Example 2.
Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the
work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time
does Y finish his work?
Solution:
Now the only reason this is trickier than the first problem is because the sequence of
events are slightly more complicated. The concept however is the same. So if our
understanding of the concept is clear, we should have no trouble at all dealing with this.
‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us
that in one hour, X will finish of the work.
‘He finishes 2/3 of the work’ This tells us that of the work still remains.
‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete of the
work.
‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be .
Therefore, rate at which Y works is .
‘In how much time does Y finish his work?’ If Y completes of the work in 1 hour, then
he will complete of the work in 40 hours.
So as you can see, even though the question might have been a little difficult to follow at
first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we
did was apply our knowledge of the concept we learnt earlier to the information in the
question in order to answer what was being asked.
Example 3.
Working together, printer A and printer B would finish a task in 24 minutes. Printer A
alone would finish the task in 60 minutes. How many pages does the task contain if
printer B prints 5 pages a minute more than printer A?
Solution:
This problem is interesting because it tests not only our knowledge of the concept of word
problems, but also our ability to ‘translate English to Math’
‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us
that A and B combined would work at the rate of per minute.
‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of
 per minute.
At this point, it should strike you that with just this much information, it is possible to
1
12
1
3
1
3
1
12
∗ =1
10
1
12
1
120
1
120
1
3
1
24
1
60
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calculate the rate at which B works: Rate at which B works = .
‘B prints 5 pages a minute more than printer A’ This means that the difference between
the amount of work B and A complete in one minute corresponds to 5 pages. So, let us
calculate that difference. It will be
‘How many pages does the task contain?’ If of the job consists of 5 pages, then the 1
job will consist of pages.
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten
hours longer to produce 660 sprockets than machine B. Machine B produces 10% more
sprockets per hour than machine A. How many sprockets per hour does machine A
produce?
Solution:
The rate of A is sprockets per hour;
The rate of B is sprockets per hour.
We are told that B produces 10% more sprockets per hour than A, thus
--> --> the rate of A is sprockets per hour.
As you can see, the main reason the 'tough' problems are 'tough' is because they test a
number of other concepts apart from just the ‘work’ concept. However, once you manage
to form the equations, they are really not all that tough.
And as far as the concept of ‘work’ word problems is concerned – it is always the
same!


age
to form the equations, they are really not all that tough.
And as far as the concept of ‘work’ word problems is concerned – it is always the
same!

− =1
24
1
60
1
40
− =1
40
1
60
1
120
1
120
= 600(5∗1)
1
120
660
t+10
660
t
∗ 1.1 =660
t+10
660
t
t = 100 = 6660
t+10
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Distance, Speed Time
https://gmatclub.com/forum/distance-speed-time-87481.html
 'Distance/Speed/Time' Word Problems Made Easy
created by: sriharimurthy
edited by: bb, walker, Bunuel
NOTE: In case you are not familiar with translating word problems into equations please
go through this post first : https://gmatclub.com/forum/word-problem ... 87346.html
What is a ‘D/S/T’ Word Problem?
Usually involve something/someone moving at a constant or average speed.
Out of the three quantities (speed/distance/time), we are required to find one.
Information regarding the other two will be provided in the question stem.
The ‘D/S/T’ Formula: Distance = Speed x Time
I’m sure most of you are already familiar with the above formula (or some variant of it).
But how many of you truly understand what it signifies?
When you see a ‘D/S/T’ question, do you blindly start plugging values into the formula
without really understanding the logic behind it? If then answer to that question is yes,
then you would probably have noticed that your accuracy isn’t quite where you’d want it
to be.
My advice here, as usual, is to make sure you understand the concept behind the
formula rather than just using it blindly.
So what’s the concept? Lets find out!
The Distance = Speed x Time formula is just a way of saying that the
distance you travel depends on the speed you go for any length of time.
If you travel at 50 mph for one hour, then you would have traveled 50 miles. If you
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travel for 2 hours at that speed, you would have traveled 100 miles. 3 hours would
be 150 miles, etc.
If you were to double the speed, then you would have traveled 100 miles in the first
hour and 200 miles at the end of the second hour.
We can figure out any one of the components by knowing the other two.
For example, if you have to travel a distance of 100 miles, but can only go at a
speed of 50 mph, then you know that it will take you 2 hours to get there. Similarly,
if a friend visits you from 100 miles away and tells you that it took him 4 hours to
reach, you will know that he AVERAGED 25 mph. Right?
All calculations depend on AVERAGE SPEED.
Supposing your friend told you that he was stuck in traffic along the way and that
he traveled at 50 mph whenever he could move. Therefore, although practically he
never really traveled at 25 mph, you can see how the standstills due to traffic
caused his average to reduce. Now, if you think about it, from the information
given, you can actually tell how long he was driving and how long he was stuck due
to traffic (assuming; what is false but what they never worry about in these
problems; that he was either traveling at 50 mph or 0 mph). If he was traveling
constantly at 50 mph, he should have reached in 2 hours. However, since he took 4
hours, he must have spent the other 2 hours stuck in traffic!
Now lets see how we can represent this using the formula.
We know that the total distance is 100 miles and that the total time is 4 hours.
BUT, his rates were different AND they were different at different times. However,
can you see that no matter how many different rates he drove for various different
time periods, his TOTAL distance depended simply on the SUM of each of the
different distances he drove during each time period?
E.g., if you drive a half hour at 60 mph, you will cover 30 miles. Then if you speed
up to 80 mph for another half hour, you will cover 40 miles, and then if you slow
down to 30 mph, you will only cover 15 miles in the next half hour. But if you drove
like this, you would have covered a total of 85 miles (30 + 40 + 15). It is fairly easy
to see this looking at it this way, but it is more difficult to see it if we scramble it up
and leave out one of the amounts and you have to figure it out going "backwards".
That is what word problems do.
Further, what makes them difficult is that the components they give you, or ask you
to find can involve variable distances, variable times, variable speeds, or any two or
three of these. How you "reassemble" all this in order to use the d = s*t formula
takes some reflection that is "outside" of the formula itself. You have to think
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about how to use the formula.
So the trick is to be able to understand EXACTLY what they are giving you and
EXACTLY what it is that is missing, but you do that from thinking, not from the
formula, because the formula only works for the COMPONENTS of any trip
where you are going an average speed for a certain amount of time. ONCE
the conditions deal with different speeds or different times, you have to look at each
of those components and how they go together. And that can be very difficult if you
are not methodical in how you think about the components and how they go
together. The formula doesn't tell you which components you need to look at and
how they go together. For that, you need to think, and the thinking is not always as
easy or straightforward as it seems like it ought to be.
In the case of your friend above, if we call the time he spent driving 50 mph, T1;
then the time he spent standing still is (4 - T1) hours, since the whole trip took 4
hours. So we have
100 miles = (50 mph x T1) + (0 mph x [4 - T1]) which is
equivalent then to: 100 miles = 50 mph x T1
So, T1 will equal 2 hours. And, since the time he spent going zero is (4 - 2), it also
turns out to be 2 hours.
Sometimes the right answers will seem counter-intuitive, so it is really
important to think about the components methodically and systematically.
There is a famous trick problem: To qualify for a race, you need to average 60 mph
driving two laps around a 1 mile long track. You have some sort of engine difficulty
the first lap so that you only average 30 mph during that lap; how fast do you have
to drive the second lap to average 60 for both of them?
I will go through THIS problem with you because, since it is SO tricky, it will
illustrate a way of looking at almost all the kinds of things you have to think about
when working any of these kinds of problems FOR THE FIRST TIME (i.e., before you
can do them mechanically because you recognize the TYPE of problem it is).
Intuitively it would seem you need to drive 90, but this turns out to be wrong for
reasons I will give in a minute.
The answer is that NO MATTER HOW FAST you do the second lap, you can't make it.
And this SEEMS really odd and that it can't possibly be right, but it is. The reason is
that in order to average at least 60 mph over two one-mile laps, since 60 mph is
one mile per minute, you will need to do the whole two miles in two minutes or less.
But if you drove the first mile at only 30, you used up the whole two minutes just
doing IT. So you have run out of time to qualify.
To see this with the d = s*t formula, you need to look at the overall trip and break
it into components, and that is the hardest part of doing this (these) problem(s),
because (often) the components are difficult to figure out, and because it is hard to
see which ones you need to put together in which way.
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In the next section we will learn how to do just that.
Resolving the Components
When you first start out with these problems, the best way to approach
them is by organizing the data in a tabular form.
Use a separate column each for distance, speed and time and a separate row for the
different components involved (2 parts of a journey, different moving objects, etc.).
The last row should represent total distance, total time and average speed for these
values (although there might be no need to calculate these values if the question
does not require them).
Assign a variable for any unknown quantity.
If there is more than one unknown quantity, do not blindly assign another variable
to it. Look for ways in which you can express that quantity in terms of the quantities
already present. Assign another variable to it only if this is not possible.
In each row, the quantities of distance, speed and time will always satisfy d
= s*t.
The distance and time column can be added to give you the values of total
distance and total time but you CANNOT add the speeds.
Think about it: If you drive 20 mph on one street, and 40 mph on another street,
does that mean you averaged 60 mph?
Once the table is ready, form the equations and solve for what has been
asked!
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Warning: Make sure that the units for time and distance agree with the units for the rate.
For instance, if they give you a rate of feet per second, then your time must be in seconds
and your distance must be in feet. Sometimes they try to trick you by using the wrong
units, and you have to catch this and convert to the correct units.
A Few More Points to Note
Motion in Same Direction (Overtaking): The first thing that should strike you
here is that at the time of overtaking, the distances traveled by both will be the
same.
Motion in Opposite Direction (Meeting): The first thing that should strike you
here is that if they start at the same time (which they usually do), then at the point
at which they meet, the time will be the same. In addition, the total distance
traveled by the two objects under consideration will be equal to the sum of their
individual distances traveled.
Round Trip: The key thing here is that the distance going and coming back is the
same.
Now that we know the concept in theory, let us see how it works practically,
with the help of a few examples.
Note for tables : All values in black have been given in the question stem. All values in
blue have been calculated.
Example 1.
To qualify for a race, you need to average 60 mph driving two laps around a 1-mile long
track. You have some sort of engine difficulty the first lap so that you only average 30
mph during that lap; how fast do you have to drive the second lap to average 60 for both
of them?
Solution:
Let us first start with a problem that has already been introduced. You will see that by
clearly listing out the given data in tabular form, we eliminate any scope for confusion.
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In the first row, we are given the distance and the speed. Thus it is possible to calculate
the time.
Time(1) = Distance(1)/Speed(1) = 1/30
In the second row, we are given just the distance. Since we have to calculate speed, let
us give it a variable 'x'. Now, by using the 'D/S/T' relationship, time can also be expressed
in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = 1/x
In the third row, we know that the total distance is 2 miles (by taking the sum of the
distances in row 1 and 2) and that the average speed should be 60 mph. Thus we can
calculate the total time that the two laps should take.
Time(3) = Distance(3)/Speed(3) = 2/60 = 1/30
Now, we know that the total time should be the sum of the times in row 1 and 2. Thus we
can form the following equation :
Time(3) = Time(1) + Time(2) ---> 1/30 = 1/30 + 1/x
From this, it becomes clear that '1/x' must be 0.
Since 'x' is the reciprocal of 0, which does not exist, there can be no speed for
which the average can be made up in the second lap.
Example 2.
An executive drove from home at an average speed of 30 mph to an airport where a
helicopter was waiting. The executive boarded the helicopter and flew to the corporate
offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip
took three hours. Find the distance from the airport to the corporate offices.
Solution:
Since we have been asked to find the distance from the airport to the corporate office
(that is the distance he spent flying), let us assign that specific value as 'x'.
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Thus, the distance he spent driving will be '150 - x'
Now, in the first row, we have the distance in terms of 'x' and we have been given the
speed. Thus we can calculate the time he spent driving in terms of 'x'.
Time(1) = Distance(1)/Speed(1) = (150 - x)/30
Similarly, in the second row, we again have the distance in terms of 'x' and we have been
given the speed. Thus we can calculate the time he spent flying in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = x/60
Now, notice that we have both the times in terms of 'x'. Also, we know the total time for
the trip. Thus, summing the individual times spent driving and flying and equating it to
the total time, we can solve for 'x'.
Time(1) + Time(2) = Time(3) --> (150 - x)/30 + x/60 = 3 --> x = 120 miles
Answer : 120 miles
Note: In this problem, we did not calculate average speed for row 3 since we did not need
it. Remember not to waste time in useless calculations!
Example 3.
A passenger train leaves the train depot 2 hours after a freight train left the same depot.
The freight train is traveling 20 mph slower than the passenger train. Find the speed of
the passenger train, if it overtakes the freight train in three hours.
Solution:
Since this is an 'overtaking' problem, the first thing that should strike us is that the
distance traveled by both trains is the same at the time of overtaking.
Next we see that we have been asked to find the speed of the passenger train at the time
of overtaking. So let us represent it by 'x'.
Also, we are given that the freight train is 20 mph slower than the passenger train. Hence
its speed in terms of 'x' can be written as 'x - 20'.
Moving on to the time, we are told that it has taken the passenger train 3 hours to reach
the freight train. This means that the passenger train has been traveling for 3 hours.
We are also given that the passenger train left 2 hours after the freight train. This means
that the freight train has been traveling for 3 + 2 = 5 hours.
Now that we have all the data in place, we need to form an equation that will help us
solve for 'x'. Since we know that the distances are equal, let us see how we can use this
to our advantage.
From the first row, we can form the following equation :
Distance(1) = Speed(1) * Time(1) = x*3
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From the second row, we can form the following equation :
Distance(2) = Speed(2) * Time(2) = (x - 20)*5
Now, equating the distances because they are equal we get the following equation :
3*x = 5*(x - 20) --> x = 50 mph.
Answer : 50 mph.
Example 4.
Two cyclists start at the same time from opposite ends of a course that is 45 miles long.
One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after
they begin will they meet?
Solution:
Since this is a 'meeting' problem, there are two things that should strike you. First, since
they are starting at the same time, when they meet, the time for which both will have
been cycling will be the same. Second, the total distance traveled by the will be equal to
the sum of their individual distances.
Since we are asked to find the time, let us assign it as a variable 't'. (which is same for
both cyclists)
In the first row, we know the speed and we have the time in terms of 't'. Thus we can get
the following equation :
Distance(1) = Speed(1) * Time(1) = 14*t
In the second row, we know the speed and again we have the time in terms of 't'. Thus
we can get the following equation :
Distance(2) = Speed(2) * Time(2) = 16*t
Now we know that the total distance traveled is 45 miles and it is equal to the sum of the
two distances. Thus we get the following equation to solve for 't' :
Distance(3) = Distance(1) + Distance(2) --> 45 = 14*t + 16*t --> t = 1.5 hours
Answer : 1.5 hours.
Example 5.
A boat travels for three hours with a current of 3 mph and then returns the same distance
against the current in four hours. What is the boat's speed in calm water?
Solution:
Since this is a question on round trip, the first thing that should strike us is that the
distance going and coming back will be the same.
Now, we are required to find out the boats speed in calm water. So let us assume it to be
'b'. Now if speed of the current is 3 mph, then the speed of the boat while going
downstream and upstream will be 'b + 3' and 'b - 3' respectively.
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In the first row, we have the speed of the boat in terms of 'b' and we are given the time.
Thus we can get the following equation :
Distance(1) = Speed(1) * Time(1) = (b + 3)*3
In the second row, we again have the speed in terms of 'b' and we are given the time.
Thus we can get the following equation :
Distance(2) = Speed(2) * Time(2) = (b - 3)*4
Since the two distances are equal, we can equate them and solve for 'b'.
Distance(1) = Distance(2) --> (b + 3)*3 = (b - 3)*4 --> b = 21 mph.
Answer : 21 mph.
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Overlapping Sets
https://gmatclub.com/forum/overlapping-sets-144260.html
ADVANCED OVERLAPPING SETS
Some hard GMAT quantitative questions will require you to know and understand the formulas for set theory, presenting three sets and
asking various questions about them. There are two main formulas to solve questions involving three overlapping sets. Consider the
diagram below:
FIRST FORMULA
.
Let's see how this formula is derived.
When we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are counted twice
and section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and subtract section g
TWICE (again to count section g only once).
In the formula above, , where AnB means intersection of A and B (sections d,
and g), AnC means intersection of A and C (sections e, and g), and BnC means intersection of B and C (sections f, and g).
Now, when we subtract (d, and g), (e, and g), and (f, and g) from , we are subtract sections d, e, and f
ONCE BUT section g THREE TIMES (and we need to subtract section g only twice), therefor we should add only section g, which is
intersection of A, B and C (AnBnC) again to get .
SECOND FORMULA
.
Notice that EXACTLY (only) 2-group overlaps is not the same as 2-group overlaps:
Elements which are common only for A and B are in section d (so elements which are common ONLY for A and B refer to the elements
which are in A and B but not in C);
Elements which are common only for A and C are in section e;
Elements which are common only for B and C are in section f.
Let's see how this formula is derived.
Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither
sum of 2 −group overlaps = AnB + AnC + BnC
AnB AnC BnC A + B + C
Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither
Total = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗ (all three) +Neither
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Again: when we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are
counted twice and section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and
subtract section g TWICE (again to count section g only once).
When we subtract from A+B+C we subtract sections d, e, and f once (fine) and next we
need to subtract ONLY section g ( ) twice. That's it.
Now, how this concept can be represented in GMAT problem?
Example 1:
Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30
are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers
are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all
three teams. How many workers are there in total?
Translating:
"are placed on at least one team": members of none =0;
"20 are on the marketing team": M=20;
"30 are on the Sales team": S=30;
"40 are on the Vision team": V=40;
"5 workers are on both the Marketing and Sales teams": MnS=5, note here that some from these 5 can be the members of Vision team
as well, MnS is sections d an g on the diagram (assuming Marketing = A, Sales = B and Vision = C);
"6 workers are on both the Sales and Vision teams": SnV=6 (the same as above sections f an g);
"9 workers are on both the Marketing and Vision teams": MnV=9.
"4 workers are on all three teams": MnSnV=4, section 4.
Question: Total=?
Applying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members:
.
Answer: 74. Discuss this question
HERE.
Example 2:
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three
academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22
students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students
sign up for exactly two clubs, how many students sign up for all three clubs?
Translating:
"Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic
clubs": Total=59, Neither=0 (as members are required to sign up for a minimum of one);
"22 students sign up for the poetry club": P=22;
"27 students for the history club": H=27;
"28 students for the writing club": W=28;
"6 students sign up for exactly two clubs": (sum of EXACTLY 2-group overlaps)=6, so the sum of sections d, e, and f is given to be 6,
(among these 6 students there are no one who is a member of ALL 3 clubs)
Question:: "How many students sign up for all three clubs?" -->
Apply second formula: -->
 --> .
Answer: 6. Discuss this question
HERE.
sum of EXACTLY  2 −group overlaps
AnBnC
Total = M + S + V − (MnS + MnV + SnV ) +MnSnV + Neither = 20 + 30 + 40 − (5 + 6 + 9) + 4 + 0 = 74
PnHnW = g =?
Total = P + H + W − (sum of EXACTLY  2 −group overlaps) − 2 ∗PnHnW + Neither
59 = 22 + 27 + 28 − 6 − 2 ∗x + 0 x = 6
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Example 3:
Of 20 Adults, 5 belong to A, 7 belong to B, and 9 belong to C. If 2 belong to all three organizations and 3 belong to exactly
2 organizations, how many belong to none of these organizations?
Translating:
"20 Adults": Total=20;
"5 belong to A, 7 belong to B, and 9 belong to C": A=5, B=7, and C=9;
"2 belong to all three organizations": AnBnC=g=2;
"3 belong to exactly 2 organizations": (sum of EXACTLY 2-group overlaps)=3, so the sum of sections d, e, and f is given to be 3,
(among these 3 adults there are no one who is a member of ALL 3 clubs)
Question:: Neither=?
Apply second formula: -->
 --> .
Answer: 6. Discuss this question
HERE.
Example 4:
This semester, each of the 90 students in a certain class took at least one course from A, B, and C. If 60 students took A,
40 students took B, 20 students took C, and 5 students took all the three, how many students took exactly two courses?
Translating:
"90 students": Total=90;
"of the 90 students in a certain class took at least one course from A, B, and C": Neither=0;
"60 students took A, 40 students took B, 20 students took C": A=60, B=40, and C=20;
"5 students took all the three courses": AnBnC=g=5;
Question:: (sum of EXACTLY 2-group overlaps)=?
Apply second formula: -->
 --> .
Answer: 20. Discuss this question
HERE.
Example 5:
In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 10 owned
rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?
Translating:
"60 people own cats, dogs, or rabbits": Total=60; and Neither=0;
"30 people owned cats, 40 owned dogs, 10 owned rabbits": A=30, B=40, and C=10;
"12 owned exactly two of the three types of pet": (sum of EXACTLY 2-group overlaps)=12;
Question:: AnBnC=g=?
Apply second formula: -->
 --> .
Answer: 4. Discuss this question
HERE.
Example 6:
When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E)
had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she
compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw
that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name
listed more than once, how many names will be on the combined roster?
Translating:
"E had 26 names, M had 28, and S had 18": E=26, M=28, and S=18;
"E and M had 9 names in common, E and S had 7, and M and S had 10": EnM=9, EnS=7, and MnS=10;
Total = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither
20 = 5 + 7 + 9 − 3 − 2 ∗ 2 +Neither Neither = 6
Total = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither
90 = 60 + 40 + 20 −x − 2 ∗ 5 + 0 x = 20
Total = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither
60 = 30 + 40 + 10 − 12 − 2 ∗x + 0 x = 4
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"4 names were on all 3 rosters": EnMnS=g=4;
Question:: Total=?
Apply first formula: -->
 --> .
Answer: 50. Discuss this question HERE.
Example 7:
There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a
course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses
and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?
Translating:
"There are 50 employees in the office of ABC Company": Total=50;
"22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course"; A=22, B=15, and
C=14;
"Nine of the employees have taken exactly two of the courses": (sum of EXACTLY 2-group overlaps)=9;
"1 employee has taken all three of the courses": AnBnC=g=1;
Question:: None=?
Apply second formula: -->
 --> .
Answer: 10. Discuss this question
HERE.
Example 8 (hard):
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked
product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the
products, what percentage of the survey participants liked more than one of the three products?
Translating:
"85% of those surveyed liked at least one of three products: 1, 2, and 3": Total=100%. Also, since 85% of those surveyed liked at
least one of three products then 15% liked none of three products, thus None=15%;
"5% of the people in the survey liked all three of the products": AnBnC=g=5%;
Question:: what percentage of the survey participants liked more than one of the three products?
Apply second formula:
Total = {liked product 1} + {liked product 2} + {liked product 3} - {liked exactly two products} - 2*{liked exactly three
product} + {liked none of three products}
 --> , so 5% liked exactly two products. More than one product liked those who liked
exactly two products, (5%) plus those who liked exactly three products (5%), so 5+5=10% liked more than one product.
Answer: 10%. Discuss this question
HERE.
Example 9 (hard):
In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play
Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many
students play exactly two of these sports?
Translating:
"In a class of 50 students...": Total=50;
"20 play Hockey, 15 play Cricket and 11 play Football": H=20, C=15, and F=11;
"7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football": HnC=7, CnF=4, and HnF=5. Notice that
"7 play both Hockey and Cricket" does not mean that out of those 7, some does not play Football too. The same for Cricket/Football
and Hockey/Football;
"18 students do not play any of these given sports": Neither=18.
Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither
Total = 26 + 28 + 18 − (9 + 7 + 10) + 4 + 0Total = 50
Total = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + None
50 = 22 + 15 + 14 − 9 − 2 ∗ 1 +None None = 10
100 = 50 + 30 + 20 −x − 2 ∗ 5 + 15 x = 5
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Question:: how many students play exactly two of these sports?
Apply first formula:
{Total}={Hockey}+{Cricket}+{Football}-{HC+CH+HF}+{All three}+{Neither}
50=20+15+11-(7+4+5)+{All three}+18 --> {All three}=2;
Those who play ONLY Hockey and Cricket are 7-2=5;
Those who play ONLY Cricket and Football are 4-2=2;
Those who play ONLY Hockey and Football are 5-2=3;
Hence, 5+2+3=10 students play exactly two of these sports.
Answer: 10. Discuss this question
HERE.
Example 10 (hard DS question on three overlapping sets):
A student has decided to take GMAT and TOEFL examinations, for which he has allocated a certain number of days for
preparation. On any given day, he does not prepare for both GMAT and TOEFL. How many days did he allocate for the
preparation?
(1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days.
(2) He prepared for either GMAT or TOEFL on 14 days
We have: {Total} = {GMAT } + {TOEFL} - {Both} + {Neither}. Since we are told that "on any given day, he does not prepare for
both GMAT and TOEFL", then {Both} = 0, so {Total} = {GMAT } + {TOEFL} + {Neither}. We need to find {Total}
(1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days --> {Total} - {GMAT } = 10 and {Total} - {TOEFL} =12.
Not sufficient.
(2) He prepared for either GMAT or TOEFL on 14 days --> {GMAT } + {TOEFL} = 14. Not sufficient.
(1)+(2) We have three linear equations ({Total} - {GMAT } = 10, {Total} - {TOEFL} =12 and {GMAT } + {TOEFL} = 14) with
three unknowns ({Total}, {GMAT }, and {TOEFL}), so we can solve for all of them. Sufficient.
Just to illustrate. Solving gives:
{Total} = 18 - he allocate total of 18 days for the preparation;
{GMAT } = 8 - he prepared for the GMAT on 8 days;
{TOEFL} = 6 - he prepared for the TOEFL on 6 days;
{Neither} = 4 - he prepared for neither of them on 4 days.
Answer: C. Discuss this question
HERE.
Example 11 (disguised three overlapping sets problem):
Three people each took 5 tests. If the ranges of their scores in the 5 practice tests were 17, 28 and 35, what is the
minimum possible range in scores of the three test-takers?
A. 17
B. 28
C. 35
D. 45
E. 80
Consider this problem to be an overlapping sets problem:
# of people in group A is 17;
# of people in group B is 28;
# of people in group C is 35;
What is the minimum # of total people possible in all 3 groups? Clearly if two smaller groups A and B are subsets of bigger group C (so
if all people who are in A are also in C and all people who are in B are also in C), then total # of people in all 3 groups will be 35.
Minimum # of total people cannot possibly be less than 35 since there are already 35 people in group C.
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Answer: C.
P.S. Notice that max range for the original question is not limited when the max # of people in all 3 groups for revised question is
17+28+35 (in case there is 0 overlap between the 3 groups).
Answer: C. Discuss this question HERE.
____________________________________________________________________________________________________________
For more questions on overlapping sets check our Question Banks
Problem Solving Questions on Overlapping Sets
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GMAT Club Math Book
Probability
https://gmatclub.com/forum/probability-87244.html
 PROBABILITY
Definition
A number expressing the probability (p) that a specific event will occur, expressed as the
ratio of the number of actual occurrences (n) to the number of possible occurrences (N).
A number expressing the probability (q) that a specific event will not occur:
Examples
Coin
There are two equally possible outcomes when we toss a coin: a head (H) or tail (T).
Therefore, the probability of getting head is 50% or and the probability of getting tail is
50% or .
All possibilities: {H,T}
Dice
p = n
N
q = = 1 − p(N−n)
N
1
2
1
2
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There are 6 equally possible outcomes when we roll a die. The probability of getting any
number out of 1-6 is .
All possibilities: {1,2,3,4,5,6}
Marbles, Balls, Cards...
Let's assume we have a jar with 10 green and 90 white marbles. If we randomly choose a
marble, what is the probability of getting a green marble?
The number of all marbles: N = 10 + 90 =100
The number of green marbles: n = 10
Probability of getting a green marble:
There is one important concept in problems with marbles/cards/balls. When the first
marble is removed from a jar and not replaced, the probability for the second marble
differs ( vs. ). Whereas in case of a coin or dice the probabilities are always the
same ( and ). Usually, a problem explicitly states: it is a problem
with replacement or
without replacement.
Independent events
Two events are independent if occurrence of one event does not influence occurrence of
other events. For n independent events the probability is the product of all probabilities of
independent events:
p = p1 * p2 * ... * pn-1 * pn
or
P(A and B) = P(A) * P(B) - A and B denote independent events
1
6
p = = =n
N
10
100
1
10
9
99
10
100
1
6
1
2
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Example #1
Q:There is a coin and a die. After one flip and one toss, what is the probability of getting
heads and a "4"?
Solution: Tossing a coin and rolling a die are independent events. The probability of
getting heads is and probability of getting a "4" is . Therefore, the probability of
getting heads and a "4" is:
Example #2
Q: If there is a 20% chance of rain, what is the probability that it will rain on the first day
but not on the second?
Solution: The probability of rain is 0.2; therefore probability of sunshine is q = 1 - 0.2 =
0.8. This yields that the probability of rain on the first day and sunshine on the second
day is:
P = 0.2 * 0.8 = 0.16
Example #3
Q:There are two sets of integers: {1,3,6,7,8} and {3,5,2}. If Robert chooses randomly
one integer from the first set and one integer from the second set, what is the probability
of getting two odd integers?
Solution: There is a total of 5 integers in the first set and 3 of them are odd: {1, 3, 7}.
Therefore, the probability of getting odd integer out of first set is . There are 3 integers
in the second set and 2 of them are odd: {3, 5}. Therefore, the probability of getting an
odd integer out of second set is . Finally, the probability of of getting two odd integers
is:
Mutually exclusive events
Shakespeare's phrase "To be, or not to be: that is the question" is an example of two
mutually exclusive events.
Two events are mutually exclusive if they cannot occur at the same time. For n mutually
exclusive events the probability is the sum of all probabilities of events:
p = p1 + p2 + ... + pn-1 + pn
or
P(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events
Example #1
Q: If Jessica rolls a die, what is the probability of getting at least a "3"?
Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The
probability of each outcome is 1/6. The probability of getting at least a "3" is:
1
2
1
6
P = ∗ =1
2
1
6
1
12
3
5
2
3
P = ∗ =3
5
2
3
2
5
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Combination of independent and mutually exclusive
events
Many probability problems contain combination of both independent and mutually
exclusive events. To solve those problems it is important to identify all events and their
types. One of the typical problems can be presented in a following general form:
Q: If the probability of a certain event is p, what is the probability of it occurring k times
in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)
Solution: All events are independent. So, we can say that:
 (1)
But it isn't the right answer. It would be right if we specified exactly each position for
events in the sequence. So, we need to take into account that there are more than one
outcomes. Let's consider our example with a coin where "H" stands for Heads and "T"
stands for Tails:
HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3
heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In
our general question, probability of occurring event k times in n-time sequence could be
expressed as:
 (2)
In the example with a coin, right answer is
Example #1
Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the
probability of raining on exactly 2 days in a 7-day period?
Solution: We are not interested in the exact sequence of event and thus apply formula
#2:
A few ways to approach a probability problem
There are a few typical ways that you can use for solving probability questions. Let's
consider example, how it is possible to apply different approaches:
Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly
chosen to form a committee, what is the probability that the committee includes both Bob
and Rachel?
Solution:
P = + + + =1
6
1
6
1
6
1
6
2
3
= ∗ (1 − pP ′ pk )n−k
P = ∗ ∗ (1 − pCn
k pk )n−k
P = ∗ ∗ = ∗C8
3 0.53 0.55 C8
3 0.58
P = ∗ ∗C7
2 0.42 0.65
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1) combinatorial approach: The total number of possible committees is . The
number of possible committee that includes both Bob and Rachel is .
2) reversal combinatorial approach: Instead of counting probability of occurrence of
certain event, sometimes it is better to calculate the probability of the opposite and then
use formula p = 1 - q. The total number of possible committees is . The number
of possible committee that does not includes both Bob and Rachel is:
 where,
 - the number of committees formed from 6 other people.
 - the number of committees formed from Rob or Rachel and one out of 6 other
people.
3) probability approach: The probability of choosing Bob or Rachel as a first person in
committee is 2/8. The probability of choosing Rachel or Bob as a second person when first
person is already chosen is 1/7. The probability that the committee includes both Bob and
Rachel is.
4) reversal probability approach: We can choose any first person. Then, if we have
Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have
neither Rachel nor Bob as first choice, we can choose any person out of remaining 7
people. The probability that the committee includes both Bob and Rachel is.
Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what
is the probability that none of them are married to each other?
Solution:
1) combinatorial approach:
 - we choose 3 couples out of 5 couples.
 - we chose one person out of a couple.
 - we have 3 couple and we choose one person out of each couple.
 - the total number of combinations to choose 3 people out of 10 people.
2) reversal combinatorial approach: In this example reversal approach is a bit shorter
N = C8
2
n = 1
P = = =n
N
1
C8
2
1
28
N = C8
2
m = + 2 ∗C6
2 C6
1
C6
2
2 ∗ C6
1
P = 1 − = 1 −m
N
+2∗C6
2 C6
1
C8
2
P = 1 − = 1 − =15+2∗6
28
27
28
1
28
P = ∗ = =2
8
1
7
2
56
1
28
P = 1 − ( ∗ + ∗ 1) = =2
8
6
7
6
8
2
56
1
28
C5
3
C2
1
(C2
1 )3
C10
3
p = = =
∗(C5
3 C2
1 )3
C10
3
10∗8
10∗3∗4
2
3
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and faster.
 - we choose 1 couple out of 5 couples.
 - we chose one person out of remaining 8 people.
 - the total number of combinations to choose 3 people out of 10 people.
3) probability approach:
1st person: - we choose any person out of 10.
2nd person: - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: - we choose any person out of 6=10-4(two couples from previous choices).
Probability tree
Sometimes, at 700+ level you may see complex probability problems that include
conditions or restrictions. For such problems it could be helpful to draw a probability tree
that include all possible outcomes and their probabilities.
Example #1
Q: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If
the first ball is not green, she takes the second ball (without replacing first) and she wins
if the two balls are white or if the first ball is gray and the second ball is white. What is the
probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?
Solution: Let's draw all possible outcomes and calculate all probabilities.
Now, It is pretty obvious that the probability of Julia's win is:
C5
1
C8
1
C10
3
p = 1 − = 1 − =
∗C5
1 C8
1
C10
3
5∗8
10∗3∗4
2
3
= 110
10
8
9
6
8
p = 1 ∗ ∗ =8
9
6
8
2
3
P = + ∗ + ∗ =4
7
2
7
1
6
1
7
2
6
2
3
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Tips and Tricks: Symmetry
Symmetry sometimes lets you solve seemingly complex probability problem in a few
seconds. Let's consider an example:
Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel.
How many ways it can be done ?
Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly
1/2 of all possible ways:
N = ∗ = 101
2 P 5
2
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Practice Questions
Easy:
1. https://gmatclub.com/forum/in-a-box-of- ... 07395.html
2. https://gmatclub.com/forum/among-a-grou ... 34499.html
3. https://gmatclub.com/forum/the-probabil ... 44730.html
4. https://gmatclub.com/forum/xavier-yvonn ... 65822.html
5. https://gmatclub.com/forum/carol-purcha ... 21728.html
6. https://gmatclub.com/forum/two-integers ... 43451.html
7. https://gmatclub.com/forum/raffle-ticke ... 36563.html
8. https://gmatclub.com/forum/sixty-percen ... 43900.html
9. https://gmatclub.com/forum/in-a-set-of- ... 68915.html
10. https://gmatclub.com/forum/a-committee- ... 44443.html
Medium:
1. https://gmatclub.com/forum/if-x-is-to-b ... 67089.html
2. https://gmatclub.com/forum/a-gardener-i ... 99822.html
3. https://gmatclub.com/forum/a-contest-wi ... 38710.html
4. https://gmatclub.com/forum/in-a-certain ... 69115.html
5. https://gmatclub.com/forum/a-committee- ... 81051.html
6. https://gmatclub.com/forum/two-thirds-o ... 10059.html
7. https://gmatclub.com/forum/a-deck-of-ca ... 21976.html
8. https://gmatclub.com/forum/set-s-consis ... 24713.html
9. https://gmatclub.com/forum/john-throws- ... 43960.html
10. https://gmatclub.com/forum/a-fair-die-w ... 84968.html
Hard:
1. https://gmatclub.com/forum/a-couple-dec ... 68730.html
2. https://gmatclub.com/forum/mary-and-joe ... 86407.html
3. https://gmatclub.com/forum/a-box-contai ... 27699.html
4. https://gmatclub.com/forum/two-dice-are ... 26477.html
5. https://gmatclub.com/forum/the-table-ab ... 05918.html
6. https://gmatclub.com/forum/minimum-of-h ... 73869.html
7. https://gmatclub.com/forum/kate-and-dav ... 97177.html
8. https://gmatclub.com/forum/when-an-unfa ... 98795.html
9. https://gmatclub.com/forum/artificial-i ... 21978.html
10. https://gmatclub.com/forum/a-deck-of-ca ... 21977.html
GMAT Club Math Book
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GMAT Club Math Book
Combinations
https://gmatclub.com/forum/combinations-87345.html
 COMBINATORICS
created by: walker
edited by: bb, Bunuel
Definition
Combinatorics is the branch of mathematics studying the enumeration, combination, and
permutation of sets of elements and the mathematical relations that characterize their
properties.
Enumeration
Enumeration is a method of counting all possible ways to arrange elements. Although it is
the simplest method, it is often the fastest method to solve hard GMAT problems and is a
pivotal principle for any other combinatorial method. In fact, combination and permutation
is shortcuts for enumeration. The main idea of enumeration is writing down all possible
ways and then count them. Let's consider a few examples:
Example #1
Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible
to arrange marbles in a row?
Solution: Let's write out all possible ways:
Answer is 6.
In general, the number of ways to arrange n different objects in a row
Example #2
Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible
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to arrange marbles in a row if blue and green marbles have to be next to each other?
Solution: Let's write out all possible ways to arrange marbles in a row and then find only
arrangements that satisfy question's condition:
Answer is 4.
Example #3
Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible
to arrange marbles in a row if gray marble have to be left to blue marble?
Solution: Let's write out all possible ways to arrange marbles in a row and then find only
arrangements that satisfy question's condition:
Answer is 3.
Arrangements of n different objects
Enumeration is a great way to count a small number of arrangements. But when the total
number of arrangements is large, enumeration can't be very useful, especially taking into
account GMAT time restriction. Fortunately, there are some methods that can speed up
counting of all arrangements.
The number of arrangements of n different objects in a row is a typical problem that can
be solve this way:
1. How many objects we can put at 1st place? n.
2. How many objects we can put at 2nd place? n - 1. We can't put the object that already
placed at 1st place.
.....
n. How nany objects we can put at n-th place? 1. Only one object remains.
Therefore, the total number of arrangements of n different objects in a row is
N = n ∗ ( n − 1) ∗ ( n − 2). . . .2 ∗ 1 = n!
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Combination
A combination is an unordered collection of k objects taken from a set of n distinct
objects. The number of ways how we can choose k objects out of n distinct objects is
denoted as:
knowing how to find the number of arrangements of n distinct objects we can easily find
formula for combination:
1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects
((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter.
Permutation
A permutation is an
ordered collection of k objects taken from a set of n distinct objects.
The number of ways how we can choose k objects out of n distinct objects is denoted as:
knowing how to find the number of arrangements of n distinct objects we can easily find
formula for combination:
1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the
order of remained (n-k) objects doesn't matter.
If we exclude order of chosen objects from permutation formula, we will get combination
formula:
Cn
k
=Cn
k
n!
k!(n−k)!
P n
k
=P n
k
n!
(n−k)!
=
P n
k
k! Cn
k
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Circular arrangements
Let's say we have 6 distinct objects, how many relatively different arrangements do we
have if those objects should be placed in a circle.
The difference between placement in a row and that in a circle is following: if we shift all
object by one position, we will get different arrangement in a row but the same relative
arrangement in a circle. So, for the number of circular arrangements of n objects we
have:
Tips and Tricks
Any problem in Combinatorics is a counting problem. Therefore, a key to solution is a way
how to count the number of arrangements. It sounds obvious but a lot of people begin
approaching to a problem with thoughts like "Should I apply C- or P- formula here?".
Don't fall in this trap: define how you are going to count arrangements first, realize that
your way is right and you don't miss something important, and only then use C- or P-
formula if you need them.
Resources
Walker's post with Combinatorics/probability problems:
Combinatorics/probability
Problems
R = = ( n − 1)!n!
n
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Practice Questions
Easy:
1. https://gmatclub.com/forum/there-are-8- ... 34582.html
2. https://gmatclub.com/forum/there-are-5- ... 88100.html
3. https://gmatclub.com/forum/a-certain-un ... 68638.html
4. https://gmatclub.com/forum/the-diagram- ... 44271.html
5. https://gmatclub.com/forum/in-a-tournam ... 55939.html
6. https://gmatclub.com/forum/how-many-way ... 47677.html
7. https://gmatclub.com/forum/in-how-many- ... 05506.html
8. https://gmatclub.com/forum/a-certain-ba ... 98055.html
9. https://gmatclub.com/forum/how-many-dif ... 54633.html
10. https://gmatclub.com/forum/if-a-person- ... 24843.html
Medium:
1. https://gmatclub.com/forum/the-letters- ... 20320.html
2. https://gmatclub.com/forum/team-a-and-t ... 41461.html
3. https://gmatclub.com/forum/the-subsets- ... 08256.html
4. https://gmatclub.com/forum/each-in-the- ... 95162.html
5. https://gmatclub.com/forum/a-company-th ... 68427.html
6. https://gmatclub.com/forum/clarissa-wil ... 42452.html
7. https://gmatclub.com/forum/in-the-table ... 66572.html
8. https://gmatclub.com/forum/if-a-code-wo ... 26652.html
9. https://gmatclub.com/forum/how-many-two ... 28130.html
10. https://gmatclub.com/forum/john-has-12- ... 07307.html
Hard:
1. https://gmatclub.com/forum/pat-will-wal ... 68374.html
2. https://gmatclub.com/forum/five-integer ... 05923.html
3. https://gmatclub.com/forum/in-how-many- ... 85707.html
4. https://gmatclub.com/forum/how-many-fiv ... 91597.html
5. https://gmatclub.com/forum/nine-family- ... 75071.html
6. https://gmatclub.com/forum/how-many-pos ... 55804.html
7. https://gmatclub.com/forum/how-many-odd ... 94655.html
8. https://gmatclub.com/forum/how-many-dif ... 71774.html
9. https://gmatclub.com/forum/a-committee- ... 04966.html
10. https://gmatclub.com/forum/in-how-many- ... 56400.html


forum/how-many-dif ... 71774.html
9. https://gmatclub.com/forum/a-committee- ... 04966.html
10. https://gmatclub.com/forum/in-how-many- ... 56400.html

Page 1 of 1 All times are UTC - 8 hours [ DST ]
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GMAT Club Math Book
Standard Deviation
https://gmatclub.com/forum/standard-deviation-87905.html
 STANDARD DEVIATION
Definition
Standard Deviation (SD, or STD or ) - a measure of the dispersion or variation in a
distribution, equal to the square root of variance or the arithmetic mean (average) of
squares of deviations from the arithmetic mean.
In simple terms, it shows how much variation there is from the "average" (mean). It may
be thought of as the average difference from the mean of distribution, how far data points
are away from the mean. A low standard deviation indicates that data points tend to be
very close to the mean, whereas high standard deviation indicates that the data are
spread out over a large range of values.
σ
variance =
∑( −xi xav)2
N
σ = =variance− − − − − − −√
∑( −xi xav)2
N
− − − − − − − − −
√
GMAT Club Math Book
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Properties
;
 only if all elements in a set is equal;
Let standard deviation of be and mean of the set be :
Standard deviation of is . Decrease/increase in all elements of a set by a
constant percentage will decrease/increase standard deviation of the set by the same
percentage.
Standard deviation of is . Decrease/increase in all elements of a set by a
constant value DOES NOT decrease/increase standard deviation of the set.
σ ≥ 0
σ = 0
{ }xi σ m
{ }
xi
a =σ
′ σ
a
{ + a}xi = σσ
′
GMAT Club Math Book
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if a new element is added to set and standard deviation of a new set is
, then:
1) if
2) if
3) if
4) is the lowest if
Tips and Tricks
GMAC in majority of problems doesn't ask you to calculate standard deviation. Instead it
tests your intuitive understanding of the concept. In 90% cases it is a faster way to use
just average of instead of true formula for standard deviation, and treat
standard deviation as "
average difference between elements and mean". Therefore,
before trying to calculate standard deviation, maybe you can solve a problem much faster
by using just your intuition.
Advance tip. Not all points contribute equally to standard deviation. Taking into account
that standard deviation uses sum of squares of deviations from mean, the most remote
points will essentially contribute to standard deviation. For example, we have a set A that
has a mean of 5. The point 10 gives in sum of squares but point 6 gives
only . 25 times the difference! So, when you need to find what set has the
largest standard deviation, always look for set with the largest range because remote
points have a very significant contribution to standard deviation.
Examples
Example #1
Q: There is a set . If we create a new set that consists of
all elements of the initial set but decreased by 17%, what is the change in standard
deviation?
Solution: We don't need to calculate as we know rule that decrease in all elements of a
set by a constant percentage will decrease standard deviation of the set by the same
percentage. So, the decrease in standard deviation is 17%.
Example #2
Q: There is a set of consecutive even integers. What is the standard deviation of the set?
(1) There are 39 elements in the set.
(2) the mean of the set is 382.
Solution: Before reading Data Sufficiency statements, what can we say about the
question? What should we know to find standard deviation? "consecutive even integers"
y { }xi {{ }, y}xi
σ
′
> σσ
′
|y − m| > σ
= σσ
′ |y − m| = σ
< σσ
′
|y − m| < σ
σ
′
y = m
| − |xi xav
(10 − 5 = 25)2
(6 − 5 = 1)2
{67, 32, 76, 35, 101, 45, 24, 37}
GMAT Club Math Book
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means that all elements strictly related to each other. If we shift the set by adding or
subtracting any integer, does it change standard deviation (average deviation of elements
from the mean)? No. One thing we should know is the number of elements in the set,
because the more elements we have the broader they are distributed relative to the
mean. Now, look at DS statements, all we need it is just first statement. So, A is
sufficient.
Example #3
Q: Standard deviation of set is 18.3. How many elements
are 1 standard deviation above the mean?
Solution: Let's find mean:
Now, we need to count all numbers greater than 42+18.3=60.3. It is one number - 76.
The answer is 1.
Example #4
Q: There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form
a new set B by adding 2 more elements to the set A. What two elements will decrease the
standard deviation the most?
A) 9 and 3
B) -3 and 3
C) 6 and 1
D) 4 and 5
E) 5 and 5
Solution: The closer to the mean, the greater decrease in standard deviation. D has 4
(equal our mean) and 5 (differs from mean only by 1). All other options have larger
deviation from mean.
Normal distribution
It is a more advance concept that you will never see in GMAT but understanding statistic
properties of standard deviation can help you to be more confident about simple
properties stated above.
In probability theory and statistics, the normal distribution or Gaussian distribution is a
continuous probability distribution that describes data that cluster around a mean or
average. Majority of statistical data can be characterized by normal distribution.
{23, 31, 76, 45, 16, 55, 54, 36}
m = = 4223+31+76+45+16+55+54+36
8
GMAT Club Math Book
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covers 68% of data
 covers 95% of data
 covers 99% of data
m − σ < x < m + σ
m − 2σ < x < m + 2σ
m − 3σ < x < m + 3σ
GMAT Club Math Book
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Practice Questions
Easy:
1. https://gmatclub.com/forum/the-data-set ... 76313.html
2. https://gmatclub.com/forum/which-of-the ... 31485.html
3. https://gmatclub.com/forum/a-certain-ch ... 43982.html
4. https://gmatclub.com/forum/if-d-is-the- ... 93979.html
5. https://gmatclub.com/forum/for-a-certai ... 28661.html
6. https://gmatclub.com/forum/a-certain-li ... 59504.html
7. https://gmatclub.com/forum/the-standard ... 99221.html
8. https://gmatclub.com/forum/the-resident ... 83362.html
9. https://gmatclub.com/forum/the-arithmet ... 29117.html
10. https://gmatclub.com/forum/the-mean-and ... 98248.html
Medium:
1. https://gmatclub.com/forum/a-certain-ch ... 43982.html
2. https://gmatclub.com/forum/set-s-consis ... 51019.html
3. https://gmatclub.com/forum/a-vending-ma ... 93351.html
4. https://gmatclub.com/forum/which-of-the ... 62530.html
5. https://gmatclub.com/forum/set-j-consis ... 06956.html
6. https://gmatclub.com/forum/the-table-ab ... 20477.html
7. https://gmatclub.com/forum/if-a-certain ... 30542.html
8. https://gmatclub.com/forum/a-researcher ... 34893.html
9. https://gmatclub.com/forum/a-certain-li ... 03989.html
Hard:
1. https://gmatclub.com/forum/if-m-is-a-ne ... 26819.html
2. https://gmatclub.com/forum/set-q-consis ... 55781.html
3. https://gmatclub.com/forum/which-of-the ... 00345.html
4. https://gmatclub.com/forum/the-list-sho ... 91966.html
5. https://gmatclub.com/forum/a-researcher ... 59754.html
6. https://gmatclub.com/forum/e-is-a-colle ... 99774.html
7. https://gmatclub.com/forum/set-a-consis ... 10876.html
8. https://gmatclub.com/forum/if-j-k-m-n-a ... 72294.html
9. https://gmatclub.com/forum/which-of-the ... 90680.html
10. https://gmatclub.com/forum/the-list-abo ... 93102.html


forum/if-j-k-m-n-a ... 72294.html
9. https://gmatclub.com/forum/which-of-the ... 90680.html
10. https://gmatclub.com/forum/the-list-abo ... 93102.html

Page 1 of 1 All times are UTC - 8 hours [ DST ]
GMAT Club Math Book
101


MANHATTAN PREP
GMAT Advanced Quant
GMAT STRATEGY GUIDE
This supplemental guide provides in-depth and comprehensive
explanations of the advanced math skills necessary for the highest-level
performance on the GMAT.
GMAT® is a registered trademark of the Graduate Management
Admissions Council™. Manhattan Prep is neither endorsed by nor
aﬀiliated with GMAC.


Table of Contents


GMAT Advanced Quant
Cover
Title Page
Copyright
Instructional Guide Series
Letter
Introduction
In This Chapter...
A Qualified Welcome
Who Should Use This Book
Try It Yourself
The Purpose of This Book
An Illustration
Learning How to Think
Plan of This Book
Solutions to Try-It Problems
Part 1: Problem Solving and Data Sufficiency Strategies
Chapter 1 Problem Solving: Advanced Principles
In This Chapter...
Chapter 1 Problem Solving: Advanced Principles
Principle #1: Understand the Basics
Principle #2: Build a Plan
Principle #3: Solve—and Put Pen to Paper
Principle #4: Review Your Work
Problem Set
Solutions
Chapter 2: Problem Solving: Strategies & Tactics
In This Chapter...
Chapter 2 Problem Solving: Strategies & Tactics
Advanced Strategies
Advanced Guessing Tactics
Problem Set
Solutions
Chapter 3: Data Sufficiency: Principles
In This Chapter...
Chapter 3 Data Sufficiency: Principles
Principle #1: Follow a Consistent Process


Principle #2: Never Rephrase Yes/No as Value
Principle #3: Work from Facts to Question
Principle #4: Be a Contrarian
Principle #5: Assume Nothing
Problem Set
Solutions
Chapter 4: Data Sufficiency: Strategies & Tactics
In This Chapter...
Chapter 4 Data Sufficiency: Strategies & Tactics
Advanced Strategies
Advanced Guessing Tactics
Summary
Common Wrong Answers
Problem Set
Solutions
Part 2: Strategies for All Problem Types
Chapter 5 Pattern Recognition
In This Chapter...
Pattern Recognition Problems
Sequence Problems
Units (Ones) Digit Problems
Remainder Problems
Other Pattern Problems
Problem Set
Solutions
Chapter 6: Common Terms and Quadratic Templates
In This Chapter...
Chapter 6 Common Terms and Quadratic Templates
Common Terms
Quadratic Templates
Quadratic Templates in Disguise
Problem Set
Solutions
Chapter 7: Visual Solutions
In This Chapter...
Chapter 7 Visual Solutions
Representing Objects with Pictures


Rubber Band Geometry
Baseline Calculations for Averages
Number Line Techniques for Statistics Problems
Problem Set
Solutions
Chapter 8: Hybrid Problems
In This Chapter...
Pop Quiz!
Hybrid Problems
Identify and Sequence the Parts
Where to Start
Minor Hybrids
Problem Set
Solutions
Part 3: Practice
Chapter 9 Workout Sets
In This Chapter...
Workout Set 1
Workout Set 1 Answer Key
Workout Set 1 Solutions
Workout Set 2
Workout Set 2 Answer Key
Workout Set 2 Solutions
Workout Set 3
Workout Set 3 Answer Key
Workout Set 3 Solutions
Workout Set 4
Workout Set 4 Answer Key
Workout Set 4 Solutions
Workout Set 5
Workout Set 5 Answer Key
Workout Set 5 Solutions
Workout Set 6
Workout Set 6: Answer Key
Workout Set 6 Solutions
Workout Set 7
Workout Set 7 Answer Key


Workout Set 7 Solutions
Workout Set 8
Workout Set 8 Answer Key
Workout Set 8 Solutions
Workout Set 9
Workout Set 9: Answer Key
Workout Set 9 Solutions
Workout Set 10
Workout Set 10 Answer Key
Workout Set 10 Solutions
Workout Set 11
Workout Set 11 Answer Key
Workout Set 11 Solutions
Workout Set 12
Workout Set 12 Answer Key
Workout Set 12 Solutions
Workout Set 13
Workout Set 13 Answer Key
Workout Set 13 Solutions
Workout Set 14
Workout Set 14 Answer Key
Workout Set 14 Solutions
Workout Set 15
Workout Set 15 Answer Key
Workout Set 15 Solutions
Workout Set 16
Workout Set 16 Answer Key
Workout Set 16: Answers and Explanations
mba Mission
mba Mission
Go Beyond Books. Try A Free Class Now.
Prep Made Personal


Acknowledgements
A great number of people were involved in the creation of the book you are holding.
Our Manhattan Prep resources are based on the continuing experiences of our instructors and
students. The overall vision for this edition was developed by Chelsey Cooley, who determined
what new areas to cover and who wrote all of the problems that are new to this edition.
Chelsey served as the primary author of this edition and Emily Meredith Sledge was the primary
editor; Emily also served as the primary author of the first edition of this guide. Mario Gambino
managed production for the many—and quite complicated—images that appear in this guide.
Matthew Callan coordinated the production work for this guide. Once the manuscript was done,
Naomi Beesen and Ben Ku edited and Cheryl Duckler and Stacey Koprince proofread the entire
guide from start to finish. Carly Schnur designed the covers.
Retail ISBN: 978-1-5062-4993-3
Retail eISBN: 978-1-5062-4994-0
Course ISBN: 978-1-5062-4995-7
Course eISBN: 978-1-5062-4996-4
Copyright © 2020 Manhattan Prep, Inc.
ALL RIGHTS RESERVED. No part of this work may be reproduced or used in any form or by any
means—graphic, electronic, or mechanical, including photocopying, recording, taping, web
distribution—without the prior written permission of the publisher, MG Prep, Inc.
GMAT® is a registered trademark of the Graduate Management Admission Council. Manhattan Prep
is neither endorsed by nor aﬀiliated with GMAC.


GMAT® STRATEGY GUIDES
GMAT All the Quant
GMAT All the Verbal
GMAT Integrated Reasoning and Essay
STRATEGY GUIDE SUPPLEMENTS
Math Verbal
GMAT Foundations of Math
GMAT Advanced Quant
GMAT Foundations of Verbal


January 7, 2020
Dear Student,
Thank you for picking up a copy of Advanced Quant. I hope this book
provides just the guidance you need to get the most out of your GMAT
studies.
At Manhattan Prep, we continually aspire to provide the best instructors
and resources possible. If you have any questions or feedback, please do
not hesitate to contact us.
Email our Student Services team at gmat@manhattanprep.com or give us
a shout at 212-721-7400 (or 800-576-4628 in the United States or Canada).
We try to keep all our books free of errors, but if you think we’ve goofed,
please visit manhattanprep.com/GMAT/errata.
Our Manhattan Prep Strategy Guides are based on the continuing
experiences of both our instructors and our students. The primary author
of the this edition of the Advanced Quant guide was Chelsey Cooley and
the primary editor was Emily Meredith Sledge. Project management and
design were led by Matthew Callan and Mario Gambino. I’d like to send


particular thanks to instructors Stacey Koprince and Ben Ku for their
content contributions.
Finally, we are indebted to all of the Manhattan Prep students who have
given us excellent feedback over the years. This book wouldn’t be half of
what it is without their voice.
And now that you are one of our students too, please chime in! I look
forward to hearing from you. Thanks again and best of luck preparing for
the GMAT!
Sincerely,
Chris Ryan
Executive Director, Product Strategy
www.manhattanprep.com/gmat 138 West 25th Street, 7th Floor, New
York, NY 10001 Tel: 212-721-7400 Fax: 646-514-7425


Introduction


In This Chapter...
A Qualified Welcome
Who Should Use This Book
Try It Yourself
The Purpose of This Book
An Illustration
Learning How to Think
Plan of This Book
Solutions to Try-It Problems


Introduction
A Qualified Welcome
Welcome to GMAT Advanced Quant! In this venue, we decided to be a
little nerdy and call the introduction Chapter 0. A er all, the point (0, 0) in
the coordinate plane is called the origin, isn’t it? (That’s the first and last
math joke in this book.)
Unfortunately, we have to qualify our welcome right away, because this
book isn’t for everyone. At least, it’s not for everyone right away.


Who Should Use This Book
You should use this book if you meet the following conditions:
If you match this description, then please turn the page.
You have achieved a scaled score of at least 47 (out of 51) on the Quant
section of either the Manhattan Prep practice test or the oﬀicial practice
computer-adaptive test (CAT).
You have worked through the Manhattan Prep All the Quant guide,
which covers all of the topics and strategies you need for the Quant
section, or you have worked through similar material from another
company. This material should include the following:
Algebra
Fractions, Decimals, Percents, and Ratios
Geometry
Number Properties
Word Problems
You are already comfortable with the core principles in these topics.
You want to raise your performance to a scaled score of 49 or higher.
You want to become a significantly smarter test-taker.


If you don’t match this description, then you will probably find this book
too diﬀicult at this stage of your preparation.
For now, you are better oﬀ working on topic-focused material, as found in
the All the Quant guide, and ensuring that you have mastered that material
before you return to this book.


Try It Yourself
Throughout the chapters of this guide, you’ll see Try-It problems—
problems designed to test your skills on certain aspects of GMAT problems.
Take a look at the following three Try-It problems, which are very diﬀicult.
They are at least as hard as any real GMAT problem—probably even harder.
Go ahead and give these problems a try. You should not expect to solve any
of them in two minutes. In fact, you might find yourself completely stuck. If
that’s the case, switch gears. Do your best to eliminate some incorrect
answer choices and take an educated guess.
Try-It #0-1
A jar is filled with red, white, and blue tokens that are equivalent
except for their color. The chance of randomly selecting a red
token, replacing it, then randomly selecting a white token is the
same as the chance of randomly selecting a blue token. If the
number of tokens of every color is a multiple of 3, what is the
smallest possible total number of tokens in the jar?
 9(A)
12(B)
15(C)
18(D)


Try-It #0-2
Arrow
 , which is a line segment exactly 5 units long with an
arrowhead at A, is constructed in the xy-plane. The x- and y-
coordinates of A and B are integers that satisfy the inequalities 0 ≤
x ≤ 9 and 0 ≤ y ≤ 9. How many different arrows with these -
properties can be constructed?
Try-It #0-3
In the diagram to the right, the value
of x is closest to which of the
following?
21(E)
 50(A)
168(B)
200(C)
368(D)
536(E)
(A)
2(B)
(C)
(D)
1(E)


(Note: This problem does not require any non-GMAT math, such as
trigonometry.)


The Purpose of This Book
This book is designed to prepare you for the most diﬀicult math problems
on the GMAT.
So…what is a diﬀicult math problem, from the point of view of the GMAT?
A diﬀicult math problem is one that most GMAT test-takers get wrong
under exam conditions. In fact, this is essentially how the GMAT measures
diﬀiculty: by the percent of test-takers who get the problem wrong.
So what kinds of math questions do most test-takers get wrong? What
characterizes these problems? There are two kinds of features:
Topical nuances or obscure principles
These topical nuances are largely covered in the Extra sections of
the Manhattan Prep All the Quant guide. This book includes many
problems that involve topical nuances. However, the exhaustive
theory of divisibility and primes, for instance, is not repeated here.
1.
Connected to a particular topic
Inherently hard to grasp or unfamiliar
Easy to mix up


Complex structures
Complex structures are essentially disguises for simpler content.
These disguises may be diﬀicult to pierce. The path to the answer is
twisted or clouded somehow.
To solve problems that have simple content but complex structures,
you need approaches that are both more general and more
creative. This book focuses on these more general and more
creative approaches.
The three problems on the previous page have complex structures;
the solutions are a bit later in this chapter. In the meantime, take a
look at another problem.
2.
May use simple principles in ways that aren’t obvious
May require multiple steps
May make you consider many cases
May combine more than one topic
May need a flash of real insight to complete
May make you change direction or switch strategies along the
way


An Illustration
Give this problem a whirl. Don’t go on until you have spent a few minutes
on it—or until you have figured it out.
Try-It #0-4
What should the next number in this sequence be?
1 2 9 64 __
Note: This problem is not exactly GMAT-like, because there is no
mathematically definite rule. However, you’ll know when you’ve solved the
problem. The answer will be elegant.
This problem has very simple content but a complex structure.
Researchers in cognitive science have used sequence-completion
problems such as this one to develop realistic models of human thought.
Here is one such model, simplified but practical.
Top-Down Brain and Bottom-Up Brain
To solve the sequence-completion problem above, you need two kinds of
thinking:


You might even say that you need two types of brain.
The top-down brain is your conscious self. If you imagine the contents of
your head as a big corporation, then your top-down brain is the CEO,
responding to input, making decisions, and issuing orders. In cognitive
science, the top-down brain is called the executive function. Top-down
thinking and planning is indispensable to any problem-solving process.
But the corporation in your head is a big place. For one thing, how does
information get to the CEO? And how preprocessed is that information?
The bottom-up brain is your preconscious processor. A er raw sensory
input arrives, your bottom-up brain processes that input extensively before
it reaches your top-down brain.
For instance, to your optic nerve, every word on this page is just a lot of
black squiggles. Your bottom-up brain immediately turns these squiggles
into letters, joins the letters into words, summons relevant images and
concepts, and finally serves these images and concepts to your top-down
brain. This all happens automatically and swi ly. In fact, it takes eﬀort to
interrupt this process. Also, unlike your top-down brain, which does things
one at a time, your bottom-up brain can easily do many things at once.
How does all this relate to solving the sequence problem above?


Each of your brains needs the other one to solve diﬀicult problems.
Your top-down brain needs your bottom-up brain to notice patterns, sniﬀ
out valuable leads, and make quick, intuitive leaps and connections.
But your bottom-up brain is inarticulate and distractible. Only your top-
down brain can build plans, pose explicit questions, follow procedures,
and state findings.
Imagine that you are trying to solve a tough murder case. To find all the
clues in the woods, you need both a savvy detective and a sharp-nosed
bloodhound.
Your top-down brain is the
detective.
Be organized, fast, and flexible to crack
the case.
Your bottom-up brain is the
bloodhound.
To solve diﬀicult GMAT problems, try to harmonize the activity of your two
brains by following an organized, fast, and flexible problem-solving
process.
Organized
You need a general step-by-step approach to guide you. One such
approach, inspired by expert mathematician George Pólya, is Understand,
Plan, Solve (UPS):


You may never have thought explicitly about steps 1 and 2 before. It may
have been easy or even automatic for you to Understand easier problems
and to Plan your approach to them. As a result, you may tend to dive right
into the Solve stage. This is a bad strategy. Mathematicians know that the
real math on hard problems is not Solve; the real math is Understand and
Plan.
Fast
Speed is important for its own sake on the GMAT, of course. What you may
not have thought as much about is that being fast can also lower your
stress level and promote good process. If you know you can solve quickly,
then you can take more time to comprehend the question, consider the
given information, and select a strategy. To this end, make sure that you
can complete calculations and manipulations fairly rapidly so that you can
aﬀord to spend some time on the Understand and Plan stages of your
problem-solving process. A little extra time invested up front can pay oﬀ
handsomely later.
Flexible
To succeed against diﬀicult problems, you sometimes have to “unstick”
yourself. Expect to run into brick walls and encounter dead ends. Returning
to first principles and to the general process (e.g., making sure that you
fully Understand the problem) can help you back up out of the mud.
Let’s return to the sequence problem and play out a sample interaction
between the two brains. The path is not linear; there are several dead ends,
as you would expect. This dialog will lead to the answer, so don’t start
reading until you’ve given the problem a final shot (if you haven’t already
Understand the problem first.1.
Plan your attack by adapting known techniques in new ways.2.
Solve by executing your plan.3.


solved it). The top-down brain is labeled TD; the bottom-up brain is labeled
BU.
1 2 9 64 __ TD: “Okay, let’s Understand this thing. At a glance, they’ve
given me an increasing list of numbers, and they want me to
find the number that “should” go in the blank, whatever
“should” means. What’s a good Plan? Hmm. No idea. Stare at
the numbers given?”
BU notices that 9 = 32 and 64 = 82. Likes the two squares.
1 2 32 82 __ TD: “Write in the two squares.”
1 2 32 82 __ BU notices that 1 is a square, too.
sq no sq sq __ TD: “Are they all perfect squares? No, since 2 isn’t.”
BU doesn’t like this break in the pattern.
1 2 32 26 — TD: “Wait, back up. What about primes, factoring all the way. 8
= 23, and so 82 = (23)2 = 26.”
BU notices 6 = 2 × 3, but so what?
1 2 32 (23)2 __ TD: “Let’s write 26 as (23)2. Anything there?”
BU notices lots of 2’s and 3’s, but so what?
TD: “Okay, keep looking at this. Are the 2’s and 3’s stacked
somehow?”


BU notices no real pattern. There’s 2−3−2 twice as you go
across, but so what? And the 1 is weird by itself.
1 2 9 64 __ TD: “No good leads there. Hmm...time to go back to the
original and try taking diﬀerences.”
BU notices no pattern. The numbers look even uglier.
1 2 9 64 __ TD: “Hmm. No good. Go back to original numbers again. What’s
going on there?”
BU notices that the numbers are growing quickly, like squares
or exponentials.
12 2 32 82 __ TD: “Must have something to do with those squares. I should
look at those again.”
BU notices a gap on the le , among the powers.
1? 21 32 82 __ TD: “How about looking at 2. Write it with exponents: 2 = 21. -
Actually, 1 doesn’t have to be 12. One can be to any power and
still be 1. The power is a question mark.”
BU notices 21 then 32. Likes the counting numbers.
BU really wants 1, 2, 3, 4 somehow.
1? 21 32 4?? __ TD: “Try 4 in that last position. Could the last term be 4
somehow?”
BU likes the look of this. 8 and 4 are related.
1? 21 32 43 __ TD: “64 is 4 to the what? 42 = 16, times another 4 equals 64, so
1 7 55


it’s 4 to the third power. That fits.”
BU is thrilled: 1, 2, 3, 4 below and 1, 2, 3 up top.
10 21 32 43 __ TD: “Extend le . It’s 10. Confirmed. The bases are 1, 2, 3, 4, etc.,
and the powers are 0, 1, 2, 3, etc.”
BU is content.
10 21 32 43 54 TD: “So the answer is 54, which is 252, or 625.”
Your own process was almost certainly diﬀerent in the details. Also, your
internal dialog was very rapid—parts of it probably only took fractions of a
second to transpire. A er all, you think at the speed of thought.
The important thing is to recognize how the bottom-up bloodhound and
the top-down detective worked together in the case above. The TD
detective set the overall agenda and then pointed the BU bloodhound at
the clues. The bloodhound did practically all the “noticing,” which in some
sense is where all the magic happened. But sometimes the bloodhound
got stuck, so the detective had to intervene, consciously trying a new path.
For instance, 64 reads so strongly as 82 that the detective had to actively
give up on that reading.
There are so many possible meaningful sequences that it wouldn’t have
made sense to apply a strict recipe from the outset: “Try X first, then Y, then
Z…” Such an algorithm would require hundreds of possibilities. Should
you always look for 1, 2, 3, 4? Should you never find diﬀerences or prime


factors because they weren’t that useful here? Of course not! A computer
can rapidly and easily apply a complicated algorithm with hundreds of
steps, but humans can’t. (If you are an engineer or programmer, maybe
you wish you could program your own brain, but so far, that’s not
possible!)
What humans are good at, though, is noticing patterns. The bottom-up
brain is extremely powerful—far more powerful than any computer yet
built.
As you gather problem-solving tools, the task becomes knowing when to
apply which tool. This task becomes harder as problem structures become
more complex. But if you deploy your bottom-up bloodhound according to
a general problem-solving process such as Understand, Plan, Solve, then
you can count on the bloodhound to notice the relevant aspects of the
problem—the aspects that tell you which tool to use.
You can break down Understand, Plan, Solve into several discrete steps:
Understand Glance at the problem briefly:
does anything stand out?
Read the problem.
Jot down any obvious formulas or
numbers.
Plan Reflect on what you were given:
what clues might help tell you
how to approach this problem?


Organize your approach:
choose a solution path.
Solve Work the problem!
You’ll get lots of practice using the UPS process throughout this guide.


Learning How to Think
This book is intended to make you smarter.
It is also intended to make you scrappier.
That description encompasses two main ideas: employing GMAT strategies
as well as textbook solution methods and knowing when to let go.
If you have traditionally been good at paper-based standardized tests, then
you may be used to solving practically every problem the “textbook” way.
Problems that forced you to get down and dirty—to work backwards from
the choices, to estimate and eliminate—may have annoyed you.
A major purpose of this book is to help you learn to choose the best GMAT
approach. On the hardest Quant problems, the textbook approach is o en
not the best GMAT approach.
Unfortunately, advanced test-takers are sometimes very stubborn.
Sometimes they feel they should solve a problem according to some
theoretical approach. Or they fail to move to Plan B or C rapidly enough, so
they don’t have enough time le to execute that plan. In the end, they
might wind up guessing purely at random—and that’s a shame.


GMAT problems o en have back doors—ways to solve that don’t involve
crazy computation or genius-level insights. Remember that in theory,
GMAT problems can all be solved in two minutes. By searching for the back
door, you might avoid all the bear traps that the problem writer set out by
the front door!
In addition to learning alternative solution methods, you also need to learn
when to let go. As you know, the GMAT is an adaptive test. If you keep
getting questions correct, the test will keep getting harder…and harder…
and harder…
At some point, there will appear a monster problem, one that announces “I
must break you.” In your battle with this problem, you could actually lose
the bigger war—even if you ultimately conquer this particular problem.
Maybe it takes you eight minutes, or it beats you up so badly that your
head starts pounding. This will take its toll on your score.
This will happen to everyone, no matter how good you are at the GMAT.
Why?
The GMAT is not an academic test, though it certainly appears to be.
Business schools are primarily interested in whether you’re going to be an
eﬀective businessperson. Good businesspeople are able to assess a
situation rapidly, manage scarce resources, distinguish between good
opportunities and bad ones, and make decisions accordingly.
The GMAT wants to put you in a situation where the best decision is, in fact,
to guess and move on, because business schools are interested in learning


whether you have the presence of mind to recognize a bad opportunity
and the discipline to let it go.
Show the GMAT that you know how to manage your scarce resources (time
and mental energy) and that you can recognize and cut oﬀ a bad
opportunity.


Plan of This Book
The rest of this book has three parts:
Part One: Problem Solving and Data Suﬀiciency
Strategies
Chapter 1: Problem Solving: Advanced
Principles
Chapter 2: Problem Solving: Strategies &
Tactics
Chapter 3: Data Suﬀiciency: Principles
Chapter 4: Data Suﬀiciency: Strategies &
Tactics
Part Two: Strategies for All Problem Types Chapter 5: Pattern Recognition
Chapter 6: Common Terms & Quadratic
Templates
Chapter 7: Visual Solutions
Chapter 8: Hybrid Problems
Part Three: Practice Workouts 1−16: Sixteen sets of 10
problems each
The four chapters in Part I focus on principles, strategies, and tactics
related to the two types of GMAT math problems: Problem Solving (PS) and
Data Suﬀiciency (DS). The next four chapters, in Part II, focus on techniques


that apply across several topics but are more specific than the approaches
in Part I.
Each of the eight chapters in Part I and Part II contains the following:
Many of these problems will be GMAT-like in format, but many will not.
Part III contains sets of GMAT-like Workout problems, designed to exercise
your skills as if you were taking the GMAT and seeing its hardest problems.
Several of these sets contain clusters of problems relating to the chapters
in Parts I and II, although the problems within each set do not all resemble
each other in obvious ways. Other Workout problem sets are mixed by both
approach and topic.
Note that these problems are not arranged in order of diﬀiculty. Also, you
should know that some of these problems draw on advanced content
covered in the Manhattan Prep All the Quant guide.
Try-It Problems embedded throughout the text
Problem Sets at the end of the chapter


Solutions to Try-It Problems
If you haven’t tried to solve the first three Try-It problems in the Try It
Yourself section at the beginning of this chapter, then go back and try them
now. Think about how to get your top-down brain and your bottom-up
brain to work together like a detective and a bloodhound. Come back
when you’ve tackled the problems, even if you don’t get to an answer (in
this case, do make a guess).
In these solutions, we’ll outline sample dialogs between the top-down
detective and the bottom-up bloodhound.
Try-It #0-1
A jar is filled with red, white, and blue tokens that are equivalent
except for their color. The chance of randomly selecting a red
token, replacing it, then randomly selecting a white token is the
same as the chance of randomly selecting a blue token. If the
number of tokens of every color is a multiple of 3, what is the
smallest possible total number of tokens in the jar?
 9(A)
12(B)
15(C)
18(D)


SOLUTION TO TRY-IT #0-1
… jar is filled with red, white, and blue tokens …
chance of randomly selecting …
TD: “I need to Understand this
problem first. There’s a jar, and it’s
got red, white, and blue tokens in it.”
BU notices “chance” and
“randomly.” That’s probability.
TD: “All right, this is a probability
problem. Now, what’s the situation?”
BU notices that there are two
situations.
… chance of randomly selecting a red token, replacing
it, then randomly selecting a white token is the same
as the chance of randomly selecting a blue token.
TD: “Let’s rephrase. In simpler
words, if I pick a red, then a white,
that’s the same chance as if I pick a
blue. Jot that down. Okay, what
else?”
… number of tokens of every color is a multiple of 3 … BU doesn’t want to deal with this
“multiple of 3” thing yet.
… smallest possible total number of tokens in the jar? TD: “Okay, what are they asking
me?”
BU notices “smallest possible total
number.” Glances at answer choices.
They’re small, but not tiny. Hmm.
21(E)


TD: “Let’s Reflect for a moment to
figure out a Plan. How can I
approach this? How about algebra—
if I name the number of each color,
then I can represent each fact and
also what I’m looking for. Okay, I use
R, W, and B. Make probability
fractions. Multiply red and white
fractions. Simplify algebraically.”
BU is now unsure. No obvious path
forward.
The chance of randomly selecting a red token,
replacing it, then randomly selecting a white token is
the same as the chance of randomly selecting a blue
token …
TD: “Let’s start over conceptually.
Reread the problem. Can I learn
anything interesting?”
BU notices that blues are diﬀerent.
TD: “How are blues diﬀerent? Hmm.
Picking a red, then a white is as likely
as picking a blue. What does that
mean?”
BU notices that it’s unlikely to pick a
blue. So there aren’t many blues
compared to reds or whites.
Fewer blues than reds or whites
B < R and B < W
TD: “Are there fewer blues? Yes.
Justify this. Focus on the algebraic
setup.”


In the very first equation above, each fraction on the
le is less than 1, so their product is even smaller.
The denominators of the three fractions are all the
same.
So the numerator of the product (B) must be smaller
than either of the other numerators
(R and W ).
BU notices fractions less than 1. All
positive.
TD: “Two positive fractions less than
1 multiplied together give an even
smaller number.”
TD: “Yes, there are fewer blues.”
BU is quiet.
If the number of tokens of every color is a multiple of 3,
what is the smallest possible total number of tokens in
the jar?
Neither R nor W can equal 3 (since B is smaller than
either).
Let R = W = 6.
(A) and (B) are out now. The smallest possible total is
now 15.
TD: “Time to go back and reread the
rest of the problem.”
BU again notices “multiple of 3,” also
in answer choices. Small multiples.
TD: “Change of Plan: Algebra by itself
isn’t getting me there. What about
plugging in a number? Try the most
constrained variable: B. Since it’s the
smallest quantity, but still positive,
pretend B is 3. Execute this
algebraically. Divide by RW.”
BU likes having only two variables.
TD: “Need to test other numbers.
Apply constraints I know—B is the
smallest number. Rule out answer
choices as I go.”
TD: “6 and 6 don’t work, because the
right side adds up to larger than 1.


Let R = 6 and W = 9.
 (C) is out too. Try the next
possibility.”
BU doesn’t like breaking the
symmetry between R and W. They
seem to be alike.
TD: “Does it matter whether R = 6
and W = 9 or the other way around?
No, it doesn’t. One is 6, the other is 9.
Plug in and go.”
TD: “This works. The answer is 3 + 6
+ 9 = 18.”
The correct answer is (D). Let’s look at another pathway—one that moves
more quickly to the back door.
ALTERNATIVE SOLUTION TO TRY-IT #0-1
… chance of randomly selecting … BU notices “chance.” BU doesn’t like probability.
TD: “Oh man, probability. Okay, let’s make sense
of this and see whether there are any back doors.
That’s the Plan.”
… the number of tokens of every color is
a multiple of 3 …
BU notices that there are only limited possibilities
for each number.
TD: “Okay, every quantity is a multiple of 3. That
simplifies things. There are 3, 6, 9, etc., of each
color.”


A jar is filled with red, white, and blue
tokens …
BU is alert—what about 0?
TD: “What about 0? Hmm…the wording at the
beginning assumes that there actually are tokens
of each color. So there can’t be 0 tokens of any
kind.”
(A) 9 (B) 12 (C) 15 (D) 18 (E) 21 TD: “Now let’s look at the answer choices.”
BU notices that they’re small.
(A) 9 TD: “Try plugging in the choices. Let’s start at the
easy end—in this case, the smallest number.”
BU notices 9 = 3 + 3 + 3.
Select a red:

Select a white:

, which is not "select a
blue"
TD: “The only possible way to have 9 total tokens
is to have 3 reds, 3 whites, and 3 blues. So…does
that work? Plug into probability formula.”
(A) 9 (B) 12 (C) 15 (D) 18 (E) 21
(B) 12
Select a red:
TD: “No, that doesn’t work. This is good. Knock out
(A). Let’s keep going. Try (B).”
BU notices 12 = 3 + 3 + 6.
TD: “Only way to have 12 total is 3, 3, and 6. Which
one’s which? Picking a red and then a white is the
same as picking a blue, so the blue should be one
of the 3’s. Let’s say red is 3 and white is 6.”


Select a white:
  , which is not “select a
blue”
(A) 9 (B) 12 (C) 15 (D) 18 (E) 21
(C) 15
TD: “That doesn’t work either. Knock out (B). Keep
going.”
BU notices 15 has a few options.
Select a red:

Select a white:
, which is not "select a
blue"
,
which is not "select a blue"
(A) 9 (B) 12 (C) 15 (D) 18 (E) 21
(D) 18
, which IS "select a blue"
TD: “I can make 15 by 3, 6, and 6 or by 3, 3, and 9.
Try 3−6−6; make blue the 3.”
TD: “Nope. What about 3−3−9.”
TD: “Not this one either.”
TD: “Knock out (C). Try (D).”
TD: “Maybe 3–6–9 first. Make blue the 3.”
TD: “That’s it! Answer’s (D).”


Many people find this second approach less stressful and more eﬀicient
than the textbook approach. In fact, there is no way to find the correct
answer by pure algebra. Ultimately, you have to test suitable numbers.
Try-It #0-2
Arrow
 , which is a line segment exactly 5 units long with an
arrowhead at A, is constructed in the xy-plane. The x- and y-
coordinates of A and B are integers that satisfy the inequalities 0 ≤
x ≤ 9 and 0 ≤ y ≤ 9. How many different arrows with these
properties can be constructed?
SOLUTION TO TRY-IT #0-2
BU notices “xy-plane.”
TD: “Let’s Understand first. This is a coordinate-
plane problem. Sketch it out. Put in boundaries
as necessary.”
BU wonders where this is going.
How many diﬀerent arrows with these TD: “What is the question asking for again?
 50(A)
168(B)
200(C)
368(D)
536(E)


properties can be constructed? Reread the question.”
BU wonders which properties.
… exactly 5 units long with an arrowhead
at A … the x- and y-coordinates of A and B
are to be integers that satisfy the
inequalities 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9.
TD: “What are the properties of the arrows
supposed to be again? Each arrow is 5 units
long.”
BU notices “integers” and “coordinates” and
pictures a pegboard.
TD: “Reflect. The tip and the end of the arrow
have to touch holes in the pegboard exactly.
Okay. The Plan is to start counting. How to
Organize?”
BU imagines many possible arrows. Brute force
can’t be the right way forward. The arrows can
point in all sorts of diﬀerent ways.
TD: “Let’s simplify the Plan. Let’s focus on just
one orientation of arrows—pointing straight up.
Draw this situation. How many places can the
arrow be?”
BU wants to go up & down, then right & le .
In one column, there are 5 positions for the
arrowhead: y = 5, 6, 7, 8, or 9. That’s the
same as 9 − 5 + 1, by the way.
TD: “Count the positions in one column, then
multiply by the number of columns. Be careful
to count endpoints.”


There are 10 identical columns: x = 0
through x = 9. 5 × 10 = 50 possible positions
for the arrow pointing straight up.
50 × 2 = 100 possible positions for the
arrow if it points straight up or to the right.
TD: “Great. I’ve Solved one part. Other
possibilities?”
BU notices the square is the same vertically as
horizontally. Go right.
TD: “I get the same result for arrows pointing
right. 50 more positions. Is that it? Am I done?”
50 × 4 = 200 possible positions
BU wonders about “down” and “le .”
TD: “These arrows can point straight down or
straight le , too. Those would have the same
result. So there are 50 positions in each of the
four directions. Calculate at this point and
evaluate answers. Eliminate (A) and (B).”
Answer seems to be (C). BU is suspicious: somehow too easy.
TD: “Tentative answer is (C), but I'm not done.”
BU wonders about still other ways for the arrows
to point.
Three up, four across:
 TD: “Could the arrows be at an angle?”
BU notices that the arrow is 5 units long,
associated with 3−4−5 triangles.


TD: “3−4−5 triangles. Yes. Put the arrow as the
hypotenuse of a 3−4−5 triangle. How can this be
done? Try to place the arrow. Remember the
reversal. Looks like there are four ways if I go 3
up and 4 across: up right, up le , down right,
down le .”
BU is happy. This is the trick.
Four up, three across:
 TD: “Likewise, there must be four ways if I go 4
up and 3 across: again, up right, up le , down
right, and down le . By the way, the answer
must be (D) or (E).”
Three up, four across, pointing up to the
right:
There are 7 positions vertically for the
arrowhead (9 − 3 + 1) and 6 positions
horizontally (9 − 4 + 1), for a total of 7 × 6 =
42 positions.
8 × 42 = 336 possible positions at an angle.
In total, there are 200 + 336 = 536 positions.
TD: “Now count just one of these ways. Same
ideas as before. Be sure to include endpoints.”
BU notices the symmetry. The 3 up, 4 across is
the same as the 4 up, 3 across, if you turn the
square.
TD: “Each of these angled ways will be the same.
There are 8 ways to point the arrow at an angle.
Finish the calculation and confirm the answer.”
The correct answer is (E). There isn’t much of an alternative to the
approach above. With counting problems, it can o en be very diﬀicult to
estimate the answer or work backwards from the answer choices.


Try-It #0-3
In the diagram to the right, the value of x is closest to which of the
following?
SOLUTION TO TRY-IT #0-3
TD: “Okay, let’s Understand this. Redraw the
figure. The problem wants the value of x. Now…
how about a Plan?”
BU notices this is an isosceles triangle, because
there are two sides labeled x. How about the
two equal angles?
180° − 45° = 135°.
Divide 135° equally across the two missing
angles. So each angle is 67.5°.
TD: “Figure out the two missing angles. Use the
180° rule.”
BU doesn’t recognize this triangle.
(A)
2(B)
(C)
(D)
1(E)


TD: “Hmm...here’s a Plan: add a perpendicular
line to make right triangles. Drop the line from
the top point. I'll label corners while I'm at it.
Now fill in angles.”
BU notices 45−45−90 and is happy.
TD: “Use the 45−45−90 to write expressions for
its sides. Then
 can be split up into two
pieces, and I can set up the Pythagorean
theorem.”
BU feels that this process is kind of ugly.
TD: “Let’s push through. Write the Pythagorean
theorem for the small triangle on the le , using
the
 as the hypotenuse.”
BU thinks this equation is really ugly.
TD: “Push through. Expand the quadratic and
simplify.”
BU doesn’t like the square root on the bottom.
TD: “Multiply by
 to get rid of it on the -
bottom of the fraction.”


BU has no idea how to take the square root of
this.
TD: “Neither do I. Let’s try estimating. If x2 is
about 3.5, then the square root must be a bit
less than 2 (since the square root of 4 is 2). 182 is
324 and 192 is 361, so the answer is around 1.8
or 1.9.”
TD: “Answer (A) is about 3.5; that matches the
squared value, not the square root. The answer
needs to be less than 2, so (B) is also wrong.
is about 1.7. Answers (D) and (E) are too small,
so the answer is (C).”
The correct answer is (C).
The method you just saw is algebraically intensive, and so your bottom-up
bloodhound might have kicked up a fuss along the way. Sometimes, your
top-down brain needs to ignore the bottom-up brain. Remember, when
you’re actually taking the GMAT, you have to solve problems quickly—and
you don’t need to publish your solutions in a mathematics journal. What
you want is to get the correct answer as quickly and as easily as possible. In
this regard, the solution above works perfectly well.
2 +
(A)
2(B)
(C)
(D)
1(E)