GMAT Quantitative Problems: Defining the Negative Space


GMAT blog, GMAT probability questions, GMAT combinations questions, GMAT quantitative sectionTake a look at the picture with this blog. It’s an iconic optical illusion. Stare at it—what do you see?  The picture is called the Great Wave off Kanagawa, painted by Katsushika Hokusai, a Japanese artist famed for his brilliant compositions. This drawing is of a wave, of course, but do you see the other wave, the reverse wave in the sky?

This image utilizes negative space. You take the whole frame, the great big rectangle, you block out that actual image—and what remains is, in its own right, an interesting picture.

You’ve seen this on the GMAT, of course. Images like this occur frequently on the Quantitative section:

GMAT blog, GMAT geometry problem

 

 

 

To find the area of the shaded region, we need to subtract the area of the smaller inner circle from the large outer circle—the difference is the area of the ring.

But the concept extends beyond simple pictures and geometry questions. Probability problems sometimes operate on a similar principle, subtracting an easy-to-find probability of failure from 1, the total of all probabilities, the “whole frame,” as it were. Once you’ve subtracted all the failures, then whatever remains, the  “negative space,” must be the chances of success!

The GMAT is a test of critical thinking. It tests your ability to find the most effective path to the solution. Sometimes, you’ll pick numbers, sometimes you’ll do the math directly, sometimes you’ll guess strategically. And sometimes, you’ll define the negative space around the answer, and solve that way. Today’s problem of the day is best solved by that principle. You can’t figure out how many arrangements follow the rule in the question stem. But you can find out how many arrangements don’t follow the rule, and subtract it from the total. Good luck!

 

Question

Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in

a single row of six chairs. If Betsy cannot sit next to Emily, how many different

arrangements of the six children are possible?

(A) 240

(B) 480

(C) 540

(D) 720

(E) 840

 

Solution:

Step 1: Analyze the Question

We have to arrange six children in six chairs, but two of

the children can’t sit together. We’re asked to calculate the

number of different arrangements of children.

 

Step 2: State the Task

We’ll calculate the number of possible arrangements of the

children. Then, we’ll subtract the number of ways Betsy can

sit next to Emily.

 

Step 3: Approach Strategically

The possible number of arrangements of six elements is

6! = 6 x 5 x 4 x 3 x 2 x 1= 720.

 

Now we’ll have to calculate the number of ways that would

violate the question stem by putting Betsy next to Emily.

If we number our seats from left to right, there are 5 ways

they can sit together if Betsy is on the left and Emily is on

the right:

 

Seats 1 & 2

Seats 2 & 3

Seats 3 & 4

Seats 4 & 5

Seats 5 & 6

 

And there are 5 more ways if Emily is on the left and Betsy

is on the right, for a total of 10. Now, for any one of those

10 ways, the four remaining children can be seated in 4!

ways: 4! = 4 x 3 x 2 x 1 = 24. So we need to subtract

24 x 10 = 240 ways that have Betsy and Emily sitting

together from our original total of 720: 720 – 240 = 480.

 

Answer choice (B).

 



Tough GMAT Probability Questions


GMAT blog, flipping a coin, probability of heads or tailsTackling some of the tougher GMAT probability questions efficiently relies on both steady practice and your ability to make two key decisions well. First, you will need to quickly and accurately assess the total number of possible outcomes (the denominator of your probability equation). Second, within a multitude of possible approaches, you will need to determine the most efficient route to calculate the number of desired outcomes (the numerator of your probability equation).

With the clock ticking away on your GMAT CAT, figuring out the total number of possibilities can be time-consuming and fraught with room for error. For instance, if a question asks about the probability of getting at least 2 heads on 5 coin tosses, you could sit there all day writing out possibilities:

HHTTT

HTHTT

HTTHT

So forth and so on. I know I got dizzy with the possibilities just writing those three out.  There is a better and more efficient way. For every coin that you toss there are 2 possibilities. You can think of the total possibilities like a permutation problem.

__2__    __2__   __2__   __2__  __2_

1st             2nd          3rd         4th         5th

Just like in a GMAT permutations question when we are trying to determine the total number of codes possible or 4-digit numbers, we would multiply these individual probabilities together. Therefore, there are 2x2x2x2x2 = 2^5 = 32 total possibilities.

Next, we need to look at the numerator  (desired outcomes). We want to find the all of the possibilities that have at least 2 heads, which means that we could have 2 heads, 3 heads, 4 heads, or 5 heads.  To do so, we would need to count all of the different ways that these possibilities could be arranged. Again, we find ourselves in a situation that will be time-consuming and fraught with error. Instead of going down this path, remember that the sum of the probabilities of a complete set of mutually exclusive possible outcomes is 1.  Thus, as is often the case on “at least” probability questions, we can look for those options that are restricted. Then we only have to count the options that have 1 or 0 heads.

TTTTT

HTTTT

THTTT

TTHTT

TTTHT

TTTTH

There are only 6 of those, instead of the 26 possibilities the other way.

Finally, we can either subtract 6/32 from 1 in order to remove all of the restricted possibilities from 1 or we can subtract 6 from 32 and use the result as the desired possibilities. Either way, the answer is 26/32, which you can reduce down to 13/16.

Let’s look at another to make sure we have this down.

Question

A test has 5 multiple-choice questions. Each question has 4 answer options (A,B,C,D). What is the probability that a student will choose “B” for at least four questions if she leaves no questions blank?  Pause a moment and try it for yourself first.

Step 1: Total number of possibilities

There are 5 questions and each has 4 possibilities, so our total possibilities would be 4x4x4x4x4 = 4^5 = 1024

Step 2: Approach Desired Strategically

Here there are far more possibilities for 0, 1, 2, or 3 “B’s,” so let’s get a total for 4 or 5 “B’s”.

All B’s — B,B,B,B,B

Four B’s — A,B,B,B,B – B,A,B,B,B – B,B,A,B,B – B,B,B,A,B — B,B,B,B,A

C,B,B,B,B – B,C,B,B,B – B,B,C,B,B – B,B,B,C,B – B,B,B,B,C

D,B,B,B,B – B,D,B,B,B – B,B,D,B,B – B,B,B,D,B – B,B,B,B,D

3 x 5 = 15 because we can repeat the same pattern for each letter other than B

We can also calculate the total possibilities of 4 B’s by calculating the possibilities for each “no-B” position.

No-B,B, B, B, B = 3x1x1x1x1 = 3

B, No-B, B, B, B = 1x3x1x1x1 = 3

B, B, No-B, B, B = 1x1x3x1x1 = 3

B, B, B, No-B, B = 1x1x1x3x1 = 3

B, B, B, B, No-B = 1x1x1x1x3 =3

A total of 15 possibilities with 4 B’s in the mix

 

That gives a total of 16 different ways that a student can choose at least 4 B’s here.

16/1024 = 1/64 as our final probability.

 

Keep these two decisions in mind each time that you approach a tough probability question on the GMAT quantitive section. You don’t have to write out all of the possible outcomes in order to tackle these on test day!

The language of probability can take a while to learn, especially if you are unfamiliar or out of practice with it to start.  Post your questions below, and we can help you get on track.



GMAT Quantitative Section: Probability Translation


GMAT blog, GMAT prep, GMAT quantitative section, GMAT practice problem, GMAT probabilityTranslating word problems into algebra is a staple skill of GMAT test-takers, one that underlies countless problems in practice and on Test Day. But some challenging translations occur as part of probability and combinatorics problems. That’s because a pair of the most basic words in the English language, “And” and “Or,” suddenly become overburdened with mathematical significance.

“And” is the simpler of the two. When “And” represents independent choices—cases in which one option or arrangement has no impact on the other choice—just multiply the outcomes. For instance:

“The number of ways to purchase three board games and two video games” is an independent choice. The board games we pick have no impact on the video games we pick. So, to translate: [The number of ways to purchase three board games] × [the number of ways to select two video games]. Of course, we’d need the combination formula to find actual values—but we’d know what to do with those values once we got them.

 “Or” is a little more complicated. It’s confusing even in conversation, after all—if I say that you can have cake or ice cream for dessert, can you have both if you want? When you CAN have both, you can treat the problem similarly to an overlapping sets problem. But in most cases on the GMAT, the “Or”s will be mutually exclusive—for instance, if you want to know the odds of drawing a heart or a diamond out of a deck of cards, there is no card that is both a heart or diamond.

A mutually exclusive OR can be translated as a “plus.” That’s all you have to do. So:

“The probability of drawing a heart or a diamond from a deck of cards,” which is the odds of one of two mutually exclusive events occurring, translates to: [The probability of drawing a heart] + [The probability of drawing a diamond].

Today’s problem of the day hinges on those same ideas. Read carefully—you’re solving for the odds of one of two outcomes (an OR), but each of those two outcomes is the specific result of two independent events (an AND). Be systematic in your translation, and I’m sure you’ll get the right result.

Post your answers below before you read the solution, and we can go over them…

 

Question:

Each person in Room A is a student, and  1/6 of the students in Room A are

seniors. Each person in Room B is a student, and  5/7 of the students in Room

B are seniors. If 1 student is chosen at random from Room A and 1 student is

chosen at random from Room B, what is the probability that exactly 1 of the

students chosen is a senior?

(A) 5/42

(B) 37/84

(C) 9/14

(D) 16/21

(E) 37/42

 

 

Solution:

Step 1: Analyze the Question

This is a complex question, but it can be broken down into

simple steps. As with any probability question, we must first

consider all of the scenarios in which the desired outcome

can be true. In this question, there are two different ways

in which exactly one of two students chosen is a senior.

Either (i) a senior is chosen from Room A and a non-senior

is chosen from Room B or (ii) a non-senior is chosen from

Room A and a senior is chosen from Room B.

 

Step 2: State the Task

Determine the probabilities of the two scenarios above and

add them together.

 

Step 3: Approach Strategically

Let’s start with (i) and find the probability that a senior is

chosen from Room A and a nonsenior is chosen from Room B.

The probability that the student chosen from Room A is a

senior is 1/6 .

 

The probability that the student chosen from Room B is not

a senior is 1- 5/7=2/7

 

So the probability that the student chosen from Room A

is a senior and the student chosen from Room B is not a

senior is (1/6) x (2/7) = 2/42 .

 

Let’s not simplify this yet, because we can expect that the

probability we will find when working with (ii) will also

have a denominator of 42.

 

Now let’s work with (ii). Let’s find the probability that a

nonsenior is chosen from Room A and a senior is chosen

from Room B.

 

The probability that the student chosen from Room A is not

a senior is 1 – 1/6 = 5/6 .

 

The probability that the student chosen from Room B is a

senior is 5/7 .

 

So the probability that the student chosen from Room A

is a not a senior and the student chosen from Room B is a

senior is (5/6) x (5/7) = 25/42 .

 

Now we sum the total desired outcomes. The probability

that exactly one of the students chosen is a senior

is (2/42) + (25/42) = 27/42 = 9/14 .

 

(C) is correct.



Three GMAT Challenges


GMAT blog, GMAT study plan, GMAT goal, MBA applicationsPiecing together the time to study for the GMAT can be challenging.  In today’s blog, I’m going to talk about three students (whose names I’m changing to protect their identities).  Each had a major obstacle to studying, and each overcame it in a different way. I hope these students’ examples can help some of you reach your GMAT and MBA goals.

 

Case Study 1: Vincent, the Entrepreneur

The Challenge: Vincent was a busy man when I was tutoring him. His schedule was very flexible—his main source of income was a business that he started and ran himself—but he was distracted at all hours by emails and phone calls related to his work.

The Solution: Vincent needed a time and place where he could study in peace.

Because of his flexible work schedule, it was easier for Vincent to find time than it is for some other students. He dedicated a daily block of time to studying, and had the discipline to stick with it—though as his tutor, I was standing by ready to make sure he stuck with it if he got distracted! 

Vincent had a harder time finding the space he needed to study. Local coffee shops were noisy, and didn’t have reliable internet connections for his CATs. But fortunately, there was a quiet study space regularly available in the local public library. Not only did that let him work in peace, but it also forced him to turn off his cell phone and disconnect from the world.

Ultimately, Vincent got a 700—though he didn’t quite reach his goal, he significantly improved his score, posting a result that combined with his entrepreneurial experience to make a top-tier-worthy application.

 

Case Study 2: Brandon, the Financier

The Challenge: Brandon had a lot of things to cope with. He was a long time out of college, so his writing and grammar skills were rusty (especially since he was a non-native, though fluent, speaker of English). Moreover, though he worked with numbers quite a bit in his job at a bank,the GMAT quantitative section proved challenging since he seldom had to do algebra, let alone geometry or probability. 

Brandon had a relatively easy work schedule and a strong work ethic, and he was able to make consistent, steady progress across the board. But after 4 weeks and 60 points of improvement, he was exhausted and burnt out. 

The Solution: Brandon and I sat down to start working on his applications. 

This was something that had to get done, so it was a good use of time—but for Brandon, it was also a welcome relief from the constant effort of GMAT studying, especially when rusty fundamentals meant nothing was coming easily Working on the applications boosted his confidence, since seeing awesome application essays reminded him that he was a strong candidate already, and his test score was just the final piece of the puzzle. And finally, writing application essays with questions like “What are your goals at business school?” restored his focus on why he was studying for the test in the first place! 

After spending a few days writing and revising application material, Brandon was ready and energized to get back to GMAT studying—and his practice test scores kept rising.

 

Case Study 3: Sally, the Management Consultant

The Challenge: Sally was working as a consultant while taking my class. She worked 70-80 hours/week during her busy periods and 50 hours/week at slower times. She spent most of her work week away from home. And perhaps most frustrating of all, Sally’s subordinates were studying for the GMAT on every train ride to and from their work site. She wanted to study with them, but didn’t think she’d be able maintain the respect necessary to manage them—especially since some of them were outscoring her on practice tests! 

The result: Sally decided not to take the test.

I realize this might not seem like an inspiring outcome, but it’s actually quite brilliant. The GMAT is not something that fits everyone’s schedule at any given time—it’s a major commitment. Forcing yourself to take a test you’re not ready for is just going to put a mediocre score on your record for the next five years. And more importantly, there is more than one path to success. Sally’s hard work has earned her a raise to a pay grade normally reserved for MBA’s! She’s hoping that with a few more years of such progress, she’ll be able to achieve her long-term career goals through an executive MBA program, which will be a better fit for her busy, hardworking lifestyle. I look forward to helping her again when that time comes.



2 Dice and the GMAT. What’s the relationship?


GMAT BlogWhile my students certainly accept that they have to take the GMAT, I often hear complaints about the applicability of GMAT topics – especially math – to real world situations.  My usual response to this observation is that the GMAT is using math questions to test critical thinking skills that business schools consider essential.

However, some GMAT problems have surprising real world applications outside the realm of business.  The sample problem below is an example of one such question.  Specifically, it can help you win at craps.

If you want to try the problem on your own, skip down to it now and then return here – part of this discussion will give you clues about the correct answer.

For those of you not familiar with craps, the basic play is fairly straightforward.  You roll two dice and if the sum of the result on each die is 7 or 11 you win.  If the sum of the results are 2, 3 or 12 you lose.  If you roll any other result the game progress to a second step that we do not need to concern ourselves with for purposes of this discussion.

Someone who is not familiar with probability might look at those rules and think, “this game is not fair – I have three ways to lose, but only two ways to win.”  But, by understanding probability we can see that craps is actually the most fair game in the casino.

The key is that every result does not have an equal probability of occurring.  To role a 2, the first die must roll a 1 and the second die must also role a 1.  Likewise, to roll a 12, both dice must come up with a 6.  A three is slightly more likely – the first die can be a 1 and the second a 2 or the first die can be a 2 and the second a 1.  This gives us four ways to lose in our initial roll.

Rolling an 11 is just as likely as rolling a 3.  You can roll a 5 on the first die and a 6 on the second or a 6 on the first and a 5 on the second.  However, we can roll a 7 in many ways.  We can roll a 1 first and a 6 second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a 6 second, or a 6 first and a 1 second.  Thus, we have six ways to make a seven when we roll our dice.

This means that we have eight ways to win and only four ways to lose.  While, the rest of the game pushes the odds back into the casino’s favor, on that initial roll we are twice as likely to win as lose.

This concept of desired versus undesired outcomes will show up on many probability questions on the GMAT – give the problem below a try to see how.

Problem:

What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1 through 6 on each side?

(A)      1/12

(B)      1/6

(C)      2/7

(D)      1/3

(E)      1/2

Solution:

When confronted with a probability question, always remember two points.  First, probability is just desired outcomes divided by possible outcomes, so you will need to find this ratio.  Second, if more than one event occurs, “and” means multiply and “or” means add.

In this problem, we want to roll a combined 7.  We can accomplish this in a number of ways.  We can roll a 1 with our first die and a 6 with our second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a 2 second, or a 6 first and a 1 second.

This gives us a total of 6 desired outcomes.  Next, we will need to find our possible outcomes.  Since we have 6 possible outcomes for the first die and 6 possible outcomes for the second die we should multiply 6 x 6 = 36 possible outcomes.  Also note, that this is an “and” situation – we roll the first AND second die – so we want to multiply rather than add.

Finally, we put desired outcomes over possible outcomes, to get 6/36 as the probability of rolling a combined 7.  6/36 simplifies to 1/6, which is answer (B).