Word Problems Print

Approximately 50 pages. Use "save as" to download this chapter to your computer. If a graphic does not appear, press the "refresh" button.

To download section click button or click on “File Save as..” in the upper left-corner of your browser

The GRE is unlikely to give you a simple algebraic equation to solve like:

10?

The GRE never asks questions that are simple and straightforward. Instead, expect questions that require you to convert complex written statements into variables:

If you tripled Adam's age, he would be double Frank's age. Frank is currently 10 years old.

This question translates into:

You have to painstakingly convert the written language into your own algebraic equation. It is easy to make errors here, just be sure to double check yourself by Plugging In your solution or Backsolving from the answer choices.

Word Problems with Undefined Variables

These questions typically ask for increases or decreases in a given amount without any specificity.


	800score Tip Many algebra word problems contain several variables and it is helpful on these questions to track the variables by using names that correspond to the items in the question. For example, if the question is about John, use the variable "j " so that you don't get confused.

How to translate GRESpeak into equations:
Example	Word	What to do?	As equation
The sum of John's age and Steven's is 19.	sum	Addition	j + s = 19
The difference between Todd, the older brother, and Sandra, the younger sister, is 5 years.	difference	Subtraction. Note that Todd is older, so it is Todd minus Sandra.	t - s = 5
The product of my age and 12 is 144.	product	Multiplication	a × 12 = 144
The difference between Todd's age and Sandra's age is 5 years.	difference	Subtraction & absolute value (you don't know who is older).	\|t - s\| = 5
Six less than my age is 22.	less than	Subtraction	a - 6 = 22
The total of my pocket change and 10 dollars is $11.33.	total	Addition	c + 10 = 11.43
10 more than my age equals 23.	more than	Addition	10 + y = 23
Four times my age is 48.	times	Multiplication	4 × y = 48

Much of the challenge in this word problem translation process is not injecting errors. Plug In(picking numbers) and Backsolving are integral to most algebra problems of this sort. These techniques allow you to check your conversion to make sure it is valid.

Example (easy)

Steven is 12 years older than Mary. 3 years ago, Steven was 5 times as old as Mary.

How old is Mary?

Solution

Mary's age = m
Steven's age = s

Translate the GRESpeak

Translation	Statement
s = 12 + m	Steven is 12 years older than Mary
s - 3 = 5(m - 3) s - 3 = 5m - 15 s = 5m - 12	3 years ago, Steven was 5 times as old as Mary.
12 + m = 5m - 12 m = 5m - 24 -4m = -24 m = 6	Now we have two equations, simply solve: Substitute s = 12 + m for s in the equation s = 5m - 12 Mary is 6 years old

Example (medium)

Ethan is as much older than Harry as Harry is older than Candice. Five years ago Ethan's age was double what the age difference between what his and Harry's will be 15 years from now. How old is Candice?

Solution

Ethan's age = e
Harry's age = h
Candice's age = c

Translate the GRESpeak

Translation	Statement
e - h = h - c	First Statement: Ethan is as much older than Harry as Harry is older than Candice
e - 5 = 2[(e +15) - (h +15)]	Second Statement: Five years ago Ethan's age was double what the age difference between what his (Ethan) and Harry's will be 15 years from now.
e - 5 = 2[(e +15) - (h +15)] e - 5 = 2[(e +15 - h -15)]	Now we have two equations, simply solve.
e - 5 = 2(e - h)	The +15 and -15 cancel each other out.
e - 5 = 2e - 2h	Distribute 2(e - h)
-e = -2h + 5	add -2e to both sides and +5 to isolate e
e = 2h - 5	multiply both sides by -1

e - h = h - c e = 2h - c	Set the first statement equal to e by adding 2h to each side.
-e = -2h + 5 add e = 2h - c 0 = - c + 5 c = 5 : Candice is 5 years old	multiply e = 2h - 5 by - 1

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 by letting x, or some other letter, represent the quantity that we wish to determine.

The 5-Step Process for Word Problems:

Quickly read the question and the answer choices to get a feel for what the question is specifically asking.
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). Stop at important points and jot down any pertinent information.
Translate the equation to paper and translate the question into an expression with variables.
If you get a mental block or don't see a shortcut, use Backsolving or Plug In (take numbers and feed them into the question--either answer choices or choose numbers).
If that does not work, start eliminating answers that are outside of the Ballpark; guess and move on.

The GRE uses ƒ(x) to describe a function

A function describes a relationship between one or more inputs and one output.

If ƒ(x) = 2x, then:
ƒ(2) = 4
ƒ(3) = 6

Note: For every ƒ(x) there is one unique value. So, for example, you can't have a ƒ(x) that corresponds to absolute value because ƒ(+2) would result in two answers: -2 and +2.

More interesting functions:

ƒ(x) = 2x/(4 - x)

What does ƒ(x +1) = ?

To solve this, simply plug in (x + 1) in for the function. This is just like Substitution for solving multiple variables.

	2(x + 1)	Substitute (x +1) in for x

	4 - (x + 1)

	2x + 2

	3 - x

Manipulating Functions

Manipulate functions as you would any number.

Note: that if you have f(x)/g(x) and g(x) = 0, then that isn't a real number because zero is in the denominator.

Given f(x) = 3x + 2 and g(x) = 4 - 5x, find:

f(x) - g(x) = [3x + 2] - [4 - 5x ] = 3x + 5x + 2 - 4 = 8x - 2
f (x) + g (x) = [3 x + 2] + [4 - 5x] = 3x - 5x + 2 + 4 = -2x + 6
f(x) × g(x) = (3x + 2)(4 - 5x) = 12x + 8 - 15x² - 10x = -15x² + 2x + 8

Compound Functions

RULE #1: go INSIDE, then go OUTSIDE

f(x) = x²

g(x) = x + 2

f(g(2)) = ?

First, take g(2) because it is on the inside
(2) + 2 = 4

Then take f(4), because it is the outside
f(4) = 16

The #$@#! 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 (easy)

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

Solution

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

Example (easy)

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 (hard)

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). After we have solved what is in the parenthesis, the second step is to do x # (what is in the parenthesis). 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 for 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: Remember order of operations when doing these.)

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 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

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) = 1 # (3 # 2) = 2

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)

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 infer that we got the right answer.


	Double Checking 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.

An ordered list of numbers is called a sequence, and each individual number is a term. Here is a simple sequence of consecutive even integers.

2, 4, 6, 8, 10.…

Here you may assume the next number of the sequence is 12, but often you can't make these assumptions.

Sequence Formula
Sequences use "subscript" which is small font at the foot of a letter. Think of this as a mechanism that generates integer numbers and runs them through a function like a computer.

a_n= 5 + n	Here the subscript n is defined by the rule 5 + n. So this sequence will be a series of multiples of integers + 5 starting with 1. 6, 7, 8, 9, 10, 11, 12...
a_n= 5n	Here the subscript n is defined by the rule 5n. So this sequence will be a series of multiples of 5: 5, 10, 15, 20, 25, 30...

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.


	Trick Question Alert Mr. GRE likes to fool you by making a list of numbers like 3, 5, 7, _ Guess what is next in this sequence? No, it is not 9, it is 11. You thought it was consecutive odd numbers, but it was actually consecutive primes. On Data Sufficiency questions, be keenly aware of the limitations when calculating sequences.

Arithmetic Progressions

Progressions where there is a constant distance between each term.

How would I calculate the 100th unit in this series?
2, 5, 8, 11, 14, 17, 20........

What an easy question?... Let's get started...
2(1), 5(2), 8(3), 11(4), 14(5), 17(6), 20(7), 23(8), 26(9), 29(10).... wow, this is fun!

The GRE obviously does not expect you to make a mad scramble to add up to the 100th term in a sequence. Instead, here is a formula for completing the problem easier:

a_n= a₁+ (n - 1)d

_n	is the term in the series.
a	is the value of the term in the series.
d	is the difference between each number.
n	is the number of terms.
a_n	This is the final number in the series.

Back to our our 100th number project... Let's plug in the numbers.

_n	100	is the term in the series
a		is the value of the term in the series
d	3	is the difference between each term. Just subtract them to get this.
a₁	2	the first number

a₁₀₀ = 2 + (99)3

a₁₀₀ = 299

Isn't that preferable to counting 100 terms?

Sum of Arithmetic Progressions

Try this question out:

Calculate the SUM of the first 100 terms in this series:

2, 5, 8, 11, 14, 17, 20........

What an easy question? Let's get started...

(1)2, (2)5 + 2 = 7, (3)8 + 7 = 15, (4)11 + 15 = 26, (5)14 + 26 =40, ... wow, this is fun!

Wait a second? Why do I always fall for for these GRE tricks? Do I have the word "sucker" stamped on my forehead? Surely there must be an easier way. Indeed there is!

n terms = n ((a₁ + a_n)/2)

n /_n	number of terms in series
a_n	The final term in the set. calculated using a_n= a₁+ (n - 1)d
a₁	first term in the series

Notice how similar this formula is to the one for adding consecutive integers (e.g. the sum of all the integers from 1 to 100). The concept here is the same: you take the average value of the numbers and then multiply it by the number of numbers.

Plug-In 100 for n, 2 for A and 299 for a_n.

n /_n	100	number of terms in series
a_n	299	a_n= a₁+ (n - 1)d (we calculated this above to be: 299.)
a₁	2	first term in the series

Plug these values into the sum formula:
n terms = n ((a₁ + a_n)/2)

n terms = 100((2 + 299)/2)

n terms = 100(150.5)

n terms = 15,050

Geometric Progression
While in an arithmetic sequence the difference between consecutive terms is always the same, in a geometric progression, the quotient of consecutive terms is what remains constant. The common ratio of these sequences is represented by the variable r.

The formula is:

a_n= a₁, a₁r¹, a₁r², a₁r³, a₁r⁴, a₁r⁵

3, 6, 12, 24, 48, 96 ....

Here r = 2, a = 6

a	6	Number of terms in series
r	2	This is the quotient between the terms. So the 3rd term divided by the 2nd term is r.

Translated:
a_n = 3, (3 × 2 = 6), (3 × 2² = 12), (3 × 2³ = 24), (3 × 2⁴ = 48), (3 × 2⁵ = 96)

Calculate the given value of a term:
What is the 6th term in this series?

3, 6, 12, 24, 48 ....

a_n= a₁r^n-1

a₁	3	First term in the series.
a_n	6	We want term 6.
r	2	Divide 24/12 or 96 by 48 and you get 2
r^n-1	32	2⁵

Sub in the values:

a₆= (3)(2⁵)

a₆= 96

Discover Patterns
If a GRE sequence or progression question seems to require more than 5 minutes to do, then it is likely that there is a shortcut or a pattern to solve the question much more quickly. There are few things Mr. GRE relishes more than the "time trap". This is where you get bogged down on a question for minutes at a time, entirely oblivious to the shortcut that could have solved the question in 30 seconds.

Common patterns include repeating of the last digit.
3¹= 3
3²= 9
3³= 27
3⁴= 81
3⁵= 243
3⁶= 729

There is a pattern corresponding with the last digit (3) for every 4th power.

Examples of Common Progression Types
Be able to recognize these progressions when you see them

Progressions	What it is?
0, 3, 6, 9, 12, 15…	X_n= 3 + n	Arithmetic Progression
1, -3, -6, -9, -12…	X_n= -3n	Multiplicative Progression
64, 32, 16, 8, 4…	n, n × 1/2, n × 1/4, r = 1/2	Geometric Progression
1, -3, 9, -27, 81...	n, n × -3, n × 9, n × -27, r = -3	Negative Geometric Progression
1, 2, 3, 5, 8, 13...	Sum of numbers. Add term 1 to term 2 to get term 3.	Misc. Progression

Sample problems

Example 1

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

Solution

(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 × 4 = -8, this is the fifth term in the sequence. 4 × -8 = -32, this is the sixth term. -8 × -32 = 256, this is the seventh term, and the correct answer, choice D.

Example 2

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

Solution

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 3

What is the next term of the sequence -3, 6, -12, 24,…?

(A) -48
(B) 48
(C) -64
(D) 64
(E) -144

Solution

This is a multiplicative progression generated by multiplying each term by the constant value of -2 in order to generate the next term. (-3 × -2 = 6; 6 × -2 = -12, etc.) To generate the next term, multiply 24 by -2 to get -48, choice (A) is the correct answer.

Example 4

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

Solution

This is one of the more complex sequences in which each subsequent term is derived from both arithmetic and multiplicative processes. Like the 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:

?	?	?	?	14
A	B	C	D	E

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) × 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) × 2) + 2
14 = ((n - 1) × 2) + 2

Subtracting 2 from both sides, we get:
12 = ((n - 1) × 2)

Dividing both sides of the equation by 2, we get:
6 = n - 1
Therefore, position D is 6.

?	?	?	6	14
A	B	C	D	E

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:

-1	0	2	6	14
A	B	C	D	E

Now we can see that B = 0, choice (C) is the correct answer.

A. Percentages

The word percent is a fraction whose denominator is 100. For example 26% is equivalent to the fraction 26/100. To change a decimal number to a percent, we simply multiply by 100%:

0.32 × 100% = 32%

If a percentage is given, move the decimal two places to the left to express its equivalent decimal form. The number 0.321 is equivalent to 32.1%.

You can convert a fraction to a decimal by dividing the fraction then moving the decimal point two spaces over and adding a percent sign:
34/100 = .34 = 34%.

Correspondingly, when you multiply something by a percentage, move the decimal point two spaces to the left.

500 × 20% = 500 × 20 = 10,000, then move the decimal two to left to 100%
(of course you could have quickly realized that 20% = 1/5 and solved in your head to get 100.)

10% multiplication trick:

To get 10% of anything, simply slide the decimal point over by 1.
3456 × 10% = 345.6

Example

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

Fraction / Percent Issues

To convert from a percent to a decimal (and vice versa)
You must take a decimal point and insert it into the percent number two spaces from the right:
80% = 0.80, while 225% equals 2.25.
To convert back, move the decimal two places over and add a percent sign. 0.8 becomes 80%
Percent to a fraction
Take the percentage number and place it as a numerator with 100 as the denominator. If you have 70% of something, then you can convert it into the fraction as 70/100.
Fraction to a percent or decimal
Divide the numerator by the denominator. (See the chart below of common conversions).
1/5 equals 5 divided into 1, producing the decimal 0.2.
To make this a percentage, multiply by 100% to get 20%

Common Decimal/Fraction Conversions

1 = 100%	1/3 = 33.33	1/4 = 25%	1/5 = 20%	1/6 = 16.66%	1/7 = 14.28%
1/2 =50%	2/3 = 66.66	3/4 = 75%	2/5 = 40%	5/6 = 83.66 %	2/7 = 28.56%
			3/5 = 60%		3/7 = 42.85%
			4/5 = 80%		4/7 = 57.14%
					5/7 = 71.42%
					6/7 = 85.71%

1/8 = 12.5%	1/9 = 11.1%	1/10 = 10%	1/11 = 9.09%	1/12 = 8.3%
3/8 = 37.5%	2/9 = 22.2%	3/10 = 30%	2/11 = 18.18%	5/12 = 41.7%
5/8 = 62.5%	4/9 = 44.4%	7/10 = 70%	3/11 = 27.17%	7/12 = 58.3%
7/8 = 87.5%	5/9 = 55.6%	9/10 = 90%	4/11 = 36.36%	11/12= 91.7%
	7/9 = 77.7%		5/11 = 45.45%
	8/9 = 88.8%		6/11 = 54.54%
			7/11 = 36.36%
			5/11 = 45.45%
			6/11 = 54.54%
			7/11= 63.63%
			8/11 = 72.72%
			9/11 = 81.81%
			10/11 = 90.90%

Cross Multiplication with Fractions

Use cross-multiplication to solve for percentages. Multiply the upper left by the lower right and set it equal the the upper right multiplied by the lower left.

Numbers		Percentage
Part Original/Whole	=	Part 100%

What is 30% of 210?

_N_
210

30%
100%

N (100%) = (30%)(210)
n = 63

Increase and Decrease of Percentages

Percent increase: If the price of a$10 item increases by 10%, the new price is the original $10 plus 10% of the $10 original (or 110% or the original). "Increase" may go by other names such as "markup" or "wholesale" (cost from factory) to "retail" (cost to public).
Percent decrease: If the price of an item decreases by 10%, the new price is the original $10 minus 10% of the $10 original (or 90% of the original). "Decrease" may go by other names such as "sale".


	How good is a 100% increase? An increase of 100% is the same as doubling something. So 10 increased by 100% is 20 because you are adding 100% of 10 to 10 to get 20. Re-read this if you didn't get it the first time.

Example

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

= 0.2

Convert to a percentage by moving the decimal point two places to the right and adding a percent sign:
% decrease = 20%

if you cross multiply:

Numbers

Percentage

	40

	50

	n

	100%

40(100%) = 50n
20% = n

Discounts & Markup

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:

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

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

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

An employee is to mark up a piece of jewelry 120%. If its wholesale cost is $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

Numbers		Percentage
_n_ 100	=	120 100%

n(100%) = 50n
20% = n

Example

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 × 100 = 66%

Example (easy)

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

P - 0.4P = 240

0.6P = 240
divide both sides by .6

therefore, P = $400

Example (easy)

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

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 (medium)

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

Taking a Percent of a Percent

What happens if you have to take a percentage of something and then modify it again? You multiply the existing total by the new total.

Example

If a stock valued at $100 increases by 10% during the first year, then increases again by another 20% in the second year, what is the stock's final value?

Solution

100 × 110% = 110

then for the second year:

110 × 120% = $132

Sometimes on these double percent questions, you should just Plug In 100 as a test number to get a handle on the quantities.

Example

Joe's portfolio lost 80% of its value, then it gained 10% of its value back. What was his final loss?

Solution

Here you can just substitute 100 for the original value to help solve the question. It goes from 100 to 20. And then add 10% to 20 to get 22.

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.

Example (easy)

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
I =	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 (easy)

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, so you have to add 12 to convert months to years. The annual interest income is $3,600 × (12) so that:

I = PiT

3600(12) = 400,000(i)

You don't need time because you are solving for a rate. You are solving for the i value.

I =	3600(12)

	400,000(i)

0.108 =	(3.6(3)(4))

	4(100)

0.108 =

10.8%

Example 5

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, so 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 (earning $240).

The fourth $1,000 will earn $160 and be worth $1,160. The last will earn $80 and be worth $1,080. The progression is $1,400, $1,320, $1,240, $1,160, $1,080.

The money in the account is the total of those five numbers:
$1,400 + $1,320 + $1,240 + $1,160 + $1,080 = $6,200

Compound Interest Formulas

P(1 + r/n)^nt
P = principle
r = rate
n = number of times per year
t = number of years

Example (easy)

$100 invested over 1 year with semi-annual interest of 4% will generate how much?

Solution

Insert in the formula P(1 + r/n)^nt

Substitute 100 for P, r = .04, n = 2 (because there are two periods in one year).

100 × (1 + .04/2)⁽²⁾⁽¹⁾ = 100 × (1.02)² = $104.04

You made $4.04

A ratio is a fraction that compares two numbers. The ratio of x to y is written as x : y. It is a relationship of part to part.

RATIOS COMPARE PART TO PART, NOT PART TO WHOLE.

If the Yankees win 100 out of 169 games, what is their ratio of wins to losses?

If they win 100 games, they lost 69 games. So the win/loss ratio is 100:69. The fraction of games won would be 100/169.

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.

Proportions

A proportion states that two ratios are equal. Two ratios involve four numbers: two numerators and two denominators. You can solve for one of these numbers 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 multiplied by the other: x = cy (where c is a constant).

Automobile company profits are generally proportional to economic growth.

They are inversely proportional (or indirectly proportional) if one is a constant divided by the other: x = c/y, or equivalently, xy = c.

Your ability to hit a baseball is inversely proportional to how nearsighted you are.

The interest rate a company pays is inversely proportional to its credit rating.

The unemployment rate is inversely proportional to GDP growth

To decide if two quantities are directly or inversely proportional, we ask the question, "Do the quantities both increase or decrease together or does one increase while the other decreases?" If they both increase or decrease at the same rate, 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

	x₁

	y₂

	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

	y₁

	x₂

	y₂

To solve problems involving proportions, 3 of the 4 numbers will usually be known and the problem will be to calculate the fourth.

Example 1

Calculate x if 4 :15 = 16 : x

Solution

The equation is written in a more obvious form:

Cross multiply to get 4x = 16(15) therefore, x = 60

Example 2

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 = 4/1 = 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.

Also, in a ratio, the sum of the ratio (in this case 5) will be multiplied by some number in order to get it to the real total. In this problem, 5 is multiplied by 8 to get 40. You can now use the multiplier for each part of the ratio to find out their actual value. In this case, x, or 4, is multiplied by 8 to get 32. And y is multiplied by 8 to get 8. In this way, if you know the ratio and one "real" value, you can find the multiplier and figure out the rest of the values in the equation.

Example 3

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

12,000 = 2.5x
x = 4,800 miles

Example 4

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.

Example 5

If 100 dollars can buy .07 grams of a rare radioactive material, how many grams can you buy with 10⁶ dollars?

(A) 7
(B) 70
(C) 700
(D) 7000
(E) 70000

Solution

Set up the question as a proportion

100/.07 = (10⁶)/x

cross multiply and solve for x. We get:

100x = .07(10⁶)
x = .07(10⁴) = 700

The correct answer is C.

The distance, D, that the object will travel in time, T, depends on the velocity, V. It is expressed mathematically as:

D = VT

NOTE: Units must match. 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 usually be in kilometers per hour (kph).

Example (easy)

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

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

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

On the sketch above points A, B, C, and D denote different towns. The sketch shows the ABCDA route of a car as well as the average speeds of the car on each leg. The car made no stops in any of the towns. Answer questions 4 and 5 based on the above information.

Example 4

If the driver wants to repeat his ABCDA trip, what must be his speed on leg DA so that the trip would last 3 hours? (Assume that the speed on other legs remains unchanged).

(A) 35 km/h
(B) 42 km/h
(C) 45 km/h
(D) 54 km/h
(E) 60 km/h

Solution

The duration of the actual journey was 30/90 + 50/50 + 60/60 + 40/40 = 3 1/3 hours. To complete the trip in 3 hours, the driver will have to finish the last leg in 2/3 hours. To do so, he has to go at 40/(2/3) = 60 km/h (E).

Example 5

If on each leg the car went 25% slower than it actually did, how much longer would the trip have lasted?

(A) 20 minutes
(B) 30 minutes
(C) 45 minutes 30 seconds
(D) 50 minutes
(E) 66 minutes 40 seconds

Solution

If on each leg the car had gone 25% slower, we can either multiply the average speed of the entire trip by .25 and subtract it from the original OR multiply by .75 to get the new speed. In the original problem, the car drove 180 km in 200 minutes, which gives us an average speed of 0.9 km/m, which can also be written as 9/10 km/m.

We now multiply this number by .75 since it is going 75% the original speed, and we get (9/10 × 75/100) = 27/40 km/m as the new speed.

We now plug this into the equation D = VT, where distance stays the same (180km) and the new velocity is 27/40 km/m.

D = VT
180 = (27/40km/m)(T)
(180) (40/27) = T
266 2/3 minutes = T

2/3 Minutes converts to 40 seconds. The new time is now 266 minutes and 40 seconds. This is 66 minutes and 40 seconds longer than the original time (E) is the correct answer.

Example

Bill takes x hours to run y miles. On Monday, Bill will run in a marathon that is z miles long. How long will he take to finish?

(A) xy
(B) y/z
(C) x/(zy)
(D) (xz)/y
(E) (zy)/x

Solution

The first part of the question tells us Bill's speed.

Speed = Distance/Time = y/x

Use this equation again using Bill's speed (y/x) and the distance of the marathon (z).

y/x = z/Time

In order to solve for time, we cross multiply and get (time)(y) = (x)(z). When we divide each side by y to solve for time, we get (xz)/y = time.

Or, if this is more work than you want to do on the test day, simply Plug-In values for x, y, and z.

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 the question 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 usually equals one completed job.

Examples of Work Questions
	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 1

It takes 3 men 8 hours to paint a house. How long will it take 5 men to paint the same house?

Solution

W = RT

The per hour rate at which each man works is:
R = 1/(3 × 8) = 1/24 houses per hour

The rate R for 5 men is (5R). The work (W) is 1 house. Our equation gives us:
1 = (5/24)(T)
T = 24/5 = 4.8 hours or 4 hours and 48 minutes.

NOTE: 0.8 hours is 0.8 × 60 = 48 minutes.

Example 2

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), which will cancel out the denominators

40 × 60 =	40 × 60 × T

	40

+	40 × 60 × T

	60

2400 =

60T

40T

2400 =

100T

2400 =

100T

24 =

Example 3

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

W is set to 1. 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:

W = RT

1 = 1/20(8) + R(8)
To solve for R, multiply by 20:

20 = 8 + 20R(8)
12 = 8(20)R
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.

Sets

A Set describes a group of items called an element or a member. Sets are usually marked off by these types of brackets {}. All the mammals of the world may be a set, of which dogs and cats are members, but lizards are a member of the reptiles group.

In mathematical terms, you can have a set of odd numbers less than 10 such as {1, 3, 5, 7,9} and a set of prime numbers less than 10 {2, 3, 5, 7} and the overlap is {3, 5, 7}.

Overlapping Sets

When a question relates to overlapping groups, try diagramming the problem with overlapping circles. This will make it easy to account for the overlap.

In group questions, you can add the two groups which represent "either" and subtract the group which represents "both" to get a total number of items in the two sets.

Example 1

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

alone
alone
overlap

total of 23 players
Also, 20 + 10 - 7(both) = 23

Greatest Possible Range Questions

When questions ask for a possible range, be sure to examine the lowest and highest possible scenarios and then multiply or add the possibilities if needed.

Example 2

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 cabinet. 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 3

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).

Tables, Charts, and Graphs (Data Interpretation)

NOTE: Data Interpretation questions are rare on the GRE. Read this section last.

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 difficult to read or understand.

Here are some things to remember when doing problems based on Data Interpretation:

Take your time and read carefully. Understand what you are being asked to do before you begin figuring.
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.
Check the units required. Be sure that your answer is in thousands, millions or whatever the question calls for.
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.
Be careful in computing problems asking for percentages.
1. Remember that to convert a decimal into a percent you must multiply it by 100. For example, 0.04 is 4%.
2. 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.
3. Remember that if quantity x is greater than quantity y, and the question asks what percent quantity x is of quantity y, then the answer must be greater than 100 percent.

Example Set: Table Chart

Examples 1-5 are based on this Table Chart. (Easy)

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
	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

Example 1

How many games did the team win during the first seven weeks?

(A) 32
(B) 29
(C) 25
(D) 21
(E) 50

Solution

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.

Example 2

What percent of the games did the team win?

(A) 75%
(B) 60%
(C) 58%
(D) 29%
(E) 80%

Solution

Choice C is correct. The team won 29 out of 50 games or 58%.

Example 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

Solution

Choice E is correct. The seventh week was the only week that the team lost more games than it won.

Example 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

Solution

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.

Note: if you read this as 5 out of 23, you mis-read the chart column for total games played versus games that week.

Example 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

Solution

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: Interpreting Graphs

Answer the following questions based on the graph above.

Example 6

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

Solution

Reading from the second 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).

Example 7

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

Solution

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).

Example 8

In which year did earnings increase by the greatest percentage over the previous year?

(A) 96
(B) 97
(C) 98
(D) 99
(E) 2000

Solution

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).

Example 9

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

Solution

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).

The charts above show exchange rate dynamics during the week. Answer questions 10-13 using the data provided by the charts.

Example 10

On Tuesday Mr. Smith bought 200 British Pounds using US Dollars. On Wednesday he exchanged these 200 British Pounds back into US Dollars. What profit did Mr. Smith make? (assume zero transaction cost).

(A) $5
(B) $8
(C) $10
(D) $15
(E) $20

Solution

On Tuesday, 1 pound cost 1.7 dollars. Thus, Mr Smith spent 1.7 × 200 = 340 dollars. On Wednesday, Mr Smith earned 1.75 × 200 = 350 dollars. Therefore, Mr Smith made a profit of 350 – 340 = 10 dollars. Or, you can just compare the exchange rates and see that 1.75 – 1.70 = 0.05, so Mr. Smith makes 5 cents off each Pound he buys and sells. Therefore: 200 × 0.05 = 10 (C).

Example 11

What was the price of one US Dollar in terms of Euro on Sunday at the end of the week?

(A) 0.67
(B) 0.75
(C) 0.80
(D) 1.20
(E) 1.33

Solution

B. At the end of the week 1 pound cost 1.6 dollars or 1.2 euros. This gives that 1.6 dollars were equivalent to 1.2 euros. Therefore, 1 dollar was equivalent to 1.2/1.6 = 3/4 = 0.75 euros.

Example 12

On the basis of these charts alone, by how much did the US Dollar depreciate against the Euro during the seven-day period?

(A) 5.00%
(B) 5.50%
(C) 6.25%
(D) 7.50%
(E) 8.25%

Solution

C. At the end of last week on Sunday one dollar cost 1.4/1.75 = 4/5 = 0.8 euros. At the end of this week one dollar cost 1.2/1.6 = 3/4 = 0.75 euros. The dollar depreciated by 0.05 euros which is 0.05/0.8 × 100% = 25/4% = 6.25%.

Questions 13-16 are based on the following chart. Two series of casting one die produced the above results. The bars indicate frequency of occurrence of each die face in the first series (blue bars) and in the second series (red bars).

Example 13

What is the average score of the second series?

(A) 2.0
(B) 2.5
(C) 3.0
(D) 3.333…
(E) 4.111…

Solution

The average score =	(sum of scores)

	(number of throws)

The average score =	[(1 × 3) + (2 × 6) + (3 × 4) + (4 × 4) + (5 × 1) + (6 × 2)]

	[3 + 6 + 4 + 4 + 1 + 2]

The average score =	60

	20

The average score = 3, (C)

Example 14

What is the median score of the first series?

(A) 3.0
(B) 3.5
(C) 4.0
(D) 4.5
(E) 5.0

Solution

C. To find the median score we have to arrange all scores of the first series in ascending order: (1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6).

Now we can compute the median as:

median =	(10th score + 11th score)

	2

median =	(4 + 4)

	2

median =	8

	2

median = 4, (C) is the correct choice.

Example 15

Assume that the die is fair (lands on each of its faces with equal probability). What is the probability that after one more throw in the first series the average score of the first series will increase?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

Solution

B. The current average score of the first series = (1 × 1 + 2 × 4 + 3 × 3 + 4 × 3 + 5 × 4 + 6 × 5)/(1 + 4 + 3 + 3 + 4 + 5) = 4. The new throw has to produce either 5 or 6 to increase the average. The probability that 5 or 6 will fall = 2/6 = 1/3

Earnings	Number of Shares Outstanding

The left chart above shows company's earnings for every month between April and August. The right chart reflects the number of shares outstanding. Using the information provided above, answer questions16 -18.

Example 16

How much greater were the company's earnings in June compared to those in April?

(A) 150%
(B) 175%
(C) 200%
(D) 250%
(E) 300%

Solution

The percentage increase in earnings between April and June = (earnings in June – earnings in April)/(earnings in April) × 100% = (5,000,000 – 2,000,000)/2,000,000 × 100% = 3/2 × 100% = 150% (A) is the correct answer.

Example 17

By how much did Earnings per Share grow from May to June?

(A) $2.50
(B) $3.33
(C) $5.00
(D) $6.67
(E) $10.00

Solution

Earnings per Share in May = $3,000,000/180,000 = $300/18 = $16.666

Earnings per Share in June = $5,000,000/250,000 = $500/25 = $20.

Thus, Earnings per Share grew by $20 – $16.666 = $3.333… or approximately $3.33. The best answer is B.

Example 18

Management predicts that earnings will decline by 5% from August to September. How many shares does the company need to buy back so that the Earnings per Share ratio does not change?

(A) 3,000
(B) 4,000
(C) 5,500
(D) 6,000
(E) 7,500

Solution

August Earnings per Share ratio = $3,000,000/150,000 = $20. Expected earnings in September = $3,000,000 × 0.95 = $2,850,000. Let x denote the number of company's shares in September. Then, $2,850,000/x = $20. From this equation x = 142,500. The company has to buy out 150,000 – 142,500 = 7,500 shares. Capital stock (buy backs) is not counted in the basic EPS, but may be counted in the diluted EPS. Many companies publish both numbers in their annual reports. (E) is the correct answer.

On the diagram above the bars indicate the production capacity of a car manufacturing plant while red bars indicate the actual production. Answer questions 19-21 using information provided by the diagram.

Example 19

If the trend of linear growth continues for both the production and the production capacity, in which year will the actual production equal the capacity?

(A) 2005
(B) 2007
(C) 2007
(D) 2008
(E) 2009

Solution

Each year the capacity grows by 5000 units while the production grows by 6000 units. Thus, each year the gap between the capacity and the actual production is being cut by 1000 units. If this trend continues, the gap will completely disappear by 2007. (C) is the correct answer.

Example 20

If the cost of production of one car was $10000 in 2002 and $15000 in 2004, what was the percentage increase in gross production costs from 2002 to 2004?

(A) 290%
(B) 350%
(C) 380%
(D) 410%
(E) 510%

Solution

Production costs in 2002 = $10,000 × 5000 = $50 million. Production costs in 2004 = $15,000 × 17,000 = $255 million. Thus, the percentage increase in production costs = ($255 mln – $50 mln)/($50 mln) × 100% = 205/50 × 100% = 410%. (D) is the correct choice.

Example 21

All cars manufactured in 2003 were sold for $10,000 each and all cars manufactured in 2004 were sold for $20,000 each. What was the percentage increase in sales revenue from 2003 to 2004 (approximately)?

(A) 166%
(B) 178%
(C) 198%
(D) 209%
(E) 255%

Solution

In 2003 the company earned $10,000 × 11,000 = $110 million. In 2004 the company earned $20,000 × 17,000 = $340 million. The percentage increased in revenue = ($340 mln – $110 mln)/($110 mln) × 100% = 23/11 × 100%. This is slightly greater than 200%. Among the listed choices, choice (D) is the most appropriate answer.


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 GRE because it is a more precise term than Average.
Range		is the difference between the highest and lowest numbers in a set.
Median		If all the numbers in a set are arranged in ascending or descending order, the middle number is the median. If there is an even number of numbers, then the median is half-way between the two middle numbers. If there is an odd number of numbers, it is just the middle number.
Average vs. 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 1

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(because there were 10 students):

Average =	20 + 25 + 3 + 30 + 35 + 40 + 60 + 80 + 90

	10

Average =

The median is, for the case of an even number of entries, the average of the two middle numbers when arranged in order:

20 + 25 + 3 + 30 + 35 + 40 + 60 + 80 + 90

Median =	30+ 35

	2

Median =

32.5

Example 2 (hard)

Solution

First, we need to arrange the elements in ascending order. Because - < x < 0, 2x and x are negative and 2x < x. For the same reason, x² and -x are positive and x² < -x < We do not know whether < -2x but it is not important as long as we know that -x and x² are the third and the fourth largest elements in the set respectively. So, the resorted set will look like (2x, x, x², -x, , -2x) or like (2x, x, x², -x, -2x, ). By definition, Median is:

the middle element of a sorted list if there is an odd number of values in the list;
the average of the two middle elements of the sorted set if there is an even number of values in the list.

In our case, the median = (factor out the x's in the numerator).
The answer is C.

Conclusion of Word Problems Chapter.