Tài liệu Master the Gre 2010 - Part 24 pptx

GEOMETRIC SEQUENCESIn a geometric sequence of numbers, each term is a constant multiple of the preceding one; in other words, the ratio between any term and the next one is constant.. In

GEOMETRIC SEQUENCES

In a geometric sequence of numbers, each term is a constant multiple of the preceding

one; in other words, the ratio between any term and the next one is constant The

multiple (or ratio) might be obvious by examining the sequence For example:

In the geometric sequence 2, 4, 8, 16, , you can easily see that the

constant multiple is 2 (and the ratio of each term to the next is 1:2)

In the geometric sequence 1, 23, 9, 227, , you can easily see that the

constant multiple is 23 (and the ratio of each term to the next is 1:23)

Once you know the multiple (or ratio), you can answer any question asking for an

unknown term—or for either the sum or the average of certain terms

25 In a geometric sequence, each term is a constant multiple of the preceding

one If the third and fourth numbers in the sequence are 8 and 216,

respectively, what is the first term in the sequence?

(A) 232

(B) 24

(C) 2

(D) 4

(E) 64

The correct answer is (C) The constant multiple is 22 But since you need to

work backward from the third term (8), apply the reciprocal of that multiple

twice The second term is ~8!S21

2D5 24 The first term is ~24!S21

2D5 2

26 In a geometric sequence, each term is a constant multiple of the preceding

one What is the sum of the first four numbers in a geometric sequence

whose second number is 4 and whose third number is 6?

(A) 16

(B) 19

2

3

(E) 20

2 In other words, the ratio

of each term to the next is 2:3 Since the second term is 4, the first term is 4 3

2

35

8

3 Since the third term is 6, the fourth term is 6 3

3

25

18

2, or 9 The sum of the four terms 58

31 4 1 6 1 9 5 21

2

3. You can also solve geometric sequence problems by applying a special formula But

ALERT!

You can’t calculate the average of terms in a geometric sequence by averaging the first and last term in the sequence: The progression is geometric, not arithmetic You need to add

up the terms, then divide by the number of terms.

r 5 the constant multiple (or the ratio between each term and the preceding one), a 5

the first term in the sequence, n 5 the position number for any particular term in the

sequence, and T 5 the particular term itself:

ar (n 2 1)5 T You can solve for any of the formula’s variables, as long as you know the values for the other three Following are two examples:

If a 5 3 and r 5 2, then the third term 5 (3)(2)25 12, and the sixth term 5 (3)(2)5

5 (3)(32) 5 96

If the sixth term is 21

16and the constant ratio is

1

2, then the first term (a) 5 22:

aS1

2D5

5 2 1 16

aS1

32D 5 21

16

a 5S21

16D~32! 5 22 The algebra is simple enough—but you need to know the formula, of course

27 In a geometric sequence, each term is a constant multiple of the preceding

one If the first three terms in a geometric sequence are 22, x, and 28,

which of the following could be the sixth term in the sequence?

(A) 232 (B) 216 (C) 16 (D) 32 (E) 64 The correct answer is (E) Since all pairs of successive terms must have the

same ratio, 22

x

28 By the cross-product method, x

2

5 16, and hence x 5 64.

For x = 4, the ratio is 4

225 22 Applying the formula you just learned, the sixth term would be (22)(22)5

5 64 For x 5 24, the ratio is24

225 2 The sixth term would be (22)(2)55 264

PERMUTATIONS

A permutation is an arrangement of objects in which the order (sequence) is

important Each arrangement of the letters A, B, C, and D, for example, is a different permutation of the four letters There are two different ways to determine the number

of permutations for a group of distinct objects

List all the permutations, using a methodical process to make sure you don’t

overlook any For the letters A, B, C, and D, start with A in the first position, then

list all possibilities for the second position, along with all possibilities for the third

and fourth positions (you’ll discover six permutations):

Placing B in the first position would also result in 6 permutations The same

applies to either C or D in the first position So, the total number of

permuta-tions is 6 3 4 5 24

Use the following formula (let n 5 the number of objects) and limit the number of

terms to the counting numbers, or positive integers:

Number of permutations 5 n(n 2 1)(n 2 2)(n 2 3) (1).

The number of permutations can be expressed as n! (“n” factorial) Using the

factorial is much easier than compiling a list of permutations For example, here’s

how to determine the number of arrangements (permutations) of the four letters

A, B, C, and D: 4! 5 4(4 2 1)(4 2 2)(4 2 3) 5 4 3 3 3 2 3 1 5 24

28 Five tokens—one red, one blue, one green, and two white—are arranged in

a row, one next to another If the two white tokens are next to each other,

how many arrangements according to color are possible?

(A) 12

(B) 16

(C) 20

(D) 24

(E) 30

The correct answer is (D) The two white tokens might be in positions 1 and 2,

2 and 3, 3 and 4, or 4 and 5 For each of these four possibilities, there are 6

possible color arrangements (3!) for the other three tokens (which all differ in

color) Thus, the total number of possible arrangements is 4 3 6, or 24

COMBINATIONS

A combination is a group of certain objects selected from a larger set The order of

objects in the group is not important You can determine the total number of possible

combinations by listing the possible groups in a methodical manner For instance, to

determine the number of possible three-letter groups among the letters A, B, C, D,

and E, work methodically, starting with A as a group member paired with B, then C,

then D, then E Be sure not to repeat combinations (repetitions are indicated in

parentheses here):

TIP

You can shortcut common factorial calculations by memorizing them: 3! 5 6, 4! 5

24, and 5! 5 120.

ALERT!

Notice that each parenthetical combination backtracks to an earlier letter.

Be sure you don’t repeat any combination and make sure you don’t backtrack to an earlier object.

A, B, C (A, C, B) (A, D, B) (A, E, B)

A, B, D A, C, D (A, D, C) (A, E, C)

Perform the same task assuming B is in the group, then assuming C is in the group (all combinations not listed here repeat what’s already listed)

B, C, D

B, C, E

B, D, E

C, D, E

The total number of combinations is 10

29 How many two-digit numbers can be formed from the digits 1 through 9, if

no digit appears twice in a number?

(A) 36 (B) 72 (C) 81 (D) 144 (E) 162 The correct answer is (B) Each digit can be paired with any of the other 8

digits To avoid double counting, account for the possible pairs as follows: 1 and 2–9 (8 pairs), 2 and 3–9 (7 pairs), 3 and 4–9 (6 pairs), and so forth The total number of distinct pairs is 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 5 36 Since the digits

in each pair can appear in either order, the total number of possible two-digit numbers is 2 3 36, or 72

Here’s something to consider: You can approach combination problems as probability

problems as well Think of the “probability” of any single combination as “one divided by” the total number of combinations (a fraction between zero (0) and 1) Use whichever method is quickest for the question at hand

PROBABILITY

Probability refers to the statistical chances of an event occurring or not occurring By

definition, probability ranges from zero (0) to 1 (Probability is never negative, and it’s never greater than 1.) Here’s the basic formula for determining probability:

Probability 5 number of ways the event can occur

total number of possible occurrences

30. If you randomly select one candy from a jar containing two cherry

candies, two licorice candies, and one peppermint candy, what is the

probability of selecting a cherry candy?

1

1 3

3 5

The correct answer is (D) There are two ways among five possible

occurrences that a cherry candy will be selected Thus, the probability of

selecting a cherry candy is2

5.

To calculate the probability of an event not occurring, just subtract the probability of

the event occurring from 1 So, referring to the preceding question, the probability of

not selecting a cherry candy is3

5 (Subtract

2

5from 1.)

On the GRE, a tougher probability question would involve this basic formula, but it

would also add a complication of some kind It might require you to determine any of

the following:

• certain missing facts needed for a given probability

• probabilities involving two (or more) independent events

• probabilities involving an event that is dependent on another event

For these three types of probability questions, which we’ll examine next, don’t try to

“intuit” the answer Probabilities involving complex scenarios such as these are often

greater or less than you might expect

Missing Facts Needed for a Given Probability

In this question type, instead of calculating probability, you determine what missing

number is needed for a given probability To do so, just plug what you know into the

basic formula and solve for the missing number

31 A piggy bank contains a certain number of coins, of which 53 are dimes

and 19 are nickels The remainder of the coins in the bank are quarters If

the probability of selecting a quarter from this bank is1

4, how many quarters does the bank contain?

(A) 30

(B) 27

(C) 24

(D) 21

(E) 18

The correct answer is (C) On its face, this question looks complicated, but it’s

really not Just plug what you know into the probability formula Let x 5 the

TIP

The probability of an event not

occurring is 1 minus the probability that it will occur.

and let x 1 72 5 the total number of coins (the fraction’s denominator) Then solve for x (use the cross-product method to clear fractions):

1

45

x

x 1 72

x 1 72 5 4x

72 5 3x

24 5 x

Probability Involving Two or More Independent Events

Two events are independent if neither event affects the probability that the other will occur (You’ll look at dependent events next.) On the GRE, look for either of these two scenarios involving independent events:

The random selection of one object from each of two or more groups The random selection of one object from a group, then replacing it and selecting

again (as in a “second round” or “another turn” of a game)

In either scenario, the simplest calculation involves finding the probability of two events both occurring All you need to do is multiply together their individual prob-abilities: (probability of event 1 occurring) 3 (probability of event 2 occurring) 5 (probability of both events occurring)

For example, assume that you randomly select one letter from each of two sets: {A,B} and {C,D,E} The probability of selecting A and C 51

23

1

3, or

1

6.

To determine the probability that three events will all occur, just multiply the third

event’s probability by the other two To calculate the probability that two events will

not both occur, subtract from 1 the probability of both events occurring.

32. From a group of seven students, one student is called on at random to answer a question Then one of the same seven students is called on at random to answer another question What is the probability that the same student will NOT be called on to answer BOTH questions?

1

6 7

48 49

The correct answer is (E) You must first calculate the chances of

picking a particular student twice by multiplying together the two

individual probabilities for the student: 1

73

1

7 5

1

49 The probability of not picking the same student twice is 1 2 1

49= 48

49.

Probability Involving a Dependent Event

Two distinct events might be related in that one event affects the probability of the

other one occurring—for example, randomly selecting one object from a group, then

selecting a second object from the same group without replacing the first selection.

Removing one object from the group increases the odds of selecting any particular

object from those that remain

You handle this type of problem as you would any other probability problem:

Cal-culate individual probabilities, then combine them

33. In a random selection of two people from a group of five—A, B, C, D,

and E—what is the probability of selecting A and B?

1

1 10

1 20

The correct answer is (C) You need to consider each of the two

selections separately In the first selection, the probability of selecting

either A or B is 2

5 But the probability of selecting the second of the two is 1

4, because after the first selection only four people remain from whom to

select Since the question asks for the probability of selecting both A and B

(as opposed to either one), multiply the two individual probabilities: 2

53

1 4

5 2

205

1

10.

You can also approach probability problems like Question 33 (above) as combination

problems For Question 33, here are all the combinations:

• A and either B, C, D, or E (4 combinations)

• B and either C, D, or E (3 combinations)

• C and either D or E (2 combinations)

• D and E (1 combination)

There are 10 possible combinations, so the probability of selecting A and B is 1 in 10

ALERT!

Strategies such as plugging in test numbers, working backward, and sizing up answer choices don’t work for most probability questions.

SUMMING IT UP

• Although the types of questions reviewed in the early sections of this chapter are the most basic of the math problems you’ll encounter on the GRE Quantitative Reasoning section, don’t underestimate how useful they’ll be as building blocks for solving more complex problems

• Certain fraction-decimal-percent equivalents show up more frequently than others on the GRE If you have time, memorize the standard conversions to save yourself time on the actual exam

• Percent change questions are typical on the GRE Quantitative Reasoning section,

so be ready for them

• As with fractions, you can simplify ratios by dividing common factors

• Review the definitions of arithmetic mean, median, mode, and range, so you’re better equipped to solve such problems on the exam

• Many arithmetic sequence questions ask for the average or sum of a sequence You may be able to “shortcut” the addition instead of calculating the average of a long sequence of evenly spaced integers

• Memorizing common factorial combinations will save you time when you encounter permutation questions on the GRE

• Work methodically on combination questions to avoid backtracking to an earlier object

• It’s wise not to try “intuiting” the answers to probability questions Many of these problems are too complex to arrive at an accurate answer this way

Theory and Algebra

OVERVIEW

In this chapter, you’ll first broaden your arithmetical horizons by dealing with

numbers in more abstract, theoretical settings You’ll examine the following

topics:

• The concept of absolute value

• Number signs and integers—and what happens to them when you apply

the four basic operations

• Factors, multiples, divisibility, prime numbers, and the “prime

factor-ization” method

• The rules for combining exponential numbers (base numbers and

“powers”) using the four basic operations

• The rules for combining radicals using the four basic operations

• The rules for simplifying terms containing radical signs

Then you’ll review the following basic algebra skills:

•

.

• Solving a system of two equations with two variables—by substitution and by addition-subtraction

• Recognizing unsolvable linear equations when you see them

• Applying simple algebraic functions

• Handling algebraic inequalities Later you’ll learn how the GRE tests your algebra skills through the use of “story” problems, which involve real-world scenarios

BASIC PROPERTIES OF NUMBERS

First let’s review the basics about integers, number signs (positive and negative), and prime numbers Make sure you’re up to speed on the following definitions, which you’ll need to know for this chapter as well as for the test:

origin) on the real number line The absolute value of x is indicated as |x| By definition, a number’s absolute value cannot be negative—that is, less than zero (0)

3 } Except for the number zero (0), every integer is either positive or negative and either even or odd

for a product of n.

positive factors: 1 and the number itself In other words, a prime number is not divisible by (a multiple of) any positive integer other than itself and 1

Number Signs and the Four Basic Operations

The four basic operations are addition, subtraction, multiplication, and division Be sure you know the sign of a number that results from combining numbers using these operations Here’s a table that includes all the possibilities (a “?” indicates that the sign depends on which number has the greater absolute value):

~1! 1 ~1! 5 1

~2! 1 ~2! 5 2

~1! 1 ~2! 5 ?

~2! 1 ~1! 5 ?

~1! 2 ~2! 5 ~1!

~2! 2 ~1! 5 ~2!

~1! 2 ~1! 5 ?

~2! 2 ~2! 5 ?

~1! 3 ~1! 5 1

~1! 3 ~2! 5 2

~2! 3 ~1! 5 2

~2! 3 ~2! 5 1

~1! 4 ~1! 5 1

~1! 4 ~2! 5 2

~2! 4 ~1! 5 2

~2! 4 ~2! 5 1

GRE problems involving combining numbers by addition or subtraction usually incor-porate the concept of absolute value, as well as the rule for subtracting negative numbers

ALERT!

The factors of any integer n

include 1 as well as n itself.

Zero (0) and 1 are not prime

numbers; 2 is the first prime

number.