The post The Speedy Road to Insufficiency appeared first on .

]]>Let’s take a look at the following Official Guide DS problem:

*If *p* and *q* are positive integers and *pq* = 24, what is the value of *p*?*

*(1) *q*/6 is an integer*

*(2) *p*/2 is an integer*

It just so happens that this is a value-based question: When we’re asked “what is the value of *p*?”, we’re being asked to provide a single value for p. If a statement or combination of statements cannot give us a single value, then it must be insufficient.

So, before you go all crazy trying to set up equations, think about what I like to call “The Speedy Road to Insufficiency.” What does that mean? Well, in a nutshell, it means that when approaching a statement or combination of statements, you should treat it as “insufficient until proven sufficient.” In other words, go into it trying to demonstrate insufficiency. Why? Well, in short, because it’s faster. How so? Well, in order to prove insufficiency, all we have to do is find two different possible values for *p* that satisfy the statement(s).* * If we can quickly locate two such values, we don’t have to do any more work. We know the statement *must* be insufficient.

Can we do this for Statement (1)? Well, if *q*/6 is an integer and *pq* = 24, then *q *could be 6, 12, or 24, and the corresponding values of *p* would be 4, 2, and 1, respectively. We just found three possible values of *p*. Guess what? We’re done with Statement (1). Definitely insufficient. And in truth, you could have stopped the moment you realized *p *could be 4 or 2. Two possible values of *p* are enough to demonstrate insufficiency.

What about Statement (2)? Well, if *p*/2 is an integer and *pq* = 24, then *p* could be 2, 4, 6, 8, 12, or 24. Definitely not just one value of *p*, so we know Statement (2) is insufficient.

Now, I’ll just give this away and tell you that Statements (1) and (2) are still insufficient when combined. But before reading the next paragraph, see if you can spot the speediest way to determine that.

Did you find it? Well, notice that Statements (1) and (2) both say that *p *could be 2 or 4. That’s it…don’t do any more work! You’ve got two possible values of *p*, and thus you’ve shown insufficiency.

This strategy also works for “Yes/No” questions, which include phrases such as “Is x odd?” or “Is y > 2?” These questions don’t ask for a specific value but instead ask you to answer “yes” or “no” to a specific question. But the strategy is the same: ”Insufficient until proven sufficient”. If you can show quickly that the answer could be either “yes” or “no”, you’ve shown insufficiency and can move on. See if you can apply the strategy to this official problem:

*If *x ≠ -y*,* *is (*x-y*) / (*x+y*) > 1 ?*

*(1) x > 0*

Don’t just post your solution! Also include the way you made things speedy.

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]]>The post Intuition and Common Sense on Geometry DS Questions appeared first on .

]]>Here’s what I mean: Suppose I ask you the question “What is x?” I then give you a statement that says “2,346.456 x + 87,234 = 912,347π”. Is the statement sufficient?

“Yes,” you’d say (hopefully!).

“But do you know what x is?” I’d respond back.

And what would you say? Yup, you got it: ”Who cares?!”

All we care about is the *ability* to find x. We don’t care what x actually is. Because the statement gives us a simple linear equation with only one variable, we have the ability to find the value of that variable, and that’s all that matters.

This seems a pretty elementary point in and of itself, but it’s one that many students often forget when out in the field tackling tough DS problems. Specifically on Geometry DS problems, the temptation can be to plug numbers into formulas and tackle the problem as if it were a PS question. But often times, this is completely unnecessary and a tremendous waste of time. Many times, you can solve Geometry DS problems intuitively using common sense and simple logic. But it all hinges on the ability to identify exactly what information you need.

Let’s take a look at the following official DS problem:

*A circular tub has a band painted around its circumference, as shown above. What is the surface area of this painted band?*

*(1) x = 0.5*

*(2) The height of the tub is 1 meter*

Stop! Don’t write any formulas! It’s great if you know the formula for the volume of a right circular cylinder, and that might come in handy on PS problems and maybe a more intricate DS problem. But let’s take a moment to think about what information is really necessary here. We want to know what the surface area of that band is. Ask yourself: What’s keeping us from knowing that? What’s missing?

Well, we can’t very well know the surface area if we don’t know how wide the cylinder is. What determines how wide it is? Radius! And if we know radius, we also know circumference. But is that enough? Nope. We also don’t know how high x is. So the two missing pieces of info can be boiled down to: ”x = ? and r = ?”

This makes perfect sense when you think about it. How can you know the surface area of something if you don’t know its dimensions? In this case, the two dimensions are the circumference around the cylinder (which can be determined by radius) and the height of the band, and we need both to get the surface area.

Now that we’ve figured out intuitively what information we need, let’s look at the statements:

Statement (1) gives us the value of x. Great…nothing about the radius, though. Insufficient.

Statement (2) gives us the height of the entire cylinder. Great…nothing about either the radius or the value of x. Insufficient.

Statements (1) and (2) together give us the height of the entire cylinder and the value of x. Awesome….where’s the value of the radius? Still nowhere to be found. Answer: E.

We didn’t write down a single equation, and yet we still got out of the problem quickly and with the correct answer.

Now, try your intuitive skills on this other official DS geometry problem. Remember, try to do it without equations! Use your common sense! And post your step-by-step intuitive solutions in the comments!

*The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with *k* identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high what is the value of *k*?*

*(1) Each of the cans has a radius of 4 centimeters.*

*(2) Six of the cans fit exactly along the length of the carton.*

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]]>The post The AWA Isn’t That Important… So Can I Blow It Off? appeared first on .

]]>But many of you also probably know that the essay portion is not nearly as consequential as the multiple-choice portions. The essays are *not* part of your 800 composite score and are instead scored on a separate scale out of 6.0 in decrements of 0.5. Generally, admissions committees use the essay scores to judge whether or not you actually wrote your application essays to the school. If you write a fantastic admissions essay filled with prose worthy of a Pulitzer but get a 2.0 AWA score, the admissions committee’s going to be suspicious.

But in the end, your composite score out of 800 weighs much more heavily on the committee’s collective mind. It’s easy, then, to come up with arguments for blowing off the essay and focusing the bulk of your attention on the multiple-choice sections. But is this really wise?

One time, a student told me that he was going to take the GMAT a second time. The first time he took the test, he didn’t achieve his desired composite score. But he got a perfect 6.0 on the AWA. So in his mind, this justified the following strategy: “I’m going to completely skip the essays the next time I take the test. It’s not important anyway.”

Several students have suggested similar strategies to me, and I understood where they were coming from. Skipping the essays outright makes the test shorter and allows you to conserve some mental energy for the multiple-choice sections. But consider the following scenario: You’re on the admissions committee of a prestigious business school. You’re reviewing an applicant, and you see that he got a 6.0 AWA on the GMAT on his first try, but got a 0.0 the second time. You don’t like the sight of the 0.0, and although you recognize that the applicant has the capability to score a 6.0, it’s readily apparent (at least in your mind) that the applicant feels cocky or lax enough to blow a whole section of the test off and just assume that it won’t matter.

Now, it may be true that some individual committees won’t care as much. But are you willing to take that risk? What if you’re applying to a competitive school that requires you to present every advantage you can? Allowing a 0.0 to show up on your transcript doesn’t seem very prudent.

Besides, does the one hour of essay-writing really drain you of so much energy that it makes a statistically significant difference on your performance for the rest of the test? You might assume the answer is yes. But it’s a pretty tenuous generalization. If you took the GMAT hundreds of times with the AWA and hundreds of times without, maybe we’d have some statistical data to go on. But as is, assuming the AWA will take away your mojo so much that your score will drop by a significant margin is unfounded speculation at best.

Another thing to keep in mind: some students actually like the essays, because they allow you to shake off initial nerves. You get the least important part of the test out of the way first, and by the time you’re done, you’ve settled into a groove and gotten used to your surroundings. For some students, this allows them to transition into the multiple-choice sections more relaxed and clear-minded. This might be the case for you too.

So all in all, you might in theory be able to get away with blowing off the essays. But all things considered, it’s not a risk worth taking, and a poor or nonexistent AWA score will not look good to admissions committees. The essays are the least consequential part of the test, so investing just a little bit of time and energy in them should not have a great effect on your composite score.

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]]>The post Tips for Comparing Fractions on the GMAT appeared first on .

]]>But is that always true?

Well, in this case, sure, it’s true. But what if I asked to compare, for example, -1/3 and -1/4? Now, it would be incorrect to say the number with the larger denominator is smaller. Now that we’re on the negative side of the number line, everything is reversed. -1/4 is actually larger than -1/3.

The confusion is only compounded when we get to fractions involving exponents. Let’s say *b* is a proper fraction (i.e. a number between 0 and 1). It’s easy enough to see that, for example, *b^5* will be less than *b^3*. For example, (1/2)^5 = 1/32 is less than (1/2)^3 = 1/8.

But students often extend this to a general rule and say something like, “The higher the exponent, the smaller the fraction.” Of course, this isn’t always true. If *b* were a negative proper fraction (i.e. between -1 and 0), then *b^5* would actually be larger than *b^3* (e.g. -1/32 > -1/8).

So how to work around this confusion? Simply put: Get rid of the terms “larger” and “smaller”. They’ll only cause trouble, because the rules switch once you go from positive to negative or vice versa. Instead, think about the fractions in terms of *their relationship to zero*.

Let’s look at our first example. We found that 1/4 is smaller than 1/3, because the fractions are both positive. Then we found that -1/4 is larger than -1/3, because the fractions are both negative. What’s the common thread? In each case, the number with the larger denominator is *closer to zero*! You can see this on a number line:

<—–(-1/3)–(-1/4)—–0—–(+1/4)–(+1/3)—–>

Only after you’ve figured out which fraction is closer to zero should you deal with “bigger” or “smaller”. Once you’ve determined that -1/4 is closer to zero than -1/3, you can easily figure out that -1/4 is larger, because being closer to zero on the negative side indicates a larger value. You could also view this as the *absolute value* getting smaller and smaller.

This proves especially helpful when dealing with exponents of negative fractions. Let’s use -1/2 as our example. As we increase the integer powers, weird stuff starts to happen:

(-1/2)^1 = -1/2

(-1/2)^2 = +1/4

(-1/2)^3 = -1/8

(-1/2)^4 = +1/16

Notice that the terms are now alternating between positive and negative values. But the common thread is that the terms always get closer to zero. In other words, the *absolute value* gets smaller and smaller as the exponents get larger. You can cut out a lot of confusion by sticking to the principle of distance from zero.

Let’s say you’re told that *b* is a non-zero value between -1 and 1, meaning that *b* is either a negative or positive proper fraction. If you’re asked to compare *b^5* and *b^3*, you can’t say that either one is always smaller or larger, since they could be either positive or negative. But you *can* say that the fraction with the higher exponent (*b^5*) will always be closer to zero. So if you are then given information saying that b is negative, you will know that *b^5* is larger (i.e. closer to zero on the negative side).

They probably won’t be so generous as to ask you to compare* b^5* and *b^4*. In that case, *b^5* is negative, since we’re taking an odd power of a negative number, and *b^4 *is positive, since we’re taking an even power of a negative number. But if they brought in absolute values and asked you to compare *|b^5|* and *|b^4|*, you could easily apply the principle we just discussed! Whether *b* is positive or negative, it must be true that *|b^5| < |b^4|*, because a higher exponent on a fraction will always move the fraction closer to zero, meaning its absolute value will decrease.

Hopefully this will alleviate some of the confusion about fractions as you go forward with your GMAT studies!

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]]>The post Filling in Your Chart: A Data Sufficiency Primer appeared first on .

]]>*A certain zoo has 288 mammals, 25 percent of which are female. What percent of the mammals in the zoo were born at the zoo?*

*(1) The number of male mammals that were born at the zoo is three times the number of female mammals who were not born at the zoo.*

*(2) The number of male mammals that were not born at the zoo is three times the number of male mammals that were born at the zoo.*

In that post, I mapped out exactly what the chart should look like:

Notice that I included not only each category, but also its total, which will end up being just as important. So once you have your chart, start filling it in appropriately.

The prompt tells us that there are 288 total animals, 25 percent of which are female. We are then asked what percent of the mammals are born at the zoo. Fill in all the information, and your chart should look like this:

Notice three very important things:

1. I used a question mark to very clearly denote the quantity I’m looking for — the TOTAL of the “Born” column.

2. I filled in EVERYTHING I could deduce from the given information. If I know that 25% of 288 is the female row, then 75% of 288 must be the male row. Don’t stop just because you’ve filled in everything mentioned! See if you can take things a step further.

3. This is Data Sufficiency; I don’t care what 25% of 288 is. I’m not going to waste time calculating it if I don’t need to. I’m more concerned that it is a DEFINITE QUANTITY, and thus I have a number for that box.

This brings up an interesting strategic point. If you really wanted to, you could just put a check mark (✓) in the box instead of “25% of 288″. That’s because you don’t care as much about the quantity itself; the most important thing is that SOME DEFINITE QUANTITY is in the box, and thus that box is accounted for.

If you follow that pattern, your chart would look like this:

Now let’s look at Statement 1. It says that the number of male mammals that were born at the zoo is three times the number of female mammals who were not born at the zoo. Okay, so we are given a ratio instead of some hard numbers. So how can we label the appropriate boxes? If you said variables, then you are right! If you represent the number of females not born at the zoo as *x*, then the male/born quantity will be *3x*. The end result should look like this:

Now, is this sufficient? Well no, because there’s no possible way to establish what quantity will replace the question mark. The chart makes this easy to see. Eliminate answer choices A and D.

So what about Statement 2? It says that the number of male mammals that were not born at the zoo is three times the number of male mammals that were born at the zoo. We can use the same process of assigning variables, but choose a different variable besides *x*, so we don’t get confused. You should get something like the following:

Is there anything we can deduce? Well yeah, because we have only one variable representing the male row. If we add y and 3y together, we get a definite number (as indicated by the check mark), so that means we can find y and thus the values in each of those boxes:

Is this sufficient? Well no, because there’s still no way to get a definite quantity for the question mark box.

Now we’ve narrowed down the answer choices to C and E. So which is it? Well… you tell me :).

I’ll leave the final step of combining the information to you guys, and you can leave an explanation in the comments. Good luck!

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]]>The post To Venn or Not to Venn? Depends on the Question. appeared first on .

]]>I’ve found that students like to ask because every student has a personal preference. Some don’t want to deal with Venn Diagrams because labeling them can get confusing at times. Others stay away from charts, either because they find charts more confusing or because they prefer the compactness of the Venn diagram.

So in essence, what the student really wants me to do is either justify a natural inclination (“I just find charts easier”) or tell them that this inclination is completely wrong and needs to be squelched immediately (“I really like Venn Diagrams, but I always get them wrong, so should I switch to charts?”).

As with so many things in life, the answer lies somewhere in between. Neither a chart nor a Venn diagram is inherently bad. But here’s the rub: When should you use one versus the other?

Let’s take a look at the following Official Guide Data Sufficiency question:

*If 75 percent of the guests at a certain banquet ordered dessert, what percent of the guests ordered coffee?*

*(1) 60 percent of the guests who ordered dessert also ordered coffee.*

*(2) 90 percent of the guests who ordered coffee also ordered dessert.*

Here, we only have two groups (coffee and dessert), and they overlap. In this case, the Venn Diagram is definitely the way to go, because you can very easily see a visual representation of the overlap and separation:

Even if the question dealt with three foods (e.g. coffee, dessert, entree) instead of just two, a Venn diagram would still be the way to go, because everything fits under one basic category: food. Even if there are different overlaps between foods (e.g. coffee and dessert, dessert and entree, entree and dessert), there is the possibility of a single overlap between all three.

On the other hand, let’s take a look at the following Knewton question:

*A certain zoo has 288 mammals, 25 percent of which are female. What percent of the mammals in the zoo were born at the zoo?*

*1. The number of male mammals that were born at the zoo is three times the number of female mammals who were not born at the zoo.*

* *

* *

*2. The number of male mammals that were not born at the zoo is three times the number of male mammals that were born at the zoo.*

Think about what might happen if you go after this question with Venn diagrams. You’d say, “Okay, so, I’ll create a Venn diagram for ‘male’ and ‘born’. But wait… I also need one for ‘male’ and ‘not born’. Wait, so I need two Venn diagrams?? And I haven’t even considered the females yet?! How am I supposed to solve this?? This is too hard!!! I give up.”

Before you are from this question untimely ripped, let’s recall the words of the inimitable George Carlin: “Calm down…have some dip!” If you’re struggling that much to set up a problem, you’re likely going about it in an inefficient way. The big problem here is that if you consider only the split of male/born, you consider none of woman/born. There are several overlaps from two completely separate categories (gender and birthplace), so Venn diagrams don’t make much sense.

But with a chart, things get MUCH more manageable:

Aside from demonstrating my basic online chart-making skills :), this figure works out much better, because it accounts for all the overlaps. And notice that we also leave space for all the totals, which will be just as important.

The big takeaway is: If you have several categories (e.g. male/female, born/not born) and these categories overlap with each other, a chart is definitely the way to go.

So how do we go about solving these two problems? Ah, that’s for you to find out :). Take a stab at them on your own, and feel free to post your solutions in the comments. Bonus points to those who create charts and/or diagrams and post links for us to see!

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]]>The post Two Shortcut Formulas for Set Problems appeared first on .

]]>Let’s say we consider all the athletes at a school. We consider those on the football team and those on the hockey team. Some athletes are on both teams, and some athletes are on neither team. Let’s assign the following variables:

F = number of athletes on the football team (including those who also play hockey)

H = number of athletes on the hockey team (including those who also play football)

B = number of athletes who play both football and hockey

N = number of athletes who play neither football nor hockey

T = total number of athletes at the school.

The following formula would apply:

F + H – B + N = T

This applies to any pair of overlapping sets. You add the total number in each group (including the overlap in both cases), then you subtract the number in the overlap, add the number in the neither group, and that gets you the overall total.

In general, you could think of it as:

**(Total in first group) + (Total in second group) – (overlap) + neither = Overall total**

Now, it’s important to keep in mind that the total for each group *includes* the overlap and not just those elements that are *only football* or *only hockey*.

“All that is great,” you might say. But then you’d add, “What about the really annoying problems that involve not two but *three* overlapping sets?” Well, there’s even a formula for that situation. In general, if you have three overlapping groups, the formula would be:

**(Total in Group 1) + (Total in Group 2) + (Total in Group 3) – (Overlap of 1 and 2) – (Overlap of 1 and 3) – (Overlap of 2 and 3) – [2 * (Overlap of 1, 2, and 3)] + (total in none of the groups) = Overall total**

*Very important reminder*: When we say “Total in Group 1″, we mean **everything** in Group 1, including the elements in the overlaps. It does *not* mean those elements in Group 1 only!

If you’re curious about why we subtract the overlaps, it’s essentially because in adding up the totals of each group, we count the overlaps more than once. For example, in adding the totals of Groups 1 and 2, the overlap between them gets counted twice, so we must subtract out one of them. When we add the totals of all three groups, the overlap of all three groups gets counted 3 times, whereas it should only be counted once; that’s why we subtract that overlap twice.

Now, the formula is a great tool, but as you probably suspected, having it memorized is just a first step. The difficulty of GMAT problems lies not in complex formulas and calculations, but in the quirkiness of their constructions and setups. Try your hand at these two problems to see what I mean. The first is an Official Guide problem, and the second is one of my own. As usual, feel free to post your step-by-step solution in the comments. Good luck!

*1. Of the 200 students at college T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from*

*A) 20 to 50
B) 40 to 70
C) 50 to 130
D) 110 to 130
E) 110 to 150*

*2. In a particular neighborhood of 150 households, some households have no electronic devices. The rest have some combination of televisions, laptops, and stereos. 75 households have televisions, 50 have laptops, and 20 have all three devices. If 45 households have exactly two of the three devices, and the number of households that have stereos is four times the number of households that have none of the three devices, how many households have stereos?*

*A) 84
B) 88
C) 92
D) 96
E) 100*

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]]>The post More Fun with Weighted Averages appeared first on .

]]>If you think I have another official GMAT problem handy, you are right . Feel free to give it a shot before reading my explanation.

*Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?*

*(A) $752
(B) $755
(C) $765
(D) $773
(E) $775*

You could certainly solve this problem the standard way: his first 6 paychecks average $750, meaning that they total $750*6 = $4500, then you’d calculate the total of the remaining paychecks and find an overall average. But instead, let’s have some fun and try to short-circuit the problem using weighted averages:

According to the problem, 6 of Manfred’s paychecks averaged $750 and the other 20 averaged $780. First of all, notice that the $780 group carries much more weight, because there are 20 paychecks versus only 6 in the other group. This means the overall average will be closer to $780 than to $750. Guess what? That means right away, we say goodbye to answers A, B, and C. So if you end up having to guess, you’ve got a 50-50 shot. Not bad!

Next, we’re relating 6 paychecks to 20 paychecks. That’s not an easy ratio to use for weighted averages. But let’s fudge things a bit and pretend that there are 6 paychecks in the $750 group and 18 in the $780 group.

In that case, we would have an 18:6 = 3:1 ratio in favor of the $780 group, which is a much cleaner ratio for dealing with weighted averages. Here’s the cool thing: numerically, that means that the overall average would be 3 times closer to $780 than to $750. Represented on a number line, it would look something like this,

750———–o———–o———–X———–780

where the X marks our final average. Again, think of the relative sizes of the groups as “weights” that pull the overall average away from the average of 750 and 780. Since there are three times as many entries in the 780 group, the overall average is pulled towards 780 and ends up three times closer to 780 than to 750.

Now, notice that the distance of our whole number line is 30. That distance of 30 is divided into 4 equal segments; thus, each marking is 30/4 = 7.5 apart. That means our “fudged” final average (X) is 780-7.5 = 772.5.

Of course, now we have to readjust for the actual problem. There are 20 entries in the $780 group, not 18. But those extra two entries mean the average is bumped up just a tiny smidge from 772.5, and thus we know that D is the final answer!

In addition, you could eliminate E, because an average of $775 would be too neat for the numbers involved. $780-$775 = $5, and $775-$750 = $25, which means that there would be 5 times as many entries in the $780 group. Obviously, it’s not that neat, since the actual ratio is 20:6 = 10:3.

**TAKEAWAY:** When you see a problem involving multiple sets of numbers, each of which has a unique average and each of which has a different number of entries (“weight”), you should immediately think of weighted averages and see if this conceptual approach can get you to the answer faster.

Okay, now I have a tricky problem for you. But this time, try it on your own! And remember, try to use the strategies we just talked about in this post. Feel free to list your step-by-step solutions in the comments, and be sure to apply the concepts behind weighted averages!

*A certain balloon is packaged only in small cases of 40 balloons or in large cases of 100 balloons. A certain stock room has a total of 70 cases of these balloons, and the average number of balloons for each case is 88. A supervisor plans to remove some of the larger cases from the stock room without removing any of the smaller cases. How many of these larger cases must be removed such that the average number of balloons for the remaining boxes is reduced to 70?*

*A) 28
B) 32
C) 35
D) 40
E) 42*

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]]>The post What the Heck is a “Weighted Average”? appeared first on .

]]>Let’s say one group of numbers has an average of 4, and a second group of numbers has an average of 6. We cannot just average 4 and 6 and conclude that the overall average of all the numbers in the two groups is 5. There could be more numbers in one group than in another, and thus the two groups would have different “weights.”

For example, if the first group has 1,000,000 numbers while the second has only 1 number, the first group is weighted much more heavily, and thus the overall average will be much closer to 4 than to 6. (When illustrating general principles, outlandish examples always do the trick )

On the GMAT, this manifests itself in several ways, but as with so many GMAT topics, the *conceptual understanding* is often much more important than the ability to do raw calculations. As such, you can bet that weighted average questions will show up in Data Sufficiency!

Hey, what do you know?! I just happen to have a sample GMATPrep problem right here! Feel free to give it a shot before reading my explanation:

*Each employee on a certain task force is either a manager or a director. What percentage of the employees are directors?*

*(1) The average salary for a manager is $5,000 less than the average salary of all employees.*

*(2) The average salary for a director is $15,000 greater than the average salary of all employees.*

*A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked.*

And now, the breakdown: It’s easy enough to narrow the answer choices down to C and E; each statement gives information on one of the two groups, but not the other. When looking at the statements together, it’s very tempting to think that the answer is E, since you are given only averages and not absolute numbers of directors or managers. However, the question asks about the *percentage* of employees who are directors, and percentages can be gotten without absolute numbers.

Believe it or not, the specific numbers involved give away the exact percentages. How do we know? Weighted averages! The difference between the average salary of a manager and the overall employee average ($5,000) is one-third that between the average salary of a director and the overall employee average ($15,000), and yet they end up “balancing each other” to get that overall average. That means that the $5,000 group must carry more weight. And since the two dollar amounts are in a 3:1 ratio, that means that there are 3 times as many managers as directors!

As such, we can conclude that 75% of the employees are managers, and 25% are directors. Final answer: C.

If you recall my previous post on averages, this makes sense. If the average director’s salary adds an amount $15,000 greater than the average, it will take *three* amounts of $5,000 below the average to balance things back to the overall average. So for every one director, there must be three managers.

**Again, notice that we didn’t have to do a single calculation. Our conceptual understanding of weighted averages is what bailed us out.**

Next week, I’ll show you how a conceptual understanding of weighted averages can even help you on Problem Solving questions!

In the meantime, I have another DS problem for you. But this time, try it on your own! Post your step-by-step solutions in the comments, and **be sure to apply the concepts behind weighted averages!**

*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.*

*A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked.*

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]]>But if the GMAT is more about reasoning than speed calculations, then why am I writing a post on quick mental math? Well, it’s true that any GMAT quant question can be solved without quick mental math (and certainly without a calculator). But that doesn’t mean the test doesn’t reward you if you can speed up the process. And by “speed up,” I don’t mean cutting corners or being less rigorous. I mean recognizing that there are ways to make calculations easier and more manageable.

Take averages. You’re probably familiar with the basic way to find averages: you add up all the elements in a set of numbers, then divide by the number of terms, and the result is the average. For example, let’s take the set (3, 7, 15, 9, 6). To calculate the average, we’d take 3+7+15+9+6 = 40, then divide that 40 by the number of elements (5) to get an average of 8.

There’s an important principle at work here: no matter what the values of those 5 elements are, they will always have an average of 8 if they sum to 40. Thus, if we change (3,7,15,9,6) to (4,6,15,9,6), the average does not change, because the numbers still sum to 40. This is shown in the “rearranged average formula,” which simply states that the sum of a group of numbers is equal to that group’s average multiplied by the number of terms in the set (Sum = Avg * # terms).

This leads to an incredibly helpful mental math trick. If you can change the individual numbers in the set such that they get closer to the average, while at the same time making sure the sum doesn’t change, you can hone in on the exact average much much quicker.

Let’s use our original set of five numbers (3, 7, 15, 9, 6) to demonstrate:

To start, let’s take the average of individual pairs. For example, 3 and 7 have 5 in the middle, and 9 and 15 have 12 in the middle. We can therefore change the list from (3, 7, 15, 9, 6) to (5, 5, 12, 12, 6). Notice that the two sets still have the same average, because we’ve changed neither the sum of the numbers nor the number of elements in the set.

We now have two 5s and two 12s. Let’s break those down. The average of 5 and 12 is 8.5, so we can change (5, 5, 12, 12, 6) to (8.5, 8.5, 8.5, 8.5, 6).

Obviously, the average of the set will not be 8.5, since you have that 6 thrown in. But on a Problem Solving question, you could possibly approximate; since 6 is not far off from 8.5, you know the average will be slightly less than 8.5, so if only one answer choice is slightly less than 8.5, you know it will be your winner.

But for the more precise, let’s finish what we started! Now comes another cool mental math trick related to averages. If all the elements were 8.5, then of course the average would be 8.5. But what effect does that 6 really have on the entire set? Well, it’s as if we had all 8.5s and then reduced the entire sum of the set by 2.5 (since we changed one 8.5 to a 6). If we spread that 2.5 difference evenly across the entire set of five numbers, each number would be reduced by 2.5 / 5 = 0.5. Therefore, it’s as if we had all 8.5s to start with and then reduced each one by 0.5. Guess what… that means the final average is 8.5 – 0.5 = 8. Done!

Another way you can quickly find an average is to use a single number as a “focal point.” This works well if the numbers are somewhat close together, but far enough apart that the average might not be immediately obvious. For example, if I gave you the list (100, 95, 85, 110, 150) and asked you to find the average, you could pick 100 as your focus and record how much less or greater each other number is than 100:

100 95 85 110 150

(+0) (-5) (-15) (+10) (+50)

Add up all the comparisons to get a net difference: -5 + -15 + 10 + 50 = +40. Again, this is the conceptual equivalent of having five 100s and adding 40 to the entire sum. So, if we spread that 40 evenly across the set, each term would increase by 40/5 = 8 from the “focal point” of 100. And *voilà*! That means the average is 100+8 = 108.

Next week, I’ll extend this to a discussion of weighted averages, which are *very* common on the GMAT!

In the meantime, try to apply these principles to the following official GMAT problem:

*Ada and Paul received their scores on three tests. On the first test, Ada’s score was 10 points higher than Paul’s score. On the second test, Ada’s score was 4 points higher than Paul’s score. If Paul’s average (arithmetic mean) score on the three tests was 3 points higher than Ada’s average score on the three tests, then Paul’s score on the third test was how many points higher than Ada’s score?*

*A) 9 B) 14 C) 17 D) 23 E) 25*

* * *

**Answer to the problem from last week’s post:**

*Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce (5/4)*w* widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?*

*A) 4 B) 6 C) 8 D) 10 E) 12*

The first step is to realize that if it takes 3 days for the two machines to produce *(5/4)w* widgets, we can multiply both quantities by 4/5 to find that it takes 12/5 days to produce *w* widgets. We do this because it’s easier to think of the amount of work as simply *w* widgets rather than *(5/4)w* widgets.

Now, let’s recall the formula we came up with a formula for combined time:

If Machine X takes x days to complete the job on its own, and Machine Y takes y days to complete the job on its own, then the two machines working together at their constant rates would take (xy)/(x+y) days to complete the job. If you’d like to review how we get that quantity, go back to last week’s post. According to what we just found, that quantity is 12/5, so:

(xy)/(x+y) = 12/5

Now, here’s where you can get REALLY clever! The prompt says that Machine X takes 2 days longer than Machine Y, so if you really wanted to, you could substitute x-2 for y in the equation above and solve for x.

However, there’s a very sneaky shortcut: Notice that we’re trying to find the number of days it takes Machine X to produce 2w widgets. All the answer choices are even numbers, which means that the number of days it takes Machine X to produce w widgets (i.e. half of whatever the answer is) is an integer! So really, we just need to find integer values of x and y that are 2 apart and that satisfy the equation!

(xy)/(x+y) = 12/5

Since x+y is an integer, it will have to be a multiple of 5, according to the ratio. There’s no way x+y could equal 5 if x and y are integer values 2 apart. You can show that with substitution or quick testing cases. But could x+y=10, in which case xy would have to be 24? Sure! x=6 and y=4 are numbers for which x is two greater than y and both x and y satisfy the equation! Therefore, it takes Machine X 6 days to produce w widgets, which means that it takes 12 days to produce 2w widgets.

Final answer: E

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