How to Answer GMAT Critical Reasoning Questions Involving Experiments

QuestioningThere are certain themes that crop up in Critical Reasoning questions so often that it’s worthwhile to treat these problem types as their own sub-categories. One category that shows up with greater frequency in each new edition of the Official Guide is one that I’ll christen, “The tainted experiment.”

The logic of these arguments is always rooted in the notion that we can only trust the results of the experiment if we have a legitimate control group, and there aren’t any other confounding variables that we’ve failed to account for. Spoiler alert: typically in GMAT questions, we will find such confounding variables tainting the experiment’s predictive value.

Imagine, for example, that you’re testing a drug designed to alleviate headaches. You have two groups of subjects: a control group that takes a placebo and an experimental group that receives the drug. The results of the experiment show that the control group has a higher rate of headaches than the group receiving the medication. Time to rejoice, notify the delighted shareholders, and move this drug to market as quickly as possible? Well, maybe.

But now imagine that the control group consisted largely of stressed-out, sleep-deprived college students living near construction sites, and the experiment group consisted of retired yoga instructors. Suddenly we’ve got other variables to contend with. Yes, it’s possible that the effectiveness of the drug is what accounts for the differential in headache incidence between the two groups. But it’s just as likely that other environmental factors are responsible. A good experiment would have controlled for these factors.

The upshot: whenever you see a question that involves an experiment with a control group, always ask yourself if there are variables that the experimenters have failed to account for.

Here’s a good example of such an argument:

In Colorado subalpine meadows, nonnative dandelions co-occur with a native flower, the larkspur. Bumblebees visit both species, creating the potential for interactions between the two species with respect to pollination. In a recent study, researchers selected 16 plots containing both species; all dandelions were removed from eight plots; the remaining eight control plots were left undisturbed. The control plots yielded significantly more larkspur seeds than the dandelion-free plots, leading the researchers to conclude that the presence of dandelions facilitates pollination (and hence seed production) in the native species by attracting more pollinators to the mixed plots. 

Which of the following, if true, most seriously undermines the researchers’ reasoning? 

A) Bumblebees preferentially visit dandelions over larkspurs in mixed plots.
B) In mixed plots, pollinators can transfer pollen from one species to another to augment seed production. 
C) If left unchecked, nonnative species like dandelions quickly crowd out native species. 
D) Seed germination is a more reliable measure of a species’ fitness than seed production.
E) Soil disturbances can result in fewer blooms, and hence lower seed production. 

This is a classic experiment argument. There are two populations: plots that contain both dandelions and larkspurs, and plots that have had all the dandelions removed, and thus contain only larkspurs. We’re told that the plots containing both types of flowers produced more larkspur seeds than the plots containing only larkspurs, thus validating the contention that the presence of dandelions has a positive benefit on larkspur seed yields.

Fortunately, the GMAT is pretty predictable. If we’re trying to weaken the conclusion derived from an experiment comparing two populations – a control group and an experimental group – we’re looking for a confounding variable. The initial hypothesis is that the presence of dandelions promotes seed production in larkspurs. An alternative hypothesis is that an environmental factor we haven’t yet considered accounts for the differential in larkspur seed production in the two groups, so that’s what we’re on the lookout for when we examine each of the answer choices.

A) Which flower bees prefer sheds no light on the validity of the experiment. A is out.

B) This answer option would be entirely consistent with the hypothesis that dandelions promote larkspur seed production. We’re trying to weaken the argument. B is also out.

C) This answer choice makes no sense. We’ve already been told that the plots containing both types of flower produce more larkspur seeds – we never want to contradict a premise. C is no good.

D) This tells us nothing about whether it is the presence of dandelions that’s helping promote larkspur seed production. D gets kicked to the curb.

E) If removing the dandelions disrupts the soil, perhaps it’s the disrupted soil, rather than the absence of dandelions, that accounts for the lower larkspur production in the plots where the dandelions have been removed. We’ve got our confounding variable – E is the answer.

Takeaway: On Critical Reasoning questions on the lookout for the tainted experiment. If you’re trying to weaken an argument regarding an experiment containing a control group and an experimental group, the key will be determining which answer choice provides a confounding variable, and thus, an alternative explanation for the conclusion given.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

Quarter Wit, Quarter Wisdom: The 3-Step Method to Solving Complex GMAT Algebra Problems

Quarter Wit, Quarter WisdomIf you have been practicing GMAT questions for a while, you will realize that not every question can be solved using pure algebra, especially at higher levels. There will be questions that will require logic and quite a bit of thinking on your part.  These questions tend to throw test-takers off – students often complain, “Where do I start from? Thinking through the question takes too much time!” Unfortunately, there is no getting away from such questions.

Today, let’s see how to handle such questions step-by-step by looking at an example problem:

N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?

(A) 29
(B) 49
(C) 58
(D) 113
(E) 131

This is not a simple algebra question, where we are asked to make equations and solve them.

We are given 6 digits: 1, 2, 3, 6, 7, 8. Each digit needs to be used to form two 3-digit numbers. This means that we will use each of the digits only once and in only one of the numbers.

We also need to minimize the difference between the two numbers so they are as close as possible to each other. Since the numbers cannot share any digits, they obviously cannot be equal, and hence, the smaller number needs to be as large as possible and the greater number needs to be as small as possible for the numbers to be close to each other.

Think of the numbers  of a number line. You need to reduce the difference between them. Then, under the given constraints, push the smaller number to the right on the number line and the greater number to the left to bring them as close as possible to each other.

STEP 1:
The first digit (hundreds digit) of both numbers should be consecutive integers – i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3** (the difference between the latter will certainly be more than 100).

We get lots of options for hundreds digits: (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**). All of these options could satisfy our purpose.

STEP 2:
Now let’s think about what the next digit (the tens digit) should be. To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1, if possible) and the tens digit of the smaller number should be as large as possible (8, if possible). So let’s not use 1 or 8 in the hundreds places and reserve them for the tens places instead, since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?

Let’s try to make a pair of numbers in the form of 2** and 3**. We need to make the 2** number as large as possible and make the 3** number as small as possible. As discussed above, the tens digit of the smaller number should be 8 and the tens digit of the greater number should be 1. We now have 28* and 31*.

STEP 3:
Now let’s use the same logic for the units digit – make the units digit of the smaller number as large as possible and the units digit of the greater number as small as possible. We have only two digits left over – 6 and 7.

The two numbers could be 287 and 316 – the difference between them is 29.

Let’s try the same logic on another pair of hundreds digits, and make the pair of numbers in the form of 6** and 7**. We need the 6** number to be as large as possible and the 7** number to be as small as possible. Using the same logic as above, we’ll get 683 and 712. The difference between these two is also 29.

The smallest of the given answer choices is 29, so we need to think no more. The answer must be A.

Note that even if you try to express the numbers algebraically as:

N = 100a + 10b + c
M = 100d + 10e + f

a lot of thought will still be needed to find the answer, and there is no real process that can be followed.

Assuming N is the greater number, we need to minimize N – M.

N – M = 100 (a – d) + 10( b – e) + (c – f)

Since a and d cannot be the same, the minimum value a – d can take is 1. (a – d) also cannot be negative because we have assumed that N is greater than M. With this in mind, a and d must be consecutive (2 and 1, or 3 and 2, or 7 and 6, etc). This is another way of completing STEP 1 above.

Next, we need to minimize the value of (b – e). From the available digits, 1 and 8 are the farthest from each other and can give us a difference of -7. So b = 1 and e = 8. This leaves the consecutive pairs of 2, 3 and 6, 7 for hundreds digits. This takes care of our STEP 2 above.

(c – f) should also have a minimum value. We have only one pair of digits left over and they are consecutive, so the minimum value of (c – f) is -1. If the hundreds digits are 3 and 2, then c = 6 and f = 7. This is our STEP 3.

So, the pair of numbers could be 316 and 287 – the difference between them is 29. The pair of numbers could also be 712 and 683 – the difference between them is also 29.

In either case, note that you do not have a process-oriented approach to solving this problem. A bit of higher-order thinking is needed to find the correct answer.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving the Pouring Water Puzzle

Quarter Wit, Quarter WisdomSome time back, we came across a GMAT Data Sufficiency word problem question based on the “pouring water puzzle”. That made us think that it is probably a good idea to be comfortable with the various standard puzzle types. From this week on, we will look at some fundamental puzzles to acquaint ourselves with these mind benders in case we encounter them on test day.

Today, we will look at the popular “pouring water puzzle”. You may remember a similar puzzle from the movie Die Hard with a Vengeance, where Bruce Willis and Samuel L. Jackson had to diffuse a bomb by placing a 4 gallon jug of water on a set of scales.

Here is the puzzle:

You have a 3- and a 5-liter water container – each container has no markings except for that which gives us its total volume. We also have a running tap. We must use the containers and the tap in such a way that we measure out exactly 4 liters of water. How can this be done?

Don’t worry that this question is not written in a traditional GMAT format! We need to worry only about the logic behind the puzzle – we can then answer any question about it that is given in any GMAT format.

Let’s break down what we are given. We have only two containers – one of 3-liter and the other of 5-liter capacity. The containers have absolutely no markings on them other than those which give us the total volumes, i.e. the markings for 3 liters and 5 liters respectively. There is no other container. We also have a tap/faucet of running water, so basically, we have an unlimited supply of water. Environmentalists may not like my saying this, but this fact means we can throw out water when we need to and just refill again.

Now think about it:

STEP 1: Let’s fill up the 5-liter container with water from the tap. Now we are at (5, 0), with 5 being the liters of water in the 5-liter container, and 0 being the liters of water in the 3-liter container.

STEP 2: Now, there is nothing we can do with this water except transfer it to the 3-liter container (there is no other container and throwing out the water will bring us back to where we started). After we fill up the 3-liter container, we are left with 2 liters of water in the 5-liter container. This brings us to (2, 3).

STEP 3: We gain nothing from transferring the 3 liters of water back to 5-liter container, so let’s throw out the 3 liters that are in the 3-liter container. Because we just threw out the water from the 3-liter container, we will gain nothing by simply refilling it with 3 liters of water again. So now we are at (2, 0).

STEP 4: The next logical step is to transfer the 2 liters of water we have from the 5-liter container to the 3-liter container. This means the 3-liter container has space for 1 liter more until it reaches its maximum volume mark. This brings us to (0, 2).

STEP 5: Now fill up the 5-liter container with water from the tap and transfer 1 liter to the 3-liter container (which previously had 2 liters of water in it). This means we are left with 4 liters of water in the 5-liter container. Now we are at (4, 3).

This is how we are able to separate out exactly 4 liters of water without having any markings on the two containers. We hope you understand the logic behind solving this puzzle. Let’s take a look at another question to help us practice:

We are given three bowls of 7-, 4- and 3-liter capacity. Only the 7-liter bowl is full of water. Pouring the water the fewest number of times, separate out the 7 liters into 2, 2, and 3 liters (in the three bowls).

This question is a little different in that we are not given an unlimited supply of water. We have only 7 liters of water and we need to split it into 2, 2 and 3 liters. This means we can neither throw away any water, nor can we add any water. We just need to work with what we have.

We start off with (7, 0, 0) – with 7 being the liters of water in the 7-liter bowl, the first 0 being the liters of water in the 4-liter bowl, and the second 0 being the liters of water in the 3-liter bowl – and we need to go to (2, 2, 3). Let’s break this down:

STEP 1: The first step would obviously be to pour water from the 7-liter bowl into the 4-liter bowl. Now you will have 3 liters of water left in the 7-liter bowl. We are now at (3, 4, 0).

STEP 2: From the 4-liter bowl, we can now pour water into the 3-liter bowl. Now we have 1 liter in the 4-liter bowl, bringing us to (3, 1, 3).

STEP 3: Empty out the 3-liter bowl, which is full, into the 7-liter bowl for a total of 6 liters – no other transfer makes sense [if we transfer 1 liter of water to the 7-liter bowl, we will be back at the (4, 0, 3) split, which gives us nothing new]. This brings us to (6, 1, 0).

STEP 4: Shift the 1 liter of water from the 4-liter bowl to the 3-liter bowl. We are now at (6, 0, 1).

STEP 5: From the 7-liter bowl, we can now shift 4 liters of water into the 4-liter bowl. This leaves us with with 2 liters of water in the 7-liter bowl. Again, no other transfer makes sense – pouring 1 liter of water into some other bowl takes us back to a previous step. This gives us (2, 4, 1).

STEP 6: Finally, pour water from the 4-liter bowl into the 3-liter bowl to fill it up. 2 liters will be shifted, bringing us to (2, 2, 3). This is what we wanted.

We took a total of 6 steps to solve this problem. At each step, the point is to look for what helps us advance forward. If our next step takes us back to a place at which we have already been, then we shouldn’t take it.

Keeping these tips in mind, we should be able to solve most of these pouring water puzzles in the future!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Why Logic is More Important Than Algebra on the GMAT

QuestioningOne common complaint I get from students is that their algebra skills aren’t where they need to be to excel on the GMAT. This complaint, invariably, is followed by a request for additional algebra drills.

If you’ve followed this blog for any length of time, you know that one of the themes we stress is that Quantitative Reasoning is not, primarily, a math test. Though math is certainly involved – How could it not be? – logic and reasoning are far more important factors than conventional mathematical facility. I stress this in every class I teach. So why the misconception that we need to hone our algebra chops?

I suspect that the culprit here is the explanations that often accompany official GMAC questions. On the whole, they tend to be biased in favor of purely algebraic solutions.  They’re always technically correct, but often suboptimal for the test-taker who needs to arrive at a solution within two minutes. Consequently, many students, after reviewing these solutions and arriving at the conclusion that they would not have been capable of the hairy algebra proffered in the official solution, think they need to work on this aspect of their prep. And for the most part it isn’t true.

Here’s a good example:

If x, y, and k are positive numbers such that [x/(x+y)]*10 + [y/(x+y)]*20 = k and if x < y, which of the following could be the value of k? 

A) 10
B) 12
C) 15
D) 18
E) 30

A large percentage of test-takers see this question, rub their hands together, and dive into the algebra. The solution offered in the Official Guide does the same – it is about fifteen steps, few of them intuitive. If you were fortunate enough to possess the algebraic virtuosity to solve the question in this manner, you’d likely chew up 5 or 6 minutes, a disastrous scenario on a test that requires you to average 2 minutes per problem.

The upshot is that it’s important for test-takers, when they peruse the official solution, not to arrive at the conclusion that they need to solve this question the same way the solution-writer did. Instead, we can use the same simple strategies we’re always preaching on this blog: pick some simple numbers.

We’re told that x<y, but for my first set of numbers, I like to make x and y the same value – this way, I can see what effect the restriction has on the problem. So let’s say x = 1 and y = 1. Plugging those values into the equation, we get:

(1/2) * 10 + (1/2) * 20  = k

5 + 10 = k

15 = k

Well, we know this isn’t the answer, because x should be less than y. So scratch off C. And now let’s see what the effect is when x is, in fact, less than y. Say x = 1 and y = 2. Now we get:

(1/3) * 10 + (2/3) * 20  = k

10/3 + 40/3 = k

50/3 = k

50/3 is about 17. So when we honor the restriction, k becomes larger than 15. The answer therefore must be D or E. Now we could pick another set of numbers and pay attention to the trend, or we can employ a bit of logic and common sense. The first term in the equation x/(x+y)*10 is some fraction multiplied by 10. So this term, logically, is some value that’s less than 10.

The second term in the equation is y/(x+y)*20, is some fraction multiplied by 20, this term must be less than 20. If we add a number that’s less than 10 to a number that’s less than 20, we’re pretty clearly not going to get a sum of 30. That leaves us with an answer of 18, or D.

(Note that if you’re really savvy, you’ll recognize that the equation is a weighted average. The coefficients in the weighted average are 10 and 20. If x and y were equal, we’d end up at the midway point, 15. Because 20 is multiplied by y, and y is greater than x, we’ll be pulled towards the high end of the range, leading to a k that must fall between 15 and 20 – only 18 is in that range.)

Takeaway: Never take a formal solution to a problem at face value. All you’re seeing is one way to solve a given question. If that approach doesn’t resonate for you, or seems so challenging that your conclusion is that you must purchase a host of textbooks in order to improve your formal math skills, then you haven’t absorbed what the GMAT is really about. Often, the relevant question isn’t, “Can you do the math?” It’s, “Can you reason your way to the answer without actually doing the math?”

*Official Guide question courtesy of the Graduate Management Admissions Council.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

GMAT Tip of the Week

A Logical Approach To Sentence Correction

(This is one of a series of GMAT tips that we offer on our blog.)

There is probably no more illogical discipline in academia than English grammar, in which seemingly the only rule is that there are generally several exceptions to each rule. Alas, as English has become a primary language of the business world, the GMAT requires examinees to be proficient in a number of grammatical devices through the Sentence Correction style of question, which comprises 1/3 of the Verbal section of the exam. Fear not (or, maybe fear slightly), however, as the GMAT focuses on a limited scope of the English language, and the Veritas Prep verbal syllabus teaches those particular skills thoroughly.

Because of the seemingly-illogical nature of the English language, it would seem illogical that some of the more-difficult GMAT Sentence Correction questions can be answered using logic. Say, for example, that you narrowed your answer choices down to the following:

B) Gary mistook the car, which was backfiring, for a gunshot.

D) Gary mistook the sound of a backfiring car for a gunshot.

While neither may appear to have a distinct grammatical error, answer choice B is simply illogical – a gunshot is a sound (or event) while a car is an object, and therefore one could not logically mistake the car for the gunshot. D, which correctly equates the sound of the car with the sound of the gunshot, is an accurate sentence, and is thus correct.

How can you use this to your advantage? Consider this — it doesn’t behoove business schools to admit students based upon their knowledge of obscure and complicated English language idioms, but it does make sense for schools to admit students who can think logically. While certain elements of grammar are distinctly testable, and you need to understand them thoroughly, if you don’t see an opportunity to employ one of the major error types, your time may be better spent by considering the logic of the remaining options than by worrying over tiny constructions of idiomatic grammar.

For more GMAT prep tips and resources, visit Veritas Prep.