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 GRE 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
J. Progressions

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 amont 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

     A sequence is a set of numbers arranged in a definite order. The numbers in the sequence are terms. An example of a sequence is 9, 6, 3, 0, - 3,..., where 9 is the first term, 0 is the 4th term, and the three dots mean that the sequence continues indefinitely.

      An arithmetic progression is a sequence in which each term is derived from the preceding term by adding or subtracting a fixed number called the common difference. The example 9, 6, 3, 0, -3,... is an arithmetic progression in which (-3) is the common difference. The progression -3, 0, 3, 6, 9,... is an arithmetic progression in which 3 is the common difference. If the first term and the common difference are given, the arithmetic progression can be written.

Example 14

The first term of an arithmetic progression is 5 and the common difference is 4. What is the sum of the first six terms?

Solution
The sequence is 5, 9, 13, 17, 21, 25,…

The sum is 5 + 9 + 13 + 17 + 21 + 25 = 90

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   =    6 
15       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

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 1999. 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 1999, 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
1999   $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%
1999	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
1999	.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.

Example 30
If a # b = a + b, then what is 2 # 3?

 

Solution
2 # 3 would equal 2 + 3, or 5.

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)

So, we are now dealing with

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.)


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)

 

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 GRE 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 GRE 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 GRE. 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.

<<go back to table of contents