800Score GMAT Guide
Chapter 6 B: Word Problems
Section 1:  Strategy
Section 2:  Age
Section 3:  Functioins
Section 4:  Sequences
Section 5:  Percent
Section 6:  Interest
Section 7:  Ratio & Proportion
Section 8:  Distance, Rate & Time
Section 9:  Work
Section 10:  Sets
Section 11:  Data Interpretation
Section 12:  Mean, Median & Mode
Section 1: Strategy

The Process for Word Problems

Underneath the convoluted language in GMAT word problems are really just simple math concepts. It is easy to get lost in the words and fail to see the manageable math equations.

The GMAT rarely asks questions that are simple and straightforward. Expect problems that require you to convert complex written statements into variables and equations.

If you tripled Adam's age, he would be double Frank's age. Frank is currently 9 years old.

This translates into the equation:
 3a = 9 2

The most effective strategy for solving a word problem is to express the question as an equation or relationship where x, or some other letter, represents the quantity that you need to find.

 800score Tip Many algebra word problems contain several variables. It is helpful to track the variables by using letters that correspond to the items in the question. For example, if the question is about Adam, use the variable "a."

The 5 Steps

Read and re-read the question until it makes sense. You might need to rephrase the question into your own words, or break the question into simpler parts. Just make sure you know exactly what the question is asking.

As you read the question, be on the lookout for tricks, traps or anything unusual.

Look at the answers before working on a solution. The GMAT often makes questions appear more complicated than they really are, so the answers can clarify your understanding of the question. For example, reading the answers can prevent you from solving for the wrong variable. You can also see how exact your answer needs to be. Can you estimate and use rounding? Can you factor then cancel?

The GMAT often requires you to convert units, so reading the answers will help ensure you use the right units.

3) Define Variables and Relationships

Look for words that define relationships and actions, such as "twice as many", "combined," "gives," or "greater than." More examples of words that show relationships and actions are in the next section and throughout this chapter.

4) Choose a Technique and Apply It

Your goal when taking the GMAT is to pick the right answers on a multiple-choice exam. You are not being graded on style. It doesn't matter if you find the answer following the usual routine. In fact, the fastest way to an answer often does not involve using a traditional method, but rather a method geared specifically to the GMAT.

Each question may have several possible approaches, so you need to develop an ability to recognize a fast and effective way to solve different question types. Once you have chosen a technique, run the numbers and follow your plan. If it is taking too long, see if you can find shortcuts on the math or change your technique. Go back to Step 1 and reread the question and answers, if necessary.

1. Plow
2. Don't Do That Math!
3. Backsolving
4. Plug-In
5. Ballpark
6. Experiment
7. Pattern

5) Eliminate Choices

Picking the correct answer choice may not require a complete solution to the problem. Eliminating all the wrong answers is the same as finding the right answer.

If you don't know an answer, you CANNOT skip the question and return to it later. But don't just guess a random answer, make an educated guess.

Use the process of elimination to decrease the number of possible answers choices, even if you can't find the exact answer. Use logic to rule out wrong answers. With every answer choice you can eliminate, you increase your odds of guessing correctly.

You have to guess aggressively on the GMAT. When a question is clearly too tough, make an educated guess and move on. Don't get stuck on one problem. Take the time you would have wasted on that problem and use it on a question you know how to solve.

The techniques in the next section will give you tools you can use to eliminate wrong answer choices.

How to Eliminate Choices

Video Courtesy of Kaplan GMAT

The examples below illustrate The 5 Steps.

Example

A library is having a book sale. Hardcover books cost \$4 each and paperbacks cost \$2 each. Lisa spent \$28 for 8 books. How many hardcover books and how many paperback books did she purchase?

Solution

The question is looking for two numbers: the numbers of hardcover and paperback books.

Define the variables and write equations based on the given information. Let h be the number of hardcovers and p be the number of paperbacks.

She bought 8 books, so h + p = 8.

Since hardcover books cost \$4, the total cost of h hardcovers is 4h.
Paperback books cost \$2, so the cost of p paperbacks is 2p.
She spent \$28, so 4h + 2p = 28.

h + p = 8
4h + 2p = 28

 Method 1: Substitution Method 2: Addition Solve for h: h = 8  p Substitute: 4h + 2p = 28 4(8  p) + 2p = 28 32  4p + 2p = 28 -2p = -4 p = 2 paperback books Substitute: h + p = 8 h + 2 = 8 h = 6 hardcover books Multiply the first equation by 4: 4h + 4p = 32 Subtract: 4h + 4p = 32  [4h + 2p = 28] 2p = 4 p = 2 paperback books Substitute: h + p = 8 h + 2 = 8 h = 6 hardcover books

Example

There are 15 marbles. There are twice as many white marbles as the number of green and blue marbles combined. One fifth of the marbles are green. How many marbles of each color are there?

Solution

Define the variables. Let w be the number of white marbles, g the number of green marbles and b the number of blue marbles.

Write an equation based on each sentence of the question.

15 = w + g + b
w = 2(g + b)
g = (1/5)(15)
g = 3

Substitute g = 3 into both of the other equations, then simplify.

 15 = w + 3 + b 12 = w + b w = 2(3 + b) w = 6 + 2b

Substitute w = 6 + 2b into the first equation.

12 = 6 + 2b + b
6 = 3b
b = 2

Substitute g = 3 and b = 2 into the original first equation.

15 = w + 3 + 2
w = 10

There are 3 green marbles, 2 blue marbles and 10 white marbles.

Example

The area of a triangle is 15 and the base is 10. What is the height of the triangle?

Solution
Use the formula
 A = 1 bh 2

 15 = 1 (10)h 2

15 = 5h
h = 3

Section 2: Age

Like all word problems, age problems require writing an equation using the words that define the relationships and actions. Age problems just use more of these words than other types of word problems.

The key is to calmly translate the words into manageable equations. Often each sentence requires a new equation.

The words that translate to the equals sign include "is," "was" and "equals."

Common words for other actions and relationships are defined in the following list.

 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's age and his younger sister Sandra's age is 5 years. difference subtraction (since Todd is older, it is Todd minus Sandra.) t – s = 5 The difference between Todd's age and Sandra's age is 5 years. difference subtraction with absolute value (you don't know who is older) |t – s| = 5 John's age is double Henry's age. double multiplication j = 2h Four years ago, Kim was the same age as Bob is now. ago subtraction k  4 = b Six less than my age is 22. less than subtraction a  6 = 22 The total of Mario's age and my age is 35. total addition m + a = 35 Ten more than my age equals 23. more than addition 10 + a = 23 Four times my age is 48. times multiplication 4a = 48 The product of my age and 12 is 144. times multiplication a × 12 = 144

Correctly translating the words into your own algebraic equation is critical, and it is easy to make mistakes. You can double check yourself by Plugging In your solution or Backsolving from the answer choices.

Example

Steven is 12 years older than Mary. Three years ago, Steven was 5 times as old as Mary.
How old is Mary?

Solution

Define the variables.

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

There are 2 variables, so you need 2 equations.

 Sentence Equation Steven is 12 years older than Mary. s = m + 12 Three years ago, Steven was 5 times as old as Mary. s  3 = 5(m  3) Simplify the equation. s  3 = 5m  15 s = 5m  12

Use the 2 equations and solve.

s = m + 12   and   s = 5m  12
m + 12 = 5m  12
4m = 24
m = 6         Mary is 6 years old.

The question only asks for Mary's age, so you dont need to solve for Steven's age. Reading the answers before solving the question would remind you that you are only solving for one age.

Example

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 age and Harry's will be 15 years from now.
How old is Candice?

Solution

Define the variables.

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

There are 3 variables, but you only need to solve for one variable, c. So you need 3 or possibly just 2 equations.

 Sentence Equation Ethan is as much older than Harry as Harry is older than Candice. e  h = h  c Simplify the equation. e = 2h  c Break down each phrase of the next sentence. Five years ago, Ethan's age e  5 was = double Χ 2 difference between  (subtraction) what Ethan's age and Harry's will be 15 years from now e + 15 and h + 15 Write the complete equation. e  5 = 2[(e + 15)  (h + 15)] Simplify the equation. e  5 = 2[e + 15  h  15] e  5 = 2[e  h] e  5 = 2e  2h e = 2h  5

Use the 2 equations and solve.

e = 2h  c
e = 2h  5

2h  c = 2h  5

c = 5         Candice is 5 years old.

In this case you only need two equations to solve for one of three unknowns – the GMAT tests your thinking, not just your math.

Section 3: Functions

Functions

A function ƒ(x) describes a relationship between one or more inputs and one output. On the GMAT, you can simply think of a function as an instruction for how to treat a particular variable or expression.

If ƒ(x) = 2x, then:

ƒ(2) = 4
ƒ(3) = 6

Example

 Let ƒ(x)   = 2x 4 – x

(a) What does ƒ(3) equal?

(b) What does ƒ(x + 1) equal?

Solution

To solve, simply substitute the given value or expression into the function.

(a) Substitute 3 wherever you see x in the function.

 ƒ(3)   = 2(3) = 6 = 6 4 – 3 1

(b) Substitute (x + 1) wherever you see x in the function.

 ƒ(x + 1)   = 2(x + 1) = 2x + 2 = 2x + 2 4 – (x + 1) 4 – x – 1 3 – x

Combining Functions

You can combine functions using any operation.

Example

Let ƒ(x)  =  3x + 2  and  g(x)  =  4 – 5x

(a) Find ƒ(x) – g(x).

(b) Find ƒ(x) + g(x).

(b) Find ƒ(x) × g(x).

Solution

(a) ƒ(x) – g(x)  =  (3x + 2)  (4  5x)  =  3x + 2  4 + 5x  =  8x  2

(b) ƒ(x) + g(x)  =  (3x + 2) + (4  5x)  =  3x + 2 + 4  5x  =  -2x + 6

(b) ƒ(x) × g(x)  =  (3x + 2) Χ (4  5x)  =  12x  15x2 + 8  10x  =  -15x2 + 2x + 8

Composite Functions

Another way of combining functions is with a composite function. This means the functions are nested, so you apply one function to find a value, then apply a second function to that value.

It is important to follow the order of operations, doing the inside function first, then the outside function.

Example

Let ƒ(x)  =  x2  and  g(x)  =  x + 2

(a) Find ƒ(g(x)).

(b) Find g(ƒ(x)).

Solution

(a) First evaluate g(2) since it is the inner function.
g(x) = x + 2, so g(2) = 2 + 2 = 4
Then apply ƒ(x) = x2
(4) = 42 = 16

(b) First evaluate ƒ(2) since it is the inner function.
ƒ(x) = x2, so ƒ(2) = 22 = 4
Then apply g(x) = x + 2
g(4) = 4 + 2 = 6

Variety of Symbols

On some questions, the functions wont use the standard ƒ(x) or g(x) format. Instead, they will use symbols, including #, & and ♣.
The symbols can be confusing, but just treat them the same as any other function and these questions will be easy. Simply plug the numbers into the function.

Example

Let a # b = a + b.

(a) Find 2 # 3.

(b) Find (2 # 3) # 2.

Solution

(a) You can think of this function as addition.
2 # 3 = 2 + 3 = 5

(b) First evaluate (2 # 3) since it is the inner function.
2 # 3 = 2 + 3 = 5
Then apply the same function, addition, again.
(2 # 3) # 2 = 5 # 2 = 7

Example

Let a @ b = ab  and  a & a = a2.

(a) Find a @ (a & b).

(b) Find (a @ a) & a.

Solution

(a) First evaluate (a & b) since it is the inner function.
Since a & a = a2, you can think of this function as multiplication.
a & b = ab

Then apply a @ b = ab. Again, you can think of this function as multiplication.
a @ ab = a2b.

(b) First evaluate (a @ a) since it is the inner function.
Since a @ b = ab, you can think of this function as multiplication.
a @ a = a2

Then apply a & a = a2. Again, you can think of this function as multiplication.
a2 & a = a2 × a = a3

Example

For the numbers x, y, z, the function # is defined as x # y = xy  x.

Find x # (y # z).

Solution

Look for the function rule. Essentially, this function takes the first number, multiplies it by the second number, and then subtracts the first number.

First evaluate (y # z) since it is the inner function.
Since x # y = xy  x, substitute y for x and z for y.
So y # z = yz  y.

Now substitute yz  y into the function.
x # (y # z) = x # (yz  y)

Apply the function rule:
take the first number: x
multiply it by the second number: (yz  y)
then subtract the first number: x

x # (yz  y) = x Χ (yz  y)  x = xyz  xy  x

Depending on the answer choices, you may need to factor out the x.
xyz  xy  x = x(yz  y  1)

Checking

One method to check your answer is to Plug In some numbers.

Choose some numbers for the variables. Let x = 1, y = 3 and z = 2.

x # (y # z) = 1 # (3 # 2)

Now use these numbers and apply the rule x # y = xy  x.

Do the inner function first and evaluate (3 # 2).
Since x # y = xy  x, substitute 3 for the first variable and 2 for the second variable.
x # y = xy  x becomes (3)(2)  3 = 3
So 1 # (3 # 2) becomes 1 # 3.

Since x # y = xy  x, substitute 1 for the first variable and 3 for the second variable.
x # y = xy  x becomes (1)(3)  1 = 2

Check your answer by using the values x = 1, y = 3 and z = 2.
x # (y # z) = xyz  xy  x becomes (1)(2)(3)  (1)(3)  1 = 6  3  1 = 2.

Since these values match the variables in the expression you found, you can infer that your expression is correct.

 800score Tip When doing complex functions or algebra, plugging in numbers can help you better understand the equation and check your answer. However, this technique takes time, so use it judiciously.

Section 4: Sequences

• Sequence
• Writing the Rules
• Arithmetic Sequence
• Arithmetic Series
• Geometric Sequence
• Other Sequences

Sequence

A sequence is an ordered list of numbers. Each number contained in a sequence is called a term.

A sequence is defined by an equation or rule. The rule can be applied to each term of a sequence to generate the next term in the sequence. The ellipsis (...) at the end of a sequence means the sequence continues using the same rule.

The key to solving sequence problems is to determine the relationship between the terms.

Example

What is the next term in the sequence?

(a)  2, 4, 6, 8, ...

(b)  1, 2, 3, 5, ...

Solution

(a) 2, 4, 6, 8, ...
Here you can probably assume the next number of the sequence is 10, but always be careful with snap assumptions on sequence problems.

(b) 1, 2, 3, 5, ...
Here you cannot be sure of the next term. If the sequence is odd integers, the next term would be 9. If the rule is each term is the sum of the previous 2 terms, the next term could be 8.

 Don't Assume The GMAT likes to trick you with a sequence like: What is the next term in this sequence?     3, 5, 7, ____, ... You may immediately think "consecutive odd numbers" and assume the next term is 9. But it could be "consecutive primes" and the next term would be 11. On Data Sufficiency questions in particular, be careful not to make assumptions about the rule in a sequence.

Writing the Rules

The key to solving sequence problems is to determine the relationship between the terms in the sequence. The rule can be applied to each term of a sequence to generate the next term in that sequence.

There is a common set of variables used in the equation or rule for a sequence.

Subscripts are used to give the position-number of a term.

an represents the value of the nth term.

d is the difference between terms.

r is the ratio between terms.

So for the sequence 2, 4, 6, 8, 10:

 value 2 4 6 8 10 term & number a1 = 2 a2 = 4 a3 = 6 a4 = 8 a5 = 10 pattern a1 = 2 a2 = a1 + 2 = 4 a3 = a2 + 2 = 6 a4 = a3 + 2 = 8 a5 = a4 + 2 = 10

Example

Write the first 4 terms of the sequence: an = n + 4

Solution

Substitute the term number for n in the equation that describes the sequence.

The first term a1 would be a1 = 1 + 4 = 5.

The second term a2 would be a2 = 2 + 4 = 6.

The third term a3 would be a3 = 3 + 4 = 7.

The fourth term a4 would be a4 = 4 + 4 = 8.

Example

Write the rule for the sequence:   5, 10, 15, 20, 25, ...

Solution

Look for a pattern that uses the position number n.

The first term a1 = 5 = 1 × 5.

The second term a2 = 10 = 2 × 5.

The third term a3 = 15 = 3 × 5.

The fourth term a4 = 20 = 4 × 5.

So the rule for the sequence is that the nth term is n × 5.

an = 5n

In an arithmetic sequence the difference between consecutive terms is constant. You add (or subtract) the same amount to go from one term to the next. From the Example above, an = n + 4 is an arithmetic sequence.

There is a general form for the formula of an arithmetic sequence. This formula allows you to calculate the value of any term in an arithmetic sequence with just the value of the first term and the difference.

arithmetic sequence    an = a1 + (n – 1)d

Example

Find the 100th term in this sequence:    2, 5, 8, 11, 14, ...

Solution

Sometimes the easiest way to find a term is to just do the calculations.

 term a1 a2 a3 a4 a5 a6 a7 a8 a9 value 2 5 8 11 14 17 20 23 26

But for 100 terms, there has to be a better way.

From looking at the first 5 terms, you can see the difference is 3. Use the formula for arithmetic sequence with a1 = 2 and d = 3.

a100 = 2 + (100 – 1)(3) = 2 + (99)(3) = 299

The 100th term is 299.

Example

The fifth term in a sequence of numbers is 19. Each term after the first term in the sequence is 3 less than the term immediately preceding it. What is the second term in the sequence?

(A) 10
(B) 13
(C) 28
(D) 30
(E) 31

Solution

A good technique to visualize this short sequence is to use blanks.

 ____ ____ ____ ____ _19_ ____ a1 a2 a3 a4 a5 a6

You 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!

19 is 3 less than the term preceding it. So a4 – 3 = a5 = 19, and a4 = 22.

Continue to fill in the blanks.

 _31_ _28_ _25_ _22_ _19_ ____ a1 a2 a3 a4 a5 a6

So the second term a2 = 28. The correct answer choice is (C).

A series is the sum of the terms in a sequence.

sequence:  5, 10, 15, 20
series:         5 + 10 + 15 + 20 = 50

There is a formula for arithmetic series. The formula allows you to calculate the sum of the arithmetic sequence using the number of terms and the values of the first and last terms.

The sum of an arithmetic sequence is the mean of the first and last terms multiplied by the number of terms.

arithmetic series
 Sn = n ( a1 + an ) 2

Notice how similar this is to the formula for adding consecutive integers, such as the sum of all the integers from 1 to 100. The concept here is the same: take the mean of the terms and multiply it by the number of terms.
(Note: This is covered further in Chapter 5B Section 2: Consecutive Numbers.)

Example

Find the sum of the first 100 terms in this series:    2 + 5 + 8 + 11 + 14 + ...

Solution

With this many terms, there must be a shortcut.

The first step is to calculate the value of the 100th term. An Example above does this calculation:

Use the formula for arithmetic sequence with a1 = 2 and d = 3.
a100 = 2 + (100 – 1)(3) = 2 + (99)(3) = 299     The 100th term is 299.

So the first term is 2, the 100th term is 299, and there are 100 terms.

 S100 = 100 ( 2 + 299 ) = 100 ( 301 ) = 100(150.5) = 15,050 2 2

In a geometric sequence the ratio between consecutive terms is constant. You multiply by the same number to go from one term to the next. From an Example above, an = 5n is a geometric sequence where the ratio is 5.

There is a formula for a geometric sequence that allows you to calculate the value of any term in a geometric sequence with just the value of the first term and the ratio.

geometric sequence     an = a1 × r n – 1

Example

Find the 10th term in the sequence:    3, 6, 12, 24, ...

Solution

First find the ratio between terms.

 6 = 2 3

 12 = 2 6

 24 = 2 12

Use r = 2 and a1 = 3 in the formula.

a10 = 3 × 2101 = 3 × 29 = 3 × 512 = 1,536           The 10th term is 1,536.

Example

Find the 7th and 8th terms in the sequence:    -729, 243, -81, 27, ...

Solution

First find the ratio between terms.

 243 = – 1 -729 3

 -81 = – 1 243 3

 27 = – 1 -81 3

So r = -1/3.

Method 1

You could use a1 = -729 in the formula and look for the 7th and 8th terms. But if you use a4 = 27 as the first term, the calculations will be easier. You will look for the fourth and fifth terms after 27.

a4 = 27 × (-1/3)4–1 = 27 × (-1/3)3 = 27 × (-1/27) = 27 ÷ -27 = -1
For (-1/3)3 remember negative × negative × negative = negative.

a5 = 27 × (-1/3)5–1 = 27 × (-1/3)4 = 27 × (1/81) = 27 ÷ 81 = 1/3
For (-1/3)4 remember negative × negative = positive.

Method 2

Another technique for this short sequence is to use blanks. This makes it easier to see the alternating positive and negative terms.

 _-729_ _243_ _-81_ _27 _ _-9 _ _3 _ _-1 _ _1/3_ a1 a2 a3 a4 a5 a6 a7 a8

So the 7th and 8th terms are -1 and 1/3.

Example

Except for the first two terms, every term in the sequence 1, -2, -2, 4,  is the product of the two immediately preceding terms. What is the seventh term of this sequence?

(A) -8
(B) 32
(C) -32
(D) 256
(E) -256

Solution

Method 1

Rather than a ratio between terms, there is a rule. Though you could write the rule for this short sequence, it is easier to use blanks or a list.

a1 = 1
a2 = -2
a3 = -2
a4 = 4
a5 = -2 Χ 4 = -8
a6 = 4 Χ -8 = -32
a7 = -8 Χ -32 = 256

The correct answer choice is (D).

Method 2

Use the strategy of reading the answers first. You can see that there are two basic pieces of information you need to find: whether the number is positive or negative, and how big it is.

First find the pattern for positive and negative without doing the calculations:
positive, negative, negative, positive, negative, negative, positive.

If you quickly calculate the fifth term, just looking at the change between the fourth term and the fifth terms shows you the value increases quickly. From that, you can infer that the largest value will be correct. The largest positive value is 256, which is (D).

Some sequences don't have a simple rule, as shown in the previous Example. If a GMAT sequence seems to require 5 minutes to complete, there almost certainly is a shortcut that solves the question more quickly.

Look for patterns. Some patterns will be arithmetic or geometric sequences. Some patterns are functions. Others will be less visible rules.

Examples of Common Sequence Types

 Terms What it is? Sequence 0, 3, 6, 9, 12, ... an = n + 3 arithmetic sequence 1, -3, 9, -27, 81, ... an = a1 × (-3)n–1: r = -3 geometric sequence 64, 32, 16, 8, 4, ... an = a1 × (1/2)n–1: r = 1/2 geometric sequence 1, 2, 3, 5, 8, 13, ... 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, ... pattern: add the two previous terms to get next term (Fibonacci sequence) 1, 2, 4, 7, 11, 16, ... 1 + 1 = 2, 2 + 2 = 4, 4 + 3 = 7, ... pattern: an = an–1 + (n – 1)

Example

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 a more complex sequence, considering that each new term is derived from both adding and multiplying.

A good technique to visualize this short sequence is to use blanks. The fifth term, E, equals 14.

 ____ ____ ____ ____ _14_ A B C D E

Translate the rule into a function. Look at each phrase.

 the value is → starting with an 2 more than → add 2 the product of → multiplication the previous integer → the previous term, an–1 and 2 → multiply by 2

So the current term an is 2 plus (2 times the previous term an–1).
an = 2 + 2an–1

You have E = a5 = 14. Use this function and apply it to E to get values for the terms D, then C, then B.

an = 2 + 2an–1
 E = 2 + 2D and E = 14 14 = 2 + 2D 12 = 2D D = 6 D = 2 + 2C and D = 6 6 = 2 + 2C 4 = 2C C = 2 C = 2 + 2B and C = 2 2 = 2 + 2B 0 = 2B B = 0

 ____ _0_ _2_ _6_ _14_ A B C D E

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

Note: Don't waste time finding the value for A.

Section 5: Percent

• Percent & Conversions
• Multiplying by Percent
• Percent Increase and Decrease
• Discounts and Markups
• Sale Prices
• Mixture Problems
• Taking a Percent of a Percent
• Common Values

Percent & Conversions

The word percent means per 100. It is a fraction whose denominator is 100. For example, 26% is equivalent to the fraction 26/100.

A percentage is used to compare values. It can describe the relationship of a part to a whole, or it can describe a magnitude of change.

Decimal to Percent
To change a decimal number to a percentage, multiply by 100; this moves the decimal point 2 places to the right. Then add the percent symbol.

0.32 = 32%        0.07 = 7%        1.75 = 175%

Percent to Decimal
To change a percentage to a decimal, divide by 100; this moves the decimal point 2 places to the left. Then remove the percent symbol.

25% = 0.25        90% = 0.90 = 0.9        32.1% = 0.321        100% = 1

Fraction to Decimal
You can convert a fraction to a decimal by dividing the numerator by the denominator. Sometimes a quicker method is to change the fraction so it has 100 as the denominator.

 21 = 0.21 100
 43 = 86 = 0.86 50 100
 13 = 13 ÷ 16  =  0.8125 16

Decimal to Fraction
To convert decimal to a fraction, move the decimal point 2 places to the right and use 100 as the denominator. Then simplify the fraction. If there is still a decimal in the numerator, move the decimal point to the right as needed, and add a 0 to the denominator for each place value.

 0.55 = 55 = 11 100 20
 2.18 = 218 = 109 = 2 9 or 2.18 = 2 18 = 2 9 100 50 50 100 50
 0.123 = 12.3 = 123 100 1000

Note: There is a chart at the end of this section with common fraction and decimal conversion values.

Fraction and Percent
To change a percentage to a fraction, change the percentage to a decimal, then the decimal to a fraction.

 34% = 0.34 = 34 = 17 100 50

Correspondingly, to change a fraction to a percentage, change the fraction to a decimal, then the decimal to percent.

 18 = 2 = 0.2 = 20% 90 10

Example

Convert 4% into a decimal and a fraction in lowest terms.

Solution

To convert 4% into a decimal, move the decimal point two places to the left. This requires adding a leading 0.
4% = 0.04

To express 4% as a fraction, put it over a denominator of 100, and then simplify the fraction.
 4 = 4 = 1 100 4 × 25 25

Multiplying by a percentage is a way to find a "part" of the original. Multiplying by a percentage is the same as multiplying by the decimal equivalent.

15% of 40 = 15% × 40 = 0.15 × 40 = 0.6                       8% of 25 = 8% × 25 = 0.08 × 25 = 2

quick trick: Multiplying by 10%

To get 10% of anything, simply slide the decimal point one place to the left.
3456 Χ 10% = 345.6

For 20%, slide the decimal point one place to the left, then multiply by 2.
3456 Χ 20% = 345.6 Χ 2 = 691.2

The same pattern hold true for 30%, 40%, 60%, 70%, 80% and 90%. But for 50%, simply divide by 2.

Example

What is 30% of 210?

Solution

You are trying to find the "part" of 210 that is equal to 30%.

 30% of 210 = 210 Χ 30% Change to multiplication. = 21 Χ 3 Slide the decimal point one place to the left, then multiply by 3. = 63

30% of 210 is 63.

You may also be given two numbers (a part and a whole) and asked to find the percentage of the whole represented by the part. One way to do this is to use a proportion, which sets 2 fractions equal to each other. Change the given numbers to a fraction and set it equal to a fraction with denominator 100. Solve for the numerator and that will be the percentage.

Example

What percent of 24 is 9?

Solution

You are trying to find the percent that matches the given fraction.

 part = percent so 9 = x whole 100 24 100

Method 1

900 = 24x       Use cross multiplication.

 x = 900 = 3 × 3 × 4 × 25 = 3 × 25 = 75 = 37.5 24 2 × 3 × 4 2 2

Method 2

Start by simplifying the fraction.
 9 = 3 24 8

 Since you know 1 = 0.125, you can quickly calculate 3 = 3(0.125) = 0.375 = 37.5% 8 8

Percent Increase and Decrease

Percentages can be used to describe the magnitude of change. To find the percentage of change, you compare the new amount to the original amount.

Percent increase: If the price of a \$30 item increases by 10%, the new price is the original \$30 plus 10% of the \$30 original, \$33. This is 110% or 1.1 times the original price. "Increase" may go by other names such as the "markup" from "wholesale" (cost from the factory) to "retail" (cost to the public).

Percent decrease: If the price of an item decreases by 10%, the new price is the original \$30 minus 10% of the \$30 original, or \$27. This is 100% – 10% = 90% or 0.9 times the original price. "Decrease" may go by other names such as "sale" or "discount."

 How much is a 100% increase? An increase of 100% is the original plus 100% of the original, which is the same as doubling or multiplying by 2. So 30 increased by 100% is 60 because you are adding 100% of 30 to 30.

Example

If the price of a stock drops from \$60 to \$45, what is the percentage decrease?

Solution

You are trying to find the percent decrease based on the original value and the new value.

 change = percent so new – original = percent original 100 original 100

 45 – 60 = x 60 100

 -15 = x 60 100
The negative tells you this is a decrease. You can leave it out of the rest of the calculations as long as you remember to use the correct terminology (e.g., decreased or dropped).
 15 = 1 = 25 =  25% 60 4 100
For this fraction, it is easier to simplify than cross-multiply.

The price dropped by 25%.

Example

The monthly cost of cable internet service went from the introductory price of \$39.99 a month to \$50.99 a month. What was the percentage increase from the introductory price?

Solution

The original price here is the introductory price.

 change = percent which is new – original = percent original 100 original 100

 50.99 – 39.99 = x 39.99 100
 11 = x 39.99 100

39.99x = 1100     (Round and use 40 instead of 39.399)
x = 27.5

The price increased by 27.5%.

A percentage can be used to apply the same proportion of change to multiple values.

Discount is the decrease in price of an item when the price is decreased by a certain percentage.

Markup is the increase in price when the cost of an item is increased by a certain percentage. The following examples illustrate this concept.

For markups and discounts, calculate:
 new – original = percent original 100

If the value is negative, it is a discount. If the value is positive, it is a markup.

Example

A pair of aerobic shoes was priced \$115 and is now discounted to \$69. What is the percentage discount?

Solution

 new – original = percent original 100
 69 – 115 = x 115 100
 -46 = x 115 100

The negative tells you this is a decrease. You can leave it out of the rest of the calculations.

 46 = 2 × 23 = 2 = 40 = x 115 5 × 23 5 100 100

x = 40

The discount is 40%.

Example

A pair of aerobic shoes is purchased at wholesale for \$69 and sold by the store for \$115. What is the percentage markup?

Solution

 new – original = percent original 100
 115 – 69 = x 69 100
 46 = x 69 100

Notice the original price here is \$69, not \$115.

 46 = 2 × 23 = 2 = 67 = x 69 3 × 23 3 100 100

x = 67

The discount is 67%.

Example

An employee is to mark up the price of a piece of jewelry by 120%. If its wholesale cost was \$110, what will be its selling price?

Solution

Notice that the price is being marked up by 120%, not to 120%.

The amount of the markup is 120% of \$110 so it's 1.2 × 110 = \$132.
The selling price is then the original price plus the markup, so \$110 + \$132 = \$242.

Another way is to calculate the selling price is 120% + 100% = 220%, so 2.2 × \$110 = \$242

Example

A college bookstore purchases trade books on a 30% margin, i.e., it purchases a trade book for 30% less than its retail price. What is the percentage markup from the wholesale price?

Solution

The wholesale price is the retail price minus 30% of the retail price.

wholesale = retail  (30% of retail) = (100%  30%) Χ retail = 70% Χ retail

So the wholesale price is 70% of the retail price.

But don't stop there. It is a common GMAT trick to require you to use the result of one part of a question to get the final answer.

This question is asking for the markup from wholesale to retail.
To make calculations easier, use \$100 rather than a variable as the retail price of a trade book. Then the wholesale is 70% Χ retail = 70% Χ \$100 = \$70.

 new – original = percent original 100
 retail – wholesale = markup wholesale 100
 100 – 70 = markup 70 100

 30 = 3 = 42.8 = markup 70 7 100 100

So the markup from wholesale to retail is 43%.

Example

Find the number of residents in a city if 20% of them, or 6200 people, ride bicycles.

Solution

Let R be the number of residents. Translate the words in an equation.

20% of R is 6200 so 0.2R = 6200

 R = 6200 = 6200 0.2 2 10
Change the decimal to a fraction. To divide by a fraction, multiply by the reciprocal.

 R = 6200 × 10 = 62,000 = 31,000 2 2

The city has 31,000 residents.

In determining sale prices, be careful not to mix up the amount taken off the original price with the new, sale price.

Example

Kathy buys a bike for \$240 after a 40% markdown. What was the original price?

Solution

Since the markdown is 40%, the purchase price is 60% of the original price.
100%  40% = 60%
purchase price = 60% of original price

Let x be the original price.

0.6x = 240

 x = 240 = 240 = 240 × 10 = 400 0.6 6 6 10

The original price was \$400.

Example

A sweater, originally \$80, is on sale for 25% off. What is the sale price?

Solution

Since the markdown is 25%, the purchase price is 75% of the original price.
100%  25% = 75%
sale price = 75% of original price

Let s be the sale price.

0.75(80) = s
 s = 0.75 × 80 = 3 × 80 = 60 4

The sale price is \$60.

Example

Kent pays 20% tax on income between \$10,000 and \$20,000 and 30% on income over \$20,000. The first \$10,000 is tax-free. If he paid \$14,000 in taxes, what was his income?

Solution

Let Kent's income be k. Divide his income into the tax brackets.

For \$0 to \$10,000, there is no tax.

The \$10,000 between \$10,000 to \$20,000 is taxed at 20%.

Income beyond \$20,000 is taxed at 30%. So,

10,000 (0%) + 10,000 (20%) + (k  20,000)(30%) = 14,000
0 + 2000 + 0.3(k  20,000) = 14,000
2000 + 0.3k  6000 = 14,000
0.3k = 18,000
 k = 18,000 × 10 = 60,000 3

Kent's income was \$60,000.

Mixture Problems

Mixture Problems

Example

How many gallons of pure water must be added to 100 gallons of a 4% saline solution to produce a 1% saline solution?

Solution

Let x be the gallons of pure water to be added.

In 100 gallons of a 4% saline solution, there are

0.04(100) = 4 gallons of salt.

In the 1% solution, the total number of gallons will be 100 + x. The amount of salt will remain constant at 4 gallons.

0.01(100 + x) = 4
1 + 0.01x = 4
0.01x = 3
x = 3/0.01 = 300

300 gallons of water need to be added to produce a 1% saline solution.

Taking a Percentage of a Percentage

What happens if you take a percentage of a number and then take a percentage of that new value? You are just multiplying the new total by a second percentage. This is a common GMAT trick.

Example

If the price of a stock starts 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 price?

Solution

In the first year, \$100 × 110% = \$110.

For the second year, \$110 × 120% = \$132.

The stock price would be \$132.

Notice that this is not the same price change as 10% + 20% = 30%.
A 30% increase would have been resulted in a \$130 price.

Fractions and Percentile

Example

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

Solution

Use \$100 for the original value to help solve the question.

The portfolio lost 80%.
100%  80% = 20%, so \$20 was left.

That \$20 gained back 10% of its value, so \$20 Χ 110% = \$22.

So 22% is the percentage of the original investment that he still has.

Subtract 100%  22%. The final percentage loss is 78%.

Notice that this is not 100%  80% + 10% = -70%, or a 70% loss.

Example

A backpack was marked for sale at 30% off, and an additional 20% was taken off at the register. If the original price was \$50, what was the final purchase price?

Solution

The discount is not 30% + 20% = 50%.

30% off means you pay 70%.
Another 20% off means paying 80% of the 70%.
\$50 Χ 0.8 Χ 0.7 = \$50 Χ 0.56 = \$28

A way to check this is to calculate the dollar amount coming off the price.
30% off of \$50 = 0.3 Χ \$50 = \$15. So the purchase price before getting to the register is \$50  \$15 = \$35.
20% off of \$35 = 0.2 Χ \$35 = \$7
So the final purchase price is \$35  \$7 = \$28.

Common Conversions

 1 = 100% 1⁄10 = 10% 1⁄5 = 2⁄10 = 20% 1⁄3 = 33.33 3⁄4 = 75% 3⁄10 = 30% 2⁄5 = 4⁄10 = 40% 2⁄3 = 66.66 1⁄2 = 50% 7⁄10 = 70% 3⁄5 = 6⁄10 = 60% 1⁄4 = 25% 9⁄10 = 90% 4⁄5 = 8⁄10 = 80%

For many other denominators, you can remember the first value then multiply to get other values.
For example, if you remember that 1/8 = 0.125, then 3/8 = 3 Χ 0.125 = .375

 1⁄6 = 16.66% 1⁄7 = 14.28% 1⁄8 = 0.125 = 12.5% 1⁄9 = 11.1% 1⁄11 = 9.09% 1⁄12 = 8.3% 5⁄6 = 83.66% 2⁄7 = 28.56% 3⁄8 = 0.375 = 37.5% 2⁄9 = 22.2% 2⁄11 = 18.18% 5⁄12 = 41.7% 3⁄7 = 42.85% 5⁄8 = 0.625 = 62.5% 4⁄9 = 44.4% 3⁄11 = 27.27% 7⁄12 = 58.3% 4⁄7 = 57.14% 7⁄8 = 0.875 = 87.5% 5⁄9 = 55.6% 4⁄11 = 36.36% 11⁄12 = 91.7% 5⁄7 = 71.42% 7⁄9 = 77.7% 5⁄11 = 45.45% 6⁄7 = 85.71% 8⁄9 = 88.8% 6⁄11 = 54.54% 7⁄11 = 36.36% 8⁄11 = 72.72% 9⁄12 = 81.81% 10⁄11 = 90.90%

Section 6: Interest

Interest is money earned (or paid) on an investment (or loan).

There are two types of interest that appear on the GMAT.

Simple interest is calculated based only on the principal (the original amount).

Compound interest is calculated based on the principal plus all previously accrued interest.

Simple Interest

The formula for calculating simple interest is:

I = P r t

The amount of interest  I  is the dollars earned or paid to the investor.

The principal  P  is the amount of money invested.

The interest rate  r  is the annual interest rate expressed as a decimal value.

The time period  t  is measured in years.

Example

A student invests \$1,000 at 10% simple interest for the summer (3 months). How much interest does the student earn?

Solution

Use the simple interest formula.

I = Prt

The principal is \$1,000. The interest rate of 10% per year is 0.10. The time is 3 months, so 3/12 of a year.

 I = 1,000(0.10)( 3 ) 12

 I = 100( 1 ) = 25 4

The student earned \$25.

Example

A professor retires with a retirement fund of \$400,000. If her monthly interest income is \$3,600, what is the annual simple interest rate?

Solution

The principal is \$400,000. The \$3,600 interest is earned per month, so the time is 1/12.

Use the simple interest formula and find r.

I = Prt

 3600 = 400,000r( 1 ) 12

 r = 3600 × 12 = 0.108 400,000

The interest rate is 10.8%.

Example

At the beginning of each year, an investment of \$1000 is added to an account earning simple interest of 8%. How much money is in the account after 5 years?

Solution

First, note that \$1000 is added EACH year, so \$5000 is invested.

Simple interest means that no interest is paid on accrued interest, only on the principal (which in this case grows by \$1000 every year).

The first \$1000 will earn interest for 5 years.
I = 1000(0.08)(5) = 400
The first \$1000 will earn \$400.
So after 5 years, the amount in the account from this first \$1000 is 1000 + 400 = \$1400.

The second \$1000 will earn interest for 4 years for a total of \$320.
The amount in the account from this \$1000 is 1000 + 320 = \$1320.

The third \$1000 will earn interest for 3 years for a total of \$240.
The amount in the account from this \$1000 is 1000 + 240 = \$1240.

The fourth \$1000 will earn \$160 and be worth \$1160.

The last \$1000 will earn \$80 and be worth \$1080.

The money in the account after 5 years is the total of these five values:

\$1400 + \$1320 + \$1240 + \$1160 + \$1080 = \$6200

Compound Interest

The formula for calculating compound interest is:

A = P (1 + rn) nt

As before, the principal P is the amount of money invested. The interest rate r is the annual interest rate expressed as a decimal value. The time period t is measured in years.

This formula gives the value of the account A, not the interest earned. The new component is n, which is how often the interest is paid or compounded per year.

Example

An account has \$1500 invested with semi-annual interest of 4%. How much will the account earn in its first year?

Solution

The principal is \$1500. The interest rate is 4%. Interest is paid twice a year, so n = 2. The time is 1 year.

A = P (1 + rn) nt

A = 1500 (1 + 0.042) 2×1

A = 1500(1.02)2 = 1500(1.0404) = 1560.60

The account now has \$1560.60. But be careful, the question asks how much was earned, which is:
1560.60 – 1500 = 60.60

The account earned \$60.60.

On the GMAT, you may encounter a question where it seems like compound interest would take too long to calculate. By looking at the answer choices you can determine if an estimate will work. Part (c) in the Example below shows how you can estimate and use logical reasoning to find the correct answer.

Example

An account has \$4000 invested and earns 5%. How much will the account be worth after one year if:

(a) the interest is simple interest

(b) the interest is compounded quarterly

(c) the interest is compounded daily

Solution

(a) The formula for simple interest is I = Prt
I = 4000(0.05)(1) = 200
The account earned \$200 in interest, so the new value is \$4200.

(b) The formula for compound interest is A = P (1 + rn) nt
The interest is compounded quarterly for a year, so n = 4 and t = 1.

A = 4000 (1 + 0.054) 4 = 4000(1.0125)4 ≈ 4000(1.05) = 4204

The new value is \$4204, so the account earned \$204 in interest.

(c) The formula for compound interest is A = P (1 + rn) nt
The interest is compounded daily for a year, so n = 365 and t = 1.

A = 4000 (1 + 0.05365) 365 = 4000(1.00014)365 ≈ 4000(1.0524) = 4210

The new value is \$4210, so the account earned \$210 in interest.

If this was a multiple-choice question, you could look at the answer choices to see how accurate the answer needed to be. Lets say the answer options are:

(A) \$200    (B) \$210    (C) \$4182    (D) \$4200    (E) \$4210

You could easily calculate that the value with simple interest would be \$4200. Since you know that the amount with compound interest must be greater, the only possible answer choice is (E) and there is no need to do that calculation.

Section 7: Ratio & Proportion

• Ratio
• Proportion
• Inversely Proportional

Ratio

A ratio is used to compare two quantities.

The ratio of numbers a and b can be expressed as:

• the ratio of a to b
• a is to b
• a : b
•  a b

A fraction always compares a part to a whole. A ratio can compare a part to a part, or a part to a whole.

example
If there are 2 oranges and 3 apples, the ratio of oranges to apples is 2:3.
The ratio and fraction of oranges to the total number of fruits is 2/5.

Example

A baseball team won 98 out of the season's 162 games.

(a) What is the ratio of wins to losses?

(b) What is the ratio of wins to games?

Solution

(a) If the team won 98 games, they lost 64 games. So the ratio of wins to losses is 98 : 64, or 98/64.

(b) The ratio of wins to games is 98/162.

Fractions of a Sum

Example

Orange concentrate is diluted with water to make juice that is 1/5 concentrate.

(a) What is the ratio of concentrate to water?

(b) What is the ratio of water to juice?

Solution

(a) Juice that is 1/5 concentrate means that the total amount of juice is 5. One part of that total 5 is concentrate. The amount of water is total juice minus concentrate, so 5 – 1 = 4. The ratio of concentrate to water is 1 : 4.

(b) The fraction of concentrate in the total amount of juice is 1/5. Water makes up the rest of the juice, so its fraction is 4/5.

Example

The ratio of two numbers is 4:1, and their sum is 40. Find the two numbers.

Solution
 Write the ratio of the two numbers x to y as x = 4 , so x = 4y y 1

Their sum is 40, so x + y = 40.

4y + y = 40         Substitute x = 4y into x + y = 40
5y = 40
y = 8

x = 4y, so x = 32

The two numbers are 32 and 8.

Ratios can be between more than just two terms. One way to look at the relationship is that they are all multiples of one value.

example
The ratio is 1 : 2 : 5.
The total is 40.
So x + 2x + 5x = 40.

Example

Red, blue and yellow marbles in a bag have a ratio of 5 to 2 to 6. After removing the red marbles, there are 32 marbles left in the bag. How many red marbles were in the bag?

(A) 4
(B) 8
(C) 20
(D) 24
(E) 52

Solution

Use the total number of marbles in the bag and the ratio of just blue plus yellow.

blue + yellow = marbles left
2x + 6x = 32
8x = 32, so x = 4.

You can use x = 4 and the original 5 : 2 : 6 ratio to find the number of red marbles.
5x = 5 × 4 = 20
There were 20 red marbles in the bag.

The correct answer choice is (C).

Be careful you're answering the question being asked. Here, the wrong answer options include x = 4, 2x = 8 blue marbles, 6x = 24 yellow marbles, and 32 + 20 = 52 marbles originally in the bag.

Ratios

Proportion

A proportion states that two ratios are equal.

A proportion comparing two ratios can be expressed as:

• a is to b as c is to d
• a : b = c : d
•  a = c b d

You can solve for any term in a proportion: a, b, c or d.

Example

 Find the value of x. 35 = x 84 24

Solution

Method 1

35 × 24 = 84x         Cross multiply.

 x = 35 × 24 Divide both sides by 84. 84

 x = 35 × 24 = 7 × 5 × 2 × 2 × 6 = 10 Factor and cancel to simplify the calculations. 84 2 × 6 × 7

x = 10

Method 2

 35 = x Factor and simplify the fractions. 84 24

 7 × 5 = x 7 × 12 2 × 12

5/12 = x / (2 × 12)
5 = x / 2
x = 10

Example

Find the value of x.         4 : 15 = 16 : x

Solution

Write the proportion using fractions.

 4 = 16 15 x

4x = 16 × 15                                   Cross multiply.

 x = 16 × 15 = 4 × 15 = 60 Divide both sides by 4. 4

x = 60

Example

If 100 dollars can buy 0.07 grams of a rare radioactive material, how many grams can you buy with 106 dollars?

(A) 7
(B) 70
(C) 700
(D) 7000
(E) 70,000

Solution

Write the question as a proportion.

 0.07 = g 100 106

0.07 × 106 = 100g            Cross multiply.

 g = 0.07 × 102 × 104 = 7 × 104 = 7 × 102 Divide both sides by 100 and simplify. 100 102

g = 700

The correct answer choice is (C).

Inversely Proportional

Two quantities are directly proportional if one equals the other multiplied by a constant, or y = cx where c is a constant.
The Examples above have all been directly proportional.

 4 = 16 is the same as 4 × 4 = 16 15 x 15 × 4 x

 35 = x is the same as 35 = 3.5x 84 24 84 3.5 × 24

When two quantities are directly proportional, both values increase or both values decrease.

Two quantities are inversely proportional if one decreases when the other increases.
One value is equal to a constant divided by the other, or
 y = c x
which can be rewritten xy = c.

examples

The unemployment rate is inversely proportional to economic growth. Unemployment goes up when economic growth goes down.

The interest rate a company pays is inversely proportional to its credit rating. The better the credit rating, the lower the interest rate.

Example

Roger drove 90 miles in 1.5 hours. How far would he go in 2.5 hours?

Solution

This represents a direct proportion: both the distance and time increase.

 90 miles = m 1.5 hours 2.5 hours

 m = 90 × 2.5 = 90 × 5 × 0.5 = 30 × 5 = 150 1.5 3 × 0.5

Roger would drive 150 miles.

Example

Roger drove 1 hour at 60 miles per hour. If he drove at 40 miles per hour, how long will it take to go the same distance?

Solution

This represents an indirect proportion: as the speed decreases, the time increases.

60 mph × 1 hour = 40 mph × h hours
60 = 40h

 h = 60 = 1.5 40

It will take 1.5 hours. The speed is 40/60 = 2/3 of the original speed, so the time is 3/2 = 1.5 times the original time.

Mileage Calculation

Example

It takes 4 men 6 hours to repair a road. Approximately how long will it take 7 men to do the job if they work at the same rate?

Solution

This represents an indirect proportion: as the number of workers increases, the time decreases.

4 men × 6 hours = 1 job

7 men × h hours = 1 job

Set the 2 equations equal.

4 × 6 = 7h

 h = 24 =  3 3 ≈  3.5 7 7

It will take the 7 men about 3.5 hours.

Section 8: Distance, Rate & Time

The distance  d  that an object will travel is equal to its rate  r  times its time  t.

d = r t

Rate is sometimes called speed or velocity. Rates are given as ratios, such as "miles per hour" or mph and "kilometers per hour" or km/h.

The distance covered in 1 hour at 20 miles per hour is the same as the distance covered in 30 minutes at 40 miles per hour.

d = r t, so 20 mph × 1 hour = 40 mph × 30 minutes

When setting up an equation, make sure that the units match. The rates 1 mile per minute and 60 miles per hour are the same.

 1 mile × 60 minutes = 60 miles 1 minute 1 hour 1 hour

The most common methods for solving motion problems are to use a system of equations or to use a proportion.

Converting Rates and Speeds

Video Courtesy of Kaplan GMAT
Example

A biker traveled 60 miles in 2.5 hours. Find the biker's average speed.

Solution

d = r t

60 = r × 2.5

r = 60 ÷ 2.5 = 24         Divide both sides by 2.5.

r = 24

The biker's average speed was 24 miles per hour.

Example

An airplane travels 1,200 miles in 2.5 hours. How far will it travel in 10 hours?

Solution

Use a proportion.

 1200 miles = m miles 2.5 hours 10 hours

 m = 1200 × 10 = 1200 × 100 = 1200 × 4 = 4800 2.5 hours 25

The plane will travel 4,800 miles in 10 hours.

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 / yz
(D) xz / y
(E) yz / x

Solution

Use d = r t and a proportion. The question assumes the average rate will be the same for the practice and the marathon. So the ratio of r = d / t will be the same.

Taking x hours to run y miles gives time and distance. The marathon is z miles long, so that is the distance. You need to find the time it takes to run z miles.

 y miles = z miles x hours ? hours

 ? = xz y

The correct answer choice is (D).

Rate/Speed

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

The distance traveled by the officer equals the distance traveled by Philip.
Let r miles per hour be Philip's speed. Use d = r t.

distance of the police officer = 100 mph × 2 hours
distance of Philip = r mph × 2.5 hours. The head start added half an hour to Philip's time.

100 mph × 2 hours = r mph × 2.5 hours

200 = 2.5r
r = 200 ÷ 2.5 = 80

Philip was driving at 80 miles per hour.

Example

It takes 7 hours for a car to drive the 400 miles from City V to City W. The return trip takes 9 hours. Find the average speed of the car.

Solution

 Using r = d , the average rate = total distance t total time

The total distance is 2(400) = 800 miles.
The total time is 7 + 9 = 16 hours.

r = 800/16 = 50

The average speed was 50 miles per hour.

Example

Joe drove 2.5 hours at 60 miles per hour and half an hour at 35 miles per hour. What was Joe's average speed?

Solution

Using d = r t:

distance = (2.5 Χ 60) + (0.5 Χ 35) = 167.5 miles
time = 2.5 + 0.5 = 3 hours

So 167.5 = 3r, and r ≈ 56. His average speed was 56 miles per hour.

Notice that the average speed is not (60 + 35)/2 = 52.5

 The average rate = total distance total time

On the graph below, the points A, B, C and D are different towns. The dashed lines show the route of a car travelling from town A to town B to town C to town D and back to A. The graph also shows the average speeds of the car on each leg. The car made no stops in any of the towns.

Example

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? Use the graph above. 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

Use t = d / r. The original journey was
30 km at 90 km/h = 30 ÷ 90 = 1/3 hour
50 km at 50 km/h = 50 ÷ 50 = 1 hour     Note: The distance of 50 is easy to find since it is the hypotenuse of a 3 : 4 : 5 right triangle.
60 km at 60 km/h = 60 ÷ 60 = 1 hour
40 km at 40 km/h = 40 ÷ 40 = 1 hour

So the time for the original journey was 1/3 + 1 + 1 + 1 = 3 1/3 hours

To complete the trip in 3 hours, the driver needs to take 1/3 hour less. Specifically, he needs to take 1/3 hour less on the last leg, DA.
DA took 1 hour, so it needs to take 1  1/3 = 2/3 of an hour.

For the time to decrease, the speed needs to increase. Use r = d / t.

40 km in 2/3 hour = r km/h

r = 40 ÷ (2 / 3) = 40 × (3 / 2) = 60

He will need to drive the last leg at 60 km/h. The correct answer choice is (E).

Example

On another ABCDA trip, the car went 25% slower on each leg. About how much longer did the trip take? Use the graph above.

(A) 27 minutes
(B) 40 minutes
(C) 67 minutes
(D) 240 minutes
(E) 267 minutes

Solution

Since the speed decreased, the time increased.

Since the speed was 25% slower, the new speed was 100% – 25% = 75% of the original speed.

Use d = r t. The distances are the same.

(original rate) × (original time) = (new rate) × (new time)

(original rate) × (original time) = (0.75 × original rate) × (new time)

original time = 0.75 × new time = 3/4 × new time

The original time was 200 minutes.

200 × 4/3 = new time

266.66 ≈ 267 = new time

Be careful. The question doesnt ask for the new time, it asks for the difference between the old and new times. A common trick on the GMAT is for the answer choices to include a previous calculation. Remember to read all the answer options before making your choice.

267 – 200 = 67 minutes

The correct answer choice is (C).

The Seven Steps to Rate/Distance Problems

Video Courtesy of Manhattan GMAT

Section 9: Work

Work

The amount of work W, accomplished in time T, depends on the rate R, at which the work is done. This relationship is described by the equation:

W = RT

The amount of work is often one completed job, so usually W = 1.
The time T is how long it takes to complete the entire job.
The rate R is the job divided by time.

 Rate = 1 job or jobs done time to do 1 job time

Rates

A work-rate sounds like a speed, but the unit is a job done, not a distance covered.

There are many work-rates in everyday life.
• She types 60 words per minute.
• The repairman makes 6 service calls per day.
• The hen lays 5 eggs per week.
In the formula, you need to use the unit rate.

examples of unit rates

It takes 1 tractor 10 hours to plow 1 field, so the rate for the tractor is 1/10 of a field per hour.
 1 = 1 = 1 tractors × hours 1 tractors × 10 hours 10

It takes x tractors 1 hours to plow 1 field, so the rate for the tractor is 1/x of a field per hour.
 1 = 1 = 1 tractors × hours x tractors × 1 hour x

It takes x tractors 4 hours to plow 1 field, so the rate for the tractor is 1/4x of a field per hour.
 1 = 1 = 1 tractors × hours x tractors × 4 hours 4x

It takes x tractors y hours to plow 1 field, so the rate for the tractor is 1/xy of a field per hour.
 1 = 1 = 1 tractors × hours x tractors × y hours xy

Example

A machine that folds and inserts letters into envelopes can do 3 envelopes in 8 seconds. How long will it take to do 240 envelopes?

Solution

 The work is 240 envelopes. The rate is 3 envelopes .   The time t is what you need to find. 8 seconds

Using the formula:
 240 = 3 t 8

 t = 240 × 8 = 640 3

It would take 640 seconds, or 10 minutes 40 seconds.

Review:
640 seconds = 600 seconds + 40 seconds = 10 minutes 40 seconds
 60 seconds = 60x seconds 1 minute x minutes

Work Rates

Example

It takes Jeff 30 minutes to rake the leaves in the yard, and it takes his brother Ken 45 minutes. How long would it take the two of them working together to rake the yard?

Solution

The work is 1 yard. You want to find the time t.
 Jeff's rate is 1 and Ken's rate is 1 .  So the combined rate is 1 + 1 . 30 45 30 45

Using the formula:
 1 = ( 1 + 1 ) t 30 45

t = 18          It would take them 18 minutes to rake the yard working together.

Review:
There are multiple methods to deal with the fractions in work-rate equations.

Method 1
Multiply both sides of the equation by the LCD
(least common denominator).

To find the LCD, first factor the denominators.
Then multiply each unique factor.

30 = 2 × 15
45 = 3 × 15
2 × 3 × 15 = 90
The LCD is 90.

 90 = 90( 1 + 1 ) t 30 45

 90 = ( 90 + 90 ) t 30 45

90 = (3 + 2) t

t = 18

Method 2

Factor each denominator.
 1 + 1 = 1 + 1 30 45 2 × 15 3 × 15

Multiply each fraction by 1 to get the same denominator. Keep in factored form.

 = ( 3 ) ( 1 ) + ( 2 ) ( 1 ) 3 2 × 15 2 3 × 15
 = 3 + 2 3 × 2 × 15 2 × 3 × 15

 = 5 3 × 2 × (3 × 5)
 = 1 3 × 2 × 5
 = 1 18

 1 = 1 t 18

t = 18

Example

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

The work is 1 day's invoices. You want to find the time t.
 Michelle's rate is 1 and John's rate is 1 .  So the combined rate is 1 + 1 . 40 60 40 60

Using the formula:
 1 = ( 1 + 1 ) t 40 60

t = 24          It would take them 24 minutes to do the job working together.

Review:

Method 1
multiply by the LCD

40 = 2 × 20
60 = 3 × 20
2 × 3 × 20 = 120
The LCD is 120.

 120 = 120( 1 + 1 ) t 40 60

 120 = ( 120 + 120 ) t 40 60

120 = (3 + 2) t

t = 24

Method 2

 1 + 1 = 1 + 1 40 60 2 × 20 3 × 20

 = ( 3 ) ( 1 ) + ( 2 ) ( 1 ) 3 2 × 20 2 3 × 20
 = 3 + 2 3 × 2 × 20

 = 5 3 × 2 × (4 × 5)
 = 1 3 × 2 × 4
 = 1 24

 1 = 1 t 24

t = 24

Example

Kelly and Shelley can together type a 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 work is 1 manuscript . Kelly's rate is 1/20. Let Shelley's rate be R. Time working together is 8 hours.

Using the formula:
 1 = ( 1 + R) 8 20

 1 = 1 + R 8 20
Divide both sides by 8.

 20 = 1 + 20R 8
Multiply both sides by 20.

2.5 – 1 = 20R

 R = 1.5 = 3 20 40

 So Shelley's rate is 3 , or she does 3 of the job in an hour. 40 40

Don't fall for the GMAT trick by stopping here; it doesn't answer the question.  Use Shelleys rate to find the time for her to type the entire manuscript alone:
 1 = 3 t 40

 t = 40 = 13 1 3 3

 So Shelley would take 13 1 hours, or 13 hours 20 minutes to type the manuscript alone. 3

 Review: 1 hour = 1 × 60 minutes = 20 minutes 3 3 1 hour

Example

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

Solution

 The rate that one man works is 1 = 1 = 1 house per hour. 3 men × 8 hours 3 × 8 24

 The rate that 5 men work is 5 . 24

Using the equation:

 1 = 5 t 24

 t = 24 = 4.8 5

It would take 5 men 4.8 hours or 4 hours and 48 minutes.

 Review: 0.8 hour = 0.8 × 60 minutes = 48 minutes 1 hour

Some jobs involve completing more than just one unit of work, so its not always the case that W = 1.

Example

It takes Monique 8 minutes to frost a cake and it takes Cheri 6 minutes to frost a cake. How long will it take them working together to frost 7 cakes?

Solution

The work is 7 cakes. You want to find the time t. Monique's rate is 1/8 and Cheri's rate is 1/6.
So the combined rate is 1/8 + 1/6.

Using the formula:

7 = (1/8 + 1/6)t
24(7) = 24(1/8 + 1/6)t        Multiply by the LCD.
24(7) = (24/8 + 24/6)t        Keep numbers in factored form.
24(7) = (3 + 4)t
24 = t

It will take them 24 minutes to frost 7 cakes.

Example

Jiro makes 3 sushi rolls in 15 minutes. Michiko makes 4 sushi rolls in 28 minutes. How long will it take them to make 36 rolls?

Solution

The work is 36 rolls. You want to find the time t.
 Jiro's rate is 3 = 1 and Michiko's rate is 4 = 1 .  So the combined rate is 1 + 1 . 15 5 28 7 5 7

Using the formula:
 36 = ( 1 + 1 ) t 5 7

t = 105

It would take them 105 minutes, or 1 hour 45 minutes.

Review:

105 minutes = 60 + 45 minutes = 1 hour 45 minutes

Method 1
multiply by LCD

 35 × 36 = 35( 1 + 1 ) t 5 7

 35 × 36 = ( 35 + 35 ) t 5 7

35 × 36 = (7 + 5) t

t = (35 × 36)/12 = 35 × 3

t = 105

Method 2

 36 = ( 1 + 1 )t 5 7

 36 = ( 7 + 5 ) t 7 × 5 5 × 7

 36 = ( 12 ) t 35

 t = 35 × 36 = 35 × 3 12

t = 105

Section 10: Sets

• Sets
• Number of Elements
• Using Sets
• Range of Values

Sets

A set is a group of distinct objects, called elements or members. Sets are used to look at the descriptions and interactions of the elements.

Elements are typically listed in {brackets}. The best way to visualize a set is with a Venn diagram. Use overlapping circles to organize the elements of a set.

 The set Animals contains the subset Mammals. In the set of Mammals, Dogs and Cats are subsets. Lizards are animals outside of the set Mammals, cold-bloodedly looking in. The set S = {1, 3, 5, 7, 9} is the odd integers less than 10. The set P = {2, 3, 5, 7} is the prime numbers less than 10. The intersection is composed of the numbers that are in both sets {3, 5, 7}.

Number of Elements

Counting problems are one type of set questions. These questions can ask about the number of elements in the set or subsets.

Example

After 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

Method 1

Draw a diagram. First put the overlap into the diagram. Then calculate the number in each subset.

For only playing basketball, 10  7 = 3.                  For only playing soccer, 20  7 = 13.

So the total number of players is basketball only + soccer only + both = 3 + 13 + 7 = 23.

The correct answer is option (C).

Method 2

Another method to find the total number in a set is to add the number in each subset and subtract the number in the intersection.

subset + subset  intersection = total elements in set
20 + 10  7 = 23

The correct answer is option (C).

Example

Out of 50 people at a buffet dinner, 14 chose ice cream for dessert, 25 chose chocolate cake, and 5 chose both.

(a) How many people had dessert?

(b) How many people chose not to have dessert?

(b) What percent of the people at the dinner had only cake?

Solution

Draw a diagram. First put the overlap number into the diagram. The 5 people who had both are included in both of the 2 sets. Then calculate the number in each subset.

For the number of people who only had ice cream, subtract the people who had both.
14  5 = 9
For the number of people who only had cake, 25  5 = 20.

(a) The number of people who had dessert is 20 + 5 + 9 = 34.

(b) Out of the 50 people, 50  34 = 16 people did not have dessert.

(c) The diagram shows 20 people had only cake, out of the 50 people at the dinner. So it is 20/50 = 40/100 = 40%.

Example

Collins did a survey in his class about what pets the students owned. The most popular pet was a dog, with 22 students having dogs. There were 16 students with cats. Nine students have other types of pets. For students with more than one pet, 6 have a dog and a cat, 4 have a dog and another type of pet, and 3 have a cat with another type of pet. Two students have dogs, cats and another pet. If the class has 40 students, what percentage of students have no pets?

Solution

Draw a diagram.
First put the overlap number into the diagram. Calculate the number in each subset by subtracting the numbers already in the intersection.

all 3 pets = 2 people

cat and dog = 6 – 2 = 4
dog and other = 4 – 2 = 2
cat and other = 3 – 2 = 1

cat only = 16 – 412 = 9
dog only = 22 – 422 = 14
other type only = 9 – 212 = 4

The total number of students in the circles is
9 + 4 + 14 + 1 + 2 + 2 + 4 = 36

If there are 40 students in the class, 40 – 36 = 4 students have no pets.

4 out of 40 = 4/40 = 1/10 = 10/100.
10% of the students have no pets.

Example

There are 36 vehicles in a parking lot. Ten of the vehicles are white and 22 are trucks. There are 14 cars that are not white. How many of the trucks are white?

Solution

Method 1
Draw a diagram. You are looking for the intersection of trucks and white.

The total number of vehicles is 36.
There are 14 that are neither trucks nor white. This number goes outside the circles.
The total number of trucks is 22.
The total number of white vehicles is 10.
So there are 36 vehicles = 14 neither + (22 trucks + 10 white – intersection)
36 = 14 + (22 + 10 – x)
22 = 32 – x
-10 = -x

So the intersection of trucks and white is 10. There are 10 white trucks.

Method 2
Use a table. The bottom row and the right column must each add to 36.

 white not white truck total trucks = 22 car 14 cars total white = 10 total vehicles = 36

Then fill in the numbers that make the bottom row and the right column add to 36.

 white not white truck total trucks = 22 car 14 cars 36 – 22 trucks = 14 cars total white = 10 36 – 10 white = 26 non-white vehicles total vehicles = 36

Fill in 2 more numbers using the totals for each column and row.

 white not white truck 26 total – 14 non-white cars = 12 non-white trucks total trucks = 22 car 14 total – 14 non-white = 0 white cars 14 cars total cars = 14 total white = 10 total non-white = 26 total vehicles = 36

Then use the totals to find the last number, the number of white trucks.

 white not white truck 10 white trucks 12 non-white trucks total trucks = 22 car 0 white cars 14 cars total cars = 14 total white = 10 total non-white = 26 total vehicles = 36

Example

Students were surveyed about their summer plans. In all, 175 said they were planning to work, 120 said they planned to study and 105 said they were going to relax. No students said they planned to do all 3. If 150 students said they planned to do just 2, how many students plan to do just 1?

(A) 100
(B) 150
(C) 200
(D) 250
(E) 400

Solution

Draw a diagram. You are looking for the areas that are not intersections of the plans.

Notice that the intersections are each counted twice:
The intersection of work and relax is included in the total count of work (175) and in the total count of relax (105).
So (the number of students in each set) minus twice (the sum of the intersections) is the answer.

175 + 120 + 105  2(150) = 400  300 = 100

The correct answer is option (A).

Example

In a display of jack-o'-lanterns, 20% looked friendly but had no nose. Half of the jack-o'-lanterns with noses looked friendly. If 60% of the jack-o'-lanterns in the display looked friendly, what percentage had noses?

Solution

Method 1

You are looking for the percentage of jack-o'-lanterns with noses. Use n% to represent the percent of jack-o'-lanterns with noses.

The first step is to draw a diagram. The sets are jack-o'-lanterns with noses and jack-o'-lanterns that look friendly.

The total in the 2 sets is 100% of the jack-o'-lanterns.

20% looked friendly but had no nose, so 20% is the friendly set without the intersection.

Half of the jack-o'-lanterns with noses did not look friendly. So 0.5n% represents the percentage of jack-o'-lanterns did not look friendly and the percentage that did look friendly.

So the percentage of jack-o'-lanterns who looked friendly is 20% + 0.5n% = 60%.
0.5n% = 40%
n% = 80%          80% of the jack-o'-lanterns had noses.

Method 2

Use a table. The bottom row and the right column must each add to 100%.

Start with the given information. Use n% to represent the percent of jack-o'-lanterns with noses.

 nose no nose friendly 0.5n% 20% 60% not 0.5n% total = 100%

Using just the first row of the table:
0.5n% + 20% = 60%
n% = 80%         80% of the jack-o'-lanterns had noses.

 800Score Tip: Count Inclusively When doing counting problems, always be sure to count the first and last of the elements.

Example

A fence consists of fence posts that are 1 foot wide and rails connecting each post that are 9 feet long. How many rails and how many fence posts would be needed for an 80 foot fence?

Solution

Each pair of post and rail covers 10 feet. So the fence needs 8 pairs of post and rail. But the trick here is to include the additional post that is needed at the end. The fence needs 8 rails and 9 posts.

Example

(a) How many elements are in the set of integers from 5 to 20?

(b) What is the sum of the set of integers from 5 to 20?

Solution

(a) You might think you just subtract. Last  first = 15  5 = 10. But make a list.

{5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} = 11 elements

Note: The formula for the number of consecutive integers in a sequence is (last  first) + 1, or L  F + 1.

(b) The first thing to notice is the sums of the pairings of the first and last elements of the set.
(first + last) = 5 + 15 = 20
(second + second to last) = 6 + 14 = 20
To use this pairing, you need to know the number of elements. For this set, there are 11 elements.
The sum will be (11 ÷ 2)(20) = 5.5 × 20 = 110.

Using Sets

Some Examples above use information about the set to ask an additional question. Using sets as the basis for questions about probability, averages, and combinations/permutations is a second type of GMAT question that uses sets.

Example

Using the set A = {1, 2, 3, 4}, how many arrangements can there be if 2 and 4 are not adjacent?

Solution

This is a question about permutations.

For a set with this few elements, a list is a simple method.

 1, 2, 3, 4 1, 4, 3, 2 2, 3, 4, 1 2, 1, 4, 3 2, 1, 3, 4 2, 3, 1, 4 3, 2, 1, 4 3, 4, 1, 2 4, 1, 2, 3 4, 3, 2, 1 4, 3, 1, 2 4, 1, 3, 2

There are 12 arrangements.

Example

Set M = {20, 70, 10, k, 20, 90}. The arithmetic mean is 45. What is the mode of the set?

(A) 20
(B) 30
(C) 40
(D) 45
(E) 60

Solution
• arithmetic mean:  sum of elements number of elements
• mode: the element that appears the most times in a set.
To find the value of k, use the arithmetic mean.

 45 = 20 + 70 + 10 + k + 20 + 90 6

 45 = 210 + k 6

45 × 6 = 210 + k
k = 270 – 210 = 60

With k = 60, rewrite the elements of the set in numerical order.
{10, 20, 20, 60, 70, 90}

The mode is 20, since it is the only element that appears more than once.

The correct answer is option (A).

Notice that the value of k is given as an answer option. A GMAT trick is to offer an answer option that is a number from a previous step in finding the real answer. The median is also given as an answer option. Since the number of elements is even, the median is the mean of the 2 center elements: (20 + 60)/2 = 40.

Range of Values

Range questions also refer to the elements in a set. When a question asks for a possible range, be sure to check both the lowest and highest possible values.

Example

An opened box contains 3 to 5 bottles of mushrooms. Each bottle contains 30 to 40 mushrooms. If 10% of the mushrooms are flawed, what is the range of the possible number of flawed mushrooms in the box?

Solution

First find the lowest value.
Three bottles and 30 mushrooms is the fewest number of mushrooms in the box.
3 Χ 30 Χ 10% = 90 Χ 0.1 = 9              At least 9 flawed mushrooms.

Then find the greatest value.
Five bottles and 40 mushrooms is the greatest number of mushrooms in the box.
5 Χ 40 Χ 10% = 200 Χ 0.1 = 20          At most 20 flawed mushrooms.

There are between 9 and 20 flawed mushrooms in the bottles in the box.

Example

Set A = {1, 2, 3, 4} and set B = {integers > 5}.
What is the range of values for the product of elements ab?

Solution

Find the minimum values and multiply.
The minimum value in set A is 1 and the minimum value in set B is 6, so the minimum for ab is 1 × 6 = 6.

Find the maximum values and multiply.
The maximum value in set A is 4. Set B has no maximum, so ab has no maximum.

The range values for the product of Set A and Set B is ab > 6.

Section 11: Data Interpretation

• Displaying Data
• Tables
• Circle Graphs
• Bar Graphs
• Double Bar Graphs
• Histograms
• Line Graphs

Displaying Data

Data can be displayed in many different ways, including tables and graphs. Both tables and graphs can be used in calculations to evaluate and interpret the data.

Graphs show the relationship of numbers and quantities in visual form, so they are the most common way of displaying data. By surveying a graph, you can very quickly learn about the relationship between two or more sets of information and determine if there are any trends.

The data can be about just one set of events, such as wins and losses or percentages of responses in a survey. This takes one graph. Data can also compare two different events, such as sales vs. earnings or production vs. capacity. Displaying this data can take more than one graph.

Data interpretation involves computing and approximating numerical values based on tables and graphs. GMAT questions go beyond just reading the data, requiring you to calculate averages or compare changes, for example. This type of question is usually in the Problem Solving format, and can appear in sets of 2 to 3 questions.

Tables

Tables give values that are organized but not represented visually. You may graph the data from the table to make comparisons and see trends. You may create a table from a graph to make calculations easier.

Examples 15 are based on the table below, which is a record of the performance of a baseball team for the first seven weeks of the season.

 Week Games Won Games Lost Total Games Played 1 5 3 8 2 4 4 16 3 5 2 23 4 6 3 32 5 4 2 38 6 3 3 44 7 2 4 50

Example 1

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

(A) 2
(B) 21
(C) 29
(D) 50
(E) 58

Solution

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

The correct answer is choice (C).

Example 2

What percent of the games did the team win?

(A) 21%
(B) 29%
(C) 42%
(D) 58%
(E) 72%

Solution

The team won 29 out of 50 games.
 29 = 58 = 58% 50 100

The correct answer is choice (D).

Example 3

Which week was the worst for the team?

(A) second week
(B) fourth week
(C) fifth week
(D) sixth week
(E) seventh week

Solution

Compare the numbers in the won - lost columns.
The seventh week was the only week that the team lost more games than it won.

The correct answer is choice (E).

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

Compare the numbers in the win - loss columns. Since there is more than 1 week with more wins than losses, you need to do a quick calculation.

Method 1

Compare the values of the ratio wins/losses. The greatest value will be the best week.

5/3 ≈ 1.67
5/2 = 2.5
6/3 = 2
4/2 = 2

The value for the third week is 2.5.

The correct answer is choice (B).

Method 2

Find the percentage of games won using the number of wins and the number of games each week.

5/8 = 62.5%
5/7 ≈ 70%
6/9 ≈ 67%
4/6 ≈ 67%

The team won 70% of its games in the third week.

The correct answer is choice (B).

Example 5

If there are 50 more games to play in the season, how many more games must the team win to end up winning 70% of the games?

(A) 18
(B) 21
(C) 35
(D) 41
(E) 42

Solution

To win 70% of all the games, the team must win 70 out of 100. Since the team won 29 games out of the first 50 games, it must win 70 – 29 = 41 games out of the next 50 games.

The correct answer is choice (D).

Circle Graphs

Circle graphs represent values as "slices of pie." Each part is labeled with a fraction or a percent, with the values adding to 1 or 100%.

Be careful in computing problems with percentages. Remember that to convert a decimal into a percentage you must multiply it by 100. For example, 0.04 is 4%. To divide by a percentage, you can multiply using the reciprocal of the percentage as a fraction, as shown in the Example below.

Examples 6 and 7 are based on the circle graph below, which shows the results of a survey taken of people who want a career working with animals.

Example 6

If 400 people were surveyed, how many people chose animal trainer?

(A) 36
(B) 40
(C) 45
(D) 360
(E) 364

Solution

Using the graph, 9% of the number surveyed chose animal trainer.

x = 400 × 9%
x = 400 × 0.09 = 400 × 9/100 = 36

The correct answer is choice (A).

Example 7

If 55 people surveyed chose pet store owner, how many people were surveyed?

(A) 40
(B) 110
(C) 121
(D) 250
(E) 400

Solution

Using the graph, 55 people is 22% of the number surveyed.

55 = 0.22x
 55 = 22 x 100

 x = 55 × 100 = 5 × 11 × 100 = 250 22 2 × 11

The correct answer is choice (D).

Notice that the total number of people surveyed is not the same in both questions, even though the questions use the same graph.

Bar Graphs

Bar graphs can be used to compare non-number categories by showing the number of times each category occurs.

Examples 8 and 9 are based on the bar graph below, which shows the results of a survey about favorite flavors of ice cream offered in the school cafeteria.

Example 8

What percent of the students chose flavors in the top 2 categories?

(A) 25%
(B) 35%
(C) 50%
(D) 55%
(E) 60%

Solution

The percent of students will be the number of students who chose chocolate and chocolate chip divided by the total number of students in the survey.

Estimate the number of students in each bar, then add.
20 + 5 + 26 + 23 + 13 + 20 = 107

top 2 flavors
26 + 23 = 49

49/107 ≈ 50%        Slightly less than half the students chose chocolate or chocolate chip.

The correct answer is choice (C).

Example 9

The cafeteria wants to add another flavor to the menu. Based on the survey, which flavor might be the best to add?

(A) Fudge Ripple
(B) Marshmallow
(C) Peanut Butter
(D) Orange Sherbet
(E) Peppermint

Solution

Look for common characteristics.
The top 2 flavors are chocolate and chocolate chip, with Rocky Road tied for third. The common ingredient is chocolate. So adding another ice cream with chocolate as an ingredient would be the best choice.

The correct answer is choice (A).

Double Bar Graphs

Bar graphs can show more than one set of values for each category.

Examples 1012 are based on the double bar graph below, which shows the results from rolling 2 dice: one white and one red.

Example 10

What is the median value rolled by the white die? (Note: Die is the singular of dice.)

(A) 2.5
(B) 3.0
(C) 3.5
(D) 4.0
(E) 4.5

Solution

Look at just the white bars. Since there were 20 rolls, an even number, the median will be the sum of the 2 middle values divided by 2.

1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6

Since the two values are the same, the median is 3.

The correct answer choice is (B).

Example 11

Estimate the arithmetic mean value rolled by the red die.

(A) 2.25
(B) 3.6
(C) 5.85
(D) 7.5
(E) 12

Solution

Look at just the red bars. The mean will be the sum of each face value multiplied by its frequency, then divided by the number of rolls.

sum of values:
(1 × 7) + (2 × 8) + (3 × 1) + (4 × 2) + (5 × 1) + (6 × 1) = 7 + 16 + 3 + 8 + 5 + 6 = 45

number of rolls:
7 + 8 + 1 + 2 + 1 + 1 = 20 rolls

45 ÷ 20 = 2.25

The correct answer choice is (A).

Example 12

Assume that the white die is fair (has an equal chance of landing on each of its faces). What is the probability that after one more roll the mean score will increase?

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

Solution

Look at just the white bars. The mean will be the sum of each face value multiplied by its frequency, then divided by the number of rolls.

sum of values:
(1 × 3) + (2 × 4) + (3 × 4) + (4 × 3) + (5 × 3) + (6 × 3) = 3 + 8 + 12 + 12 + 15 + 18 = 58

number of rolls:
2 + 4 + 4 + 3 + 4 + 3 = 20 rolls

58 ÷ 20 = 2.9

The current mean is 2.9. To increase the mean, the roll will need to be greater than the mean. So the roll will need to be a 3, 4, 5, or 6.

The probability of rolling a 3, 4, 5, or 6 = 4/6 = 2/3

The correct answer choice is (D).

The bars in a double bar graph can be "stacked" to show more than one set of values for each category.

Examples 1315 are based on the stacked bar graph below, which shows the production vs. the production capacity of a car manufacturing plant. The red portion shows the actual production.

Example 13

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

(A) 2007
(B) 2008
(C) 2009
(D) 2010
(E) 2011

Solution

Each year the gap between production and capacity decreased by 1000. In 2006, the gap is 3000. So in 3 years, in 2009, the gap would disappear.

The correct answer is choice (C).

Example 14

The cost of production of one car was \$10,000 in 2004. The cost was \$15,000 in 2006. What was the percent increase in gross production costs from 2004 to 2006?

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

Solution

Production costs in 2004:   \$10,000 × 5,000 = \$50 million

Production costs in 2006:   \$15,000 × 17,000 = \$255 million

The percent increase in production costs is the difference in costs divided by the original costs.

 difference in costs = 255 – 50 = 205 = 410 = 410% original cost 50 50 100

The correct answer choice is (D).

Example 15

All cars manufactured in 2005 were sold for \$10,000 each and all cars manufactured in 2006 were sold for \$20,000 each. What was the percent increase in sales revenue from 2005 to 2006?

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

Solution

In 2005, the revenue was \$10,000 × 11,000 cars = \$110 million.

In 2004 the revenue was \$20,000 × 17,000 cars = \$340 million.

The percent increase in revenue is the difference in revenue divided by the original revenue.

 difference in revenue = 340 – 110 = 230 This is slightly greater than 200%. original revenue 110 110

The correct answer choice is (D).

Histograms

A histogram is very similar to a bar graph, but has no spaces between the bars. The main difference is that histograms use continuous grouped data to show frequency trends.

Examples 16 and 17 are based on the histogram below, which shows the results of a survey on the frequency of eating out for various age groups.

Example 16

Which age group eats out for a third of a week's meals?

(A) 2534
(B) 3544
(C) 4554
(D) 5564
(E) 6574

Solution

A week has about 21 meals, so a third of a week's meals is 7.

The age interval of 3544 eats about 7 meals out per week.

The correct answer choice is (B).

Example 17

What is the arithmetic mean number of meals eaten out per week?

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

Solution

 5.5 + 7 + 3.2 + 1.4 + 2.7 = 19.8 ≈ 4 5 5

The mean number of meals eaten out per week by adults is 4.

The correct answer is choice (B).

Note: To do calculations based on data in a histogram, make sure the intervals are all equal. Here, the intervals all are 10 years, so no adjustments need to be made to the calculations.

Examples 1821 are based on the histograms below, which show the sales and earnings of a company from 2005 to 2010.

Example 18

During what two-year period did the company's earnings increase the most in absolute terms?

(A) 20052007
(B) 20062007
(C) 20062008
(D) 20072009
(E) 20082010

Solution

Using the second graph, you are looking for the greatest change between 3 adjacent bars.

The biggest increase between 2 adjacent bars is from \$5 million in 2006 to \$10 million in 2007, which is a \$5 million increase. Then the change from \$10 million in 2007 to \$12 million in 2008 is a \$2 million increase.

The two-year increase from 2006 to 2008 was \$7 million -- clearly the largest on the graph.

The correct answer choice is (C).

Example 19

During the years 2006 through 2008, what were the arithmetic mean earnings per year?

(A) \$6 million
(B) \$7.5 million
(C) \$9 million
(D) \$10 million
(E) \$27 million

Solution

You can make a quick table to organize the data from the graph.

 Year Earnings 2006 \$5 million 2007 \$10 million 2008 \$12 million

 5 + 10 + 12 = 27 = 9 3 3

The mean earnings were \$9 million per year.

The correct answer is choice (C).

Example 20

In which year did earnings decrease by the greatest percentage from the previous year?

(A) 2006
(B) 2007
(C) 2008
(D) 2009
(E) 2010

Solution

Use the second graph.

To find the percent decrease, divide the change by the original amount.
Note: Percent decrease here means the proportional change, not the numerical change. You can compare the change as a percent, fraction, or decimal.

Make a quick table to organize the data and calculations.

 Year Earnings 1-year decrease Calculation 2005 8 n/a 2006 5 37.5% (8 – 5)/8 = 3/8 = 0.375 = 37.5% 2007 10 increase 2008 12 increase 2009 11 8% (12– 11)/12 = 1/12 ≈ 8% 2010 8 27% (11– 8)/11 = 3/11 ≈ 27%

The correct answer choice is (A).

Example 21

If the company's earnings are less than 10 percent of sales during a year, then the Chief Operating Officer gets a 5% pay cut. How many times between 2005 and 2010 did the Chief Operating Officer take this pay cut?

(A) none
(B) one
(C) two
(D) three
(E) four

Solution

Use the second graph.

To find the percent decrease, divide the change by the original amount.
Note: Percent decrease here means the proportional change, not the numerical change. You can compare the change as percent, fraction, or decimal.

Make a quick table to organize the data and calculations.

 Year 10% of Sales Earnings Earnings < 10%? 2005 0.1 × 80 = 8 8 equal: no cut 2006 0.1 × 70 = 7 5 5 < 7: cut 2007 0.1 × 50 = 5 10 10 > 5: no cut 2008 0.1 × 80 = 8 12 12 > 8: no cut 2009 0.1 × 90 = 9 11 11 > 9: no cut 2010 0.1 × 100 = 10 8 8 < 10: cut

Comparing the right columns shows that earnings were less than 10 percent of sales in 2006 and 2010.

The correct answer is choice (C).

Line Graphs

One type of line graph is a broken line graph. Segments are used to join the values but the graph has multiple slopes.

Examples 2224 are based on the line graphs below. The graph on the left shows a company's earnings for each month between April and August. The graph on the right shows the number of outstanding shares.

Example 22

How much greater were the company's earnings in June than in April?

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

Solution

The percentage increase in earnings is the difference in earnings divided by the original earnings.

 difference in earnings = June earnings – April earnings = 5 – 2 = 3 = 1.5 = 150% original earnings April earnings 2 2

Since both values are in millions, you dont need trailing zeros in the calculations.

The correct answer is choice (A).

Example 23

How much did the earnings per share increase by 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 = 300 = 6 × 50 = 50 ≈ 16.67 180,000 18 6 × 3 3

earnings per share in June:

 5,000,000 = 500 = 20 250,000 25

20 – 16.67 = 3.33

The earnings per share grew by \$3.33.

The correct answer is choice (B).

Example 24

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

earnings per share in August:

 3,000,000 = 300 = 15 × 20 = 20 150,000 15 15

The expected earnings in September are 5% less than August, which means the earnings will be 95% of the August earnings.

Let x be the number of outstanding shares in September.
So the earnings per share ratio in September will be

2,850,000/x = 20
20x = 2,850,000
x = 142,500

The number of shares in August is 150,000. The number of shares in September needs to be 142,500.

150,000 – 142,500 = 7,500

The company has to buy back 7,500 shares.

The correct answer is choice (E).

Note: 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.

Examples 2527 are based on the line graphs below, which show exchange rate dynamics during one week.

Example 25

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

Method 1

On Tuesday, 1 pound cost 1.7 dollars. Mr. Smith bought 200 pounds for \$1.70 Χ 200 = \$340.

On Wednesday, Mr. Smith sold 200 pounds Χ \$1.75 = \$350 .

So Mr. Smith made a profit of \$350  \$340 = \$10.

The correct answer is choice (C).

Method 2

Compare the exchange rates. Since 1.75  1.70 = 0.05, the profit is 5 cents on each pound he bought and sold.
So 200 Χ 0.05 = \$10.

The correct answer is choice (C).

Example 26

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

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

Solution

At the end of the week 1 pound cost 1.6 dollars and 1 pound cost 1.2 euros. This means that 1.6 dollars were equivalent to 1.2 euros.

 1.6 dollars = 1 dollar 1.2 euros x euros

 x = 1.2 = 3 × 0.4 = 0.75 1.6 4 × 0.4

So, 1 dollar was equivalent to 0.75 euros.

The correct answer is choice (B).

Example 26

On the basis of these graphs alone, how much did the dollar depreciate against the euro during this seven-day period?

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

Solution

At the end of last week, the first Sunday on the graph, 1 pound cost 1.75 dollars and 1 pound cost 1.4 euros. This means that 1.75 dollars were equivalent to 1.4 euros.

 1.75 dollars = 1 dollar 1.4 euros x euros

 x = 1.4 = 2 × 0.7 = 4 = 0.8 1.75 2.5 × 0.7 5

So, 1 dollar was equivalent to 0.8 euros.

From the Example above, at the end of the week, on the second Sunday on the graph, 1 dollar was equivalent to 0.75 euros.

So the dollar depreciated by 0.05 euros.
To find the percentage decrease, divide the change by the original amount.

 0.05 = 5 = 5 = 1 = 0.0625 = 6.25% 0.8 80 5 × 16 16

The correct answer is choice (C).

Section 12: Mean, Median & Mode

• Averages
• Arithmetic Mean and Median
• Mode and Range
• Comparing Data

Averages

An average is a number used to describe a set of data. Averages give a measure of the "middle" of a set.

The most common averages are the arithmetic mean, the median and the mode. The mean is often thought of as the average, but the terms are not interchangeable. The GMAT uses the term arithmetic mean instead of average to be more precise.

Arithmetic Mean and Median

The arithmetic mean is the sum of a set of numbers divided by the number of elements in the set.

 mean = sum of the values number of values

The median is the "middle" number of a list of data. To find the median, arrange the numbers in numercial order. If the number of values is odd, the median is the middle number. If the number of values is even, the median is the mean of the two middle numbers.

The median can differ greatly from the arithmetic mean.

Using the median, half of the people in a neighborhood pay more than the median rent, and half pay less. For example, if 5 people have monthly rent of \$1100, \$900, \$1000, \$10,000 and \$3000, the median is \$1100.

The mean does not split the people into a top half and bottom half. For example, for the monthly rents given above, the mean rent is \$3200.

Compare this median of \$1100 and the mean of \$3200. If a large number of people pay very little rent and a few pay a huge amount, the mean would be quite high, but the median would be surprisingly low.

Mode and Range

Two other values can be used to describe a set.

The mode is the most frequently occuring number in a set of data. There can be more than one mode.

{1, 2, 3, 3, 4, 5}       mode = 3
{1, 2, 3, 3, 4, 5, 5}   modes = 3 and 5

The mode has the feature that it can be used to describe things other than numbers that are the most frequently occuring (such as white trucks).

The range isn't an average. It describes the spread or dispersion of a data set.
The range is the difference between the largest and smallest values.

range = (greatest value) – (least value)

Example

Find the arithmetic mean, median and range.

(a) 2, 6, 10, 20, 20, 26

(b) 5, 45, 15, 10, 5

Solution

(a)
 mean = 2 + 6 + 10 + 20 + 20 + 26 = 14 6

The number of values is even, so the median is the mean of the two middle numbers.
median = (10 + 20)/2 = 15
range = 26 – 2 = 24

(b)
 mean = 5 + 45 + 15 + 10 + 5 = 16 5

Putting the values in numerical order makes it easier to see the median and range.

5, 5, 10, 15, 45

The number of values is odd, so the median is the middle number.
median = 10
range = 45  5 = 40

Example

Victor earned scores of 70, 75, 80, 85, 85 and 90 on his first six algebra tests. What is the relationship between the averages of these scores?

(A) median < mode < mean
(B) median < mean < mode
(C) mode < median < mean
(D) mean < mode < median
(E) mean < median < mode

Solution

mean = (70 + 75 + 80 + 85 + 85 + 90)/6 = 485/6 ≈ 80.8
median = (80 + 85)/2 = 82.5
mode = 85

80.8 < 82.5 < 85, so mean < median < mode.

The correct answer is choice (E).

Example

Set P = {a, a, a, a, b, b, b}. If a < b, is the mean greater than the median?

Solution

The median is the middle value, so the median is a.

mean = (4a + 3b)/7

The question is about the relationship between the expressions.

median ? mean

7a ? 4a + 3b
3a ? 3b
a ? b

Since a < b was given and the median is a, the mean is greater than the median.

Comparing Data

Every set of numbers has a mean, median and range. These three values can be used to compare data by seeing how much the numbers differ.

Example

A politician announces at a press conference that the average yearly salary per person in the state is now \$80,000.

After her announcement, three groups of 5 people were surveyed about their salaries.

(I) \$80,000, \$80,000, \$80,000, \$80,000, \$80,000
(II) \$71,000, \$74,000, \$82,500, \$82,500, \$90,000
(III) \$15,000, \$27,000, \$33,000, \$100,000, \$225,000

Is the claim about average salaries reasonable for each group?

Solution

To solve this problem, compare the mean, median and range of each data set.

(I) The mean and median are \$80,000. The range is 0. The claim is reasonable for this group.

(II) The mean is \$80,000. The median is \$82,500. The range is \$19,000. So the mean and median support the claim. The range is somewhat high, showing the data is spread out. The claim is still somewhat reasonable for this group.

(III) The mean is \$80,000. The median is \$30,000. The range is \$210,000. The claim is not reasonable for this group.

Example

The mean age of a group of 10 firefighters is 18 years. If 2 firefighters join the group, the mean age increases by 2 years. What is the mean age of the new firefighters?

(A) 18
(B) 20
(C) 24
(D) 30
(E) 60

Solution

The mean age of the first 10 firefighters is 18, so the total of their ages is 10 × 18 = 180.

The mean age of the 12 firefighters is 18 + 2 = 20, so the total of their ages is 12 × 20 = 240.

The difference between the 2 totals is the total for the ages of the 2 new firefighters.
240 – 180 = 60

So the mean of the ages of the 2 added firefighters is 60 ÷ 2 = 30.

The correct answer is choice (C).

Arithmetic Mean

Example

Monica drove for 2 hours at 45 miles per hour and 3 hours at 60 miles per hour. What was her average (arithmetic mean) speed for the whole trip?

Solution

The formula is distance = rate × time, or d = rt. You want to find the rate.

distance = (2 × 45) + (3 × 60) = 270 miles
time = 2 + 3 = 5 hours

So 270 = 5r, and r = 54. Her average speed was 54 miles per hour.

Notice that the average speed is not (45 + 60)/2 = 52.5

The average rate = (total distance) / (total time)

Example

Set S consists of 5 consecutive integers. Set P is created by increasing the value of each element in set S by 2 times its place in S. (The value of the first element in S is increased by 2, the value of the second element by 4, and so on.) How much greater is the mean of set P than the mean of set S?

(A) 3
(B) 5
(C) 6
(D) 10
(E) 30

Solution

The question here is the change in the mean. The change in the mean will be the sum of each of the changes divided by the number of changes.

mean of changes = (2 + 4 + 6 + 8 + 10) / 5 = 6

So the mean of set P is 6 greater than the mean of set S.

The correct answer is choice (C).

Example

 Set S = { 1 , -x , x , x2 , -2x , 2x }. &nb.   If  - 1 < x < 0, what is the median of set S? 2 2

 (A) 1 – x 4 2

 (B) x2 – 1 2 4

 (C) x(x – 1) 2

 (D) 1 2x

 (E) x(x + 1) 2

Solution

Method 1

Since -1/2 < x < 0, the simplest technique is to Plug In: substitute a number to find a reasonable value. Use x = -1/4. Then the values are:

1/2

-x = -(-1/4) = 1/4

x = -1/4

x2 = (-1/4)2 = 1/16

-2x = -2(-1/4) = 1/2

2x = -1/2

Now arrange the data in numerical order. Since there are 6 elements in the set, you are looking for the 2 middle values.

-1/2, -1/4, 1/16, 1/4, 1/2, 1/2

 Using the values, the median is (1/16(1/16) + (1/4) . 2

 So using the variables, the median is x2 + (-x) = x2 – x = x(x – 1) . 2 2 2

The correct answer choice is (C).

Method 2

First arrange the elements in numerical order.

Because -1/2 < x < 0:
• 2x and x are negative and 2x < x.
• x2 and -x are positive and x2 < -x < 1/2.
It is not clear whether 1/2 < -2x. But it is not important as long as we know that -x and x2 are the third and fourth largest elements in the set (because there are six total elements in the set).

So, tSo, the elements in numerical order will either be {2x, x, x2, -x, 1/2, 2x} or {2x, x, x2, -x, 2x, 1/2}

With six elements in the set, the median will be the mean of the two middle values of the set. >
 median = x2 – x = x(x – 1) 2 2

The correct answer choice is (C).

 Example

123 × 341 = ?

A) 40,816
B) 41,913
C) 42,637
D) 42,755
E) 43,321

 Solution Don't just do the calculation.  Stop and think. Take a look at the answer choices.  Multiply the last digit by the last digit.  You get 123 x 341 or 3. The GMAT is a multiple-choice test, so pick the only answer choice ending in 3, which is (B).

 Example

 3 x 324 = ? 162 x 4

Solution

Always look to cancel.  Use factoring.

 3 × 324 = 3 × 4 × 9 × 9 = 9 = 1 1 162 × 4 3 × 6 × 9 × 4 6 2

Instead of factoring all the way, you may notice that 324 = 162 × 2.

The GMAT often provides these easy cancellation shortcuts to save you time, so you need to be able to spot them.

 Example

There are 90 students going on a field trip.  A bus carries 36 students.  How many buses are needed?

A) 2              B) 2.5           C) 3              D) 3.5           E) 4

 Solution The direct calculation is 90/36 = 2.5, which is answer option (B).  But that is way too easy.  Don't just do the math.  Stop and look for the trick.  The question asks for the number of buses.  There is no such thing as half a bus!  The correct answer is option (C).

Backsolving means inserting the answer choices into the variable(s) in the question. You can use this technique when there are variables in the question and numbers in the answer choices. (Backsolving does not apply to Data Sufficiency questions.)

By substituting the answers into the variables in the question, you can eliminate the answer choices that dont work and skip the complicated algebra of solving for the variables. Backsolving is also an effective way to double-check your algebraic solution.

 800score Tip: A fundamental weakness of multiple-choice tests is that you can Backsolve questions. GMAT questions are written with this in mind and are often designed so they cannot be backsolved. On the other hand, sometimes questions are designed so backsolving is the best technique. The GMAT is testing your resourcefulness and expects you to have Backsolving in your toolkit.

Don't jump to Backsolving too quickly. Try some calculations to decide if the problem is too complicated to solve algebraically. If after reading the question and the answer choices you are stumped, start plugging in answer choices.

Start at the middle answer, (C). The answer choices are usually arranged in ascending value from (A) to (E). So if (C) doesnt work you can move on to (D)/(E) or (A)/(B) depending on whether you need a higher or lower value.

Many GMAT questions have two distinct solution strategies. The 800score Prep Guide examples will often feature two approaches: calculations or algebra and Backsolve or Plug-In.

 Example

When the positive integer x is divided by 24, the remainder is 10. If x is divided by 8, the remainder is 2. What is the value of x?

A) 18      B) 34      C) 40      D) 49       E) 57

 Solution There are variables in the question and numbers in the answer choices, so this is a good place to backsolve. We can eliminate answer choice (C) because 40 divided by 24 has a remainder of 16, not 10. Moving to choice (B), we see that 34 divided by 24 does have a remainder of 10. Now check the second restriction. When 34 is divided by 8, it gives a remainder of 2. Choice (B) meets both restrictions, so its the correct answer. Backsolving turned a hard algebra question into an easy arithmetic question.

Plug-in means you choose your own numbers to insert into a question. This is useful when there are variables in both the question and the answer choices.

Translating the variables into numbers will also help you understand the question. Play with numbers to see if you notice any patterns or restrictions.

Make sure the numbers you choose fit the question's parameters, and pick a variety of numbers to make sure you cover all reasonable possibilities.

Try positive and negative numbers, and zero. For example, you could plug in -2, 0 and 2.

If the question doesn't specifically require "integers," make sure to try some positive and negative fractions (a common trick on the GMAT).

Many GMAT questions have two distinct solution strategies. The 800score Prep Guide examples will often feature two approaches: calculations or algebra and Backsolve or Plug-In.

When to Use Plug-In

Use plug-in for questions that have a small and finite set of numbers to check.

How many primes between 11 and 30 satisfy this statement?
Start substituting a few prime numbers. This works well on small sets of numbers because you can test a few or even all the options without spending too much time.

Plug-in also works well in true/false questions, such as: Is ab > 0? For a question like this, it is logical to test the conditions using a variety of numbers.

Plug-in can be used very effectively in Data Sufficiency questions to test if the statements are sufficient. You should use plug-in regularly on Data Sufficiency questions, just as you should use Backsolve constantly on multiple-choice questions.

 Example

If n is an even integer, which of the following must be an odd integer?

A) 3n – 2      B) 3(n + 1)       C) n – 2       D) n/3       E) n/2

 Solution This question has variables in the question and in the answer choices, so plug-in is the best technique. Try n = 2. Check (A). If n = 2, then 3n –­ 2 = 4, which is not odd. Check (B). If n = 2, then 3(n + 1) = 9. Since the target is an odd integer, this answer choice works. You can try another even number and/or use the rest of the answer options to double-check. For example, n = 2 works with choice (E) to make an odd number, but (E) is even with any other even value for n. Note: When plugging in, if you get a result that works for two or more answer choices, you must plug in another number. Use a number that is fundamentally different from the initial number, such as larger, positive or negative, zero or a fraction.

 Example

If x > y, then when is x2 > y?

I. x < 0

II. x > 0

III. x > 1

A) I only      B) II only      C) III only      D) I and II      E) I and III

 Solution Quickly reading the question, it seems like x2 > y is always true. So there must be a trick. Plug-in is the best technique. Check statement (I). Both x and y will be negative. Since any negative number squared is positive, positive x2 will be greater than any value of the negative y. So statement (I) is true. Before checking statements (II) and (III), notice how similar they are. The only difference is that statement (II) includes fractional values between 0 and 1. Check statement (II). Let x = 4 and y = 3. Checking x2 > y, 16 > 3. So (II) seems to be true. Now try a fraction. Let x = 1/2 and y = 1/3. Checking x2 > y, but 1/4 < 1/3, so statement (II) is false. In checking statement (II) we checked numbers greater than 1, so we also checked statement (III) and know it is true. Statements (I) and (III) are true, so the correct answer is option (E).

How to avoid complex algebra by Picking Numbers

Video Courtesy of Kaplan GMAT

Ballpark means estimating rather than doing the full calculation. Looking at the answers before doing any calculations will tell you how exact your calculations need to be and whether you can use ballpark.

When to Use Ballpark

• When the answer choices are spread over a wide range. You can use your rough estimate to eliminate answer choices that are out of range.
• Multiplying and dividing large numbers or dealing with fractions. Use rounding to get an estimate.
• On Data Sufficiency questions, to gauge whether a statement is sufficient.
• To double-check calculations. Make sure your solution is in the ballpark of your estimate.

 800score Secret: Do you know how the geometry questions with drawings always say "not drawn to scale"? Actually they pretty much are drawn to scale. This means that you can double-check your geometry answers by seeing if they "fit" the geometry picture. It also means you can ballpark geometry questions by looking at the drawing. If a drawing looks like an equilateral triangle (three equal sides), and you came up with three sides of 3, 12, and 14, you need to double-check your math! Deliberately making a drawing not look like the correct answer would just be going too far... even for the GMAT.

 Example

What is 36,568 divided by 12,985?

A) 2.26      B) 2.816      C) 3.08       D) 4.23      E) 5.65

 Solution Round the numbers. For 36,568, use 36 or 37. For 12,985, use 13. 36 ÷ 12 = 3 so 36 ÷ 13 is just under 3. 37 ÷ 13 is also just under 3. So the most reasonable answer is option (B) 2.816.

 Example

Which is greater: 24/51 or 26/49?

 Solution It helps when dealing with fractions to simplify to a basic fraction like 1/4, 1/2 and 3/4. 24 is less than half of 50, so 24/51 is less than 1/2. 26 is greater than half of 50, so 26/49 is greater than 1/2. So 26/49 is greater than 24/51.

 Example

If 0.303z = 2,727, then z =

A) 0.9      B) 9      C) 90       D) 900      E) 9,000

 Solution The answer choices are far apart, so use ballpark. Round the decimal: 0.303 is close to 1/3, so 1/3 of z ≈ 2,727, and therefore z ≈ 3 × 2,727. What answer could possibly be correct? You don't even have to do the math. The answer must be 9,000. There are no other answers even in the thousands. The correct answer is option (E). Note: Don't make the mistake of 1/3 of z = 2,727, so z ≈ 2,727 ÷ 3. This common error is another GMAT trick.

Under the intense pressure of test day, you can expect to forget some basic math rules. If this happens, you can take a moment to experiment with easy numbers in order to recall the rule.

Be careful — experiments can be very time consuming, so only use this technique if you are doing well on time.

 Example

Solve: 4525 ÷ 4512 = ?

 Solution You have forgotten the rule about division and exponents. Experiment using a base of 2. 24 ÷ 22 = 16 ÷ 4 = 4 = 22            Look at the exponents: 4 – 2 = 2 26 ÷ 23 = 64 ÷ 8 = 8 = 23            Look at the exponents: 6 – 3 = 3 So you subtract exponents when you divide. 4525 ÷ 4512 = 4525 – 12 = 4513

 Example

Solve: 144 + 225

 Solution You don't remember if splitting the square root works. Is √144 + 225  = √144  + √225  ? Experiment. Is √4 + 9  = √4  + √9 ? You can see that √13  ≠ 2 + 3. So to simplify you need to add, then factor. √144 + 225   = √369  = √3 x 3 x 41  = 3√41

A pattern is a repetition of numbers that results from repeating calculations. Sometimes you can find patterns in questions that otherwise seem impossible to solve.

 Example

What is the units digit of 3801?

Solution

You know you don't have time to calculate. This is a reasoning question. There must be a shortcut.

Try to find a pattern starting with smaller exponents. The pattern shows up pretty quickly.

 31 = 3 35 = 3 × 81 = 243 32 = 9 36 = 3 × 243 = 729 33 = 27 37 = 3 × 729 = 2187 34 = 3 × 27 = 81 38 = 3 × 2187 = 6561

The units digit repeats after "4 powers." The pattern is clearest for exponents that are multiples of 4.

Every exponent that is a multiple of 4 has a units digit of 1. Since 800 is a multiple of 4, the units digit of 3800 is 1. Looking at the pattern, the units digit of 3801 will be 3.

 800score Tip: When a question makes you scratch your head and think it will take an hour to solve, chances are there is a shortcut. Using Backsolve, Plug-In, Find Patterns or math shortcuts, you will likely find the easy way to solve it.

 Example

What is the sum of integers from 1 to 100, inclusive?

 Solution Look for a pattern starting with smaller groups of numbers. 1 + 2 + 3 + 4 = 1 + 2 + 3 + 4 = 5 + 5 = 10 1 + 2 + 3 + 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 + 6 = 7 + 7 + 7 = 21 First trick: Add first and last. Second trick: How many pairs are there for an even number of integers? You can check the pattern by doing one more batch of numbers. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = (1 + 10)(10/2) = (11)(5) = 55 So the sum of the integers from 1 to 100 is (1 + 100) (100/2) = (50)(101) = 5,050

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