - Odd = Odd
exg. 16 - 5 = 11 (odd)
Odd - Odd = Even
exg. 9 - 5 = 4 (even)
Even/Odd Example
If k is an odd integer, state whether each of the following is odd or even:
a) k + k + k
b) k × k × k
c) k + 2k
d) 2k × k
Solution
a) (k + k) is even. Thus (k + k) + k is an even plus an odd, which is odd.
b) k × k is odd. Thus (k × k) × k is an odd times an odd, which is odd.
c) k + 2k is an odd plus an even, which is odd.
d) 2k is even. An even times an odd is even.
Strategy 1: If you do not remember the rules, merely plug in by selecting an odd number (e.g.,
3) or an even number and perform the required operation.
Strategy 2: Get used to plugging in numbers. On most math questions, you will have to use the
either plug in strategy or backsolve.
A factor is an integer that divides another number resulting in a whole number. Consequently, a
number can be expressed as a multiple of each of its factors. The number 24 has factors 1, 2, 3,
4, 6, 8, 12, 24. Note that 24 is a multiple of any one of its factors, i.e., 24 is a multiple of 8.
The least common multiple (LCM) of several numbers is the smallest integer which is a
common multiple of the several numbers.
LCM example
Write the LCM (Least Common Multiple) of 6 and 12.
Solution
The factor 2 × 3 is used only once because it occurs in both numbers. Thus, the LCM is (2 × 3 ×
2) = 12. Or, what is the smallest number that both 6 and 12 go into? 12. 12 is the lowest number
divisible by both 6 and 12.
EXAMPLES
Example 1
Compute each of the following:
a) 2 - 3
b) 4 + 6 - 10
c) (-5 + 2)(-3)
Solution
a) 2 - 3 = -1
b) 4 + 6 - 10 = 10 - 10 = 0
c) (-5 + 2)(-3) = (-3)(-3) = 9
Example 2
State all the factors of 63.
Solution
The number 63 can be divided by 1, 3, 7, 9, 21, and 63. Hence, these are its factors.
Example 3
Write 63 as a product of prime numbers.
Solution
The number 63 is obviously not divisible by 2, but is divisible by 3. Hence,
63 = 3 × 21 = 3 × 3 × 7
Both 3 and 7 are prime numbers. We do not include 1 as a prime number.
Example 4
Find the LCM (Least Common Multiple) of the three numbers 20, 30 and 50.
Solution
Write each number as a product of prime numbers:
20 = 2 × 2 × 5
30 = 2 × 3 × 5
50 = 2 × 5 × 5
The factor 2 × 5 occurs in all three numbers; thus it is used only once.
The LCM is LCM = (2 × 5) × 2 × 3 × 5 = 300
C. Fractions
A fraction is one number divided by another number. It is division, such as 3/5.
The numerator is the top number, and the bottom number is the denominator. The denominator
represents the number of equal parts into which an entity has been divided; the numerator
represents the number of parts that are selected. For example, if a garden is divided into 5 equal
plots, 3 of the plots is 3/5 of the garden.
A fraction that has 0 as its denominator (e.g., 5/0) is infinitely large and undefined (not a real
number or an integer). If the numerator is zero, then the fraction equals zero (e.g. 0/5 = 0). If the
fraction has a numerator equal to the denominator (e.g., 5/5), the fraction is equal to 1.
A mixed number is a number that is an integer plus a fraction. The number 4 2/3 is the integer 4
plus the fraction 2/3. Any mixed number can be written as a fraction, and any fraction greater than
1 can be written as a mixed number. To express 4 2/3 as a fraction we multiply 4 × 3, add the
numerator to this product, 12 + 2 = 14, and divide by the denominator: 4 2/3 = 14/3. To convert
the fraction 17/5 into a mixed number, divide by the denominator (17 divided by 5 is 3 with 2
remaining), and add the remainder over the denominator: 17/5 = 3 2/5.
Example 5
Convert 4 5/7 into a fraction.
Solution
We multiply 4 × 7 and obtain 28. Add 5 to this and obtain 33. Put this over 7, and we find 4 5/7 =
33/7
Example 6
Convert 79/9 into a mixed number.
Solution
Divide 79 by 9 and obtain 8 with 7 remaining. Now add to the 8 the fraction 7/9 (8 7/9).
D. Equivalent Fractions
A fraction that has a common factor in both numerator and denominator is equal to the fraction
with the common factor canceled. The fraction 6/10 is equivalent to the fraction 3/5 since the
common factor 2 occurs in both numerator and denominator of 6/10. In fact, the following
fractions are all equivalent: 3/5 = 6/10 = 9/15 = 12/20. A fraction that has no common factors in
the numerator and denominator is said to be expressed in lowest terms.
A fraction with a negative numerator or denominator is equivalent to a negative fraction, that is,
-3/5 = 3/-5 = - (3/5). If both numerator and denominator are negative, the fraction is positive, that
is, -3/-5 = 3/5.
Example 7
Express 26/16 as a mixed number in lowest terms.
Solution
The mixed number is found by dividing by 16 giving 1 with 10 remaining. Hence
26/16 = 1 10/16 = 1 5/8
E. Multiplying and Dividing Fractions
To multiply fractions, cancel out any common factors that appear in both numerators and
denominators. Then multiply all numerators to form one numerator and all denominators to form
one denominator. This final fraction may then be written as a mixed number, if desired. To divide
fractions, say (x/y)/(a/b), invert the divisor (the fraction a/b) and multiply the two fractions, i.e.,
(x/y)/(a/b) = (x/y) × (b/a).
For instance,
(5/6) × (7/2) = (7× 5)/(6 × 2) = 35/12
F. Adding and Subtracting Fractions
To subtract one fraction from another, we simply add a negative fraction to a second fraction.
Consequently, the rules for adding and subtracting are the same. The first step is to write the
fractions such that each fraction has the same denominator. Then add or subtract the
numerators. Then simplify.
1/3 + 1/4 = 4/12 + 3/12 = 7/12
To write all fractions with the same denominator, a quick choice is to multiply all denominators
together. Hoever, this may give a rather large denominator. To avoid a large denominator, we
could find the least common denominator (LCD); it is the least common multiple (LCM) of all
the denominators.
G. Decimals
A decimal fraction is a fraction whose denominator is a power of 10 but not a factor of the
numerator. For example, the fraction 3/100 is written in decimal form as 0.03; the fraction
223/10,000 is 0.0223. Note that the number to the right of the decimal point is the numerator of
the decimal fraction, and the denominator is 10 raised to the power n where n is the number of
places to move the decimal point. To write 0.0031 as a decimal fraction, 31 is the numerator, and
because we moved the decimal point 4 places, the denominator is 10 = 10,000; consequently,
the decimal fraction is 31/10,000.
For decimals
.1 = one tenths
.01= one one-hundredths
.001 = one one-thousandths
Rule #1: If you multiply a decimal by 10, you would move the decimal 1 to the right; 100, move 2
to the right; 1000, move 3 to the right, and so on.
For instance, 100,000 × 0.0054 = 540.
You see that there are 5 zeros in 100,000; therefore, we moved the decimal 5 places to the right.
Rule # 2: When you divide by 10, you move the decimal once to the left; 100, move 2 to the left;
1000, move 3 to the left.
Example 8
Write 31.2/(100,000) as a decimal.
Solution
When we divide by 100,000, which is 10 , we move the decimal point 5 places to the left. Then
31.2 = 0.000312.
A mixed decimal is the sum of an integer and a decimal fraction, much like the mixed fraction.
The integer 5 added to the decimal fraction 35/100 is 5 + 35/100 = 5 + 0.35 = 5.35. The mixed
decimal, or a decimal fraction, is usually simply called a decimal. (Note: the leading 0 in 0.35 is
simply convention; it does not have mathematical significance: 0.35 = .35.)
Example 9
Write the fraction 649/100 as a decimal.
Solution
The numerator 649 can be written as the decimal "649.0". When we divide by 100, we move the
decimal point 2 places to the left so that
649/100 = 6.49
It is not necessary to write any zeros after the 9; that is, 6.490 is equivalent to 6.49.
Example 10
Express 0.075 as a fraction in lowest terms.
Solution
The decimal is expressed as a fraction .075 = 75/1000 The denominator and numerator are
factored resulting in 75/1000 = (25 × 3) / (25 × 40) = 3/40
H. Adding and Subtracting Decimals
To add or subtract decimals, we write the decimals in a column with the decimal points aligned
vertically. First, combine the decimals with a plus sign. Next, combine the decimals with a minus
sign. Then subtract the sum of the negative decimals from the sum of the positive decimals. In
performing these tasks, we add zeros to the right of the decimal point so that each number has
an entry in each column. For example, if we subtract 3.021 from 5, we write 5 as 5.000 so that
there is an entry in each of the 3 places to the right of the decimal in both numbers.
Example 11
Add 5 + 2.783 + 3.04.
Solution
Write the decimals in a column, with zeros added if none exist:
5.000
2.783
+ 3.040
10.823
Example 12
Compute 6.98 + 3.217 + 3 - 3.637
Solution
Add the positive decimals and the negative decimals:
+ 6.980
+ 3.217
+ 3.000
- 3.637
9.56
Subtract the sum of the negative decimals from the sum of the positive decimals:
13.197
- 3.637
9.560
I. Multiplying and Dividing Decimals
Multiply two decimals just like you would multiply two integers. The number of decimal places
in the product is then equal to the total of the decimal places in the two decimals.
To divide two decimals, move the decimal point in the divisor (the number doing the dividing)
to the right so that the divisor is an integer. Move the decimal point in the dividend to the right the
same number of places. Now perform the division, placing the decimal point in the answer directly
above the decimal point in the dividend.
Example 13
Compute 0.05 × 12.
Solution
Set up the multiplication as though the decimals were integers:
12 × .05 = .60
The answer must have a total of 2 decimal places: .60.
J. Averages and Medians
The average, or arithmetic mean, is the sum of a set of numbers divided by the total number
of elements in the set. Arithmetic mean is used on the GMAT because it is a more precise term
than average.
If all the numbers in a set are arranged in ascending or descending order, the middle number
is the median. The median is different from the arithmetic mean. Half of the people in a country
earn more than the median income, and half earn less. The average income does not split the
people into a top half and bottom half. For example, if 5 people have weekly incomes of $200,
$300, $500, $13000 and $6000, the median is $500, but the average is $4000. If a large number
of people earn very little and a few earn a huge amount, the average would be quite impressive,
but the median would be surprisingly low.
Example 14
Ten students on an exam scored 20, 30, 30, 25, 30, 35, 80, 60, 40, and 90. Calculate the average
and the median.
Solution
The average is the sum of all the numbers divided by 10: average = (20 + 25 + 3 × 30 + 35 + 40 +
60 + 80 + 90) / 10 = 44. The median is, for the case of an even number of entries, the average of
the two middle numbers when arranged in order: median = (30+35)/2 = 65/2 = 32.5
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Algebraic Expressions
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This chapter is divided into three parts:
I. Simplifying Rules
II. Complex Expressions with Exponents
III. Manipulating Complex Expressions
I. Simplifying Rules
A. Exponent Rules
B. Simplifying Expressions
A. Exponent Rules
This section is intended as a basic review of algebra; skim the material and review what is
necessary. When attempting to solve algebra problems, it is important that you strictly follow the
laws concerning exponents.
Rule I: Add the exponents when multiplying two powers of the same base:
33 =3
Rule II: Multiply the exponents when obtaining the power of a power:
(3 )
=3
Rule III: Subtract the exponents when dividing a power of a specified base by another power of
the same base:
3 /3 = 3
a negative exponent is the equivalent of (1/the number) raised to the power. For example, 3 = 1
/ 3 or 1 / 9.
Rule IV: The power of a product of factors is written by raising each factor to the specified power.
In general,
(abc) = a b c
Rule V: The power of a fraction is written by raising the numerator and the denominator to the
specified power. This is expressed by
(a/b) = a /b
B. Simplifying Expressions
Rule I: Perform multiplications and divisions before you perform additions and subtractions. The
expression x + 2y/3 is not the same as (x+2y)/3.
Rule II: Combine all like terms in an expression. The expression 2x + x - y + 4y is simplified by
combining the like terms resulting in 3x + 3y .
Rule III: Perform operations inside parentheses first. The expression (x + 2y)/3 is not the same
as x + 2y/3. With the parentheses present, we add first, then divide; with no parentheses, we use
Rule I and divide first, then add.
Rule IV: Eliminate inner parentheses first and the outermost parentheses last. In the expression
x(x + 2(3x + 4) -3), we remove the inner parentheses first obtaining x(x + 6x + 8 - 3); then we
combine like terms giving x(7x + 5). We may then remove the last parentheses providing 7x +
5x. Often brackets and braces are used if two or three sets of parentheses are needed.
II. Complex Expressions with Exponents
A. The Products of Monomials and/or Polynomials
B. Quadratic Equations
C. Factoring
D. Division of Algebraic Expressions
A. The Products of Monomials and/or Polynomials
To multiply a binomial (a polynomial with two terms like, x + 3) by a binomial use FOIL:
First: Mulitply the first terms of each binomial.
Outside: Multiply the outside terms (the outer extreme terms of the expression).
Inside: Multiply the inside terms (the inner extreme terms of the expression).
Last: Multiply the last terms of each binomial.
Then add them making sure to combine like terms.
Example 1
(x + 3)(x - 5) =
Solution
F: x × x = x
O: -5 × x = -5x
I : 3 × x = 3x
L: +3 × -5 = -15
Therefore, the answer is the sum of F + O + I + L:
x - 5x + 3x - 15
then combine (-5x and +3x)
x - 2x - 15
Example 2
(2a + b)(a - 2b) = 2a -4ab + ab - 2b
=2a - 3ab - 2b
B. Quadratic Equations
A quadratic equation may be written in the standard form:
ax + bx + c = 0 where a, b, and c are constants.
If either b or c is zero, the equation is relatively easy to solve. If neither b nor c is zero, we will
consider only those equations in which the quadratic expression on the left side can be factored.
Try this sample using the techniques from the section above:
x - 2x - 15
Here we do "reverse foil."
FIRST:
The first number x is easy. It must be the product of two numbers, which in this case are most
likely x and x.
(x
)(x
)
LAST:
The last number -15 is trickier:
The second number in each binomial must multiply to be 15, so we are looking for two
factors:(15, 1) and (3,5). One of the two numbers must be negative.
The middle number is -2x. The middle number is the combination of the OUTSIDE, INSIDE steps.
Note that among the factors, 3 and 5 can sum to -2x, here is how:
(x + 3)(x - 5)
OUTSIDE = -5x
INSIDE = 3x
Total= -2x
Thus,
(x + 3)(x - 5) must be the binomial solution because FOIL will produce
x - 2x - 15.
Strategy: Quadratic equations usually have two answers. For example X = 25 has two answers,
-5 and +5. The test writers often generate trick questions by exploiting the fact that many students
forget that these equations have two answers. Data sufficiency questions will often turn on this
issue.
Example 3
Solve x + 4x = 0
Solution
The left side of this equation can be factored so that we can write it as
x(x + 4) = 0.
This equation is satisfied if either factor is zero. Consequently we write,
x=0
x + 4 = 0.
The solution is x = 0, x = -4.
Example 4
Solve x - 5x + 4 = 0.
Solution
This quadratic equation has all terms present, so we expect that the left side can be factored.
(x - 4)(x - 1) = 0
We set each factor equal to zero and solve for x:
x-4=0
x-1=0
The solution is x = 4, 1.
Example 5
Solve 2x - 3 = (4 / x) + x.
Solution
First, put the equation in standard form by multiplying both sides by x.
This equation is factorable so we write
2x - 3x = 4 + x
x - 3x - 4 = 0
(x - 4)(x + 1) = 0
Set each factor equal to zero and solve for x:
x - 4 = 0; x = 4
x + 1 = 0; x = -1
The solution is x = 4, -1.
To multiply a monomial by a monomial, multiply the numerical coefficients and then follow the
laws of exponents with the same base. For example,
(2rs )(-3r s )= 2(-3)(r x r )(s x s ) = -6r s
To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial.
This is illustrated by
4x(3x - 2xy + y ) = 12x - 8x y + 4xy
To multiply two polynomials, multiply one of them by each term of the other and then combine like
terms. The following illustrates the process:
(2x - y)(2 + x - 3y) = 2x(2 + x - 3y) - y(2 + x - 3y)
= 4x + 2x - 6xy - 2y -xy + 3y
= 4x - 2y + 2x - 7xy + 3y
C. Factoring
Before we proceed with the rules of division, we need to introduce the notion of factoring. If an
algebraic expression can be written as the product of two other algebraic expressions, the two
other expressions are called factors. Sometimes there may be more than two factors. Factoring
an expression will often allow us to simplify a problem, so it is important when solving many
algebra problems.
There are several algebraic expressions that occur frequently. You should be able to recognize
their factors immediately.
Five common algebraic factoring types:
1. ax + ay = a(x+y)
2. x - y = (x + y)(x - y)
3. x + 2xy + y = (x + y)
4. x - 2xy + y = (x-y)
5. x + (a + b)x + ab = (x + a)(x + b)
Example 6
Factor 3a b - 15a b
Solution
Note that 3a b is common to both terms. Factor it out and obtain
3a b - 15a b = 3a b(b - 5a ).
The two factors are 3a b and (b - 5a ).
Example 7
Factor 4x - 9.
Solution
This is the difference of two perfect squares: (2x) and 3 . Using factoring expression No.2 (see
above), we see that
4x - 9 = (2x - 3)(2x + 3)
NOTE: No. 2 is the most common expression that you will need to factor, so be sure to recognize
it quickly.
Example 8
Factor y + 6y + 9.
Solution
The number 9 is 3 , so we try type No.3 and observe that:
y + 6y + 9 = (y + 3) .
Note that this can be factored using type No.5. What two numbers add together to give 6 and
multiply together to give 9? The answer is 3 and 3 so that
y + 6y + 9 = (y + 3)(y + 3).
Example 9
Factor r - 8r + 16.
Solution
The number 16 is 4 so we try type No.4 and observe that
r - 8r + 16 = (r - 4) .
This also can be factored using type No.5. What two numbers add together to give -8 and multiply
together to give 16? The answer is - 4 and - 4 so that
r - 8r + 16 = (r - 4)(r - 4).
D. Division of Algebraic Expressions
There are two common divisions that you may be asked to perform. The first involves dividing
an algebraic expression by a monomial, such as
(3x + 6ax + 15xa )/3x
We first recognize the numerator as a type No.1 expression and factor out 3x. The factor 3x
then cancels out with the denominator and we have
(3x + 6ax + 15xa )/3x = [3x(1 + 2ax + 5a )]/3x = 1 + 2ax + 5a
The second type of division involves divisions such as
(5x - 15x + 10) / (5x -10)
We write the denominator as 5(x -2) and the numerator as 5(x - 3x + 2). The numerator is a
type No.5 expression: What two numbers added together give - 3 and multiplied together give 2?
They are -2 and -1. Hence, we can write
[5(x - 3x + 2)] / [5(x - 2)] = [(x - 2) (x -1)]/[x - 2] = x - 1.
III. Manipulating Complex Expressions
A. Multiplying and Dividing Algebraic Fractions
B. Addition of Algebraic Fractions
C. Equations
D. Equivalent Equations
E. Linear Equations
F. Inequalities
G. Simultaneous Equations
A. Multiplying and Dividing Algebraic Fractions
When multiplying algebraic fractions, we search for common factors: factors that are common
to both numerators and denominators. These factors are canceled, and the resulting fraction
simplified, if possible. When dividing algebraic fractions, we invert the divisor (the fraction we're
dividing by) and then multiply, searching for common factors.
Example 1
Perform the division.
3a - 4 divided by 3ab - 4b
2 - 3a
4 - 9a
Solution
First, invert the divisor and express the division as the multiplication.
3a - 4 x 4 - 9a
2 - 3a
3ab - 4b
Next, factor any factorable expressions.
4 - 9a = (2 - 3a)(2 + 3a)
3ab - 4b = b(3a - 4)
The given division is now written and the common factors (in red) canceled.
3a - 4 x
2 - 3a
(2 - 3a)(2 + 3a) = 2 + 3a
b(3a - 4)
b
B. Addition of Algebraic Fractions
When adding (or subtracting) algebraic fractions, we proceed as in arithmetic:
o
o
o
Find the lowest common denominator (LCD) of the fractions.
Write each fraction using the LCD of the fractions.
Add (or subtract) the numerators. The denominator will be the LCD. Simplify the
resulting fraction.
Example 2
Express as a single fraction:
2x + 1 - x - 3
2a
3a
Solution
The lowest common denominator is 6a. Each fraction is written with its denominator as 6a. This
gives
2x + 1 - x - 3
2a
3a
= 3(2x+1) - 2(x-3)
6a
6a
= 6x + 3 - 2x - 6
6a
6a
= 6x + 3 - 2x + 6
6a
= 4x + 9
6a
C. Equations
An equation is a mathematical statement that two algebraic expressions are equal. The letters
in the equation are the unknowns. An equation may be thought of as asking the question, "What
numerical value of the unknown satisfies the equation?" (Satisfies means to make both sides
equal.) The value of the unknown is called the solution of the equation, or root of the equation.
We solve an equation by finding the numerical value of all of its roots.
There are equations that do not have real roots. For example , the equation x = -4 does not
have a real root; the square of any real number, whether positive or negative (e.g., + 2 or -2) is
positive. Hence, there is no real root to x = -4.
D. Equivalent Equations
Two equations that have the same roots are said to be equivalent. To solve an equation, we
often put the equation in a form that is more easily solved with one or both of the following
operations:
o
o
Add or subtract the same term to or from each side of an equation.
Multiply or divide each side of an equation by the same number or algebraic
expression.
E. Linear Equations (linear equations do not contain squares or square roots)
To solve a linear equation that contains fractions, first remove all fractions by multiplying both
sides by the lowest common denominator of all the fractions.
Example 3
x/3 + x/2 = 5
Solution
Multiply both sides by the LCD which is 6: (3 x 2)
2x + 3x = 30
5x = 30
x=6
Then move all terms containing the unknown to one side and all terms that do not contain an
unknown to the other side. Factor out the unknown from all terms that contain the unknown;
finally, divide each side by the coefficient of the unknown.
Example 4
Solve a(a + t) = b - bt for t
Solution
To solve for t, first remove the parentheses:
a + at = b - bt
Move "- bt" to the left side and "a " to the right side:
bt + at = b - a
We can factor t out of the two terms on the left side; we can also factor the right side to obtain t(b
+ a) = (b + a)(b - a).
Divide both sides by (b + a) and find the solution to be
t = b - a.
F. Inequalities
An inequality is simply a comparison of two quantities or expressions. The following symbols
with their meanings are used:
> is greater than
< is less than
>= is greater than or equal to
<= is less than or equal to
If we place all numbers on a number line with negative numbers to the left of zero and positive
numbers to the right of zero, then if A is to the right of B we state that A > B; if A is to the left of B
then A < B. Consequently, we conclude that -5 > -7 and -3 < 2.
-7
-5
-3
0
2
We also form inequalities using algebraic symbols. The inequality 3x + 2 > x - 6 is solved just like
an algebraic equation is solved. We subtract 2 from each side of the inequality so that 3x > x - 8.
Then subtract x from each side so that 2x > - 8. Divide by 2 and obtain x > - 4. This is the
solution. Any number greater than - 4 satisfies the inequality.
There are several rules that we must follow when manipulating inequalities:
o
o
o
o
The same number or algebraic expression can be added or subtracted from each
side of an inequality.
The same positive number (or positive algebraic expression) can multiply or
divide each side of an inequality.
Both sides of the same type of inequality can be added and the inequality
remains. (If x < y and w < z, then x + w < y + z.)
If a negative number (or negative algebraic expression) multiplies or divides each
side of an inequality, the inequality sign must be reversed. (Be sure to remember
this; it often leads to errors!)
Example 5
Solve the inequality 2x - 2 > x - 5.
Solution
The inequality is treated in much the same manner as an algebraic equation. Add 2 to each side:
2x > x - 3.
Subtract x from each side and the solution is x > -3
NOTE: Retain the same inequality symbol throughout the solution unless an operation reverses
the symbol.
Example 6
Solve the inequality 3r + 5 > 6r - 7.
Solution
First, subtract 5 from each side:
3r > 6r - 12
Next, subtract 6r from each side:
-3r > -12.
Now divide each side by (-3) and reverse the inequality. The solution is r < 4.
Rather than working with the negative signs, we could have added 7 to each side
of the original inequality to obtain 3r + 12 > 6r. Then subtract 3r from each side
and write 12 > 3r so that the solution is 4 > r. This is equivalent to the above
solution except that the symbol r is on the right side rather than the more
conventional left side.
G. Simultaneous Equations
Two unknowns in two equations are solved by either of two methods.
1. Substitution
2. Addition or Subtraction.
1. The substitution method:
a. Solve the first equation for one unknown in terms of the other unknown and substitute the
result into the second equation.
b. Solve the resulting equation for the unknown.
c. Substitute the value of this known unknown into either original equation and solve for the
second unknown.
Example 7
Solve for x and y.
x-y=2
2x + y = -5
using the substitution method.
Solution
Using the substitution method, the first equation gives
x=2+y
Substitute this into the second equation and solve for y:
2(2 + y) + y = -5
4 + 2y + y = -5
3y = -9
y = -3
Substitute this value back into the first equation and solve for x:
x+3=2
x = -1
The solution is x = -1, y = -3.
2. The addition or subtraction method:
a. Multiply one equation by a properly chosen number so that one of the unknowns has the same
coefficient in both equations.
b. Add or subtract the equations so that one of the unknowns is eliminated.
c. Solve the resulting equation for the remaining unknown.
d. Substitute the value of this known unknown into either original equation and solve for the
second unknown.
Example 8
Solve for x and y
x-y=2
2x + y = -5
This time use the addition or subtraction method.
Solution
We simply add the two equations since this will eliminate y:
x-y=2
+ 2x + y = - 5
3x
= -3
x
= -1
Substitute this value back into the first equation and solve for y:
(-1) - y = 2
-y = 3
y = -3
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Word Problems
For easier use, print out this guide
Option #1:
Print out Word Problems section to use
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chapter >>
Word Problems use simple math concepts and apply them in a contorted and complicated
manner. The usual strategy to solve a Word Problem is to express the question as a
mathematical equation letting x, or some other letter, represent the quantity that we wish to
determine.
The five step process for Word Problems:
1. Quickly read the question and the answer choices to get a feel for what the question is
specifically asking.
2. Read the question again (on the GMAT you have nearly two minutes per math question,
so there is time to spare as long as you budget it properly--read Chapter One for pacing
information).
3. Translate the equation to paper and translate the question into an expression with
variables.
4. If you get a mental block or see a shortcut, use Backsolving (take numbers and feed
them into the question--either answer choices or numbers you choose).
5. If that does not work, start eliminating answers that are outside of the ballpark; guess
and move on.
I. Review of Word Problem Concepts
A. Percentages
B. Interest, Discount, and Markups
C. Progressions
D. Uniform Motion
E. Work
F. Ratio and Proportion
G. Grouping and Counting
H. Tables, Charts, and Graphs (Data Interpretation)
I. Symbols
A. Percentages
The word percent is abbreviated by the symbol % and is a fraction whose denominator is 100.
26% is equivalent to the fraction 26/100. To change a decimal number to a percent, we simply
multiply by 100; the number 0.321 is equivalent to 32.1%. If a percentage is given, move the
decimal two places to the left to express its equivalent decimal form.
Example 1
Convert 4% into a decimal and a fraction in lowest terms.
Solution
To convert 4% into a decimal, we move the decimal point two places to the left:
4% = 0.04
To express 4% as a fraction, we divide by 100:
4/100 = 1/25
Hence,
4% = 0.04 = 1/25
Example 2
If the price of a stock falls from $50 to $40, what is the percentage of decrease?
Solution
First, subtract the numbers resulting in the decrease: 50 - 40 = 10. Then divide by the original
amount:
(50 - 40) / 50 = 10 / 50 = .2
Convert to a percentage by moving the decimal point two places to the right:
% decrease = 20%
Example 3
An employee is to mark up a piece of jewelry 120%. If it cost $100, what should its selling price
be?
Solution
The amount of the markup is 1.2 × 100= $120
The selling price is then $100 + $120 = $220
Example 4
A college bookstore purchases trade books on a 40% margin, i.e., it purchases a trade book for
40% less than its retail price. What is the percentage markup based on its wholesale price?
Solution
Since the retail price is not given, the percentage markup that we seek must be the same for all
trade books. Therefore, let the retail price of a trade book be $100 (rather than the symbol x).
Then the bookstore's purchase price is
100 - 100 × 0.4 = 100 - 40 = $60
If a book sells for $100 and costs $60, its percentage markup is
%markup = (100 - 60) / 60 × 100 = 40 / 60 x 100 = 66%
Example 5
Kathy buys a bike for $240 after a 40% markdown. What was the original price?
Solution
Let P be the original price. Then
P - P × 0.4 = 240
0.6P = 240
divide both sides by .6
therefore, P = $400
Example 6
Find the number of residents of a city if 20% of them, or 6,200 people, ride bicycles.
Solution
Let R be the number of residents. The equation that represents the verbal statement is
0.2R = 6,200. R = 6200/.2 = 62000/2 = 31,000 people.
Example 7
Kent pays 20% taxes on income between $10,000 and $20,000 and 30% on income over
$20,000. The first $10,000 is tax free. If he pays $14,000 in taxes, what was his income?
Solution
Let Kent's income be L. Then the total tax is
0.2(20,000 - 10,000) + 0.3(L - 20,000) =14,000
2,000 + 0.3L - 6,000 = 14,000
0.3L = 14,000 + 4,000 = 18,000
L = 18,000/.3 = $60,000
Example 8
How many gallons of pure water must be added to 100 gallons of a 4% saline solution to provide
a 1% saline solution?
Solution
Let x be the gallons of pure water to be added. There are 100 × 0.04 = 4 gallons of salt and 96
gallons of pure water in a 4% saline solution. The total number of gallons will be x + 100. The
amount of salt will remain constant. Hence,
0.01(x + 100) = 4
0.01x + 1 = 4
0.01x = 3
x = 3/.01 = 300 gallons
B. Interest, Discount, and Markups
The interest, I ,earned on the amount, P, of money invested depends on the interest rate, i,
and the time, T, the money is invested. This is represented by the equation
I = PiT
The interest would be the dollars earned (or paid), the interest rate is always the annual
interest rate (unless otherwise stated), and the time is measured in years. Simple interest means
that the interest, I, is determined using the total time period, e.g. 10 years, rather than
compounding the interest, that is, adding the interest, I, to the amount, P, after each year.
Discount is the percent reduced on the price of an item. Markup is the amount of increase
when the cost of an item is increased a certain percent. The following examples will illustrate this
concept.
For markups and discounts, calculate:
New Price - Original
Original
If the value is negative, that is the amount of the discount. If the number is positive, that is the
amount of the markup
Example 9
A student invests $1,000 at 10% for the summer (3 months). How much interest does the student
earn?
Solution
The interest is calculated to be
I = PiT
= 1000(0.10)(3/12)= $25
We have expressed the 10% interest rate as 0.10 and the 3 months as 3/12 of a year since the
interest rate is assumed to be an annual rate.
Example 10
A professor retires with a retirement fund of $400,000. If she is paid monthly interest of $3,600,
what is the interest rate?
Solution
The interest rate is assumed to be an annual rate. The annual interest income is $3,600 (12) 50
so that
I = PiT
3600(12) = 400,000(i)
I = 3600(12) / 400,000= (3.6(3)(4))/4(100) = 10.8/100 = 0.108
or 10.8%
Example 11
A pair of aerobic shoes is marked $120 and is discounted to $90. What is the percent discount?
Solution
The percent discount is based on the initial cost. It is
% discount = ((120-90)/120) × 100
30/120 × 100 = 25%
Example 12
A pair of running shoes is purchased at wholesale for $90 and is sold to a store for $120. What is
the percent markup?
Solution
The percent markup is based on the original cost. It is
% markup ((120 - 90) / 90) × 100
= 30 / 90 × 100 = 33 1/3%
Example 13
An investment of $1,000 is placed into a particular account at the beginning of each year at a
simple interest of 8%. How much money is in the account after 5 years (no compounded
interest)?
Solution
First, note that 1,000 is placed in EACH year, that is 5,000 is invested.
The first $1,000 will earn interest for 5 years for a total of $80 × 5 = $400. Its value will be $1400.
The second $1000 will earn interest for 4 years for a total of $320. Its value will be $1,320. The
third $1,000 will be worth $1,240. The progression is $1,400, $1,320, $1,240, $1,160, $1,080.
The money in account is the total of those five numbers:
$1,400 + $1,320 + $1,240 + $1,160 + $1,080 = $6,200
C. Progressions
J. Progressions
Sequences and Progressions
An ordered list of numbers is called a sequence. For example, a sequence of positive even
numbers would be:
2, 4, 6, 8, 10, …
Within a sequence, each individual member is called a term. The terms are defined by their
position in the sequence. For example, in the above sequence, 2 is the first term, 4 is the second
term, and 6 is the third term. The ellipsis symbol (…) indicates that the sequence continues
beyond the terms that are written. For the above sequence, the next term would be 12, then 14,
and so on. Even though these two terms (and those beyond) are not written out, we know that
they are terms within the sequence because the ellipsis indicates that the sequence continues
forever (to infinity).
A common kind of sequence problem that you may encounter on the GMAT is one in which you
are given a few terms of a sequence and asked to define the next term. In the above example,
given that the sequence was defined as consisting of positive even numbers, it was easy to
deduce that the next two terms in the sequence are 12 and 14. Often however, sequences will be
more complicated. Given more complex sequences, how can you determine the next term in a
sequence of numbers?
The key to solving these problems is to determine the relationship between the terms in the
sequence that you are given. This relationship can be described in terms of a progression, a
function or manipulation that can be applied to each individual term of a sequence that will
generate the next term in that sequence.
The types of progressions typically found on the GMAT can often be described as being either
arithmetic or multiplicative. Let's now look at examples of each of these.
Arithmetic progressions
Simply stated, in an arithmetic progression, a fixed amount is added to each term in order to
generate the next term. An important consequence of this is that the difference between any two
consecutive terms will remain constant for the entire sequence.
Here is an example of an arithmetic sequence:
0, 3, 6, 9, 12, 15,…
What is the constant value, or fixed amount, being added to each term to generate the next? We
can figure this out by subtracting any term in the sequence from the next term. If we subtract 6
from 9, the difference is 3. We see that this difference is the same if we subtract 9 from 12, 12
from 15, 0 from 3, or 3 from 6. Once you determine the difference between the terms, it is easy to
generate the next terms in the sequence:
0, 3, 6, 9, 12, 15, 18, 21, …
It may also be helpful for some problems to translate the sequence into a general form. In this
case, we could represent this as n, n+3, (n+3) +3, … Keep in mind, however, that this is useful
only in thinking about the general trend, that is, what is the nature of the relationship between the
terms. It does not, however, substitute for the actual sequence, since it gives us no information as
to the starting point. The general equation representation of n, n+3, (n+3) +3, … could represent
0, 3, 6, 9,… or it could represent 1, 4, 7, 10,… or it could be 0.5, 3.5, 6.5, 9.5, …, etc. Keeping
that caveat in mind, when solving problems involving more complex sequences, it is often useful
to jot down the general formula describing that sequence.
Let's look at another example. What are the next two terms of the following sequence?
2, 4.5, 7, 9.5, 12, 14.5, …
It's usually easiest to start with the first terms in the series, as those are typically the smaller
numbers, and thus easier to manipulate quickly. If we subtract 2 from 4.5, we get 2.5. If we
subtract 4.5 from 7, the difference is also 2.5. Going through all the given pairs of terms in the
sequence, we can confirm that this relationship holds true, that each term can be generated
through the addition of 2.5 to the preceding term. (Note: if this relationship were not to hold true
for each pair of terms, we would decide that this is not an arithmetic progression, and we would
then test whether it were instead a multiplicative progression, described below.) Once we
determine the constant value of 2.5, it is straightforward to generate the next two terms in the
sequence, 17 and 19.5. In this case, the general form of the sequence would be n, n + 2.5,…
Let's try another sequence. What are the next two terms in the following sequence?
19, 12, 5, -2, -9, …
Again, to determine the next two terms, we first must determine the constant value that is the
difference between pairs of terms. If we subtract 19 from 12, we have a difference of -7. If we
subtract 12 from 5, again we have -7, and likewise for the remaining terms of the sequence. Once
we know this value, we can quickly ascertain the next two terms, -16 and -23. How could we write
this in a general form? n, n + (-7),… This example should have also demonstrated to you another
important feature of arithmetic progressions: the constant values may be either positive or
negative. Keep this in mind when solving these problems.
Multiplicative Progressions
In a multiplicative progression, the ratio of consecutive terms is constant. Rather than adding a
constant value to a term in order to generate the next term as we do for arithmetic progressions,
for multiplicative progressions, we multiply each term by a constant value to generate the next
term. For example, let's consider the following sequence:
1, -3, 9, -27, 81, …
First, let's determine the constant value separating sets of consecutive terms. Depending on your
personal preference, you can think about it in terms of division or multiplication. What number
must we multiply 1 by in order to get -3 as a product? Alternatively, -3 divided by what number
results in 1? The answer (in both cases) is -3. Next, confirm that the constant value can be
multiplied to each of the terms in the sequence to generate the next (that is, -3 x -3 = 9; 9 x -3 = 27; and -27 x -3 = 81). In general form, this would be n, n x (-3),…
Let's look at another example of a geometric progression:
64, 32, 16, 8, 4, …
What is the next term in this sequence? First, determine the constant value. 64 multiplied by what
number results in a product of 32? Or, what number must 32 be divided by to result in 64? The
answer is 0.5, or ½. Again, confirm that each term can be multiplied by this constant value to
generate the following term. Once you have established that you are correct in identifying this as
a geometric progression, you can generate the next term by multiplying the constant value of 0.5
or ½ by 4, resulting in 2. In general form: n, n x ½,…
More complex sequences
Now, not all sequences will be either arithmetic or multiplicative progressions. Some sequences
will be defined by more complex functions than either addition or multiplication alone. For
example, imagine a sequence of numbers in which each term (except the first two) is generated
by the addition of the two preceding terms:
1, 1, 2, 3, 5, 8, 13, …
This is actually a well-studied sequence of numbers called the Fibonacci series. After the first two
numbers, we can represent each term of the series as n = (n-1) + (n-2), or n is the sum of the two
preceding numbers. (By using subscripts, we are indicating the previous terms, where n-1 is the
term immediately preceding n and
n-2 is the term immediately preceding n-1.)
Alternatively, you may see a sequence in which each subsequent term is derived from both
arithmetic and multiplicative processes. For example, the following sequence begins with 2, and
all subsequent terms are generated by adding 1 to the preceding term, and then multiplying that
sum by 2:
2, 6, 14, 30, 62, …
The general representation of this, not including the first term, would be n = ((n-1) + 1) x 2, or n
equals the sum of the preceding term and 1, multiplied by 2.
For sequences of this type, it is much more difficult to determine the relationship between the
terms at first glance. However, it is rare that you would actually be asked to do this on the GMAT .
Instead, you may see a problem in which they define the relationship for you (as in giving you the
rule to add 1 to the preceding term, and then multiply the sum by 2), and then ask you to
generate the next term of the sequence.
Sample problems
Let's now look at some sample sequence problems that you might find on the actual exam.
Example 33
Except for the first two numbers, every number in the sequence 1, -2, -2, 4, … is the product of
the two immediately preceding numbers. What is the seventh term of this sequence?
(A) -8
(B) 32
(C) -32
(D) 256
(E) -256
Answer explanation: (D) We are given the rule to use in order to generate the next terms of the
sequence: multiply the two immediately preceding numbers to generate the next. We already
have the first four terms, and need to generate three more. -2 x 4 = -8, this is the fifth term in the
sequence. 4 x -8 = -32, this is the sixth term. -8 x -32 = 256, this is the seventh term, and the
correct answer, choice D.
Example 34
The fifth term in a sequence of numbers is 19 and each number after the first number in the
sequence is 3 less than the number immediately preceding it. What is the second number in the
sequence?
(A) 31
(B) 30
(C) 28
(D) 13
(E) 10
Answer explanation: (C) This is an example of an arithmetic progression in which we are given
only one term and asked to determine another. We are told that each term is 3 less than the
previous term. A good technique to use in solving this is to draw out five blanks to represent the
terms of the sequence, and fill in the fifth one, the one term we know, with 19:
We are told that each term is 3 less than the term immediately preceding it. Does that mean that
the fourth term, the one immediately preceding 19, will be 3 more or 3 less? (Be sure to read
carefully!) The fourth term will be three more than 19, or 19 + 3, or 22. In this way, we can work
backwards to generate each of the first four terms, resulting in a sequence that looks like this:
Now we can see that the second term is 28, choice C.
Example 35
What is the next term of the sequence -3, 6, -12, 24,…?
(A) -48
(B) 48
(C) -64
(D) 64
(E) -144
Answer explanation: (A) This is a multiplicative progression generated by multiplying each term
by the constant value of -2 in order to generate the next term. (-3 x -2 = 6; 6 x -2 = -12, etc.) To
generate the next term, multiply 24 by -2 to get -48, choice A.
Example 36
In a sequence of integers, A, B, C, D, E…, the value of each integer except the first is equal to
two more than the product of the previous integer and 2. If E equals 14, what is the value of B?
(A) -14
(B) -8
(C) 0
(D) 4
(E) 8
Answer explanation: (C) This is one of the more complex sequences in which each subsequent
term is derived from both arithmetic and multiplicative processes. Like that previous problems, it
may be useful to create a little diagram to hold the places for the terms as you figure them out.
We are told that the fifth term, E, equals 14, so we can fill in that information as follows:
We are told that the value of each integer is equal to two more than the product of the previous
integer and 2. A general representation of this would be n = ((n-1) x 2) + 2. It is extremely
important to be very precise in your interpretation of the description of the relationship between
the terms of the sequence. In this case, a number is multiplied by 2, then 2 is added to that
product in order to generate the next term.
Now, if we use this general formula and substitute our known term, 14, for n, we can derive the
preceding term as follows:
n = ((n-1) x 2) + 2
14 = ((n-1) x 2) + 2
Subtracting 2 from both sides, we get
12 = ((n-1) x 2)
Dividing both sides of the equation by 2, we get
6 = n-1
Therefore, position D is 6. Knowing the value of D, we can now apply the formula again to solve
for C. Once we have C, we can solve for B. Finally, we end up with the sequence as follows:
Now we can see that B = 0, choice C.
Summary for sequence problems:




If you are asked to solve for a term in the sequence, determine what the
relationship is between the terms of the sequence.
Is the sequence an arithmetic progression, multiplicative progression, or a more
complex sequence defined within the problem? If you think it's an arithmetic
progression, determine the common value and make sure that each term of the
sequence can be generated through the addition of that common value to the
preceding term. Likewise, if you think it's a multiplicative progression, determine
the common value and confirm that each term can be multiplied by this constant
value to generate the following term.
If it helps, write out a more general representation of the sequence or the formula
used to determine the next term of a sequence using n to represent any given
term.
When you are asked to solve for the value of a specific term, it may be useful to
make a little diagram to ensure that you are solving for the correct position.
D. Uniform Motion
When an object is moving at a constant speed (or velocity), the object is traveling with uniform
motion. The distance, D, that the object will travel in time, T, depends on the velocity, V. It is
expressed mathematically as
D = VT
If we desire the distance in miles, we usually express the velocity in miles per hour (mph) and
time in hours. If the distance is in kilometers, then the velocity would be in kilometers per hour
(kph).
Example 15
A biker travels 60 miles in 2.5 hours. Determine the biker's average speed.
Solution
The equation relating distance, velocity and time provides
60 = V(5/2)
Divide both sides by 5/2 to solve for V.
V = (60)2/5 = 24 miles/hour
Example 16
A car travels between two cities 400 miles apart in 7 hours. The return trip takes 9 hours. Find the
average speed of the car.
Solution
The total distance is 2(400) = 800 miles. The total time is 7 + 9 = 16 hours. The average speed is
found from D = VT:
800 = V(16)
V = 800/16 = 50 miles/hour
Example 17
A police officer, traveling at 100 miles per hour, pursues Philip who has a 30 minute head start.
The police officer overtakes Philip in two hours. Find Philip's speed.
Solution
Let x miles per hour be Philip's speed. The distance traveled by the officer equals the distance
traveled by Philip:
2 × 100 = (2 + 30/60)x
200 = (2 + 0.5)x
200 = 2.5x therefore x = 80 mph
E. Work
The amount of work, W, accomplished in time, T, depends on the rate, R, at which the work is
being accomplished. Work problems are quite similar to the problems of uniform motion. The
equation we use is
W = RT
Try to solve by determining the rate per time period (usually per hour or per minute). The rate,
R, is most often expressed as the job to do divided by the time, where W = 1 job. For example, a
tractor plows 1/10 of a field each hour; the job is one field, so the rate is 1/10 of a field per hour. If
it takes x tractors to do one job in 1 hour, then each tractor works at a rate of 1/x of the job per
hour. If it takes x tractors 4 hours to do one job, then each tractor works at one quarter of the
previous rate, or at the rate of 1/4x of the job per hour. In general, if it takes x tractors y hours to
do one job, the rate that each tractor works is 1/(x × y) of the job per hour.
Example 18
It takes 3 men 8 hours to paint a house. How long will it take 5 men to paint the same house?
Solution
The per hour rate at which each man works is
R = 1/(3 x 8) = 1/24 houses per hour
The rate for 5 men is (5R). The work is 1 house. Our equation gives us
1 = 5/24T
Therefore, T = 24/5 = 4.8 hours or 4 hours and 48 minutes.
NOTE: 0.8 hours is 0.8 × 60 = 48 minutes.
Example 19
Michelle can input a day's invoices into the computer system in 40 minutes, and John can input
the same invoices in 60 minutes. How long will it take both of them, working simultaneously, to
input the invoices?
Solution
Michelle's rate for doing the job is 1/40 of the job per minute. John's rate is 1/60 of the job per
minute. Let the time they work be T. Then the sum of the work that Michelle does and the work
that John does must equal one job:
1=(1/40)T + (1/60)T
This is most easily solved by multiplying by 40(60):
40(60) = [40(60)]/40 × T + [40(60)]/60 × T
2400 = 60T + 40T
T = 24 minutes
Example 20
Kelly and Shelley can type the manuscript in 8 hours. Kelly can type the manuscript alone in 20
hours. How long would it take Shelley to type the manuscript?
Solution
The rate that Kelly works is 1/20 of the job per hour. Let the rate that Shelley works be R. To do
one job in 8 hours we have
1 = 1/20(8) + R(8)
To solve for R, multiply by 20:
20 = 8 + 20R(8)
12 = 8(20)R
Therefore, R = 12/[8(20)] = 3/40 of the job per hour.
To type the entire manuscript alone, Shelley takes
T = W/R = 1/(3/40) = 40/3 = 13 1/3 or 13 hours and 20 minutes.
F. Ratio and Proportion
A ratio is a fraction that compares two numbers. The ratio of x to y is written as x : y or x/y.
Ratios are usually used to compare quantities of the same type, for example, the ratio of the
length of a Toyota to the length of a Cadillac. We would not form the ratio of the length of a
Toyota to the cost of a Cadillac.
A proportion states that two ratios are equal. Two ratios involve four numbers: two numerators
and two denominators. Most often, one of these four numbers is not known; it is found by
equaling the two ratios, such as
2 =
15
6
x
The unknown x is then found by cross multiplying:
2x = 15(6)
therefore, x = 45
Two quantities are directly proportional if one is a constant times the other: x = cy (where c is
a constant). They are inversely proportional (or indirectly proportional) if one is a constant divided
by the other: x = c/y, or equivalently, xy = c. If a quantity is stated to be proportional to another,
the word directly is implied, so if x is stated to be proportional to y, it means x = cy.
To decide if two quantities are directly or inversely proportional, we ask the question, "Do the
quantities both increase or decrease or does one increase while the other decreases?" If they
both increase or decrease, they are directly proportional; if one increases while the other
decreases, they are inversely proportional.
To solve an equation that represents a direct proportion, such as x = cy, we set up the
equation as
x1 = y1
x2 y2
where the subscript 1 refers to the first situation and the subscript 2 to the second situation. If the
equation results from an inverse proportion, such as xy = c, we have
x1 = y2
x2 y1
To solve problems involving proportions, 3 of the 4 numbers will usually be known and the
problem will be to calculate the fourth.
Example 21
Calculate x if 4:15 = 16:x
Solution
The equation is written in a more obvious form as
4 = 16
15 x
4x = 16(15) therefore, x = 60
Example 22
The ratio of two numbers is 4:1, and their sum is 40. Find the two numbers.
Solution
This is expressed mathematically as
x/y = 4
x + y = 40. The first equation can be written in the form x = 4y. This is substituted into the
second equation to yield 4y + y = 40. 5y = 40, therefore, y = 8.
Since x = 4y, we find that x = 4(8) = 32. The two numbers are 8 and 32.
Example 23
If an airplane travels 1,200 miles in 2.5 hours, how far will it travel in 10 hours?
Solution
This represents a direct proportion: both the distance traveled and time increase. Consequently, if
we let x = distance the airplane will travel, we have
1200 = 2.5
x
10
12,000 = 2.5x
x = 4,800 miles
Example 24
What is the ratio of 2/3 to 5/4?
Solution
The ratio is (2/3)/(5/4) which is equal to 2/3 × 4/5 =8/15.
G. Grouping and Counting
Overlapping Groups
When a question relates to overlapping groups, try diagramming the problem with overlapping
circles. This will make it easy to account for the overlap.
Example 25
If, in a certain school, 20 students play soccer, 10 play basketball, and 7 play both, how many
students play basketball, soccer or both?
(A) 20
(B) 22
(C) 23
(D) 25
(E) 29
Solution
Using the diagram above we have deduced some new facts:
3 play only basketball
13 only play soccer
7 play both
total of 23 players
Possible Range Questions
When questions ask for a possible range, be sure to examine the lowest and highest
possibilities.
Example 26
A cabinet contains 3 to 5 bottles, each of which contains 30 to 40 mushrooms. If 10 percent of the
mushrooms are flawed, what is the greatest possible number of flawed mushrooms in the
cabinet?
(A) 51
(B) 40
(C) 30
(D) 20
(E) 12
Solution
There are, at most, 5 bottles, each of which contains at most 40 mushrooms; so, there are, at
most, 5 × 40 = 200 mushrooms in the drum. Since 10 percent of the mushrooms are flawed, there
are at most 20 (20 = 10% × 200) flawed mushrooms. The answer is (D).
Count Inclusively
When doing counting problems, always be sure to count the first and last of the range of
items.
Example 27
Fence posts are being placed at 20 foot intervals along a road 1000 feet long. If the first fence
post is placed at one end of the road, how many fence posts are needed?
(A) 49
(B) 50
(C) 51
(D) 52
(E) 53
Solution
The average test taker would simply take 1000 and divide it by 20 to get 50. However, to get the
right answer, you must include the first and last choice. Since the road is 1000 feet long and the
fence posts are 20 feet apart, there are 1000/20 = 50, or 50 full sections in the road. If we ignore
the first fence post and associate the fence post at the end of each section with that section, then
there are 50 fence posts (one for each of the fifty full sections). Counting the first fence post gives
a total of 51 fence posts. The answer is (C).
H. Tables, Charts, and Graphs (Data Interpretation)
Graphs and charts show the relationship of numbers and quantities in visual form. By looking
at a graph, you can see at a glance the relationship between two or more sets of information. If
such information were presented in written form, it would be hard to read and understand.
Here are some things to remember when doing problems based on data interpretation:
1. Take your time and read carefully. Understand what you are being asked to do
before you begin figuring.
2. Check the dates and types of information required. Be sure that you are looking
in the proper columns, and on the proper lines, for the information you need.
3. Check the units required. Be sure that your answer is in thousands, millions, or
whatever the question calls for.
4. In computing averages, be sure that you add the figures you need and no others,
and that you divide by the correct number of years or other units.
5. Be careful in computing problems asking for percentages.
6. (a) Remember that to convert a decimal into a percent you must multiply it by
100. For example, 0.04 is 4%.
(b) Be sure that you can distinguish between such quantities as 1% (1 percent)
and .01% (one one-hundredth of 1 percent), whether in numerals or in words.
(c) Remember that if quantity X is greater than quantity Y, and the question asks
what percent quantity X is of quantity Y, the answer must be greater than 100
percent
Example Set #28: Table Chart
Examples 1-5 are based on this Table Chart.
The following chart is a record of the performance of a baseball team for the first seven weeks of
the season.
Games Won
Games Lost
Total No.of
Games Played
First Week
5
3
8
Second Week
4
4
16
Third Week
5
2
23
Fourth Week
6
3
32
Fifth Week
4
2
38
Sixth Week
3
3
44
Seventh Week
2
4
50
1. How many games did the team win during the first seven weeks?
(A) 32
(B) 29
(C) 25
(D) 21
(E) 50
Choice B is correct. To find the total number of games won, add the number of games won for all
the weeks, 5 + 4 + 5 + 6 + 4 + 3 + 2 = 29.
2. What percent of the games did the team win?
(A) 75%
(B) 60%
(C) 58%
(D) 29%
(E) 80%
Choice C is correct. The team won 29 out of 50 games or 58%.
3. According to the chart, which week was the worst for the team?
(A) second week
(B) fourth week
(C) fifth week
(D) sixth week
(E) seventh week
Choice E is correct. The seventh week was the only week that the team lost more games than it
won.
4. Which week was the best week for the team?
(A) first week
(B) third week
(C) fourth week
(D) fifth week
(E) sixth week
Choice B is correct. During the third week, the team won 5 games and lost 2, or it won about 70%
of the games that week. Compared with the winning percentages for other weeks, the third
week's was the highest.
5. If there are fifty more games to play in the season, how many more games must the
team win to end up winning 70% of the games?
(A) 39
(B) 35
(C) 41
(D) 34
(E) 32
Choice C is correct. To win 70% of all the games, the team must win 70 out of 100. Since it won
29 games out of the first 50 games, it must win (70 - 29) or 41 games out of the next 50 games.
Example Set #29: Interpreting Graphs
Answer the following questions based on the graph above.
1. During what two-year period did the company's earnings increase the most?
(A) 95-97
(B) 96-97
(C) 96-98
(D) 97-99
(E) 98-00
Reading from the graph, the company's earnings increased from $5 million in 1996 to $10 million
in 1997, and then to $12 million in 1998. The two-year increase from '96 to '98 was $7 million-clearly the largest on the graph. The answer is (C).
2. During the years 1996 through 1998, 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
1996 $5 million
1997 $10 million
1998 $12 million
To figure out the average, add (5 + 10 + 12)/3 = 9. The answer is (C).
3. In which year did earnings increase by the greatest percentage over the previous year?
(A) 96
(B) 97
(C) 98
(D) 99
(E) 2000
To find the percentage increase (or decrease), divide the numerical change by the original
amount.
Year
Earnings
% increase from year
before
1995
8
n/a
1996
5
decrease
1997
10
100%
1998
12
20%
1999
11
decrease
2000
8
decrease
The largest number in the right-hand column, 100%, corresponds to the year 1997. The answer is
(B).
4. If the company's earnings are less than 10 percent of sales during a year, then the Chief
Operating Officer will get a 50% pay cut. How many times between 1995 and 2000 did the
Chief Operating Officer take a pay 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).
Year
10% of sales
Earnings
is 10% of sales
greater than
earnings?
1995
.10 x 80 = 8
8
no cut
1996
.10 x 70 = 7
5
cut
1997
.10 x 50 = 5
10
no cut
1998
.10 x 80 = 8
12
no cut
1999
.10 x 90 = 9
11
no cut
2000
.10 x 100 = 10
8
cut
Comparing the right columns shows that earnings were less than 10 percent of sales in 1996 and
2000. The answer is (C).
I. Symbols
On some questions the test will create new functions. You can identify these questions by the
symbols that are used--triangles, squares, ampersand, etc.). These questions are generally easy
as long as you don't get confused by the symbols. Simply take the function and plug in the
numbers.
Look at this:
If a # b = a + b, then what is 2 # 3?
Explanation
2 # 3 would equal 2 + 3, or 5. This is a function. In english "a # b = a + b" means (pretend you are
a computer) take the first number (a) and add it to the second number (b)
Example 31
If a # b = a + b, then what is (2 # 3) # 2?
Solution
Solve inside of the parentheses first. 2 # 3 would equal 2 + 3, or 5. Then (5) # 2 = 5 + 2 or 7.
Example 32 (harder)
If for numbers x, y, z the function # is defined as
x # y = xy - x
then
x # (y # z) =
Solution
The first step to solving x # (y # z) is to solve inside the parenthesis (y # z), then after we have
solved what is in the parenthesis the second step is to do x # (what is in the parenthesis), then
the third step is to solve the equation using the symbol.
Step 1 (solve the parenthesis-- y # z)
1a) if x # y = xy - x (as stated in the question stem)
1b) then y # z = yz - y ( you get this by substituting y for x and z and y)
Step 2 (insert the parenthesis value)
the original question asks x # (y # z), we have already solved y # z, which according to 1b above
y # z = yz - y
So, in the original equation x # (y # z), substitute yz - y for y # z.
Now, x # (y # z) = x # (yz - y)
(NOTE: WHEN SOLVING QUESTIONS WITH SEVERAL NUMBERS AND OPERATIONS,
ALWAYS MULTIPLY AND DIVIDE BEFORE YOUR ADD AND SUBTRACT, FOR EXAMPLE 7
+ 3 x 2 = 13, NOT 20. DO NOT SIMPLY GO FROM LEFT TO RIGHT, MULTIPLY AND DIVIDE
BEFORE YOU ADD AND SUBTRACT.)
So, we are now dealing with
x # (yz - y)
Step 3 (apply the # to the final equation)
The # symbol means x # y = xy -x.
(Essentially, take the first number--here x--, multiply it by the second number--here y--and then
subtract the first number.
Let's apply that to the equation at the end of step 2
x # (yz - y) = x(yz-y) - x
then factor it out the x's.
= xyz - xy - x
= x(yz - y - 1)
Help, I still don't get it!
To play with this question, try inserting numbers such as 1 and 3 for x and y to see how
x # y = xy - x
or
1 # 3 = 1(3)- 1
1#3=2
So now look at
x # (y # z) =
1. Set x =1 , y =3, z = 2 and plug them in
1 # (3 # 2)
2. Do the parenthesis first (as is the rule always). To make a problem less intimidating,
break it into smaller component parts:
(3 # 2) = (3)(2) - (3)
3#2=6-3
3#2=3
3. Go back to the original question and plug in the value you solved for the parenthesis.
1 # (3 # 2), plug in the value for (3 # 2) to get
1 # (3) = (1)(3) - (1)
So, 1 # (3) = 2
4. Now that you have applied the function, you have the result
x # (y # z) =
5. Set x =1 , y =3, z = 2 and plug them in and the result is 2.
Now, for fun, look at the answer we came up with before
x # (y # z) = x(yz - y - 1)
See, now plug in x = 1, y=3, z=2 for x(yz - y - 1), what do you get
1 [(3)(2) - 3 - 1] = 2,
so we can assume we got the right answer.
When doing complex functions or algebra, plug in numbers to better
understand the equation and to check if you got the right answer. This
takes time, though, so use this judiciously.
Have any more questions or suggestions, email 800score.com
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Probability questions are becoming increasingly common. Probability questions tend to be
bundled among the difficult questions, so high scorers will commonly encounter them. If you are a
low scorer and are pressed for time, consider skipping most of the material past "Simple
Probability." This is a computer-adaptive test, and low scorers aren't likely to encounter the most
difficult probability question types.
A. Simple Probability
B. Probability of Multiple Events
C. Independent and Dependent Events
D. Mutually Exclusive Events
E. Conditional Probabilities
F. Combinations
A. Simple Probability
In general, the probability of an event is the number of favorable outcomes divided by the total
number of possible outcomes.
Probability= (# of favorable outcomes) / (# of possible outcomes)
Example 1
What is the probability that a card drawn at random from a deck of cards will be an ace?
Solution
In this case there are four favorable outcomes:
(1) the ace of spades
(2) the ace of hearts
(3) the ace of diamonds
(4) the ace of clubs.
Since each of the 52 cards in the deck represents a possible outcome,
there are 52 possible outcomes. Therefore, the probability is 4/52 or 1/13.
The same principle can be applied to the problem of determining the probability of obtaining
different totals from a pair of dice.
Example 2
What is the probability that when a pair of six-sided dice are thrown, the sum of the numbers
equals 5?
Solution
There are 36 possible outcomes when a pair of dice is thrown. Consider that if one of the dice
rolled is a 1, there are six possibilities for the other die. If one of the dice rolled a 2, the same is
still true. And the same is true if one of the dice is a 3,4,5, or 6. If this is still confusing, look at the
following (abbreviated) list of outcomes: [(1,1),(1,2),(1,3),(1,4),(1,5),(1,6); (2,1),(2,2),(2,3)…
(3,1),(3,2),3,3)… (4,1)…(5,1)…(6,1)….
The total number of outcomes is 6 × 6 = 36. Since four of the outcomes have a total of 5
[(1,4),(4,1),(2,3),(3,2)], the probability of the two dice adding up to 5 is 4/36 = 1/9.
Example 3
What is the probability that when a pair of six-sided dice are thrown, the sum of the number
equals 12?
Solution
We already know the total number of possible outcomes is 36, and since there is only one
outcome that sums to 12, (6,6--you need to roll double sixes), the probability is simply 1/36.
NOTE: The material from here on through the end of the section is dense and intended only for
medium to high scorers. Because this is a CAT (computer-adaptive test), it is relatively unlikely
that lower scorers will encounter these questions, and, if they are short of time, they are better off
putting their time into other sections.
B. Probability of Multiple Events
For questions involving single events, the formula for simple probability is sufficient. For
questions involving multiple events, the answer combines the probabilities for each event in ways
that may seem counter-intuitive. The following strategy is excellent for acquiring a better feel for
probability questions involving multiple events or for making a quick guess if time is short. We will
focus on questions involving two events.

If two events have to occur together, generally an "and" is used. Take a
look at Statement 1: "I will only be happy today if I get email and win the
lottery." The "and" means that both events are expected to happen
together.

If both events do not necessarily have to occur together, an "or" may be
used as in Statement 2, "I will be happy today if I win the lottery or have
email."
Consider Statement 1. Your chances of getting email may be relatively high compared to your
chances of winning the lottery, but if you expect both to happen, your chances of being happy are
slim. Like placing all your bets at a race on one horse, you've decreased your options, and
therefore you've decreased your chances. The odds are better if you have more options, say if
you choose horse 1 or horse 2 or horse 3 to win. In Statement 2, we have more options; in order
to be happy we can either win the lottery or get email.
The issue here is that if a question states that event A and event B must occur, you should
expect that the probability is smaller than the individual probabilities of either A or B. If the
question states that event A or event B must occur, you should expect that the probability is
greater than the individual probabilities of either A or B. This is an excellent strategy for
eliminating certain answer choices.
These two types of probability are formulated as follows:
Probability of A and B
P(A and B) = P(A) × P(B).
In other words, the probability of A and B both occurring is the product of the probability of A and
the probability of B.
Probability of A or B
P(A or B) = P(A) + P(B).
In other words, the probability of A or B occurring is the sum of the probability of A and the
probability of B (this assumes A + B cannot both occur). If there is a probabiilty of A and/or B
occuring, then you must subtract the overlap.
Look at the following examples.
Example 4
If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on
the second toss the coin lands tails?
a) 1/6
b) 1/3
c) ¼
d) ½
e) 1
Solution
First note the "and" in between event A (heads) and event B (tails). That means we expect both
events to occur together, and that means fewer options, a less likely occurrence, and a lower
probability. Expect the answer to be less than the individual probabilities of either event A or
event B, so less than ½. Therefore, eliminate d and e. Next we follow the rule P(A and B) = P(A)
× P(B). If event A and event B have to happen together, we multiply individual probabilities. ½ ×
½ = ¼. Answer c is correct.
NOTE: Multiplying probabilities that are less than 1 (or fractions) always gives an answer that is
smaller than the probabilities themselves.
Example 5
If a coin is tossed twice what is the probability that it will land either heads both times or tails both
times?
a)1/8
b)1/6
c)1/4
d)1/2
e)1
Solution
Note the "or" in between event A (heads both times) and event B (tails both times). That means
more options, more choices, and a higher probability than either event A or event B individually.
To figure out the probability for event A or B, consider all the possible outcomes of tossing a coin
twice: heads, heads; tails, tails; heads, tails; tails, heads. Since only one coin is being tossed, the
order of heads and tails matters. Heads, tails and tails, heads are sequentially different and
therefore distinguishable and countable events. We can see that the probability for event A is ¼
and that the probability for event B is ¼. We expect a greater probability given more options, and
therefore we can eliminate choices a, b and c, since these are all less than or equal to ¼. Now we
use the rule to get the exact answer. P(A or B) = P(A) + P(B). If either event 1 or event 2 can
occur, the individual probabilities are added: ¼ + ¼ = 2/4 = ½. Answer d is correct.
NOTE: We could have used simple probability to answer this question. The total number of
outcomes is 4: heads, heads; tails, tails; heads, tails; tails; heads, while the desired outcomes are
2. The probability is therefore 2/4 = ½.
The following chart summarizes the "and's" and "or's" of probability:
Probability
Formula
Expectation
P(A and B)
P(A) × P(B)
Lower than P(A) or P(B)
P(A or B)
P(A) + P(B)
Higher than P(A) or
P(B)
C. Independent and Dependent Events
The types of events that we have discussed so far are all independent events. By independent
we mean that the first event does not affect the probability of the second event. Coin tosses are
independent. They cannot affect each other's probabilities; the probability of each toss is
independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards
are independent events if you put the cards back. An example of a dependent event, one in which
the probability of the second event is affected by the first, is drawing a card from a deck but not
returning it. By not returning the card, you've decreased the number of cards in the deck by 1,
and you've decreased the number of whatever kind of card you drew. If you draw an ace of
spades, there are 1 fewer aces and 1 fewer spades. This affects our simple probability: (number
of favorable outcomes)/ (total number of outcomes. This type of probability is formulated as
follows:
If A and B are not independent, then the probability of A and B is
P(A and B) = P(A) × P(B|A)
where P(B|A) is the conditional probability of B given A.
Example 6
If someone draws a card at random from a deck and then, without replacing the first card, draws
a second card, what is the probability that both cards will be aces?
Solution
Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = 4/52 = 1/13.
Given that the first card is an ace, what is the probability that the second card will be an ace as
well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability
of A and B is 1/13 × 1/17 = 1/221. The same reasoning is applied to marbles in a jar.
Example 7
If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is
the probability that, drawing twice, you will select two red marbles if you return the marbles after
each draw?
Solution
First, let's determine the number of red and blue marbles respectively. The ratio 2:3 tells us that
the total of 30 marbles must be broken into 5 groups of 6 marbles, each with 2 groups of red
marbles and 3 groups of blue marbles. Setting up the equation 2x + 3x = 5x =30 employs the
same reasoning. Solving, we find that there are 12 red marbles and 18 blue marbles. We are
asked to draw twice and return the marble after each draw. Therefore, the first draw does not
affect the probability of the second draw. We return the marble after the draw, and therefore, we
return the situation to the initial conditions before the second draw. Nothing is altered in between
draws, and therefore, the events are independent.
Now let's examine the probabilities. Drawing a red marble would be 12/30 = 2/5. The same is true
for the second draw. Since we want two red marbles in a row, the question is really saying that
we want a red marble on the first draw and a red marble on the second draw. The "and" means
we should expect a lower probability than 2/5. Understanding that the "and" is implicit can help
you eliminate choices d and e which are both too big. Therefore, our total probability is P(A and
B) = P(A) ×. P(B) = 2/5 × 2/5 = 4/25.
Now consider the same question with the condition that you do not return the marbles after each
draw. The probability of drawing a red marble on the first draw remains the same, 12/30 = 2/5.
The second draw, however, is different. The initial conditions have been altered by the first draw.
We now have only 29 marbles in the jar and only 11 red. Don't panic! We simply use those
numbers to figure our new probability of drawing a red marble the second time, 11/29. The events
are dependent and the total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 11/29 = 132/870 =
22/145.
If you return every marble you select, the probability of drawing another marble is unaffected; the
events are independent. If you do not return the marbles, the number of marbles is affected and
therefore dependent.
D. Mutually Exclusive Events
Another type of probability deals with mutually exclusive events. What do we mean by
mutually exclusive events? And what does it mean for two events not to be mutually exclusive?
Consider the following example of drawing cards:
Example 8
What is the probability that a card selected from a deck will be either an ace or a spade?
a)2/52
b)2/13
c)7/26
d)4/13
e)17/52
Solution
First, identify events A and B and notice the "or" in between them. That means a greater
probability than either A or B individually. Therefore, we expect the answer to be greater than
4/52(ace)=1/13 or 13/52(spade)= 1/4. Eliminate a and b. The tricky part of this question lies in the
fact that when we figure probability, we are really just counting, and sometimes, we count twice.
In this case we have counted the ace of spades twice. If you don't see this, consider what the 4 in
4/52 stands for: ace of hearts, ace of diamonds, ace of clubs, ace of spades. The 13 in 13/52
stands for all the spades: 1,2,3…King, Ace(of spades). Therefore if we just combined the
probabilities by the rule for P(A or B) = P(A) + P(B) we would be over counting. We have to
subtract 1/52, the ace of spades that was counted twice. Our answer becomes 4/52 + 13/52 1/52 = 16/52 = 4/13.
Another way to think about the question is to just count aces and spades; that is, use simple
probability. There are 13 spades in a deck and 3 aces other than the ace of spades already
included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible
outcomes, or 16/52 = 4/13.
In the above example, events A and B are not mutually exclusive. Figuring the probability for
event A includes part of the probability of event B, and we must therefore subtract out this "overcounted" probability to get the correct answer.
The following example illustrates mutually exclusive events:
Example 9
What is the probability that a card selected from a deck will be either an ace or a king?
a)1/169
b)1/26
c)2/13
d)4/13
e)8/13
Solution
The question asks for either an ace or a king. Since there are four kings and four aces in a deck,
the probabilities for event A and event B are the same, 4/52 = 1/13. Our answer must be more
than this, so eliminate a and b. Do kings and aces have anything to do with each other? Is there
such a thing as an ace of kings or a king of aces? No, so we don't have to worry about having
over-counted; the events are mutually exclusive. The probability is straightforward: P(A or B) =
P(A) + P(B) = 1/13 + 1/13 = 2/13. C is correct.
Again we could have used simple probability. Count the total number of kings and aces (4+4) and
divide by the total number of cards in a deck: 8/52 = 2/13.
E. Conditional Probabilities
A conditional probability is the probability of an event given that another event has occurred.
Example 10
What is the probability that the total of two dice will be greater than 8 given that the first die is a
6?
Solution
This can be computed by considering only outcomes for which the first die is a 6. Then,
determine the proportion of these outcomes that total more than 8. All the possible outcomes for
two dice are shown in the section on simple probability. There are 6 outcomes for which the first
die is a 6: (6,1),(6,2),(6,3),(6,4),(6,5),(6,6), and of these, there are four that total more than 8. The
probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3.
1. Probability of A and B
If A and B are independent, then the probability that events A and B both occur is p(A and B) =
p(A) × p(B). In other words, the probability of A and B both occurring is the product of the
probability of A and the probability of B. What is the probability that a coin will come up with
heads twice in a row? Two events must occur: a heads on the first toss and a heads on the
second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 × 1/2
= 1/4. Now consider a similar problem: someone draws a card at random out of a deck, replaces
it, and then draws another card at random. What is the probability that the first card is the ace of
clubs and the second card is a club (any club)?
Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since
13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4.
Therefore, the probability of both events is 1/52 × 1/4 = 1/208.
What's the probability of A and B (2 of 2) if A and B are not independent? If A and B are not
independent, then the probability of A and B is p(A and B) = p(A) × p(B|A) where p(B|A) is the
conditional probability of B given A. If someone draws a card at random from a deck and then,
without replacing the first card, draws a second card, what is the probability that both cards will be
aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 =
1/13. Given that the first card is an ace, what is the probability that the second card will be an ace
as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the
probability of A and B is 1/13 × 1/17 = 1/221.
2. Probability of A or B
If events A and B are mutually exclusive, then the probability of A or B is simply:
p(A or B) = p(A) + p(B).
What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible
to get both a 1 and a 6, these two events are mutually exclusive. Therefore,
p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3
If the events A and B are not mutually exclusive, then
p(A or B) = p(A) + p(B) - p(A and B).
The logic behind this formula is that when p(A) and p(B) are added, the occasions on
which A and B both occur are counted twice. To adjust for this, p(A and B) is subtracted.
Example 11
What is the probability that a card selected from a deck will be either an ace or a spade?
Solution
The relevant probabilities are
p(ace) = 4/52
p(spade) = 13/52
The only way an ace and a spade can both be drawn is to draw the ace of spades. There is only
one ace of spades, so p(ace and spade) = 1/52. The probability of an ace or a spade can be
computed as p(ace)+p(spade)-p(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
F. Combinations
Suppose that a job has two different parts. There are m different ways of doing the first part,
and there are n different ways of doing the second part. The problem is to find the number of
ways of doing the entire job. For each way of doing the first part of the job, there are n ways of
doing the second part. Since there are m ways of doing the first part, the total number of ways of
doing the entire job is m x n. The formula that can be used is
Number of ways = m x n
For any problem that involves two actions or two objects, each with a number of choices, and
asks for the number of combinations, the above formula can be used.
Example 13
William wants a sandwich and a drink for lunch. If a restaurant has 4 choices of sandwiches and
3 choices of drinks, how many different ways can he order his lunch?
Solution
Since there are 4 choices of sandwiches and 3 choices of drinks, using the formula
Number of ways = 4(3) = 12
Therefore, the man can order his lunch 12 different ways.
Now, what if the combinations available decrease after each combination is taken?
Example 14
There are five meal options in the cafeteria of a certain school. Assuming that a different meal
must be eaten each day, and each different type of meal must be eaten once before any type of
meal can be eaten a second time, how many different orders of meals can a student eat in the
first five days?
Solution
The answer is 120. There are five types of meals, so the total number of possibilities is 5!. (the "!"
stands for factorial), or 5 x 4 x 3 x 2 x 1 = 120. Since a different meal is assigned to every day,
you must reduce the amount that you multiply by on a daily basis from 5 to 4 to 3 to 2 to 1. If you
like, assign the letters A, B, C, D and E to the five meals and see how many different orders you
can create.
What if two or more samples are chosen at a time? If we have objects a, b, c, d and want to
arrange them two at a time--that is, ab, bc, cd, etc. (We have four combinations taken two at a
time). If you have four different combinations taken two at a time, you can write this as 4 C 2,
which can be written as
4C2=4x3
2x1
Examples
8C3=8x7x6
3x2x1
10 C 4 = 10 x 9 x 8 x 7
4x3x2x1
Have any more questions or suggestions? Email 800score.com
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Note#2: You cannot calculate (1/6) (5/6) + (5/6) (1/6) which undercounts the possiblities. Simply
add combinations where it cannot be true.
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Data Sufficiency
Option #1:
Print out Data Sufficiency section to use offline (9 pages)
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View the Data Sufficiency online. Browse through the
sections of this chapter >>
This section is broken into 4 parts:
I. Introduction
II. Strategies for Solving Data Sufficiency Questions
III. Data Sufficiency Trick Questions
IV. More practice questions
I. Introduction
In this chapter, we will review strategies for the Data Sufficiency questions and go over trick
questions test designers write to fool you on these questions.
The Data Sufficiency questions (typically 1/3 of all the math questions) do not require the test
taker to find a solution. Instead, the Data Sufficiency questions require the test takers only to find
if each of the statements provides enough information for solving the question.
Data Sufficiency question instructions look like this:
Directions: In each of the problems, a question is followed by two statements containing certain
data. You are to determine whether the data provided by the statements are sufficient to answer
the question. Choose the correct answer based up on the statement's data, your knowledge of
mathematics, and your familiarity with everyday facts (such as number of minutes in an hour or
cents in a dollar). (international students: 100 cents to the dollar).
Choose choice
A) if statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not;
B) if statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not;
C) if statements (1) and (2) taken together are sufficient to answer the question, even though
neither
statement by itself is sufficient;
D) If either statement by itself is sufficient to answer the question;
E) If statements (1) and (2) taken together are not sufficient to answer the question, requiring
more data pertaining to the problem.
Note: Diagrams accompanying problems agree with information given in the questions, but may
not agree with additional information given in statements (1) and (2).
All numbers used are real numbers.
The Data Sufficiency questions are broken into the stem (the top question and two
statements). You answer the question by determining if the information in the two statements is
sufficient to answer the question.
Lets look at an example to clarify this.
(stem) What is the sum of a + b?
(statement) (1) A = 5
(statement) (2) B = 10
Explanation: Statement (1) tells you that A is 5. This is not enough information to answer the
question. Statement (2) alone is also not enough to answer the question. However, if you
combine the two statements, knowing that A=5 and B=10, then you can determine the solution to
the question
II. Strategies for Solving Data Sufficiency Questions
1. Memorize the Data Sufficiency answer choices.
The directions and answer choices for Data Sufficiency questions never change. Memorize them
so that you have no problems on test day. There is no excuse for walking into test day without
these five answer choices perfectly memorized!
A) Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not.
B) Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not.
C) Statements (1) and (2) taken together are sufficient to answer the question, even though
neither statement by itself is sufficient.
D) Either statement by itself is sufficient to answer the question.
E) Statements (1) and (2) taken together are not sufficient to answer the question, requiring more
data pertaining to the problem.
Some students confuse C and D. Read each answer choice carefully.
1. Note that A requires that B not be sufficient, and vice versa with B.
2. C stipulates that A and B cannot be able to answer the question alone. This means that
although A and B together may be able to answer, the answer is not C if either one can answer
the question alone.
3. The answer is D if both can answer the question independently, even if both can answer the
question together.
What does it mean that a statement is "sufficient"?
Sufficient does not mean that a statement is right or true, just that you can use the statements to
derive an answer. Many beginning students err and think a statement is not sufficient if it proves a
statement false.
2. Methodically progress through the two statements
It takes mental discipline to progress through the Data Sufficiency questions. The test writers
deliberately build tricks to each question. There are two basic questions that you must ask
yourself on every Data Sufficiency question:
Step 1: Can you answer the question using the information from statement (1) only?
Step 2: Can you answer the question using the information from statement (2) only?
Step 3: If the answer to both of these questions is "no," then you ask yourself a third question:
can you answer the question if you combine the information from both statements?
Example
Does 3 + x = 1?
1) x is positive
2) x is an odd number
Solution
(1) alone is sufficient, because it proves that 3 + x cannot equal 1. 3 plus a positive number
cannot equal 1. Thus, statement (1) is sufficient because it establishes that the statement is false.
(2) Statement (2) is also sufficient, because it proves 3 + x cannot equal 1. 3 plus an odd number
cannot equal 1. Therefore, it is sufficient. Since both statements are sufficient, the answer must
be D.
3. Data Sufficiency process of elimination strategies
In Step 2, as you progress through each statement, you may eliminate questions. Just solve for
one of the statements and you are halfway done.
Statement 1 is insufficient: automatically eliminate choices A and D, which require (1) to be
sufficient.
Statement 1 is sufficient: automatically eliminate choices B, C and E, which require (1) to be
insufficient.
Statement 2 is insufficient: automatically eliminate choices B and D, which require (1) to be
sufficient.
Statement 2 is sufficient: automatically eliminate choices A, C and E, which require (1) to be
insufficient.
4. Analyze questions in terms of sufficiency.
Do not think in terms of "what is the exact value," "is this true or false?" Instead, review questions
in terms of one question "is there enough information to answer the question?" Look at each
statement and ask yourself if it provides enough information to arrive at a conclusion.
III. Data Sufficiency Trick Questions
The tricks used in Data Sufficiency questions come from a narrow pool of tricks. Learn to
identify each of the tricks, and you will be in a strong position to answer each question. In
approaching each question, apply the three step method we discussed earlier:
Step (1) Look at the question stem.
Step (2) Look at each statement individually.
Step (3) Then look at both statements in combination.
It is important that you have the discipline to stick to this approach. The Data Sufficiency
questions tend to be trick questions, particularly the difficult ones, and straying from this basic
strategy will increase the chances of you being fooled.
Remember that standardized tests are based on the premise that you can separate students
into groups of ability. In order to do this, the less capable students must get questions wrong. To
make sure less capable students get low scores, the tests are deliberately designed with trick
questions specifically made to fool you.
Selected Trick Questions
AMNESIA TRICK
How many adults ride bicycles in city A if all adults in City A either ride bicycles or drive cars?
(1) 85% of the 10,000 adults in city A drive cars.
(2) 8500 adults in city A drive cars.
(A) Statement (1) is sufficient since if 8,500 drive cars then 1,500 ride bicycles. Statement (2) is
not sufficient since we do not know the total population; it cannot be assumed from (1).
The trick here is that 1 alone can answer the question. Although 1 and 2 together may answer
the question, the answer is still A. The unskilled reader will carry over the information from
statement 1 when reading statement 2 and not catch the flaw with statement 2 (it does not tell
you the population). Trick #2: note that the question doesn't tell you the total population of City A,
but the total population is not relevant since the question only asks for "Adults".
This question shows how you must have discipline and stick to the 3 step process.
(1) Read the stem
(2) Read each statement individually
(3) If both statements cannot answer the question alone, then look at both statements together.
Before you try to combine statement 1 and 2, make sure each answer can/cannot answer the
question. When you first read statement 2, temporarily forget what you read in statement 1 so
that you may evaluate if (2) alone is sufficient. Hence the name of the trick question: "Amnesia."
Get temporary amnesia after reading statement 1 and don't use statement 1's information when
you first evaluate statement 2 (because you need to see if statement 2 is sufficient alone.
DELAY TRICK
How much was a certain Babe Ruth baseball card worth in January 1991?
(1) In January 1997 the card was worth $100,000.
(2) Over the ten years 1987-1997, the card steadily increased in value by 10% each 12 months.
(1) alone is obviously insufficient. To use (2) you need to know what the card was worth at some
time between 1987 and 1997. So (2) alone is insufficient, but by using (1) and (2) together you
can figure out the worth of the baseball card in January 1991. The trick here is not to do the
calculations. If you tried to actually calculate the value in January 1991, you have fallen into the
trap. All that matters is that sufficient information is available.
The test designers make these questions to make you waste time so that you do not finish the
test on time. This is called the DELAY trick because it causes you to be delayed and lose
valuable time if you do unnecessary calculations.
BACKSOLVE
Is the two-digit integer, with digits r (first) and m(second), a multiple of 7?
(1) r + m = 13
(2) r is divisible by 3
With statements 1 and 2 we may determine that the two digit number is not a multiple of 7. Using
statement (1) Try all the two digit numbers that sum 13: 94, 85, 76, 67, 58, 49. Of those, only 49
is divisible by 7. So, using statement 1, rm may or may not be a multiple of 7; it is insufficient. (2)
Is not sufficient because there are many numbers with r that are divisible by 3 and that are
multiples of 7 (35, 63, 98). Combined, there are NO possible numbers rm that are divisible by 7
and satisfy statements 1 and 2. The answer is NO, rm is not a multiple of 7. Using statements 1
and 2 we may deduce this.
Using statement 2, however, 49 is not a multiple of 3. So, combining the two statements proves
that rm is not a multiple of 7. In other words, we've used the two statements to deduce that rm is
not a multiple of 7.
This looks like a very intimidating question. As a rule, when you encounter a highly intimidating
question such as this one, you should plug in possible answers. This question defies an algebraic
solution, so it must be solved through backsolving.
RED HERRING TRICK
Billy sells twice as many $20 tickets as Tim, and Tim sells three times as many $10 tickets as
Billy. How many tickets did Billy sell? Tickets are either $10 or $20.
(1) Tim sold a total of 35 tickets
(2) Together Billy and Tim sold 70 tickets for $1000
1 is not sufficient. Let x = the number of $20 tickets sold by Tim and y = the number of $10 tickets
sold by Billy. Then
Billy sold 2x ($20 tickets) + y ($10 tickets)
Tim sold x ($20 tickets) + 3y ($10 tickets)
(2) implies 70 = x + 2x + y + 3y and 1000 = 20(x + 2x) + 10 (y + 3y)- divide this equation by 20 to
simplify. Subtract these two equations
70 = 3x + 4y
-50 = -3x - 2y
20 = 2y may be solved for x and y and subsequently y = 10 and x = 10, Billy sold 2(10) + 10 = 30
tickets.
The trick here is that 1 is completely unnecessary and a distraction. The information in 1 may
help answer the question, but it is unnecessary; 2 can do it alone.
International students: A "red herring" is an American/English phrase for something that is a
distraction to the issue. In this case, the first statement is a distraction.
SUPER STATEMENT TRICK
What is the average (arithmetic mean) of 3x and 12z?
(1) x + 4z = 20
(2) x + z = 8
Yes, combining A and B will solve the question, but A can do it alone. The trap is C. Students will
know that the two statements together can solve the question. SUPER STATEMENT questions
involve questions where together both statements can solve a question, but carefully examined,
one statement may solve it alone.
The given information asks for the average of 3x and 12z, which is (3x + 12z) / 2, or 3(x + 4z)/2.
Statement 1 tells us the value of x + 4z, (x + 4z)/3. So you can solve the average formula directly
without using the second statement. x + 4z = 20, so 3x + 12z = 60, meaning that the average =
30. You may use statement 2 to solve the problem, but statement 1 can do it itself (thus
disqualifying choice C, which requires both 1 and 2 to be insufficient).
HINT: on difficult Data Sufficiency questions, the statements usually have more value than it
appears at first glance.
IV. More practice questions
Example 1
Is x > 4?
1) x squared = 9
2) x is an integer
Solution
(1) implies that x = +/- 3 (+/- means positive or negative). Both +3 and -3 are less than 4, so the
answer is "NO" and (1) is sufficient, that is NO, x is not greater than 4. A "NO" answer is equally
acceptable as a "YES" answer. It is only necessary that there is sufficient information to answer
the question. (2) does not provide any necessary information. The correct response is A.
Example 2
What is x - y?
1) x + y = 8
2) x - 2y = 2
Solution
(1) is not sufficient since (x - y) is the quantity desired. Likewise, (2) is not sufficient. But (1) and
(2) together provide us with 2 equations and two unknowns from which x - y can be determined.
The correct response is C. (We may solve the problem by subtracting (2) from (1): 3y = 6,
therefore y = 2 and x = 6, so that x - y = 6 - 2 = 4. This calculation is, however, unnecessary.)
Example 3
How old is Gloria?
1) Gloria's age is four times Alex's age plus Becky's age.
2) Becky was Alex's age fifteen years ago.
Solution
(1) is obviously not sufficient as is (2). Can the question be answered with (1) and (2)? Let x be
Gloria's age, y be Alex's age, and z be Becky's age. (1) states that x = 4y + z. (2) states that z 15 = y. These two equations contain three unknowns; consequently, we cannot determine x.
More information is needed and the correct response is E.
Example 4
A student group sold only donuts and GMAT books to raise funds. How many GMAT books were
sold?
1) 30% of the 90 items sold were GMAT books.
2) 63 donuts were sold.
Solution
(1) is sufficient since 30% of 90 is 0.3 x 90 = 27. (2) is not sufficient since we do not know the
total number of items sold. So the correct response is A. A note of caution: Never let information
in (1) influence your decision regarding the information in (2). In this example we cannot assume
that 90 items were sold when deciding if (2) provides sufficient information. This is the Amnesia
trick that undisciplined test takers will always fall into. Remember to look at each statement
individually before comparing the two.
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