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Tiêu đề Math Review: Number Forms, Relationships, and Sets
Trường học Peterson's
Chuyên ngành GRE Preparation
Thể loại Tài liệu
Năm xuất bản 2010
Thành phố Unknown
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Số trang 10
Dung lượng 57,28 KB

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ARITHMETIC MEAN, MEDIAN, MODE, AND RANGE Arithmetic mean simple average, median, mode, and range are four different ways to describe a set of terms quantitatively.. To find a missing num

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10. 0 , p , q , 1

p q

p q

+ +

1 1

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (B) Remember three rules, which apply to any positive

numerator p and denominator q:

If p = q, the fraction’s value remains unchanged (and is always 1).

If p q, the fraction’s value will decrease.

If p , q, the fraction’s value will increase (as given in the example).

It makes no difference that p and q are fractional values less than 1 rather than

larger numbers As long as p and q are both positive, the above three rules always

apply So in the example, if the centered inequality were 1 , p , q or 0 , p , 1

, q, it wouldn’t make any difference—quantity B would still be greater than

quantity A (If you’re not convinced, try plugging in some sample values for p

and q.)

A GRE question might ask you to alter a ratio by adding or subtracting from one or

both terms in the ratio The rules for altering ratios are the same as for altering

fractions In either case, set up a proportion and solve algebraically for the unknown

term

11 A drawer contains exactly half as many white shirts as blue shirts If four

more shirts of each color were to be added to the drawer, the ratio of white

to blue shirts would be 5:8 How many blue shirts did the drawer

originally contain?

(A) 14

(B) 12

(C) 11

(D) 10

(E) 9

The correct answer is (B) Represent the original ratio of white to blue shirts

by the fraction x

2x , where x is the number of white shirts, then add 4 to both the

numerator and denominator Set this fraction equal to 5

8(the ratio after adding

shirts) Cross-multiply to solve for x:

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x 1 4

2x 1 45

5 8

8x 1 32 5 10x 1 20

12 5 2x

x 5 6

The original denominator is 2x, or 12.

Ratios with More Than Two Quantities

You approach ratio problems involving three or more quantities the same way as those involving only two quantities The only difference is that there are more “parts” that make up the “whole.”

12 Three lottery winners—X, Y, and Z—are sharing a lottery jackpot X’s

share is1

5 of Y’s share and

1

7 of Z’s share If the total jackpot is $195,000, what is the dollar amount of Z’s share?

(A) $15,000 (B) $35,000 (C) $75,000 (D) $105,000 (E) $115,000 The correct answer is (D) At first glance, this problem doesn’t appear to

involve ratios (Where’s the colon?) But it does The ratio of X’s share to Y’s share

is 1:5, and the ratio of X’s share to Z’s share is 1:7 So you can set up the following triple ratio:

X:Y:Z 5 1:5:7

X’s winnings account for 1 of 13 equal parts (1 1 5 1 7) of the total jackpot 1

13of

$195,000 is $15,000 Accordingly, Y’s share is 5 times that amount, or $75,000, and Z’s share is 7 times that amount, or $105,000

In handling word problems involving ratios, think of a whole as the sum of its fractional parts, as in the method used to solve the preceding problem: 1

13(X’s share)

1 5

13(Y’s share) 1

7

13(Z’s share) 5 1 (the whole jackpot).

TIP

Remember: When you add or

subtract the same number

from both the numerator and

denominator of a fraction—or

from each term in a ratio—you

alter the fraction or ratio,

unless the original ratio was 1:1

(in which case the ratio is

unchanged).

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Proportion Problems with Variables

A GRE proportion question might use letters instead of numbers, to focus on the

process rather than the result You can solve these problems algebraically or by using

the plug-in strategy

13 A candy store sells candy only in half-pound boxes At c cents per box,

which of the following is the cost of a ounces of candy? [1 pound 5 16

ounces]

(A) c

a

(D)ac

8c

a

The correct answer is (D) This question is asking: “c cents is to one box

as how many cents are to a ounces?” Set up a proportion, letting x equal

the cost of a ounces Because the question asks for the cost of ounces,

convert 1 box to 8 ounces (a half pound) Use the cross-product method to

solve quickly:

c

85

x

a

8x 5 ac

x 5 ac

8

You can also use the plug-in strategy for this question, either instead of algebra

or, better yet, to check the answer you chose using algebra Pick easy numbers to

work with, such as 100 for c and 16 for a At 100 cents per 8-ounce box, 16 ounces

of candy cost 200 cents Plug your numbers for a and c into each answer choice.

Only choice (D) gives you the number 200 you’re looking for

ARITHMETIC MEAN, MEDIAN, MODE, AND RANGE

Arithmetic mean (simple average), median, mode, and range are four different ways

to describe a set of terms quantitatively Here’s the definition of each one:

Arithmetic mean (average): In a set of n measurements, the sum of the

measurements divided by n.

Median: The middle measurement after the measurements are ordered by size (or

the average of the two middle measurements if the number of measurements is

even)

Mode: The measurement that appears most frequently in a set.

Range: The difference between the greatest measurement and the least

measurement

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For example, given a set of six measurements, {8,24,8,3,2,7}:

Mean 5 4 (8 2 4 1 8 1 3 1 2 1 7) 4 6 5 24 4 6 5 4 Median 5 5 The average of 3 and 7—the two middle measurements in the set

ordered in this way: {24,2,3,7,8,8}

Mode 5 8 8 appears twice (more frequently than any other measurement) Range 5 12 The difference between 8 and 24

For the same set of values, the mean (simple average) and the median can be, but are not necessarily, the same For example: {3,4,5,6,7} has both a mean and median of 5 However, the set {22,0,5,8,9} has a mean of 4 but a median of 5

For any set of terms, the arithmetic mean (AM), also called the simple average, is the sum of the terms (a 1 b 1 c 1 ) divided by the number of terms (n) in the set:

AM 5~a 1 b 1 c 1 !

n

On the GRE, easier questions involving simple average might ask you to add numbers together and divide a sum In finding a simple average, first be sure the numbers being added are all of the same form or in terms of the same units

14 What is the average of 1

5, 25%, and 0.09?

(A) 0.18 (B) 20%

(C) 1

4

(D) 0.32 (E) 1

3

The correct answer is (A) Since the answer choices are not all expressed in the

same form, first rewrite numbers in whichever form you think would be easiest to work with when you add the numbers together In this case, the easiest form to work with is probably the decimal form So rewrite the first two numbers as decimals, and then find the sum of the three numbers: 0.20 1 0.25 1 0.09 5 0.54 Finally, divide by 3 to find the average: 0.54 4 3 5 0.18

To find a missing number when the average of all the numbers in a set is given, plug into the arithmetic-mean formula all the numbers you know—which include the average, the sum of the other numbers, and the number of terms Then, use algebra to find the missing number Or, you can try out each answer choice, in turn, as the missing number until you find one that results in the average given

ALERT!

The mean and median of the

same set of numbers can be

the same—but this is not

necessarily the case.

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15 The average of five numbers is 26 Four of the numbers are 212, 90, 226,

and 10 What is the fifth number?

(A) 16

(B) 42

(C) 44

(D) 68

(E) 84

The correct answer is (D) To solve the problem algebraically, let x 5 the

missing number Set up the arithmetic-mean formula, then solve for x:

26 5~90 1 10 2 12 2 26! 1 x

5

26 562 1 x

5

130 5 62 1 x

68 5 x

Or you can try out each answer choice in turn Start with choice (C), which

supplies the middle value, 44 The sum of 44 and the other four numbers is 106

Dividing this sum by 5 gives you 21.2—a number less than the average of 26 that

you’re aiming for So you know the fifth number is greater than 44—and that

leaves choices (D) and (E) Try out the number 68, choice (D), and you’ll obtain

the average of 26

If the numbers are easy to work with, you might be able to determine a missing term,

given the simple average of a set of numbers, without resorting to algebra Simply

apply logic

16 If the average of six consecutive multiples of 4 is 22, what is the greatest

of these integers?

(A) 22

(B) 24

(C) 26

(D) 28

(E) 32

The correct answer is (E) You can answer this question with common

sense—no algebra required Consecutive multiples of 4 are 4, 8, 12, 16, etc Given

that the average of six such numbers is 22, the two middle terms (the third and

fourth terms) must be 20 and 24 (Their average is 22.) Accordingly, the fifth term

is 28, and the sixth and greatest term is 32

Simple Average and Median

A tougher question might ask you to find the value of a number that changes an average

from one number to another

TIP

Numerical answer choices are listed either in ascending or descending order of value, so

if you’re working backward from the choices, start with choice (C), the median value.

If choice (C) is too great or too little, you’ve narrowed down the options either to choices (A) and (B) or to choices (D) and (E).

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When an additional number is added to a set, and the average of the numbers in the set changes as a result, you can determine the value of the number that’s added by applying the arithmetic-mean formula twice

17 The average of three numbers is 24 If a fourth number is added, the

arithmetic mean of all four numbers is 21 What is the fourth number?

(A) 210 (B) 2 (C) 8 (D) 10 (E) 16 The correct answer is (C) To solve the problem algebraically, first determine

the sum of the three original numbers by the arithmetic-mean formula:

24 5a 1 b 1 c

3

212 5 a + b + c

Then, apply the formula again accounting for the additional (fourth) number The new average is 21, the sum of the other three numbers is 212, and the number

of terms is 4 Solve for the missing number (x):

21 5212 1 x

4

24 5 212 1 x

8 5 x You approach arithmetic-mean problems that involve variables instead of (or in

addition to) numbers in the same way as those involving only numbers Just plug the information you’re given into the arithmetic-mean formula, and then solve the problem algebraically

18 If A is the average of P, Q, and another number, which of the following

represents the missing number?

(A) 3A 2 P 2 Q (B) A 1 P 1 Q (C) A 1 P 2 Q (D) A 2 P 1 Q (E) 3A 2 P 1 Q The correct answer is (A) Let x 5 the missing number Solve for x by the

arithmetic-mean formula:

A 5 P 1 Q 1 x

3

3A 5 P 1 Q 1 x

ALERT!

Don’t try the plug-in strategy

(working backward from the

answer choices) to solve

problems like the one here; it’s

too complex Be flexible and

use shortcuts wherever you

can—but recognize their

limitations.

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A GRE question involving arithmetic mean might also involve the concept of median.

If the question deals solely with numbers, simply do the math to calculate the average

and the median

The median value of 1

2,

2

3,

3

4, and

5 6

The arithmetic mean (average) of

1

2,

2

3,

3

4, and

5 6

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (A) First, calculate Quantity A Both columns list the

same four numbers from smallest to greatest in value So Quantity A (the

median) is the average of 2

3 and

3

4: 2

3

3

4

2

8

12

9 12 2

17 12 2

17 24

+

Next, calculate Quantity B, using the common denominator 12:

1

2

2

3

3

4

5 6

6 8 9 10 12

33 12

The mean of the four numbers is one fourth of 33

12, or

33

48 To make the

com-parison, convert Quantity A to a fraction with the denominator 48: 17

24

34 48

Since the numerator in Quantity A (34) is greater than the one in Quantity B (33),

Quantity A is greater than Quantity B

A more challenging question might involve variables instead of (or in addition to)

numbers Here’s a good example:

20. Distribution K contains four terms, represented by the four

expressions p, q, p + q, and q 2 p, where 0 , q , p.

The median of Distribution K

The arithmetic mean (simple average)

of Distribution K

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

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greatest in value among the four terms Here are the four terms, ranked from least to greatest in value:

(q 2 p), q, p, (p + q)

The median value, then, is the average (arithmetic mean) of the two middle terms

p and q:

2

Next, express the arithmetic mean (AM) of the four terms (Quantity B):

AM= q+ +p (p+q)+(qp)= q+p

4

3 4

Since both quantities are positive, you can multiply by 4 across columns to eliminate fractions, leaving the following simpler comparison:

Quantity A: 2p + 2q Quantity B: 3q + p Subtract p and 2q from both columns to further simplify the comparison: Quantity A: p

Quantity B: q

As you can easily see now, Quantity A is greater than Quantity B

STANDARD DEVIATION

Standard deviation is a measure of dispersion among members of a set Computing

standard deviation involves these five steps:

Compute the arithmetic mean (simple average) of all terms in the set

Compute the difference between the mean and each term

Square each difference you computed in step 2

Compute the mean of the squares you computed in step 3

Compute the non-negative square root of the mean you computed in step 4 For example, here’s how you’d determine the standard deviation of Distribution A: {21, 2, 3, 4}:

Arithmetic mean 521 1 2 1 3 1 4

8

45 2 The difference between the mean (2) and each term:

2 2 (21) 5 3; 2 2 2 5 0; 3 2 2 5 1; 4 2 2 5 2 The square of each difference: {32, 02, 12, 22} 5 {9, 0, 1, 4}

The mean of the squares:9 1 0 1 1 1 4

14

4 5 7 2

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A GRE question might ask you to calculate standard deviation (as in the preceding

example) Or, a question might ask you to compare standard deviations You might be

able to make the comparison without precise calculations—by remembering to follow

this general rule: The greater the data are spread away from the mean, the greater the

standard deviation.

21. Distribution A: {1, 2.5, 4, 5.5, 7}

Distribution B: {1, 3, 4, 5, 7}

The standard deviation

of Distribution A

The standard deviation of Distribution B

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The quantities are equal.

(D) The relationship cannot be determined from the information given.

The correct answer is (A) In both sets, the mean and median is 4, and the

range is 6 But the standard deviation of A is greater than that of B, because 2.5

and 5.5 are further away than 3 and 5 from the mean

22 Which of the following distributions has the greatest standard deviation?

(A) {21, 1, 3}

(B) {1, 3, 5}

(C) {0, 4, 5}

(D) {23, 21, 2}

(E) {2, 4, 6}

The correct answer is (C) Notice that in each of the choices (A), (B), and (E),

the distribution’s range is 4 But in choice (C) and choice (D), the range is 5 So

the correct answer must be either choice (C) or (D) Notice that the middle term

in choice (C), 4, is skewed farther away from the mean than the middle term in

choice (D) That means that choice (C) provides the distribution having the

greatest standard deviation

ARITHMETIC SEQUENCES

In an arithmetic sequences of numbers, there is a constant (unchanging) difference

between successive numbers in the sequence In other words, all numbers in an

arithmetic series are evenly spaced All of the following are examples of an arithmetic

sequence:

• Successive integers

• Successive even integers

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• Successive multiples of the same number

• Successive integers ending in the same digit

On the GRE, questions involving an arithmetic sequence might ask for the average or the sum of a sequence When the numbers to be averaged form an arithmetic (evenly spaced) sequence, the average is simply the median (the middle number or the average of the two middle numbers if the number of terms is even) In other words, the mean and median of the set of numbers are the same Faced with calculating the average of a long sequence of evenly-spaced integers, you can shortcut the addition

23 What is the average of the first twenty positive integers?

Express your answer to the nearest tenth

Enter a number in the box

The correct answer is (10.5) Since the terms are evenly spaced, the

average is halfway between the 10th and 11th terms—which happen to be the integers 10 and 11 So the average is 10.5 (This number is also the median.) If you take the average of the first term (1) and the last term (20), you get the same result:

1 1 20

21

2 or 10

1 2 Finding the sum (rather than the average) of an arithmetic (evenly spaced) sequence

of numbers requires only one additional step: multiplying the average (which is also the median) by the number of terms in the sequence The trickiest aspect of this type

of question is determining the number of terms in the sequence

24 What is the sum of all odd integers between 10 and 40?

(A) 250 (B) 325 (C) 375 (D) 400 (E) 450 The correct answer is (C) The average of the described numbers is

25—halfway between 10 and 40 (in other words, half the sum of 10 and 40) The number of terms in the series is 15 (The first term is 11, and the last term is 39.) The sum of the described series of integers 5 25 3 15 5 375

When you calculate the average or sum of a sequence of evenly spaced numbers, be careful counting the number of terms in the sequence For instance, the number of

positive odd integers less than 50 is 25, but the number of positive even integers less

than 50 is only 24

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