Posted by Clear Admit on December 23, 2009, at 8:00 pm
Posted in:  GMAT - Quantitative , GMAT News , GMAT Test Prep Company Series , GMAT Tips Today’s GMAT tip comes from our friends at Veritas Prep. In this article, they present the first installment of their “Think Like the Testmaker Series”:
As Veritas Prep’s Director of Academic Programs, Brian Galvin runs all of our GMAT prep courses.
Greetings, readers! In this space over the coming months, the GMAT experts at Veritas Prep want to urge you to think beyond the skills covered on the GMAT — algebra, geometry, grammar, etc.– and get inside the minds of the writers of the exam. The GMAT tests not only your academic capabilities, but also (and more importantly) your problem solving, logical reasoning, and higher-order thinking skills. To employ these skills toward success, it’s helpful to begin thinking about the GMAT as a logic puzzle or mind game — in many respects, you’re in a chess match with the author of each question, trying to anticipate the trap he is setting for you as you calculate your next move.
If you embrace this competitive challenge, you’ll put yourself in a position to better anticipate the steps necessary to ensure that you answer each question correctly, and you’ll also tend to enjoy the process more. Did you dislike Geometry class in high school, but waste countless hours of your life playing Tetris on your GameBoy or computer? Both involve the heavy use of geometrical thinking, but the challenge of Tetris engaged your mind in a way that class may not have been able to do. Think of the GMAT as a game, and you should see great results.
As an initial step in “thinking like the testmaker,” let’s consider one of the time-honored tricks of the GMAT — asking a question that could be asked in multiple ways, and providing answer choices that would answer each of those variations.
Consider the question:
A 40-foot length of rope is cut in to two sections, for which the shorter section is 1/3 the length of the longer. How much longer, in feet, is the longer section of rope than the smaller?
A) 10
B) 13.33
C) 20
D) 26.67
E) 30
Note that this question can be asked in multiple ways. If we break down the question, we’ll find that the shorter piece of rope is 10 feet long and the longer piece is 30 feet. The difference between the two is 20 feet. The question asks for the difference between the two lengths, so the correct answer is 20, or answer choice C. Please notice, however, that if you simply solved for either length of rope, you might be inclined to choose choices A or E, and that, if you mistakenly solved for 1/3 or 2/3 of the initial piece of rope, you’d end up with choices B or D.
Accordingly, the lesson here is that you need to recognize what the question specifically asks for, as the most common wrong answers on the GMAT are simply the right answers to the wrong questions. To become a more astute test taker , ask yourself in practice “How could they ask this question in a different way?” so that you can anticipate the multiple potential questions on each question you face, and remind yourself that you need to be careful when submitting your answer.
For more practice in thinking like the test maker, try a GMAT practice test and track how often you answer the question asked, rather than the question that you thought was asked!
For more information on Veritas Prep, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!
Today’s GMAT tip comes from ManhattanGMAT. This article is a follow-up to an earlier one on weighted average problems. This time, ManhattanGMAT instructor Stacey Koprince offers advice on a more difficult weighted average problem:
Earlier, we tackled a medium-level GMATPrep® weighted average question. This time, we’ll try a harder GMATPrep® weighted average question in order to test whether you learned the concept as well as you thought you did.
As we discussed, every weighted average problem I’ve seen (so far!) on GMATPrep is a Data Sufficiency question. This doesn’t mean that they’ll never give us a Problem Solving weighted average problem, but it does seem to be the case that the test-writers are more concerned with whether we understand how weighted averages work than with whether we can actually do the calculations. Last time, we focused on understanding how weighted averages work via writing some equations. We’ll try to apply that understanding to our harder problem this time, along with a more efficient solution method.
Let’s start with a sample problem. Set your timer for 2 minutes…. and… GO!
* “A contractor combined x tons of gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x?
“(1) y = 10
“(2) z = 16
There are two kinds of gravel: “10% gravel” and “2% gravel.” These are our two “sub-groups.” When the two are combined (in some unknown – for now! – amounts), we get a 3rd kind: “5% gravel.” The number of tons of “10% gravel” (x) and the number of tons of “2% gravel” (y) will add up to the number of tons of “5% gravel” (z), or x + y = z. We need to find the number of tons of “10% gravel” used in the mixture.
The problem this time throws in a new wrinkle: we’re not just trying to calculate a ratio this time. We have to have enough info to calculate the actual amount of “10% gravel” used. In the other problem, we never had to worry about the actual number of employees. We’ll have to keep that in mind to see how things might change.
This problem never mentions the word average. That’s annoying. – how are we supposed to tell that this is a weighted average problem? Basically, the problem should talk about 2 sub-groups that are combined in some way to make a 3rd overall group, or mixture of the original 2 sub-groups. The problem will often discuss these groups in terms of percentages (as this problem does) or ratios (as the other problem did). That starts to tell us that some kind of averaging is happening.
Next, we know that we’ll either have an unweighted (“normal”) average or a weighted average, so check to see whether this is a normal average. We start with 10% gravel and 2% gravel. If we mixed exactly equal amounts of each (a normal average), what would the resulting mixture be?
(10+2)/2 = 6
Does the problem say we end up with 6% gravel? No. The amount of “10% gravel” is not equal to the amount of “2% gravel.” Therefore, this is not a “normal” average; now we know we have a weighted average problem.
Let’s go back to our test of “10%” and “2%” as an unweighted average. Instead of calculating with a formula (as we did up above and last time), let’s draw it out visually. Draw a straight, horizontal line on a piece of paper. Label the left end “2” and the right end “10” to represent our two different sub-groups of gravel:
2————————–10
An average of any two numbers will always appear between those two numbers. If the average is unweighted (50% of each is used), then the average will be exactly halfway between the two:
2————6————10
If the average is weighted, then it will NOT appear exactly halfway between. It will appear closer to one end or the other, depending upon the weighting. For example, if we have all “2% gravel” and zero “10% gravel,” what’s the average? A 100% weighting of “2% gravel” will give us an “average” of 2.
2————————–10
In this problem, they tell us that the average of the mixture is 5, so the average is closer to the “2” end of the line than to the “10” end of the line. (Imagine a tug-of-war between 2 and 10.) Now we know that there’s more “2% gravel” in the mixture than “10% gravel” because the “2” end of the line has “pulled” the average closer to its end.
2———5—————10
To find the weighting: calculate the difference between the two ends of the line. In this case, the difference is 10 – 2 = 8. Determine how far the dominant end (2, in this case) has “pulled” the average: calculate how far away the weighted average appears from the other end of the line. In this case, the weighted average 5 is 5 units away from the 10 sub-group.
The “dominant” end (2) therefore has the weighting 5/8, because it has “pulled” the weighted average 5/8 of the way towards its end of the line.
The “non-dominant” end (10) is just the remaining amount of the “adds up to 1” figure (from our article last time): 3/8. The “non-dominant” end (10) has “pulled” the weighted average only 3/8 of the way toward its end of the line.
Now we know that, of the total mixture, 5/8 of it will be “2% gravel” and 3/8 of it will be “10% gravel.”
Conceptually, we want to realize that if a problem tells us the two starting points (the ends of the line) and the weighted average (the middle number), then we know we will be able to calculate the relative weightings of the two sub-groups. (This is the exact same concept that we learned on the last problem!)
Let’s examine the statements to see whether this knowledge might be useful.
Statement 1 says “y = 10,” which tells us that there are 10 tons of the “2% gravel.” If I know I have 10 tons of the “2% gravel,” and I also know that 5/8 of all of the gravel will be this “2% gravel,” then can I calculate x, the amount of “10% gravel?” Yes! 10 tons = 5/8 of the total. Divide each side by 5: 2 tons = 1/8 of the total. Multiply by 3: 6 tons = 3/8 of the total. (I don’t actually need to do this calculation on this problem because this is Data Sufficiency, but I would if I saw this on Problem Solving.) Statement 1 is sufficient; eliminate choices B, C, and E.
Statement 2 says “z = 16,” which tells us that there are 16 tons of the “5% gravel.” I also know that z represents the sum of x and y, or 8/8. If I know that 16 is 100%, or 8/8, of the amount, then I can also calculate x: 16 tons = 8/8 of the total. Divide by 8: 2 tons= 1/8 of the total. (Is this starting to look familiar?) Multiply by 3: 6 tons = 3/8 of the total. Statement 2 is also sufficient.
The correct answer is D.
We can simplify this further (for future data sufficiency questions) by saying: if we’re told the two “ends of the line” for calculating the average, as well as the overall weighted average, then we can calculate the relative weightings, or ratio, of the sub-groups (just as we did last time). If we’re also given one of the three actual amounts, then we can calculate all of the actual amounts in the problem. In this case, if we’d realized that before examining the statements, we could have asked ourselves, “Does the statement give me one of the three real amounts?” and quickly recognized that each statement is sufficient by itself.
Key Takeaways for Data Sufficiency Weighted Average Problems:
(1) Determine that you have a weighted average problem: this occurs when an average is described (even if the word “average” is not in the problem!), but that average is not a standard 1:1 or equally weighted average.
(2) Carefully write down what you were asked to solve, then determine what you know, what you don’t know, and what you would need to know in order to solve (before you look at the statements).
(3) Check the given statements to see whether you can find a “match” (that is, a statement tells you what you had already decided you would need to know in order to solve).
* GMATPrep® questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!
Today’s GMAT tip comes from our friends at ManhattanGMAT. In this article, ManhattanGMAT instructor Stacey Koprince offers advice on weighted average problems:
This week, we’re going to tackle a GMATPrep® question from the quant side of things. We’ll tackle a medium-level question this week in order to learn how to master weighted average questions in general, and in a later article, we’ll try a very hard one – just to see whether you learned the concept as well as you thought you did.
Before we begin, I want to mention that every weighted average problem I’ve seen on GMATPrep is a Data Sufficiency question. This doesn’t mean that they’ll never give us a Problem Solving weighted average problem, but it does seem to be the case that the test-writers are more concerned with whether we understand how weighted averages work than with whether we can actually do the calculations. So we’re going to work on that conceptual understanding today and then we’ll discuss a neat calculation shortcut later (built on the same principles!), just in case we do need to solve.
Let’s start with a sample problem. Set your timer for 2 minutes…. and… GO!
* ” At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?
“(1) There are 52 male employees at the company.
“(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.”
Given “a certain company,” we’re asked to determine the ratio (not a real number, just a ratio – key point!) of two subgroups that together make up all employees: males to females.
So, what do we know? We know that male employees have an average of 9.8 years of experience. We also know that female employees have an average of 9.1 years of experience. What would be useful to solve? It would be useful to know about the actual number of male and female employees. Alternatively, it would be useful to know about the relationship between the number of male employees and the number of female employees. (For example, if they told me 60% of the employees were female, then I would know the ratio of males to females was 40:60, or 2:3, even though I wouldn’t know the actual number of employees.)
Most of what we’re going to do next is just to explain how weighted averages work. Once you understand how this works, you will not actually have to do these calculations on DS questions (this will take way longer than 2 minutes!); you’ll be able to determine conceptually whether enough info was provided to solve.
In the given problem, could there be equal numbers of male and female employees? Go take a look at the problem again and see what you think.
Let’s say that there are, in fact, 50 male employees and 50 female employees. If the male employees’ average experience is 9.8 years and the female employees’ average experience is 9.1 years, then what is the average experience for the whole group? That would just be the average of 9.8 and 9.1. Is the average of those two numbers 9.3 (the “total group” average given in statement 2)? No. So now we know we’ve got a weighted average problem; in other words, the number of male employees is not equal to the number of female employees. (Bonus question: can you tell, just based on what we’ve discussed so far, whether there are more male or female employees?)
In order to understand how weighted averages work, let’s calculate a few things and let’s start by using the weighted average formula to see what happens in a case where we have equal numbers of employees (which, again, is not true for this problem – we’re just examining the concept).
We know the two sub-group averages, 9.8 and 9.1, and we’re also assuming an equal weighting of the two averages, 50:50, which simplifies to 1:1. Put that ratio, 1:1, in a form where the two parts add up to 1: ½:½. Each average gets paired with its “adds up to 1” weighting:
[(9.8)(1/2) + (9.1)(1/2)] = 9.45
Because the weightings are equal, this can be simplified to the standard average formula (below); this is why we don’t bother calculating the “adds up to 1” weighting when the weightings are equal.
(9.8 + 9.1) / 2 = 9.45
What if the weightings are not equal, though? Let’s say that there were 40 male employees and 60 female employees. Then, the ratio would be 40:60, or 2:3, and the “adds up to 1” weighting would be 2/5: 3/5. (The easiest way to determine the “adds up to 1” weighting is to first add the two parts of the given ratio, 2 and 3, to get 5. 5 becomes the denominator of both fractions and the original numbers, 2 and 3, become the numerators of each respective fraction: 2/5 and 3/5.)
The weighted average formula would become:
[(9.8)(2/5) + (9.1)(3/5)] = 9.38
Here’s the “abstract” version of this formula:
[(average #1)(a) + (average #2)(b)] = weighted average, where:
a + b = 1, and
a and b represent the relative weightings of the two sub-groups
In the given problem, we don’t know a and b, but we do know the two sub-group averages, 9.8 and 9.1, so we can write these two formulas:
9.8x + 9.1y = c
a+b = 1
We need to see whether we have enough information in the statements such that a and b could be calculated.
Statement 1 says “There are 52 male employees at the company.” That gives us an actual number for the male employees; that might be good. We want the male to female ratio, though; does this statement tell us anything about the other group, female employees? No. Not sufficient. Eliminate answers A and D.
Statement 2 says “The average number of years of experience for the company’s male and female employees combined is 9.3 years.“
That’s something we can add to one of our formulas: c, the weighted average, is 9.3:
9.8x + 9.1y = 9.3
a+b = 1
What do we have? We have two distinct, linear equations with two variables, a and b. Can we solve for a and b? Yes! Sufficient!
The correct answer is B.
We can simplify this further (for future data sufficiency questions) by saying: if we know the two sub-group averages and we know the overall weighted average, then we know we can solve for a and b, the relative weightings of the two sub-groups. (Don’t bother to write the equations, of course – it’s data sufficiency!) In this case, a:b represents the requested ratio (male:female).
Key Takeaways for Data Sufficiency Weighted Average Problems:
(1) Determine that you have a weighted average problem: this occurs when an average is discussed or could be calculated, but that average is not a standard 1:1 or equally weighted average.
(2) Carefully write down what you were asked to solve, then determine what you know, what you don’t know, and what you would need to know in order to solve (before you look at the statements). Remember that, if you have two sub-group averages and the overall average, then you can determine the relative weightings of the sub-groups.
(3) Check the given statements to see whether you can find a “match” (that is, a statement tells you what you had already decided you would need to know in order to solve).
Answer to bonus question: there are more female employees at the company because the weighted average, 9.3, is closer to 9.1 (the female employee figure) than to 9.8 (the male employee figure).
* GMATPrep® questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!
Today’s GMAT challenge question comes from our friends at ManhattanGMAT. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution:
Problem
Two positive numbers differ by 12 and their reciprocals differ by 4/5. What is their product?
(A) 2/15
(B) 48/5
(C) 15
(D) 42
(E) 60
Solution
Don’t be afraid to assign variables even when none are given in the problem. “Two positive numbers differ by 12” can be written as:
x – y = 12
And “their reciprocals differ by 4/5” can be written as:
1/y – 1/x = 4/5
(Note: Here, we’ve assigned x as the bigger of the two numbers and y as the smaller, so we’ve intuited that 1/y is the larger reciprocal and 1/x the smaller, and so arranged them in that order to write 1/y – 1/x = 4/5).
Now we have a system of two variables and two equations. Note that it is NOT necessary to solve for x and y, since we are being asked for the product, xy.
First, let’s simplify the second equation by finding a common denominator for the terms on the left:
x/(xy) – y/(xy) = 4/5
(x – y)/(xy) = 4/5
Note that the denominator is xy, which is exactly the quantity we want to find.
Since we know from the first equation that x – y = 12, substitute 12:
12/(xy) = 4/5
60 = 4xy
15 = xy
The correct answer is (C).
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!
Today’s GMAT tip comes from our friends at ManhattanGMAT. This article serves as a follow-up to a recent one on max/min questions. This time around, ManhattanGMAT instructor Stacey Koprince offers advice on an even more difficult sample GMATPrep® max/min question:
Recently, we tackled two GMATPrep® questions; if you missed that article, go read it before continuing with this one. Make sure you try the two sample problems and take the time to master the concepts before you try the super-hard question below.
Okay, this sample problem is from our own archives. Set your timer for 2 minutes…. and… GO!
Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?
(A) 40
(B) 38
(C) 33
(D) 32
(E) 30
You have an answer, right? Even if you don’t know what the answer is and you have to guess… you’ve still picked an answer, right? If you haven’t, go pick an answer before you keep reading!
As we discussed, the most important thing to notice here is the word “largest.” This one word is going to be the determining factor in how we set this problem up, right from the very beginning.
So, we have a set of 7 numbers. The average of those 7 numbers is 20; can we calculate anything from that? Yes – the sum! The basic formula for an average is A = S/n, where A is the average, S is the sum, and n is the number of items. We know A and n, so plug the two numbers in to get 20*7 = 140 for the sum.
The problem is asking us to maximize one figure: the last (and, of course, highest) number in the set. If all 7 numbers have to add up to 140, and we want to make one number as large as possible, then what do we have to do to the remaining six numbers?
We have to minimize all 6 of the remaining numbers – so, for the rest of the problem, we need to figure out how to make the other 6 numbers as small as possible.
Do we know anything about those 6 other numbers? We were told that the median is 20; what does that mean? Draw out some dashes on your scrap paper, one for each number in the set:
____ ____ ____ ____ ____ ____ ____
Now, how can we represent the fact that the median is 20? We have an odd number of terms. The median will be the middle term (the fourth, in this case) and it will actually equal 20. So add that to your diagram, along with an “x” for the term we’re supposed to maximize:
____ ____ ____ _20_ ____ ____ __x__
The problem also gives us some info about the first term:
“the smallest number in the set is 5 less than half the largest number”
Hmm. We don’t know what the largest number is, of course – that’s what the problem asks us to maximize! But we’re calling that largest number “x” so let’s write the smallest term in terms of x: (½)x – 5. Add that to the diagram:
_(½)x – 5_ ____ ____ _20_ ____ ____ __ x __
Okay, now what – what’s our goal again? Oh, right, we want to minimize everything that isn’t that last term, “x.” Okay, so what can we minimize? We can’t change the first term; that’s going to be (½)x – 5 no matter what. And we can’t change the fourth term; that’s going to be 20 no matter what. What about the second, third, fifth, and sixth terms? What are the smallest possible values for each of those?
Let’s start with the rules for writing out a bunch of numbers in order to show a median. When you write a set of numbers to find the median, the requirement is to write the numbers from smallest to largest. Let’s say that we have to write these three numbers in order: 20, 14, 18. We would write them: 14, 18, 20. Moving to the right, each number is higher than the previous number. Moving to the left, each number is lower than the previous number.
Is that all? What if we had to write these three numbers in order: 20, 14, 20? Then, we would write: 14, 20, 20. Moving to the right, the second term, 20, is higher than the first term, 14, but the third term, 20, is equal to the second term, 20. So the full rule is not that the numbers have to increase as you move to the right or decrease as you move to the left. The rule is that, as you move to the right, the number has to be equal to or higher than the number to the left. Similarly, as you move to the left, the number has to be equal to or lower than the number to the right.
So, let’s get back to our problem. We’re trying to minimize the remaining slots. What is the smallest possible value for the fifth term, keeping in mind that the number has to be equal to or higher than the fourth term? The fourth term is 20, so the smallest value for the fifth term is also 20. For the same reason, the smallest value for the sixth term is also 20.
_(½) x – 5_ ____ ____ _20_ _20_ _20_ __ x __
What about the second and third terms? The second term has to be equal to or higher than the first term, and the first term is (½)x – 5. Therefore, the smallest possible value for the second term is also (½)x – 5. For the same reason, the smallest value for the third term is also (½)x – 5.
_(½) x – 5_ _(½) x – 5_ _(½) x – 5_ _20_ _20_ _20_ __ x __
Now we have representations for all seven terms: either real numbers or variable expressions. We know the seven terms add up to 140. Time to set up an equation and solve for x!
[(½)x – 5] + [(½)x – 5] + [(½)x – 5] + 20 + 20 + 20 + x = 140
(3/2) x – 15 + 60 + x = 140
(5/2) x = 95
x = 95(2/5)
x = 38
The correct answer is B.
Key Takeaways for Max/Min Problems (same as before!):
(1) figure out what variables are “in play” (what figures we can manipulate in the problem)
(2) figure out whether each variable needs to be maximized or minimized in order to achieve the desired outcome (the thing the problem asks us to do)
(3) do the work (carefully, as always!)
Note: the key takeaways are the same as before, when we did some lower-level max/min problems. The basic process doesn’t change; we just have a bit more we need to know and a bit more we need to do on the very hard problem we did this week.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!
Today’s GMAT tip comes from the folks at test prep firm ManhattanGMAT. As a follow-up to another recent GMAT article, in this article, ManhattanGMAT instructor Stacey Koprince breaks down another example of a GMATPrep® Sentence Correction (SC) question:
This week, we’re going to analyze a particularly tough GMATPrep® Sentence Correction question.
First, set your timer for 1 minute and 15 seconds and try the problem!
“Research has shown that when speaking, individuals who have been blind from birth and have thus never seen anyone gesture nonetheless make hand motions just as frequently and in the same way as sighted people do, and that they will gesture even when conversing with another blind person.
“A) have thus never seen anyone gesture nonetheless make hand motions just as frequently and in the same way as sighted people do, and that
“B) have thus never seen anyone gesture but nonetheless make hand motions just as frequently and in the same way that sighted people do, and
“C) have thus never seen anyone gesture, that they nonetheless make hand motions just as frequently and in the same way as sighted people do, and
“D) thus they have never seen anyone gesture, but nonetheless they make hand motions just as frequently and in the same way that sighted people do, and that
“E) thus they have never seen anyone gesture nonetheless make hand motions just as frequently and in the same way that sighted people do, and”
Okay, have you got your answer? Now, let’s dive into this thing! What did you think when you read the original sentence?
This is a very tough problem; when I read the sentence the first time, I actually had to stop and try to strip the sentence down to its basic core, then figure out how the modifiers fit. Until I did that, I couldn’t go any further.
First, we have “research has shown,” a subject-verb pair. That’s the start of the core. The research has shown some things. What are those things (in simple form)? In the following sentence, the words in <brackets> are my simplification of the sentence; these words do not represent the original sentence.
“Research has shown THAT when speaking, <certain>¹ individuals nonetheless make hand motions <in a certain way>², and THAT <when speaking>³ they will gesture <in another way>4.”
1<certain> takes the place of “who have been blind from birth and have thus never seen anyone gesture.”
2<in a certain way> takes the place of “just as frequently and in the same way as sighted people do.”
3<when speaking> is implied by parallelism; this second thing is something that occurs when speaking, just as the first thing is something that occurs when speaking. This parallelism is indicated by the second instance of the word “that” and is reinforced by the pronoun “they,” which refers to the subject (individuals) of the first “that” clause.
4<in another way> takes the place of “even when conversing with another blind person.”
So what we’ve really got is:
“Research has shown THAT when speaking, <certain> individuals nonetheless make <certain> hand motions, and THAT they will gesture <in a certain way>.”
Simplify that even more:
“Research has shown THAT X, and THAT Y.” (X and Y are parallel and are both things that the research has shown.)
In the original sentence, the main word in X is “individuals” and the main word in Y is “they,” so we already have proper parallelism.
Are the other four choices also correct just at the core level of the sentence? Part of the core is not underlined: “Research has shown THAT X.” We know, then, that the Y part should be introduced with another THAT (in order to indicate that these two parts, X and Y, should be parallel). Choices B, C, and E all omit the THAT in front of Y, so they are not correct.
D also uses the core structure “and THAT Y,” so D is okay as far as that issue is concerned. How do the rest of A and D compare? A begins “have thus never seen” while D begins “thus they have never seen.” What’s the major difference? D includes the subject “they” while A omits a subject. Do we want a subject here? Now we need to dive into one of the modifiers.
“individuals who have been blind from birth and have thus never seen anyone gesture nonetheless make”
As we discussed earlier, “individuals” is a subject; the matching verb is “make”: “individuals nonetheless make <certain> hand motions.” The words in between “individuals” and “nonetheless” are modifiers – and because we have two separate modifiers connected by the word “and,” we need to make those two modifiers parallel.
“Individuals who J and K nonetheless make”
A: “Individuals who [have been blind from birth] and [have thus never seen anyone gesture] nonetheless make”
D: “Individuals who [have been blind from birth] and [thus they have never seen anyone gesture], but nonetheless they make”
So, are they both properly parallel? The J modifier is not part of the underline, so we know that the structure of K has to match the existing structure of J. J’s main construction is a verb in the present-perfect tense, so K should have the same structure. In choice A, K does begin with a present-perfect verb, but in choice D, K beings with a noun (“they”). That’s not parallel. Eliminate D.
Now, we’re down to one answer choice. The correct answer is A.
There are other ways we could have eliminated answers. For example, choices B and D both use the phrase “but nonetheless” to indicate a contrast. Each word indicates a contrast by itself, so using both words together is redundant.
There’s another split between “just as frequently and in the same way as” and “just as frequently and in the same way that.” Which one is right? The word “and” once again indicates parallelism, so there’s something parallel about the part before and the part after the “and.” Try each part individually.
“She runs just as frequently as he skis.” That’s fine. Can we say “She runs just as frequently he skis?” No – we need that second “as” after the word “frequently.” The full phrase is “just as frequently as.” So that’s why we have parallelism in this sentence! In the structure “just as frequently and in the same way as,” the second “as” applies to both parts (“just as frequently as” and “in the same way as”). We can’t use “just as frequently and in the same way than” because that would leave us with either “just as frequently” (with no second “as”) or “just as frequently than” – neither of which is correct.
The major take-aways here:
(1) when doing SC, first attack the errors that you know how to do and reuse your prior analysis as much as you can; you may not have to use all of the errors / differences in order to find the right answer!
(2) scan SC answer choices vertically to find differences; don’t read horizontally
(3) know how to recognize and properly construct noun modifiers and adverbial modifiers
(4) watch out for parallelism markers – the markers are often little words but they can make a big difference!
* GMATPrep® question courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Posted by Clear Admit on December 2, 2009, at 8:00 pm
Posted in: General , GMAT News , GMAT Test Prep Company Series , GMAT Tips Today’s GMAT tip comes from the folks at Kaplan. In this article, Kaplan teacher Brian Fruchey shares advice on best practices for using the GMAC Official Guide:
The GMAC Official Guide is a fantastic book. (So fantastic, that when students enroll in a Kaplan GMAT course, we buy a copy for them!) While the book is stuffed full of good questions, make sure that you use it appropriately. Case in point: the GMAC does not indicate the relative “toughness” of a particular question. Realize that, if you are looking to score 600 + on Test Day, you will have to supplement your studying with additional material. The following outlines strategies on how to use the Official Guide for the following sections:
1 – Background of the GMAT (Sections 1,2, and 3)
The GMAT is a different kind of test than you have probably ever taken before. The first three sections of the OG (lingo for the Official Guide) provide the background a new test taker needs. We suggest to really memorize sections 1 and 2. Knowing what the GMAC says about the test will help put the rest of the content and strategies into a framework.
Section 3 is a diagnostic test. This is a really great way to jump into the content of the GMAT. I would advise you to not focus on the timing considerations at this point. Set a couple hours aside and work through each of the question sets. The diagnostic test is divided by question type (Problem Solving vs. Data Sufficiency) – don’t let this trick you. The real GMAT is not clearly subdivided.
Finally, make sure you go over the explanations after the section. Knowing why a question is wrong is a lot better than just knowing that a question is wrong – working through your right and wrong answers is really the beginning of your GMAT preparation.
2 – Math Review (Section 4)
Ok. So you just finished a diagnostic test. You have reviewed your questions and have learned about the test. Hopefully, the quantitative section wasn’t awful. However, no matter who you are, you probably found at least a couple of the concepts difficult.
Section 4 of the OG outlines the math that you will need to know for the GMAT. While the OG does not go into extreme detail, it is very important for test takers to know what is being tested on Test Day. I would advise testers to review this material and make sure they are comfortable with each of these concepts. However, also realize that the test focuses on a couple of the concepts far more than other concepts. Your study program needs to distinguish where you should focus your energies – the OG basically introduces the concepts.
3 – Content Structure (Sections 5 – 9)
Sections 5 through 9 are the heavy content sections. The GMAT views the test through a prism of question types. I view the test differently – through a prism of competencies, since a competency can be tested through several different question types. This makes learning the GMAT that much easier when you can see there are only a few things test takers truly must know. Since the OG is a GMAC book and not a strategy-based program, you will be looking at the questions by individual question type.
Each of these content areas contains a great deal of questions as well as instructions on answering the questions. Use these questions to refine and develop your test taking ability. At the very least, before you take the real GMAT, make sure you have cleaned out all of these questions and reviewed the explanations.
4 – Analytical Writing Assessment (Section 10)
Finally, the last section of the book deals with the Analytical Writing Assessment (or AWA). Please read this section in detail to understand explicitly what the GMAT wants you to write. They are pretty clear about the structure. If you are concerned about the AWA section (the score is looked at!), make sure you practice with the sample prompts in the book. Additionally, the book outlines some suggested scores for sample essays. Before reading the samples, set a timer for 30 minutes and try to write the sample yourself. Comparing your essay to the sample ones in the book will give you a decent idea of how your score might stack up.
The OG is a great resource. It’s an essential piece of any GMAT preparation program. In a later post, I’ll talk more about how the OG differs from other resources and how you can study specific question types from past actual GMATs.
For more information on Kaplan, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Today’s GMAT tip comes from our friends at ManhattanGMAT. In this article, ManhattanGMAT instructor Stacey Koprince offers explanations to sample GMATPrep® max/min questions:
This week, we’re going to tackle two GMATPrep® questions, this time from the quant side of things. My students have been asking (really, complaining!) about maximize / minimize questions lately. A lot of students aren’t sure about the most efficient approach to these kinds of questions. We’ll tackle these two GMATPrep® questions this week in order to learn how to master max/min questions in general.
Let’s start with a sample problem. Set your timer for 2 minutes…. and… GO!
*Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
The most important thing to notice here is the word “maximum.” This one word is going to be the determining factor in how we set this problem up, right from the very beginning.
Most of the time, when we’re asked to maximize something, we will need to minimize the remaining variables in the problem. Conversely, if the problem asks us to minimize something, we will usually need to maximize the remaining variables. (There are times, though, when we will need to minimize some other variable in order to minimize the desired thing or maximize some other variable in order to maximize the desired thing – so we do need to pay attention.)
This time, they’re asking us to maximize one figure: the lightest box.
If three items have an average weight of 7, then collectively, the three items have a weight of 7×3 = 21. If three items have a median weight of 9, then the middle of the three items is actually 9. This one isn’t a variable; we can’t change this number. The first (or lightest) of the three, therefore, has to be equal to or less than 9 (because it is to the left of the median). Check the answers quickly – in this case, unfortunately, that information doesn’t help us to eliminate any answers.
If the middle box is actually 9, then we can subtract that from 21 to get the combined weight for the other two boxes. 21 – 9 = 12. So the lightest and heaviest boxes have to add up to 12.
Now, do we want to minimize or maximize the weight of the heaviest box?
The heaviest box has to be equal to or greater than 9 (because it is to the right of the median). We want to minimize the weight of this box in order to maximize the weight of the lightest box. So, the smallest possible weight for the heaviest box is 9.
If the heaviest box is minimized to 9, and the heaviest and lightest add up to 12, then the maximum weight for the lightest box is 3. The correct answer is C.
Make sense? If you’re sure you’ve got it, try this harder one. Set your timer for 2 minutes!
*A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?
(A) 10,700
(B) 10,800
(C) 10,900
(D) 11,000
(E) 11,100
In this case, there are 11 voting districts, each with some number of people. We’re asked to find the minimum possible population in the least populated district – or the smallest population that any one district could possibly have.
Let’s say that we’re going to minimize the population in District 1. Because all 11 districts have to add up to 132,000 people, we want to maximize the population in Districts 2 through 10. How can we do that? Now, we need more information from the problem:
“no district is to have a population that is more than 10 percent greater than the population of any other district”
So, if the smallest district has 100 people, then the largest district could have 110 people but can’t have any more than that. If the smallest district has 500 people, then the largest district could have 550 people but can’t have any more than that. How did we calculate those numbers? In each case, we take 10% of the original number and add that figure to the original number to give us our maximum.
In the given problem, we don’t know the number of people in the smallest district, so let’s call that x. If the smallest district is x, then calculate 10% and add that figure to x: x + 0.1x = 1.1x. So the largest district could be 1.1x but can’t be any larger than that.
We want to maximize all of the 10 remaining districts, so we should assume that all 10 districts are equal to 1.1x. As a result, we have (1.1x)(10) = 11x people in the 10 maximized districts (Districts 2 through 10), as well as our original x people in the minimized district (District 1).
The problem told us that all 11 districts add up to 132,000, so write that out mathematically:
11x + x = 132,000
12x = 132,000
x = 11,000
The correct answer is D.
Key Takeaways for Max/Min Problems:
(1) figure out what variables are “in play” (what figures we can manipulate in the problem)
(2) figure out whether each variable needs to be maximized or minimized in order to achieve the desired outcome (the thing the problem asks us to do)
(3) do the work (carefully, as always!)
* GMATPrep® questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Today’s GMAT challenge question comes from our friends at ManhattanGMAT. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution:
Problem
The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n3/3 + cn2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?
(A) 1,010
(B) 1,164
(C) 1,240
(D) 1,316
(E) 1,476
Solution
The brute-force way to solve this problem is literally to add up the first 15 positive perfect squares, from 1 to 225, inclusive. This is not necessarily completely out of bounds, given that we only have to sum up 15 numbers, all of which we should know already, and several of which are small. However, we should look for a shortcut using the formula.
Unfortunately, there is an unknown constant in the formula, but by using a small test number, we can solve for this constant. You can certainly pick n = 1, since it is a positive integer:
12 = 13/3 + c12 + 1/6
1 = 1/3 + c + 1/6
1/2 = c
If you feel uncomfortable picking n = 1, you can pick n = 2 and come to the same result almost as quickly.
Now, we plug n = 15 into the formula and solve:
12 + 22 + … + 152 = 153/3 + 152/2 + 15/6
= 15×15×15/3 + 15×15/2 + 15/6
= 15×15×5 + 15×15/2 + 5/2
= 225×5 + 225/2 + 5/2
= 1,125 + 230/2
= 1,125 + 115
= 1,240
The correct answer is (C).
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Posted by Clear Admit on November 18, 2009, at 8:00 pm
Posted in: GMAT - Verbal , GMAT News , GMAT Test Prep Company Series , GMAT Tips Today’s GMAT tip comes from the test prep firm ManhattanGMAT. In this article, ManhattanGMAT instructor Stacey Koprince analyzes an example of a GMATPrep® Sentence Correction (SC) question:
First, set your timer for 1 minute and 15 seconds and try the problem!
“The Achaemenid empire of Persia reached the Indus Valley in the fifth century B.C., bringing the Aramaic script with it, from which was derived both northern and southern Indian alphabets.
“A) the Aramaic script with it, from which was derived both northern and
“B) the Aramaic script with it, and from which deriving both the northern and the
“C) with it the Aramaic script, from which derive both the northern and the
“D) with it the Aramaic script, from which derives both northern and
“E) with it the Aramaic script, and deriving from it both the northern and”
Okay, have you got your answer? Now, let’s dive into this thing! Was everything okay in the original sentence or do you want to examine anything further?
On this one, perhaps the “, from which” in the original sentence caught your eye. This marker signifies a noun modifier. Noun modifiers are supposed to modify the closest main noun; in this case, that means the noun before the comma. That noun is the pronoun “it.” So, what is “it” referring to?
“It” is referring to the empire. So the noun modifier is telling us “the empire, from which was derived <some other alphabet>.” Can an alphabet be derived from an empire? Figuratively, perhaps, but not literally. Literally, one alphabet is derived from another alphabet – so the sentence should convey that meaning.
So, we’ve found an error in the original sentence, and we can immediately cross off choice A. Next, let’s scan the remaining answers to see whether we can reuse our new-found knowledge. Do any of the other choices repeat the error that we just found?
Answers C and D have “script, from which,” so they both fix that problem. Answers B and E change things up a bit – they introduce an “and” after the comma, so we no longer have a straight noun modifier marker.
Now, each person has to make a choice: do you want to try to figure out what’s happening with these new “and” markers in B and E? Or do you want to try to find something else? There isn’t one right answer to this question; it just depends upon your strengths and weaknesses. In this case, let’s examine B and E further.
B says “it, and from which deriving…” The “and” is a parallelism marker: X and Y. If this choice is correct, then, it should have some X and Y components that can be made parallel. Right after the “and,” we have the “from which” modifier marker, so this is the start of the Y component of the “X and Y” parallelism structure. What is the X? Ah, there’s the problem! We don’t have a parallel noun-modifier X component before the “and.” Eliminate B.
We know already that E also introduced an “and” at this point in the sentence, so let’s see if we can reuse our work from B on E. This time, we have “script, and deriving…” The word “deriving” is the start of our Y component; what is the parallel X component? Perhaps it’s the word “bringing” from the non-underlined portion? Let’s test it out.
“The empire reached the Valley, bringing with it <a script>, and deriving from it <some alphabets>.” These “comma –ing” structures are adverbial modifiers, which modify the preceding clause (subject and verb). In addition, the parallelism sets up certain expectations; for instance, when using the same pronoun in the same position in two parallel structures, the expectation is that the pronoun refers to the same noun both times. What does the first “it” refer to? Does the second “it” refer to the same noun?
No. The first “it” refers to the “empire,” while the second “it” refers to the “script.” Further, parallel structures should be able to be used independently to complete a sentence. We should be able to say: (1) “The empire reached the Valley, bringing with it a script.” (2) “The empire reached the Valley, deriving from it some alphabets.” What does the “it” refer to in the second sentence? The Valley? The empire? Neither one makes sense. Eliminate E.
So now we’ve narrowed it down to C and D, both of which use a “script, from which” construction. Scan the two choices vertically, comparing parts until you find any differences. There are only two: one uses the singular “derives” while the other uses the plural “derive,” and one includes “the” in front of both northern and southern while the other does not use “the” for either one.
The derives / derive split seems as though it should be straightforward – we just have to determine whether we need the verb to be singular or plural, right? Great. So, what is the subject that matches with this verb?
Ask yourself who or what is doing the action – what derives from what? The script derives from the alphabets? Or the alphabets derive from the script? The original sentence describes the latter scenario. Alternatively, the sentence could be written in the passive voice: “the script is derived from the alphabets.”
Compare these structures to choices C and D; do they say “is derived”? No, the GMATPrep® sentence is not written in passive voice; rather, it uses the active voice: “derive/derives.” So the subject is actually “the northern and southern alphabets,” which is plural. The verb should be the plural “derive.” Eliminate D. The correct answer, by process of elimination, is C.
This last bit of analysis also shows the biggest trap in this problem: many students will think that the subject is “script” and that the verb should therefore be “derives.” Consequently, those students will eliminate the right answer, C, and choose a very tempting wrong answer, D. Another possible trap is exhibited in choice E, something we call false parallelism: students will like the apparent parallelism between “bringing with it” and “deriving from it,” but the two are only superficially parallel. There is no parallelism between the two pronouns and the second item doesn’t make sense in the context of the entire sentence.
The major take-aways here:
(1) when doing SC, reuse your prior analysis as much as you can
(2) scan SC answer choices vertically to find differences; don’t read horizontally
(3) know how to recognize and properly construct noun modifiers and adverbial modifiers
(4) watch out for parallelism markers – the markers are often little words but they can make a big difference!
* GMATPrep® question courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Today’s sample problem comes from our friends at ManhattanGMAT. To help prepare for the exam, see if you can solve the problem first, then read on for the correct answer and explanation.
Problem
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
(A) 17/300
(B) 1/15
(C) 1/12
(D) 1/10
(E) 3/25
Solution
First, make sure that you grasp the question. To find the desired probability, we need to count the integers between 1 and 300 that fit the given constraint. (We will then divide this count by 300, to determine the final answer.)
The constraint is worded in a confusing way, so we should attempt to reword it. If necessary, put in sample numbers to make the conditions make sense. We need an “integer raised to an exponent that is an integer greater than 1.” In other words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive.
Now we need to count these perfect squares, etc. in an efficient way. Let’s start with the squares. What’s the biggest perfect square less than 300? Test numbers if necessary. 162 = 256 and 172 = 289, but 182 = 324. Thus, we have 12 through 172, for 17 perfect squares.
Now, let’s count the cubes. Leave out 13, since we’ve already counted 1 as 12. 53 = 125 and 63 = 216, but 73 = 343. Thus, we have 5 more integer powers (23 through 63, inclusive), for a cumulative total of 22.
What about the fourth powers – 24, 34, etc.? We’ve already counted any of these that matter, because they are also perfect squares: 24 = 42, 34 = 92, etc. We can leave out any higher even powers for the same reason (26 = 82, 28 = 162, etc.).
However, we must consider additional odd powers, continuing to leave out 1 raised to any power. 25 = 32, 35 = 243, but 45 = something greater than 300, as we can see by considering that 45 = 44 × 4 = 162 × 4 = 256 × 4. So we have two more integer powers (25 and 35), for a cumulative total of 24.
Seventh powers: 27 = 128, but 37 = something much greater than 300 (since 37 = 35 × 32 = 243 × 9). Be sure to stop calculating when you see that the result is outside the bounds of the problem. We have 1 more power, for a cumulative total of 25.
Ninth powers: 29 = 128 × 4 = greater than 300. So we can stop here.
Finally, we compute 25/300 = 1/12.
The correct answer is (C).
For more information on ManhattanGMAT, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at nine different firms!
Posted by Clear Admit on November 11, 2009, at 8:00 pm
Posted in: General , GMAT - Quantitative , GMAT News , GMAT Test Prep Company Series , GMAT Tips Today’s GMAT tip comes from the folks at test prep firm Manhattan Review, who have some advice on how to handle rate problems on the GMAT:
Rate problems are very prevalent on the Quantitative section of the GMAT, particularly in Problem Solving, but to a lesser extent, Data Sufficiency as well. The majority of rate problems require solid proficiency in ratio concepts and knowledge of the two primary rate formulae:
Speed Formula: time x speed = distance
Productive Work Formula: time x rate = units produced
Which formula you use has everything to do with the question, and you will need to adjust the formula accordingly. Rate problems are, by definition, word problems, so being able to translate the information in the question into mathematical form is essential.
Let’s try a practice question:
Two buses, A and B, started simultaneously from opposite ends of the same 50-mile bus route and traveled toward each other, making their regular trips. Bus A, traveling at a constant rate, completed the 50-mile trip in 2 hours; Bus B, traveling at a constant rate, completed the 50-mile trip in 1 hour. How many miles had Bus A traveled when it met Bus B?
From this information, we can determine the speed of the two buses easily:
Bus A – 50 miles/2 hrs = 25 miles/hr
Bus B – 50 miles/1 hr = 50 miles/hr
The ratio of the Bus A to Bus B is 2:1.
Now, adjust the rate formula so that “time” is the amount of time passed when the buses meet. Bus A will have traveled 25(time) and Bus B will have traveled 50(time). Logically, something may occur to you right away – the entire length of the route has been covered, by either Bus A or Bus B. That means, together, at their time of meeting, the combined distance of the two buses is 50 miles. Now, set up your equation again.
25(time) + 50(time) = 50 miles
Combine like terms: 75(time) = 50 miles
time = 2/3 hour (or 40 minutes, but do not change units halfway through a question!)
Now, to figure out the distance covered by Bus A, plug the time and speed into our formula again:
2/3 hour x 25 miles/hr = 16 2/3 miles.
The exceptional flexibility of this formula is what makes it so useful, but also what can make a rate problem much trickier than the word problems you may remember from your pre-college days. This question requires you to know that you can use combined rates without changing the formula’s format.
The weekend is a great time to get some GMAT studying done. Today we share with you a challenge problem, provided by our friends at ManhattanGMAT. To help prepare for the exam, see if you can solve the problem first, then read on for the correct answer and explanation.
Problem
How many different sets of positive square integers, each greater than 1, add up to 75?
(A) 1
(B) 4
(C) 7
(D) 10
(E) 12
Solution
First, lay out the possible numbers you can use in the sum. The positive square integers greater than 1 but less than 75 are 4, 9, 16, 25, 36, 49, and 64.
Now, let’s create a possible set and then see whether we can adjust it. It’s fairly obvious that three 25’s add up to 75, so our first set is {25, 25, 25}.
We might now recall the most famous example of the Pythagorean Theorem: 9 + 16 = 25. So we can swap out, successively, a 25 and replace it with a 9 and a 16. With sets, order does not matter, so we get three more possible sets:
{25, 25, 9, 16}
{25, 9, 16, 9, 16}
{9, 16, 9, 16, 9, 16}
This gives us 4 sets so far. However, we can now swap out 16’s for four 4’s. We can do so as follows.
One possible swap for the set with one 16:
{25, 25, 9, 4, 4, 4, 4}
Two possible swaps for the set with two 16’s:
{25, 9, 16, 9, 4, 4, 4, 4}
{25, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
And three possible swaps for the set with three 16’s:
{9, 16, 9, 16, 9, 4, 4, 4, 4}
{9, 16, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
Before going to the larger squares, we should glance over our list and see whether we can do any swaps within the sets we’ve already created, using only squares equal to 25 or less. The only swap we can do is in the last set: we can swap out nine 4’s and replace them with four 9’s:
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4} = three 9’s and twelve 4’s
becomes
{9, 9, 9, 9, 9, 9, 9, 4, 4, 4} = seven 9’s and three 4’s
We are now at a total of 11 sets, having exhausted the possibilities that only involve the squares equal to 25 or less. Are there any sets that involve larger squares?
We can quickly check:
64 can’t be in the set, because the leftover (11) can’t be formed from the sum of 9’s and/or 4’s.
49 can’t be in the set, because the leftover (26) can’t be formed from the sum of 25’s, 16’s, 9’s, and/or 4’s.
36 CAN be in the set. The leftover (39) can be written as the sum of three 9′s and three 4′s, so we get
{36, 9, 9, 9, 4, 4, 4}
Thus, the total number of different sets is 12.
The correct answer is (E).
Posted by Clear Admit on November 4, 2009, at 8:00 pm
Posted in: General , GMAT - Verbal , GMAT News , GMAT Test Prep Company Series , GMAT Tips For today’s GMAT tip, test prep firm Manhattan Review shares their advice on how to work through a sentence correction question on the GMAT:
There are three types of wrong answer when it comes to sentence corrections:
1. Well, I figured I got that one wrong. That was tricky.
2. Oh, is that so? I’ve got to study that concept, there.
3. What?!? That’s impossible! I’m calling GMAC to tell them they’ve made an error!
This post is in honor of the third type of sentence correction.
Let’s go through one that fits the category, or at least it does to me:
Among the objects found in the excavated temple were small terra-cotta effigies left by supplicants who were either asking the goddess Bona Dea’s aid in healing physical and mental ills or thanking her for such help.
(A) in healing physical and mental ills or thanking her for such help
(B) in healing physical and mental ills and to thank her for helping
(C) in healing physical and mental ills, and thanking her for helping
(D) to heal physical and mental ills or to thank her for such help
(E) to heal physical and mental ills or thanking her for such help
(This is from the 11th Official Guide, p. 654)
Now, at first glance, I knew there was something wrong – I needed to find an answer choice with better parallelism! So I chose D — perfectly parallel, and it made beautiful sense.
UNTIL I discovered that the answer was A!
I could not describe my bafflement! Until I thought it over and worked the question out again. So, here is the fruit of my labor for your benefit:
What I failed to do is test the parallelism of my answer choice in the CONTEXT of the original sentence. If you look carefully, you’ll see that the initial verb is NOT “healing” but “asking”! This is a traditional either/or structure, and all verbs must be in gerund form. With this in mind, when you look at choices B, C, D, and E, it is easy to see where they fall short:
B makes the phrase “either/and” instead of “either/or.”
C mixes gerund with infinitive.
D, my original choice, is actually the worst of them all! Not only does the infinitive break the parallel structure initiated by “asking,” but it is inappropriate to follow the word “aid” with an infinitive.
E mixes gerund with infinitive.
Just because the answer choice seems to have good parallel structure, that doesn’t mean that it will remain neatly parallel in the context of the original sentence. Good luck everybody!
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