September 15, 2012
The Wrentham Village Premium Outlets are a great place to stop for cheap brand-name clothes, and they’re a popular tourist destination for visitors to Massachusetts. Like all tourist/retail locations, they need to get people in the door. They’ve tried lot of things, but their latest gimmick has interesting implications for GMAT students. They’ve started stacking discounts.
Nearly every store in the mall has signs that say something like, “65% off, PLUS take an additional 20% off!” Moreover, a coupon book gives additional discounts—the particular store with that sign also offered 15% off purchases over a certain value.
To the unenlightened, this seems too good to be true. After all, 65% + 20% + 15% = 100%. Are we seriously to believe that the outlet store is giving away things for free?
Well, that might be a trap answer on the GMAT—and it’s a trap answer for the unwary consumer as well. But because we have been practicing GMAT quant, we know better. Even though the signs say “additional” and “plus,” we’re not really adding. 65% off means that the baseline price 35% of the retail value, and a further 20% off means we pay 80% of that discounted value. When translating from English to Math, the word “of” means “times.” So, when we take a percentage “of” a percent, we multiply; the results of the previous example are as follows:
(1 – 0.65)(1 – 0.2)(1 – 0.15) = (0.35)(0.8)(0.85) = 0.238
We end up with a 76.2% discount all told; that’s a pretty good deal, but hardly the 100% sale that some might have mistakenly expected!
When stacking percentage increases or decreases on the GMAT, you need to multiply—or, you can pick 100 and plug it into the equation. But however you solve, you cannot just add the numbers together; and you can quickly rule out any answer choice that is just a sum of the percents in the stem.
September 4, 2012
On the GMAT, there is only one correct answer to each question (How many caught the Highlander reference in the title? Be honest!).
I know, big surprise, right?
But that simple, obvious statement leads us to a powerful deduction. Some Problem Solving questions on the Quantitative section will have terms, variables, or unknowns that are unsolvable—they could take multiple values on the basis of the information in the stem. And we’re not talking Data Sufficiency here. “Not sufficient” isn’t a choice (Occasionally, “Cannot be determined” is a choice on problem solving questions. This answer is usually a trap, but you can use Data Sufficiency solving techniques to see if multiple answers are possible). So if the answer choices are numbers or proportions, and some term in the question stem is unsolvable, that undetermined x-factor can’t affect the outcome. Some ratio or mathematical step in the solution has to result in that variable “canceling out,” because otherwise the problem would have multiple correct solutions and therefore could not appear on the GMAT!
This is one of the ways that the Kaplan strategy of Picking Numbers works. Once you’ve identified an unknown that cancels out, you can plug in any value for that unknown, and be confident that your result is the right one. For instance, consider the following problem:
A runner runs downhill from point A to point B at 15 kilometers per hour, then runs uphill along the same path from point B to point A at 10 kilometers per hour. Assuming the time spent turning was negligible, what was the runner’s average speed during the round trip?
Here, the distance the runner travels is unknown and unknowable. But we’re asked for average speed. That’s a ratio; if we double the distance, we’ll double the time, and get the same speed. The answer to this question won’t change if he’s running one meter or one million!
Of course, neither one nor one million makes this problem particular easy to solve. Since every distance will give us the same average speed, we should pick the distance that takes the least effort when we plug it into the question. What’s the right distance to choose to make the arithmetic work? And, how can we use it to find the average speed?
August 27, 2012
One of the most important techniques to solving algebra problems, on the GMAT quantitative section or otherwise, is factoring. This technique, taking advantage of the “distributive property” of multiplication, lets you pull a common factor outside of a sum of terms, or to distribute it across those terms. In other words:
2x + 2y + 2z ↔ 2(x + y + z)
But did you know that the distributive property applies to grammar?
Well, not literally. But for quant experts confused by Parallelism in Sentence Correction, it can be helpful to imagine it as a distribution problem. When a sentence has a list of items, auxiliary verbs such as the “had” in “had been,” and prepositions such as “by” and “in,” can be “distributed” or “factored” across the list.
…by name, by date, or by subject ↔ …by (name, date, or subject)
Of course, just like with distribution, you have to do it right—leave out a term, you end up with nonsense!
2(x + y + z) ≠ 2x + y + 2z
Sorted by (name, date, or subject) ≠ Sorted by name, date, or by subject.
Here, we stumble across the limits of this analogy. The right and left of the mathematical inequality are different, but each could show up in a different problem. On the other hand, the right-hand sentence fragment can never be correct. But the principle remains. You can think of parallelism as algebra, following fixed, predictable rules that you can learn and manipulate. Now try the problem below to test the concept (click here for even more practice):
By the time they completed their journey, the young explorers had overcome their
fears, sharpened their survival skills, and had developed a healthy respect for
nature’s potential destructiveness.
skills, and had developed a healthy respect
skills, and developed a healthy respect
skills and a healthy respect developed
skills, developing a healthy respect
skills, all the while developing a healthy respect
Step 1: Read the Original Sentence Carefully, Looking for Errors
This sentence contains an underlined verb, which means that we must check for subject-verb
agreement and correct tense. The verb phrase “had developed” agrees with its subject, “the young
explorers.” Because the young explorers’ actions precede the completion of their journey, the use
of the past perfect tense is correct. However, the phrase “had developed a healthy respect” is part
of a list of three things the young explorers did during their journey, and we must therefore also
check for parallel structure. Here there is an error, as the second and third items on the list do not
have the same structure. The second verb on the list, “sharpened,” does not repeat the helping verb,
“had,” that is used at the beginning of the list to indicate the past perfect tense. But the third verb,
“had developed,” does repeat the word “had.” We can rule out (A).
Step 2: Scan and Group the Answer Choices
Of the remaining choices, (B) simply removes “had” but leaves the rest of the underlined segment
alone, while (D) and (E) change the verb to “developing.” Like (B), choice (C) uses “developed,” but
changes the word order.
Step 3: Eliminate Choices Until Only One Remains
Because we cannot change anything in the sentence except the underlined part, we cannot correct
the parallelism error by changing “sharpened” to “had sharpened.” Our only recourse is to change
“had developed” to “developed.” (B) does just that without changing anything else, and this will
likely turn out to be the correct answer. A quick look at (C) reveals that in changing the word order,
it destroys the parallelism of the list of three actions completely. As for (D) and (E), changing “devel-
oped” to “developing” changes the tense and again destroys the list’s parallel structure. Choice (B)
remains the correct answer.
August 20, 2012
Take 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.
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!
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?
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
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).
August 16, 2012
Tackling 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:
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.
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.
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.
August 10, 2012
We’ve already covered modifiers in GMAT sentence correction several times before. But, as one of the most common question types on the verbal section, and one of the types that requires the most finesse, there is still more to cover!
Today, I want to address a common misconception. Generally, modifiers must be placed as close as possible to the thing they modify. However, students sometimes mistake “as close as possible” for “adjacent.” Many test-takers find themselves confused when a long string of nouns, often peppered with prepositions, precedes a modifier. But as long as the modifier can be unambiguously linked to a specific part of that phrase, the sentence is grammatically correct. To illustrate, look at the following sentence, which is correct as written:
The members of parliament who attended the conference were pleased with the lush accommodations they received.
The modifier is the phrase “who attended the conference,” and the “who” follows “The members of parliament.” This setup should look suspicious, and it requires our attention—does the modifier apply to “parliament,” or to “the members of parliament”? Well in this case, “parliament” isn’t a “who”! The modifier correctly, clearly, and properly modifies “members of parliament.”
Let’s look at another one to be sure that we are clear:
The tides of the Pacific Ocean, which ebb and flow with regularity, have been the sailor’s ally for centuries.
The modifier here unambiguously refers to “tides,” and not “the Pacific Ocean,” because the plural verbs “ebb” and “flow” can’t refer to a singular Ocean. It’s also correct.
But the challenge of sentence correction lies not only in recognizing right sentences, but also in spotting errors. How could an answer choice on the GMAT get this grammar wrong? Well, for starters, the modifier could unambiguously refer to the wrong part of the noun:
Error: The cats of Hemingway House, which is characterized by extra toes on each foot, are almost as popular a tourist attraction as the House itself.
The “which” clause incorrectly describes the Hemingway House as having extra toes—not just because of the noun’s placement, but also because the verb tense is the wrong one! Of course, if the verb tense could apply to multiple parts of the noun, we’re in just as much trouble:
Error: The ruling of the High Court, which is a source of constant embarrassment for the government, has made the news once again.
This time, the singular verb “is” could refer grammatically to the “ruling” or to the “High Court”–and worse, both the ruling and the Court could reasonably be the “source of constant embarrassment!” Such ambiguity would not be considered correct on a GMAT verbal question.
Have more questions about modifiers? Post them below, and we’ll analyze them together.
August 8, 2012
Translating 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…
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?
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.
August 6, 2012
As anyone who has spent any time on GMAT Sentence Correction can tell you, the English language is complex. SC problems will frequently test idioms and tricky verb tenses, among other things. But despite a few exceptions (do you know the difference between economic and economical?), subtle shifts in the meanings of similar words aren’t usually tested in GMAT sentences. They are, however, tested on Critical Reasoning and Analytical Writing prompts.
Assumptions on the GMAT occur when the scope of discussion shifts between the evidence and the conclusion. In an earlier article, I discussed a stimulus involving burgers. One such “scope shift” in that article was that the evidence discussed cholesterol, while the conclusion discussed health in general; another involved evidence about a price reduction and a conclusion about increased consumption of burgers. Some of these are easier to spot than others, but all of them involve looking for changes in terms and terminology.
But sometimes, there is a change of meaning, even though the actual words are the same. Consider the following example:
Buddy claims he hurt his back lifting a heavy box of yogurt onto the store’s shelves. However, he was in the “diet” section of the store, stocking shelves with light yogurt. Clearly the only boxes he lifted were light; his claim for workers compensation must be a fabrication.
This argument is, of course, absurd! But if you’re locked into the GMAT mode of thinking (which is a good thing!) you might wonder why. This problem doesn’t seem to shift scope—both the evidence and the conclusion talk about the yogurt being light, right?
The key is that the author is “equivocating,” a technical term for using the same word with different meanings. “Light” here means “Diet” in the evidence but “Not Heavy” in the conclusion—that’s a pretty big gap, leading to deeply flawed reasoning. This pattern isn’t terribly common on GMAT problems, but it shows up from time to time, usually on Flaw questions. Keep your eyes peeled for words with multiple, ambiguous, or unclear meanings on the GMAT, and on today’s question of the day, an AWA prompt.
The following appeared in an internal memo for the Weekly Globe newspaper.
The proposal to reduce the celebrity section of our print edition from 6 pages weekly to 2 pages is misguided. The celebrity pages on our website average more hits per article than does any other section of our website; clearly the public is most interested in celebrity news. The proposed change would not only hurt our profits, but also betray our dedication to serving the public’s interests.
Discuss how well reasoned…
Post your analysis below, and we’ll let you know if there is anything you missed.
July 23, 2012
One of the simplest arithmetic rules is that when you divide something by itself, you get 1.
What’s 3/3? 1.
What’s x/x? 1
By the same logic, what do you get with, say, inches/inches?
That’s also equal to 1.
So let’s say that you need to find the number of seconds in 2 minutes. You can probably do this in your head! Multiply two by sixty and it’s 120 second. But have you ever stopped to wonder why that works? Well, you want seconds to remain, so you want to get rid of minutes—that means you want minutes on the top and on the bottom. You did the math instinctively, but if you had broken it down step by step it would look like this:
Minutes on top and bottom cancel:
Now, on test day you’d never go through all the steps just for something as simple as that last example. But that basic principle makes it easy to solve much more complex conversions on the GMAT. For instance, my car has a 17 gallon gas tank and gets 32 miles per gallon on the highway. If I’m highway driving at an average speed of 68 miles per hour, how long can I drive without needing to stop for gas? (Assume I don’t take bathroom breaks)
The question asks for hours. Once everything has canceled, we’ll be left with the unit we want on top–so to get hours on top, we start with the reciprocal of my speed:
Now, miles is extraneous. Since it’s on the bottom, we multiply by a proportion that puts miles on top:
Then, finally, to get rid of gallons, use the same strategy:
Most rate conversion problems in the GMAT quantitative section, however complex, function similarly. You’re just multiplying a series of fractions together—all that’s challenging is keeping the fractions the “right way” up so that you keep the units you want and cross of the units that you don’t.
Magnabulk Corp sells boxes holding d magnets each. The boxes are shipped in
crates, each holding b boxes. What is the price charged per magnet, in cents, if
Magnabulk charges m dollars for each crate?
Step 1: Analyze the Question
A complicated setup, with oodles of variables. Picking
Numbers will probably be a safe approach.
Step 2: State the Task
Our task is to calculate the price per magnet. The word per
signals a rate
Many GMAT word problems have wrong answers that can
be eliminated logically, and this is no exception. Since the
answer is the amount of money that each magnet costs,
we can be sure that the more dollars charged per crate (m),
the more money each magnet would cost. In other words,
the right answer would have to get bigger as m gets bigger.
Answers (A), (C), and (E) have m in the denominator, so
those expressions would get smaller as m gets bigger.
Those answers can be eliminated.
Step 3: Approach Strategically
We need to solve for “magnets” and “cents.”
What do we know about the number of magnets? Scanning
through the question stem, we find “Magnabulk Corp sells
boxes holding d magnets.” That means d magnets per box:
What do we know about boxes? “Boxes are shipped in
crates, each holding b boxes.” That’s b boxes per crate:
What do we know about the crates? “Magnabulk charges
m dollars for each crate.” That’s m dollars per crate:
And, of course, dollars convert to cents at the rate of
100 cents per dollar:
Then set up multiplication such that cents are in the
numerator, magnets in the denominator, and everything
That’s answer (B).