Transcript GMAT Boot Camp

Loyola MBA Program
Instructor
Nate Straight, Loyola M.B.A. ‘08
• College of Business’ Director of
Assessment; Business Statistics &
Quantitative Methods Instructor
• GMAT Score of 760, 99th percentile
• [email protected]
Materials
www.loyno.edu/~nmstraig/GMAT
Structure and Scoring
Test Structure
4 sections, taken in this order:
Analytical Writing – 30 minutes
Integrated Reasoning – 30 minutes
Math / Quantitative – 75 minutes
Verbal Reasoning – 75 minutes
Test Structure
Analytical Writing (30 mins.):
1 essay requiring analysis of the
reasoning behind a presented
argument and communication
of a critique of the argument
Test Structure
Integrated Reasoning (30 mins.):
12 multiple-choice questions
Interpretation of graphics / tables
Reasoning from multiple sources
Multiple-step/two-part analysis
Test Structure
Math / Quantitative (75 mins.):
37 multiple-choice questions
50% problem-solving
50% “data sufficiency”
Test Structure
Verbal Reasoning (75 mins.):
41 multiple-choice questions
40% sentence correction
30% reading comprehension
30% critical reasoning
GMAT Scoring
Scores use the following ranges:
Analytical Writing – 0 to 6
Integrated Reasoning – 1 to 8
Math / Quantitative – 0 to 60
Verbal Reasoning – 0 to 60
Total Score – 200 to 800
GMAT Scoring
Your GMAT Score Report:
Math
%
Verb
%
Total
%
AWA
%
IR
%
36
42
30
56
550
48
4.5
38
6
79
Your total score ↑ depends only on your
math and your verbal reasoning scores.
Loyola’s average GMAT score is a 520
Improving your Score
More Test Structure
The GMAT is taken on a computer,
one question at a time, and you can
not skip or come back to questions.
The math / quantitative and verbal
sections are “computer-adaptive”.
Computer-Adaptive
The math and verbal sections start
with medium-difficulty questions
and ramp up or down the difficulty
based on your prior performance.
Your score increases with difficulty.
Computer-Adaptive
The goal of adaptive scoring is to
gradually hone in on your level of
ability, which determines your score.
Earlier questions identify your general
ability and have a large impact on your
score. Later questions are ‘fine-tuning’.
Computer-Adaptive
800
700
600
500
400
300
200
Q1
Q2
Q3
Q4
…
Q35
Q36
Q37
Computer-Adaptive
Strategies for adaptive testing:
Early questions are very important;
spend your time getting them right.
Especially early, difficult questions
can dramatically improve your score.
Plan to never miss an easy question.
Computer-Adaptive
Strategies for adaptive testing:
Unanswered questions at the end
will decrease your score; even if not
much individually, they will add up.
If you are running out of time, use
process of elimination, then guess.
Computer-Adaptive
Strategies for adaptive testing:
Early, easy-sounding questions will
have easy answers; later, as difficulty
rises these “easy” answers are a trap.
If an answer seems obviously right,
rethink it carefully before choosing it.
Process of Elimination
“18 percent of American corn is grown in
Iowa. If the total amount of corn grown in
America is 12 billion bushels, how many of
those bushels are grown outside of Iowa?”
A. 2.2 billion
D. 11.8 billion
B. 6 billion
E. 13.2 billion
C. 9.8 billion
Process of Elimination
“A company’s sales increased 10% from
2012 to 2013, and then 20% from 2013 to
2014. By approximately what percent did
sales increase overall from 2012 to 2014?”
A. 20%
D. 32%
B. 24%
E. 40%
C. 30%
Process of Elimination
“An MBA student took 2 classes in Fall with
an average grade of 80, then 3 classes in the
Spring with an average grade of 90. What
is the student’s average grade for the year?”
A. 83.3
D. 86
B. 85
E. 88
C. 85.5
General Strategies
There is no substitute for content
knowledge. Process of elimination
and other strategies are a last resort.
It will always be quicker to solve the
problem directly than to work through
5 answers trying to find the right one.
General Strategies
There is also no substitute for knowing
your own ability level and limitations.
You should spend as little test time as
possible trying to jog your faded highschool algebra or grammar memory
bank in search of long-lost knowledge.
General Strategies
The first 15-20 seconds after reading a
problem should be spent considering
a) whether you know how to solve the
problem; b) if not, whether you can
figure out how to identify the answer.
Be fully honest in this self-reflection.
General Strategies
If you know that you do not know how
to solve a problem, and cannot even
come up with a way to find an answer,
just take your best guess and move on.
Staring at the problem will not make
you understand it; it only wastes time.
General Strategies
Practice exams are more helpful than
random study. There are 2 full-length
tests available at: http://www.mba.com/us/
the-gmat-exam/prepare-for-the-gmat-exam.aspx
Take the first test, review in detail all
the content you missed, then re-take.
Overview and Strategies
Math Overview
2 types of questions:
50% problem-solving
50% “data sufficiency”
A calculator is neither provided nor
allowed for the math section. Do not
use one while you study for the exam.
Math Overview
Problem-solving content:
Order of operations
Properties / types of numbers
Fractions, ratios, and proportions
Operations using fractions / decimals
Percentages and percent change
Math Overview
Problem-solving content:
Probability and permutations
Mean, median, and mode
Range and standard deviation
Exponents and square roots
Operations using exponents
Math Overview
Problem-solving content:
Algebra for “word problems”
“Solving for X” / plugging-in
Solving systems of equations
Using algebraic functions
Algebraic factorization
Math Overview
Problem-solving content:
Distance/rate/time problems
Work/rate/time problems
Geometry: Angles and area
Solids: Volume and surface area
Lines: Slope, intercept, and areas
Math Overview
Data sufficiency content:
All of the above concepts, focused
on the “how” and “why” of a process
rather than the “what [is the answer]”
Evaluate sufficiency of 2 statements
Math Overview
Data sufficiency example:
“What is the value of x?
2
1) x = 9
A.
B.
C.
D.
E.
2) x is negative”
Statement 1 alone is sufficient; Statement 2 alone is not
Statement 2 alone is sufficient; Statement 1 alone is not
Both statements 1 and 2 together are sufficient; neither
statement alone is sufficient to answer the question
Each statement alone is sufficient to answer the question
Statements 1 and 2 together are not sufficient
Math Strategies
The GMAT contains relatively few
“one-step” problems. Nearly every
question will require more than one
calculation or test multiple concepts.
Remember, the easiest, most obvious
answers are often set up as traps.
Math Strategies
There are two good ways to tackle a
question you don’t know how to solve:
If the answers are algebraic, plug in
a number to simplify the question
If the answers are numbers, plug
each value into the question itself.
Math Strategies
“Jeremy can run 5 miles in t minutes.
If he can run x miles in 20 minutes, which
of the following is equal to the value of x?”
A. t / 100
D. 10 / t
B. t / 10
E. 100 / t
C. t
Let’s make t = 50, for 5 “10-minute” miles.
Math Strategies
“Jeremy can run 5 miles in 50 minutes.
If he can run x miles in 20 minutes, which
of the following is equal to the value of x?”
A. 50 / 100
D. 10 / 50
B. 50 / 10
E. 100 / 50
C. 50
Let’s make t = 50, for 5 “10-minute” miles.
Math Strategies
“Jeremy can run 5 miles in 50 minutes.
If he can run x miles in 20 minutes, which
of the following is equal to the value of x?”
A. 50 / 100 = .5
D. 10 / 50 = .2
B. 50 / 10 = 5
E. 100 / 50 = 2
C. 50
= 50
Math Strategies
“Mr. Hooper’s Ice Cream Shop sells only
vanilla and chocolate. Today the ratio of
vanilla to chocolate cones sold was 2 to 3,
but if 5 more vanilla cones had been sold
the ratio would have been 3 to 4. How
many vanilla cones did he sell today?”
A. 20
B. 25
C. 30
D. 35
E. 40
Math Strategies
“… the ratio of vanilla to chocolate cones
sold was 2 to 3. If 5 more vanilla cones had
been sold the ratio would have been 3 to 4.
How many vanilla cones did he sell today?”
A. 20
B. 25
C. 30
D. 35
E. 40
If you don’t know how to solve it with
algebra, plug in the answers. Start with C.
Math Strategies
“… the ratio of vanilla to chocolate cones
sold was 2 to 3. If 5 more vanilla cones had
been sold the ratio would have been 3 to 4.
How many vanilla cones did he sell today?”
C. 30 / x = 2 / 3, so x = 45 chocolate cones
5 more than 30 is 35, so the ratio would be
35 vanilla to 45 chocolate… which isn’t 3/4.
Math Strategies
“… the ratio of vanilla to chocolate cones
sold was 2 to 3. If 5 more vanilla cones had
been sold the ratio would have been 3 to 4.
How many vanilla cones did he sell today?”
A. 20
B. 25
C. 30
D. 35
E. 40
C is out. The numbers in the ratio were too
close together, because C was small. Try E.
Math Strategies
“… the ratio of vanilla to chocolate cones
sold was 2 to 3. If 5 more vanilla cones had
been sold the ratio would have been 3 to 4.
How many vanilla cones did he sell today?”
E. 40 / x = 2 / 3, so x = 60 chocolate cones
5 more than 40 is 45, so the ratio would be
45 vanilla to 65 chocolate… so E is correct.
Specific Content Review
Order of Operations
Perform arithmetic in this order:
1. Powers and square-roots
2. Multiplication and division
3. Addition and subtraction
But, anything in parentheses should
be solved first, from the inside out.
Order of Operations
Solve:
2 3
16 + 32 − 5
A. -20
D. 200
B. 5
E. 400
C. 128
2
Powers, Exponents, Roots
Exponents follow a few basic rules:
3
2
3+2
4 ∗ 4 = 4
4
6
2
5
=4
4 ∗ 4 ∗ 4 ∗ (4 ∗ 4)
4−2
2
6 =6
=6
6∗6∗6∗6
= (6 ∗ 6)
(6 ∗ 6)
Powers, Exponents, Roots
Exponents follow a few basic rules:
2
2
5 ∗ 6 = 5 ∗ 6
2
5 ∗ 5 ∗ 6 ∗ 6 = 5 ∗ 6 ∗ (5 ∗ 6)
3
5
3
3+2
3
2
2 ∗ 3 = 2 ∗ 3
= 2∗3 ∗3
2∗2∗2 ∗ 3∗3∗3∗3∗3 =
2∗3 ∗ 2∗3 ∗ 2∗3 ∗3∗3
Powers, Exponents, Roots
Exponents follow a few basic rules:
2 3
 7
2∗3
=7
=7
7∗7 ∗ 7∗7 ∗ 7∗7
−5
=1 2
1/3
3
2
8
6
5
=
8
Powers, Exponents, Roots
General facts about exponents:
A negative number to an even power
2
will be positive : −3 = −3 ∗ −3 = 9
A negative number to an odd power
3
will be negative: −2 = −2 ∗ −2 ∗ −2
Powers, Exponents, Roots
8
4
9 /
8
1 4
9
98 ∗ 1
2
4
−3
9 ∗3
4
−3
3 ∗3
3
6
1/2 6
/ 3
/ 36 ∗ 1/2
2 2
= 3
4−3
=3
−3
∗3
1
=3
Fractions, Ratios, Proportions
A fraction is a part divided by a whole:
7 𝑚𝑎𝑙𝑒𝑠

15 𝑒𝑚𝑝𝑙𝑜𝑦𝑒𝑒𝑠
or 7 divided by 15
A ratio is one part divided by another:
7 𝑚𝑎𝑙𝑒𝑠

8 𝑓𝑒𝑚𝑎𝑙𝑒𝑠
or a ratio of 7 : 8
Fractions, Ratios, Proportions
Fractions may be written in many
different equivalent ways. To find an
equivalent fraction, multiply or divide
the top and bottom of the original
fraction by the same number:
3

5
→
3
5
∗
2
2
→
3∗2
5∗2
→
6
10
Fractions, Ratios, Proportions
A proportion is a collective term for
a set of equivalent fractions or ratios:
3

5
1

2
=
6
10
=
2
4
=
=
15
25
5
10
=
=
30
50
6
12
=
=
60
100
10
20
= 𝑒𝑡𝑐
= 𝑒𝑡𝑐
Fractions, Ratios, Proportions
You will often need to be able to “solve
a proportion” on GMAT problems, as
in the prior ice cream shop example:
“If the ratio of vanilla to chocolate
cones sold is 2 to 3 and 30 vanilla
cones were sold today, how many
chocolate cones were sold today?”
Fractions, Ratios, Proportions
“If the ratio of vanilla to chocolate is
2 to 3 and 30 vanilla cones were sold,
how many chocolate cones were sold?”
Solve this step by setting up a set of
two equivalent ratios, a proportion:
2 30
=
3
𝑥
Fractions, Ratios, Proportions
You could solve this proportion using
algebra, or you can just think of what
number we multiplied the top and
bottom of the original fraction by:
2 15 30
∗
=
3 15 𝟒𝟓
2 30
=
3
𝑥
Operations using Fractions
Simplifying fractions:
Simplifying means to divide the top
and bottom by the same number;
this process may be repeated:
220

200
→
1/2
220/10
1/2
200/10
=
22
20
→
22/2
20/2
=
11
10
Operations using Fractions
Reciprocals of fractions:
If asked to divide 1/x where x is any
fraction, the answer is to “flip” x:
1

3 5
=
5
3
Operations using Fractions
Reciprocal relationships:
If two fractions are equal, you can
flip both and they will still be equal:
220

200
=
11
10
→
200
220
=
10
11
Operations using Fractions
Addition or subtraction:
Make each fraction have the same
bottom number using a proportion,
then add/subtract the top numbers:
3

5
2
4
+ →
3∗4
5∗4
+
2∗5
4∗5
Operations using Fractions
Addition or subtraction:
Make each fraction have the same
bottom number using a proportion,
then add/subtract the top numbers:

3∗4
5∗4
+
2∗5
4∗5
→
12
20
+
10
20
→
22
20
→
11
10
Operations using Fractions
Multiplication or division:
For multiplication, multiply the 2 top
and 2 bottom numbers; for division,
flip the 2nd fraction, then multiply:
2

3
4
5
∗ →
2∗4
3∗5
→
8
15
Operations using Fractions
Multiplication or division:
For multiplication, multiply the 2 top
and 2 bottom numbers; for division,
flip the 2nd fraction, then multiply:
2 4
 /
3 5
2
3
5
4
→ ∗ →
2∗5
3∗4
→
10
12
→
5
6
Operations using Fractions
Combining numbers and fractions:
Any number x may be written as x/1,
or as any fraction that simplifies to it:
5
6
3
1
5
6
3 ∗ + 2 → ∗ +
12
6
→
15
6
+
12
6
=
27
6
Operations using Fractions
If
1
𝑥
=
3
2
solve:
1
𝑥+2
2
A. 9/64
D. 64/9
B. 16/9
E. 9/16
C. 4/9
Percent & Percent Change
A percentage is a fraction over 100:
57% =
57
100
‘x percent of y’ is x written as a fraction
over 100 multiplied by the value of Y:
35% 𝑜𝑓 400 𝑖𝑠 … ? →
35
100
∗ 400 = 120
Percent & Percent Change
Use proportions to solve percentages:
Of the 400 trucks for sale on a used
car lot, 120 have air-conditioning;
what percent of trucks have a/c?
120

400
= 𝑥%? →
120
400
=
𝑥
100
→
120
400
∗ 100
Percent & Percent Change
An “x percent change” is calculated as
the proportion of the amount of the
change over the original amount:
“If a home was purchased for $200k
and sold for $230k, by what percent
did the home appreciate in value?”
Percent & Percent Change
“If a home was purchased for $200k
and sold for $230k, by what percent
did the home appreciate in value?”
𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙
𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙
=
30
200
=
30
200
=
𝑥
100
=
𝑥
100
→
→
30
200
∗ 100 = 𝑥
15 =
𝑥
1
𝑥
𝑥
1
Percent & Percent Change
The GMAT contains many problems
that make use of multiple percentages.
Although it is possible to solve these
using proportions and/or algebra, it is
better to plug in a starting value of 100
and calculate other values as you go.
Percent & Percent Change
“60 percent of Loyola seniors are female.
If 40 percent of females and 30 percent of
males have studied abroad in this year’s
senior class, what is the total percent of
Loyola seniors that have studied abroad?”
A. 24 B. 30 C. 36 D. 42 E. 70
Percent & Percent Change
“60 percent female… 40 percent of females
and 30 percent of males studied abroad…”
A. 24
B. 30
C. 36
D. 42
E. 70
Imagine there are 100 seniors, 60 females
40 percent of 60 females is 40/100*60 = 24
30 percent of 40 males is 30/100*40 = 12
24 females + 12 males = 36/100 = 36%
Combinations & Permutations
“At a football game, 4/5 of seats in the
lower deck were sold. If 1/4 of all the
seats are in the lower deck, and 2/3 of
total seats sold, what fraction of the
unsold seats were in the lower deck?”
A.
3
20
B.
1
6
C.
1
5
D.
1
3
E.
7
15
Combinations & Permutations
A permutation (or combination) is the
number of ways to order (or select) a
number of choices from a given set.
It is best to treat these problems as a
multistep series of consecutive choices
and adjust if necessary if not ordered.
Combinations & Permutations
For an ordered set (a permutation or
“arrangement”), imagine “steps” I, II,
III, etc & multiply together the number
of choices still remaining at each step.
𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑦𝑜𝑢 𝑠𝑒𝑎𝑡
4 𝑝𝑒𝑜𝑝𝑙𝑒 𝑖𝑛 𝑎 𝑟𝑜𝑤 𝑜𝑓 5 𝑐ℎ𝑎𝑖𝑟𝑠?
Combinations & Permutations
The “steps” I, II, III, etc are the points
at which a selection is to be made.
𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑦𝑜𝑢 𝑠𝑒𝑎𝑡
4 𝑝𝑒𝑜𝑝𝑙𝑒 𝑖𝑛 𝑎 𝑟𝑜𝑤 𝑜𝑓 5 𝑐ℎ𝑎𝑖𝑟𝑠?
I
II
III
IV
= 5 ∗ 4 ∗ 3 ∗ 2 = 120 𝑠𝑒𝑎𝑡𝑖𝑛𝑔𝑠
Combinations & Permutations
For an unordered group (combination
or “selection”), first imagine the boxes
for steps I, II, III, etc, then adjust for
possible orderings of choices A, B, etc.
𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑦𝑜𝑢 𝑐ℎ𝑜𝑜𝑠𝑒
3 𝑑𝑒𝑙𝑒𝑔𝑎𝑡𝑒𝑠 𝑓𝑟𝑜𝑚 7 𝑛𝑜𝑚𝑖𝑛𝑒𝑒𝑠?
Combinations & Permutations
𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑦𝑜𝑢 𝑐ℎ𝑜𝑜𝑠𝑒
3 𝑑𝑒𝑙𝑒𝑔𝑎𝑡𝑒𝑠 𝑓𝑟𝑜𝑚 7 𝑛𝑜𝑚𝑖𝑛𝑒𝑒𝑠?
I
II
III
= 7 ∗ 6 ∗ 5 = 210 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑠
For example, delegates B, D, & G chosen:
B
D
G
Combinations & Permutations
For example, delegates B, D, & G chosen:
B
D
G
But B, D, & G could have been chosen in
many different equivalent “orderings”.
D
B
G
D
B
G
B
G
D
Combinations & Permutations
Consider how many orderings B, D, & G
could have been chosen in as a group:
i
ii
iii
= 3 ∗ 2 ∗ 1 = 6 𝑜𝑟𝑑𝑒𝑟𝑖𝑛𝑔𝑠 𝑜𝑓
′
𝑡ℎ𝑒 𝑠𝑖𝑛𝑔𝑙𝑒 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛′ 𝑜𝑓 𝐵, 𝐷, & 𝐺
Combinations & Permutations
3 𝑑𝑒𝑙𝑒𝑔𝑎𝑡𝑒𝑠 𝑓𝑟𝑜𝑚 7 𝑛𝑜𝑚𝑖𝑛𝑒𝑒𝑠:
I
II
III
= 7 ∗ 6 ∗ 5 = 210 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑠
But every delegate set is repeated 6x:
i
ii
iii
= 3 ∗ 2 ∗ 1 = 6 𝑜𝑟𝑑𝑒𝑟𝑖𝑛𝑔𝑠 𝑔𝑟𝑜𝑢𝑝
Combinations & Permutations
3 𝑑𝑒𝑙𝑒𝑔𝑎𝑡𝑒𝑠 𝑓𝑟𝑜𝑚 7 𝑛𝑜𝑚𝑖𝑛𝑒𝑒𝑠:
I
II
III
i
ii
iii
= 210 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑠 / 6 𝑜𝑟𝑑𝑒𝑟𝑖𝑛𝑔𝑠
There are 210 total selections, but
there are 6 equivalent orderings for
each set, or 210 / 6 = 35 combinations
Combinations & Permutations
“To fill a number of vacant positions,
an IT firm needs to hire 2 database
administrators from 6 applicants,
and 3 developers from 5 applicants.
How many selections could be made?”
A. 25
B. 60
C. 150 D. 180 E. 1800
Averages & Variation
“Mean, median, and mode”:
Mean (or average) is the sum of all
values divided by the count of values.
Median is the “middle” number in a
set of values when put in rank-order.
Mode is the most frequent value.
Averages & Variation
{3, 4, 6, 7, 7, 9}
Mean =
3+4+6+7+7+9
6
=
36
6
=6
Median = ½-way between 6 and 7 = 6.5
Mode = 7
Averages & Variation
The mean (or average) has 3 parts:
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑚𝑒𝑎𝑛 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
GMAT questions commonly test the
ability to apply this formula in an
unexpected “direction” or context
Averages & Variation
“A company has two sales teams, A and B,
with 4 and 6 salespeople respectively. If the
average number of monthly sales generated
by salespeople in team A is 20 sales and the
average in team B is 30 sales, what is the
overall average number of sales generated?”
Averages & Variation
“If average sales in team A (4 people) is 20
and average sales in team B (6 people) is
30, what is the overall average # of sales?”
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
To solve for the overall average, take this
formula and start filling in what you know.
Averages & Variation
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
4 𝑓𝑟𝑜𝑚 𝐴 + 6 𝑓𝑟𝑜𝑚 𝐵
The number of sales figures being averaged
is 4 from team A and 6 from team B, or 10.
Averages & Variation
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
𝑠𝑢𝑚 𝑜𝑓 𝐴 + 𝑠𝑢𝑚 𝑜𝑓 𝐵
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
10
To find the sum of values from each team,
you have to use the average of each team.
Averages & Variation
𝑠𝑢𝑚 𝑜𝑓 𝐴 + 𝑠𝑢𝑚 𝑜𝑓 𝐵
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
10
𝑠𝑢𝑚 𝑜𝑓 𝐴
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝐴 (20) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝐴 (4)
The average of A, 20, is equal to the sum of
A (unknown) divided by the 4 salespeople.
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 20 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟 4 = 𝑠𝑢𝑚 𝑜𝑓 𝐴
Averages & Variation
80 + 𝑠𝑢𝑚 𝑜𝑓 𝐵
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
10
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 30 ∗ 𝑛𝑢𝑚𝑏𝑒𝑟 6 = 𝑠𝑢𝑚 𝑜𝑓 𝐵
80 + 180 260
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
=
= 26
𝑓 10 𝑓
10
So the overall average number of sales is 26
which is different from the obvious answer.
Averages & Variation
“The average of a set of 9 numbers is 8.
Two of the numbers are 11 and 12. What
is the average of the remaining values?”
A. 4.5 B. 5
C. 5.4 D. 6
E. 7
Averages & Variation
Range and Standard Deviation:
The range of a set of values is the
highest value – the lowest value.
The standard deviation is a number
that represents an average “distance
from the mean” in the set of values.
Averages & Variation
{3, 4, 6, 7, 7, 9}
Range = 9 – 3 = 6
St. Dev. = approx. 2
Mean
↓
Data →
0
1
2
3
4
5
Dist. = 2
Dist. = 3
6
7
8
Dist.=1
Dist. = 3
9
Averages & Variation
Standard Deviation concepts:
Two sets of data with different
means do not necessarily have
different standard deviations.
Distance from the mean, not other
data points, determines the std. dev.
Averages & Variation
Set S has a mean of 10 and a std. dev. of 3. We are
going to add two numbers to Set S. Which pair of
numbers would decrease the std. dev. the most?
A. {2, 10}
B. {10, 18}
C. {7, 13}
D. {9, 11}
E. {16, 16}
Algebra Content Review
Fundamentals of Algebra
Algebra is a process that has 2 parts:
1. Being able to translate a problem
into mathematical expressions,
functions, or algebraic equations
2. Knowing how to use or modify the
resulting expressions or equations
Fundamentals of Algebra
Unknown, to-be-solved-for values
are represented by letters in algebra:
“The price of a pair of shoes is equal
to 3 times the price of a pair of jeans.”
𝑠 =3∗𝑗
Fundamentals of Algebra
You can perform the same operation,
any operation, to both sides of an
equation and it will remain equal.
𝑠 =3∗𝑗
(𝑠)
(3 ∗ 𝑗)
+2=
+2
40
40
Fundamentals of Algebra
Usually, you do this in order to simplify,
not add to, an equation with the goal of a
single letter (unknown value) on one side:
𝑥 − 5 = 2 + 3𝑥
−3𝑥 + 5
+ 5 − 3𝑥
−2𝑥
= 7
Fundamentals of Algebra
Usually, you do this in order to simplify,
not add to, an equation with the goal of a
single letter (unknown value) on one side:
−2𝑥
= 7
−2𝑥
−2
7
=
−2
Fundamentals of Algebra
Usually, you do this in order to simplify,
not add to, an equation with the goal of a
single letter (unknown value) on one side:
𝑥
7
=−
2
= −3.5
Fundamentals of Algebra
“If x is equal to 1 more than the product
of 3 and z, and y is equal to 1 less than the
product of 2 and z, then how much greater
is 2x than 3y when z has a value of 4?”
A. 1
B. 2
C. 3
D. 5
E. 6
𝑥 = 3𝑧 + 1 → 3 ∗ 4 + 1 = 13 → 2𝑥 = 26
𝑦 = 2𝑧 − 1 → 2 ∗ 4 − 1 = 7 → 3𝑦 = 21
Systems of Equations
In general, to solve for a given number
of unknown values (i.e. letters), you
need as many equations as letters.
3𝑥 + 2𝑦 = 6
5𝑥 − 𝑦 = 10
Systems of Equations
2 ways to solve a “system of equations”:
1. Solve one equation for a letter (y),
then plug into the other equation.
2. Change one equation so that if you
add it to the other equation, one
letter (y) will be cancelled out.
Systems of Equations
1. Solve for y, plug in, then solve for x:
3𝑥 + 2𝑦 = 6
Solve for y → 5𝑥 − 𝑦 = 10
−5𝑥
− 5𝑥
−𝑦 = 10 − 5𝑥
𝑦 = −10 + 5𝑥
Systems of Equations
1. Solve for y, plug in, then solve for x:
3𝑥 + 2𝑦 = 6
Solve for y → 5𝑥 − 𝑦 = 10
Systems of Equations
1. Solve for y, plug in, then solve for x:
Plug in y →
3𝑥 + 2𝑦 = 6
𝑦 = −10 + 5𝑥
Systems of Equations
1. Solve for y, plug in, then solve for x:
Plug in y →
3𝑥 + 2(−10 + 5𝑥) = 6
3𝑥 + 2(−10) + 2(5𝑥) = 6
3𝑥 − 20
+ 10𝑥
=6
13𝑥 = 26 → 𝑥 = 13 26 = 2
Systems of Equations
1. Solve for y, plug in, then solve for x:
Plug in y →
3𝑥 + 2𝑦 = 6
→ 𝑥=2
𝑦 = −10 + 5𝑥
5(2)
𝑦 = −10 + 10 → 𝑦 = 0
Systems of Equations
2. Change equations so y cancels out:
Change →
3𝑥 + 2𝑦 = 6
5𝑥 − 𝑦 = 10
2 ∗ 5𝑥 − 𝑦 = 10 ∗ 2
2 5𝑥 + 2 −𝑦 = 20
10𝑥 − 2𝑦 = 20
Systems of Equations
2. Change equations so y cancels out:
Change →
3𝑥 + 2𝑦 = 6
10𝑥 − 2𝑦 = 20
13𝑥 /13 = 26 /13
𝑥
=2
Systems of Equations
“A symphony sells 3 kinds of tickets: box
seats for $40, general admission for $20,
and student admission for $10. On a recent
night they sold 200 tickets, 40 to students,
and made $4,000 in all. How many general
admission tickets did the symphony sell?”
A. 96
B. 120 C. 140 D. 160 E. 180
Systems of Equations
“15 years ago, Adam was 3 times as old as
Bob. Today, Adam is twice as old as Bob.
How old will Adam be 5 years from now?”
A. 35
B. 45
C. 50
D. 60
𝐴 − 15 = 3 ∗ (𝐵 − 15)
𝐴 = 2𝐵
E. 65
Systems of Equations
Cautions about systems of equations:
1. If the 2 equations are equivalent to
each other, you cannot solve them.
3𝑥 + 2𝑦 = 6
6𝑥 + 4𝑦 = 12
Systems of Equations
2. On the other hand, sometimes you
don’t need 2 equations; GMAT will
put 2 variables in 1 equation, but do
so such that 1 variable cancels out.
2
15 𝐴 + 𝐵 − 6𝐵 = 30 + 3𝐵(5 − 2𝐵)
2
15𝐴 + 15𝐵 − 6𝐵 = 30 + 15𝐵 − 6𝐵
2