How’s everyone else doing on this quiz? …

How’s everyone else doing on this quiz? …

How’s everyone else doing on this quiz? …

*All this talk about the new SAT is interesting and all, but we’ve still got two years to live with the old one, so let’s get back to our regularly scheduled PWNing.*

Here’s a minorly important circle fact that I find a lot of students don’t know: **when a wheel is rolling—without slipping—it makes travels a distance of one circumference each time it makes one complete turn. **The SAT doesn’t test this often, but it’s certainly in the realm of testable concepts.

I think one of the easiest ways to understand this is to picture a roll of tape. If you hold down the loose end, and roll the tape along a table until it’s made one complete revolution, how much tape will you have unrolled? Obviously, one circumference. And how far did the roll travel? It traveled the same distance as the amount of tape it left on the table—one circumference.

Does that make sense?

It does? OK, cool. Then let’s try a tricky question together:

- In the figure above, a wheel with center
Band a radius of 12 cm is resting on a flat surface. A diameter is painted on the wheel. If the wheel begins to rotate in a clockwise direction and rolls along the surface without slipping, how far willBtravel before the painted diameter is perpendicular to the surface for the first time?

(A)2π cm

(B)4π cm

(C)8π cm

(D)12π cm

(E)16π cm

Awesome question, right? I know.

To solve it, first note that the wheel will have to rotate 30º just to get to the point where the painted diameter is parallel with the surface. That might be easier to see if you draw a line through *B* that’s parallel to to ground, like so:

Once the diameter is parallel to the surface, it’s going to have to rotate another 90º to make the diameter perpendicular to the ground. So, in total, the circle will have to rotate 120º.

As is true of many circle questions, then, this one really comes down to ratios. 120º is 1/3 of a full 360º circle. If a circle travels a full circumference when it makes a full revolution, then it will travel 1/3 of a circumference when it makes 1/3 of a revolution. So all we need to do is find the full circumference, and then find 1/3 of it.

distance traveled = 8π

Easy, right?

…Say *what* now?

This isn’t tested on the SAT all that often, but it has appeared (you’ll find an example in the Blue Book: Test 3 Section 5 Number 8) and I’ve had a bunch of kids tell me lately that they don’t remember ever learning it in school.

When you have two polynomials that equal each other, their corresponding coefficients equal each other.

IF:

*ax*^{2} + *bx* + *c* = *mx*^{2} + *nx* + *p*

THEN:

*a *=* m*

*b *=* n*

*c *=* p*

You might find it useful, in fact, when you’re presented with polynomials that equal each other, to stack them on top of each other and put circles (use your imagination because I can’t figure out how to circle things in HTML) around the corresponding coefficients:

*ax*^{2} + *bx* + *c*

=

*mx*^{2} +* nx* + *p*

(

x+ 9)(x+k) =x^{2}+ 4kx+p

- In the equation above,
kandpare constants. If the equation is true for all values ofx, what is the value ofp?

Alllllright. First, foil the left hand side:

(*x* + 9)(*x* + *k*)

= *x*^{2} + 9*x* + *kx* + 9*k*

Might look a little better like this:

= *x*^{2} + (9 + *k*)*x *+ 9*k*

Now stack up the two sides, and see what equals what:

*x*^{2} + (9 + *k*)*x *+ 9*k*

=

*x*^{2} + 4*kx* + *p*

So we know that:

9 + *k* = 4*k*

9*k* = *p*

From here, this is cake, no?

9 + *k* = 4*k*

9 = 3*k*

3 = *k*

9(3) = *p*

**27 = p**

Math is fun!

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Also check out this brutal old weekend challenge testing the same concept.

So this isn’t a *super* important thing as far as how often it appears on the SAT, but it does pop up time and again, so if you’re shooting for perfection (or close to it) you might want to pay attention. Otherwise, you can get by just fine without this little nugget (but you might as well read it, since you’re here anyway).

Do you know what *prime factorization* is? Basically, the prime factorization of a number is the way you would build that number by multiplying together only prime numbers. To find the prime factorization of a number, divide by 2 if you can. Do that as many times as you can. Once you can’t do that anymore, try dividing by 3 as many times as you can. Then by 5. Then by 7. Then by 11. I think you get the idea.

What is the prime factorization of 13728?

Whoa. Big number. Lots of people like to make trees when they do this. Let’s do that. Damn I wish you and I were in the same room with a chalkboard right now. This is going to take flippin’ forever.

See how, when I couldn’t divide by 2 anymore, I went to three, and then to 11? I knew I was done when I had two prime numbers, 11 and 13. If I multiplied all those numbers back together, I’d get 13728 again. For serious. Try it:

2 × 2 × 2 × 2 × 2 × 3 × 11 × 13 = 13728

Because sometimes the SAT asks hard questions (and if you took the October 2011 SAT, you can confirm) about the lowest multiple of two numbers that’s also perfect square. It just so happens that prime factorization is a *great* way to find a perfect square.

The prime factorization of a perfect square will contain even numbers of each prime number. Look back at the prime factorization of 13728. That’s not a perfect square. There are 5 2s, and one each of 3, 11, and 13. We can use this information to find the lowest multiple of 13728 that’s a perfect square. In order to make a prime number, we’re going to need another 2, another 3, another 11, and another 13. Yikes. That’s gonna be a big number.

2 × 2 × 2 × 2 × 2 × **2** × 3 × **3** × 11 × **11** × 13 × **13**= 11778624

Peep the **bold underlines**. Those are the factors I’ve added. My new product is huge. It’s also a perfect square. Seriously.

√11778624 = 3432

And there are no multiples of 13728 that are less than 11778624 that are perfect squares. Scout’s honor.

Glad you asked. Try this (no multiple choice — it’s a grid-in):

- If
p^{2}is a multiple of both 8 and 35, andpis a positive integer, what is the least possible value ofp?

So…yeah. Start by doing a prime factorization of 8 and 35.

8 = 2 × 2 × 2

35 = 5 × 7

Note that you have odd numbers of all three used prime factors. You’re gonna need another 2, another 5, and another 7.

2 × 2 × 2 × **2** × 5 × **5** × 7 × **7** = 19600

*p*^{2} = 19600

Confirm that 19600 is a multiple of both 8 and 35 (of course it is):

19600 ÷ 8 = 2450

19600 ÷ 35 = 560

Yes, it worked. So what’s *p*? Just take the square root of 19600!

√19600 = 140

*p* = 140

Note the tempting false shortcut: just multiply 8 by 35 and square the result. But if you do that, you get 78400 for *p*^{2} and 280 for *p*. That’s not the smallest possible *p*, as we just showed.

Like I said, you don’t see this often on the SAT, but if you’re shooting for perfection, you’ll want to know this relationship between prime factors and perfect squares.

When I posted that question writing contest a few weeks back, I thought I’d get some pretty good stuff. Because you guys are smart. And then, for a little while, not much happened. I was all </3. But then some great stuff started rolling in and I was all \(*o*)//. I bet, reading this paragraph, you’re all o_O. Right?

From Debbie Stier:

- Candy bars were passed out among 10 students. If the average (arithmetic mean) number of candy bars that the students received was 11, what is the greatest possible number of candy bars that one student could have received?

A) 11

B) 20

C) 101

D) 109

E) 110

Commenter Katie (fresh off a weekend challenge win) posted a few great grid-ins, too:

- Dan wrote a 7 digit phone number on a piece of paper. He tore the paper accidentally and the last two digits were lost. What is the max number of arrangements of two digits, using digits 0 through 9, he could use to find the correct number?
- If x + (1/x) = 4, what is the value of x
^{2}+ 1/(x^{2})? - The lengths of the side of an isosceles triangle are 30, n, n. What is the smallest possible perimeter of the triangle?
- If (a/4) + (b/8) + (c/24) = 1, what is one possible value for abc?

And of course, this great question was submitted to me by email.

- Some of these require some
*teensy*clarifications before they’re ready for prime time. - I have no solutions!

Commenters, this is where you come in.

Since you’ve been paying such close attention, you know by now that the difficulty of math questions increases as a section progresses*. On a 20 question section, you can count on #1 to be super easy, #5 to be bit tougher, #10 to require more than a modicum of thought, and #20 to be a royal pain. Duh, right? You know this. If you’ve ever taken an SAT or PSAT, it’s almost impossible not to have noticed this. But have you thought about what it means for you, the intrepid test taker?

This simple fact has 2 important implications:

- Easy questions are more important than the hard ones for your score (covered a previous post).
- You should be suspicious of “easy” answers to “hard” questions (covered here).

When you’re faced with a question that’s supposed to be more difficult, you should resist the urge to jump on an answer choice that seems immediately obvious. It’s probably best to illustrate this with an example.

- Stephen wins the lottery and decides to donate 30% of his winnings to charity. Then he decides to give 20% of what he has left to his mother. What percent of his winnings does Stephen have left for himself?

(A) 67%

(B) 56%

(C) 54%

(D) 50%

(E) 14%

This question isn’t too tough and you might not need my help to solve it, but before we get I want to ask you something.

What choice should you *not even consider*? Well, if this question was just asking you to start with 100%, and subtract 30% and 20% and end up at 50%, it wouldn’t be a #15. It would be a #5. So there’s *no flippin’ way* it’s (D). Since we know we’re later in the test, there must be something else going on here. And sure enough, we see that Stephen gives 20% of what he has left *after* he’s already donated 30% to charity.

If you’ve ever been told to watch out for “Joe Bloggs” answers, that’s what I’m talking about here. I want to be clear: “Joe Bloggs” is not a technique for answering a question correctly, and anyone who tells you otherwise hasn’t spent enough time learning about the SAT. But it *is* helpful to remember that, once you’re about halfway through a math section, the questions are supposed to require some thought, so you shouldn’t fall for answers that require no thought at all.

Put another way: *This is not a way to get questions right. This is a way not to get questions wrong.*

To get it right, as we’ll often do with percent questions, we’re going to plug in.

- Say Stephen won $100 (some lottery).
- He gives $30 to charity.
- Now he’s got $70 left, 20% of which he gives to his poor old mother.
- 20% of 70 is 14, so he gives his mom $14.
- $70 – $14 = $56, so he’s got $56 left for himself. 56% of his original winnings.
- The answer is (B).

* On a grid-in section, difficulty increases twice — once from 1-8 (the multiple choice bits) and again from 9-18 (the grid-in bits). Also, in the rare event where you get a few questions that refer to the same graph or set of equations, the first one will be very easy and the last will be very hard (like a mini-section inside a section).

So the question writing contest I proposed a few weeks back didn’t exactly explode onto the scene like I thought it would, but I still think it’s a fantastic way for you to improve your skills, so just FYI: it’s still open.

I got the question above in an email the other day. The writer prefers to remain anonymous, but I’ve given him Beta access to the Math Guide because it’s an awesome question. It incorporates circle properties, shaded regions, and has great, well-planned incorrect choices. That’s good hustle.

Source. |

One technique-able counting problem type that you might come across on the SAT is what I’ll call a “possibilities” problem*. It might involve cards (but *not* playing cards – the SAT doesn’t like those), or pictures being lined up on a wall in different orders. Your job will be to determine the number of possible outcomes given a particular scenario. Like so:

- Mike is arranging seven of his various awards and commendations on a shelf in his office. If he insists that his hard-fought Class of 1999 Math Award be placed in the center, in how many different orders could he arrange the seven items?

(A) 60

(B) 72

(C) 120

(D) 720

(E) 1440

The best way to tackle a possibilities problem like this is to draw a bunch of blanks like you’re about to play Hangman, and then start thinking, methodically, through the choices you have at every step along the way. I’m going to illustrate this process with slightly more thoroughness than you probably will on test day (you won’t need to make up award names, but I will because it’s *hilarious*):

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Class of 1999 Math Award | ||||||

Choices | 1 |

First, as I did above, you must account for any special conditions or restrictions. The Math Award must go in the middle, so there’s *only one choice* for Position 4.

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Invisible Man Award | Class of 1999 Math Award | |||||

Choices | 6 | 1 |

Once all the restrictions are accounted for, start filling in the rest of the spaces. To fill Position 1, Mike has 6 different awards to choose from. Say, for argument’s sake, he chooses the Invisible Man Award next.

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Invisible Man Award | Acne League – Most Improved | Push Ups Contest – Last Place | Class of 1999 Math Award | Spin the Bottle – Luckiest Player Ever | 5^{th} Grade Science Fair – 2^{nd} Place | Shortest Fight in School History – Loser |

Choices | 6 | 5 | 4 | 1 | 3 | 2 | 1 |

To fill the next spot, since he used up the Invisible Man Award, he has 5 awards left to choose from. Once he chooses the Acne League – Most Improved award for Position 2, he has 4 choices for Position 3. He continues this process until he’s filled all the positions.

To calculate the number of different arrangements Mike could have made, multiply the number of choices he had at every step:

6 × 5 × 4 × 1 × 3 × 2 × 1 = 720

**When solving a possibilities problem, set up the hangman blanks, then imagine yourself actually performing the task described. First take care of special conditions or restrictions, and then take care of everything else. At every step, ask yourself “How many choices do I have?****” And then ask yourself “How many choices do I have now?” And then – you guessed it – ask yourself how many choices you have ****again****. **

**Stop when you run out of choices.**

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*Aside from this footnote, I’m deliberately avoiding the terms “permutation,” “combination,” and “factorial” here. That’s not because I don’t know them; it’s because I’ve found that they manufacture more confusion than they alleviate on the SAT. Remember: The SAT is not a math test! If you prefer to solve the questions laid out here by cramming them into nPr and nCr notations in your calculator, be my guest, but don’t cry to me when you miss counting questions on the SAT.

An unfortunate truth about the SAT is that while many questions can be answered with snappy tricks (many of which can be found on these pages), not all of them can. *Most* “counting” questions (and probability questions, for that matter) fall into this category.

Yes, I’m serious. *Most*.

Basically, if you don’t see within 15 seconds or so that you’re dealing with a matching problem, or a possibilities problem where you can just set up hangman blanks and count, then you should bail on looking for shortcuts and just start listing things. That’s right. List them.

Stop complaining. Listing them isn’t “the long way”! *Sitting there with your leg shaking and your hands on your head trying to see a shortcut where there is no shortcut while time ticks away is “the long way”!*

(Grid-in)

- How many positive integers less than 100 are not divisible by 7?

OK, so maybe I lied a *little* bit: there is a small shortcut, which is to treat this like a shaded region problem and find the opposite of what they’re asking for. In other words, list all the positive integers less than 100 that *are* divisible by 7:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

I’m counting 14 there. Since there are 99 total positive integers less than 100, the answer is 99 – 14 = 85. Nice, right?

I get a lot of questions from students about listing problems like these, and their questions always end the same way: “Is there a faster way?” Not really. But if you stop worrying about a faster way and just start listing as soon as you come across this kind of question, and you count carefully, you’ll probably get through it without much of a problem.

UPDATE: For this particular problem, there *is* one other shortcut. I hesitate to even mention it because I don’t want to dilute the point that your general reaction to a non-pattern-conforming counting question should just be to start listing, but when you get “How many positive integers less than *m* are not multiples of *n*?” questions you can also follow these steps:

- Find the greatest multiple of
*n*that’s less than*m*. In this case, 98 is the greatest multiple of 7 that’s less than 100. - Divide by
*n*. In this case, 98/7 = 14. - That means there are 14 multiples of 7 less than 100.

The SAT will throw two common kinds of “counting” problems your way. I’ll handle one of them in this post. The other kind, well, I’ll get to it when I get to it. 🙂

I like to call this kind of problem a *matching* problem. It’ll usually involve a bunch of people who all need to shake hands, or a league in which every team needs to play every other team. Or a massage club where everyone has to give everyone else a back massage. I don’t know…whatever. Everyone has to touch everyone at the touching party?

…this is going nowhere. Let’s see an example.

- Each team in a kickball league plays each other team 4 times during the season. If there are 7 teams in the league, how many games long is the season?

(A) 28

(B) 56

(C) 84

(D) 112

(E) 116

What you’re going to want to do here is draw a diagram.

Arrange the letters A-G (representing the 7 teams) in a large circle. Now draw

lines connecting each letter to each other letter, carefully counting as you draw. (If you try to count after you’re done drawing, you’re going to have a pretty difficult time getting an accurate count.) The best way to go about this is to draw every line that you can that originates at

A, and then do the same for B, etc.

You’ll know you’re done when you have something that resembles a star with all its outer points connected. Like so:

The number of lines you just drew – 21, you awesome counter you – equals the number of games required for each team to play each other once. If each team has to play each other 4 times, multiply 21 by 4 to get the answer: **84!** BAM!

Pretty amazing, right? It’s just so…beautiful. No, stop crying. It’s totally inappropriate for you to be crying right now. I know it’s pretty but you need to stop. I *refuse* to move on until you stop crying.

Of course, if you want to represent the above diagram mathematically, you could say that Team A needs to play all the other teams, so it plays 6 games. Then Team B needs to play all the teams except Team A (since they already played), so that’s 5 more games. Follow that line of reasoning until its end and you get:

6 + 5 + 4 + 3 + 2 + 1 = 21

But I think the star is prettier (and easier to remember).

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Guys. It’s apparently going to break 100° today in New York. Seriously.

The prize this week for the first correct answer: You will awake in a bathtub of ice, and have no idea how you got there. Your first concern will be a suspicious scar on your abdomen, but that will quickly be replaced by relief that you are in a bathtub of ice, and not in New York City, where it is frikkin’ 100°.

A wheel is rolling in a straight line, without slipping, on a flat surface. Two points on the wheel have paint on them, and they are leaving spots on the surface as the wheel rolls (wheel is rolling from left to right in the figure). What is

x?

Put your answers in the comments. I’ll post a solution Monday.

UPDATE: You guys rock. Special congratulations to the anonymously sweltering newcomer, for getting it first. Great job.

Solution below the cut.

The key to getting this right is recognizing that when a wheel makes a revolution, if it’s not slipping, it travels exactly one circumference. So when this particular wheel makes spots on the surface 8 and 10 units apart, we know that the full circumference is 18 units. We also know that the next dot will be 8 away from the previous one, and that the angle we’re looking for corresponds to an arc length of 8.

From there, you simply must recognize the opportunity to use a part/whole ratio (see this post for more on how many difficult circle problems are actually ratio problems) to solve for *x*.

It’s not uncommon for a question or two involving three-dimensional shapes to appear on the SAT. Luckily, most of the time these questions either deal directly with the simple properties of three-dimensional shapes (like surface area and volume), or are just 2-D questions in disguise. It’s pretty rare to come across a truly difficult 3-D question — but you know I’m gonna give you some in this post because I care about you so.

Generally speaking, the SAT will give you every volume formula that you need, either in the beginning of the section (rectangular solid — *V* = *lwh*; right circular cylinder — *V* = π*r*^{2}*h*) or in the question itself in the (exceedingly) rare case where you’ll have to deal with the volume of a different kind of solid. It’s worth mentioning, though, that **the volume of any right prism* can be calculated by finding the area of its base, and multiplying that by its height.**

For example, if you needed to calculate the volume of a prism with an equilateral triangle base, you’d find the area of an equilateral triangle:

And multiply that by the height of the prism:

You almost definitely won’t need this particular formula on the SAT, but it’s nice to know how to find the volume of a right prism in general: just find the area of the base, and multiply it by the height.

Most volume questions you’ll see on the SAT will require you to deftly maneuver between the volume of a solid and its dimensions. Let’s see an example (and showcase my fresh new drawing software):

- If the volume of the cube in the figure above is 27, what is the length of
AF?

(A) 3

(B) 3√2

(C) 3√3

(D) 3√5

(E) 6

Remember that a cube is the special case of rectangular solid where all the sides are equal, so the volume of a cube is the length of one edge CUBED:

So far, so good, right? Now it’s time to do the thing that you’re going to find yourself doing for almost every single 3-D question you come across: work with one piece of the 3-D figure in 2-D.

The segment we’re interested in is the diagonal of the square base of the cube. If we look at it in 2 dimensions, it looks like this:

The diagonal of a square is the hypotenuse of an isosceles right triangle, so we can actually skip the Pythagorean Theorem here since we’re so attuned to special right triangles. *AF* = 3√2. That’s choice (B).

**The surface area of a solid is simply the sum of the areas of each of its faces**. Easy surface area problems are really easy. Trickier surface area problems will often also involve volume, like this example:

- If the volume of a cube is 8
s^{3}, which of the following is NOT a value ofsfor which the value of the surface area of the cube is greater than the value of the volume of the cube?

(A) 0.5

(B) 1

(C) 2

(D) 2.25

(E) 3

Yuuuuck. What to do? Well, to find the surface area of a solid, you need to know the areas of its faces. To find those areas, you need to know the lengths of the sides of the solid. Luckily for us, it’s pretty easy to find the lengths of the sides of this cube, since we know that the volume is 8*s*^{3}. Take the cube root of the volume to find the length of one side of the cube:

If a side of the cube is 2*s*, then the area of one face of the cube is (2*s*)^{2}, or 4*s*^{2}. There are 6 sides on a cube, so the surface area of the cube is found thusly:

6 × 4*s*^{2} = 24*s*^{2}

From here, it’s trivial to either backsolve, or solve the inequality spelled out in the question:

24*s*^{2} > 8*s*^{3}

3*s*^{2} > *s*^{3}

3 > *s*

The answer must be (E), the one choice for which the inequality is NOT true.

* Right circular cylinders and rectangular solids are both special cases of right prisms — a right prism is any prism whose top lines up directly above its bottom.

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I wanted to take it easier on you guys after last week’s hard-as-hell question, so here’s one that could actually appear on an SAT.

This week’s prize: you and your future random college roommate will have remarkably similar tastes in music. Trust me — it matters.

- After
mdays, the average (arithmetic mean) temperature for the month of June in the city of Riverside was is 87° Fahrenheit. If the temperature the next day was 99° and the average temperature for June rose to 89°, what ism?

Post your answers in the comments. I’ll put up the solution on Monday evening.

UPDATE: Excellent job, Guilherme. Solution below the cut.

To solve this question, you’re going to want…bum bum BUMMMM….the AVERAGE TABLE!!!

m days | 87° | |

1 day | 99° | 99° |

m + 1 days | 89° |

The power of the average table is that it makes it very easy to deal with sums; sums end up being the path to success for most average questions on the SAT. Above I’ve filled in all the given values (the average and sum temperature on the 1 day are, obviously, the same). We’re going to solve by filling the rest of the table in. The sum of the temperatures for the first *m* days will be 87*m*. The sum for the *m* + 1 days will be the average (89) for those days times *m* + 1, but it will ALSO be the sum of the first *m* days plus the temperature of the 1 day:

m | 87 | 87m |

1 | 99 | 99 |

m + 1 | 89 | 87 m + 99or89( m + 1) |

Important enough to repeat even though I kinda already said this: I can make two separate expressions in the right-hand corner. One comes from adding up the first two entries in the column on the right, and one comes from multiplying the first two values in the last row. If you’re not following the way this table works, please see my original post on the table, which should answer your questions.

Here’s the thing: now I can solve for *m* because I have two expressions containing *m*, which equal each other!

87*m* + 99 = 89(*m* + 1)

87*m* + 99 = 89*m* + 89

87*m* + 10 = 89*m*

10 = 2*m*

5 = *m*

Oh hell yes.

Like the average (or, as some say, the *arithmetic mean*), the median and the mode are useful properties of a set of numbers and can give statisticians great at-a-glance insight into the nature of copious data. When the SAT gets its hands on them, though, they are usually stripped of any analytical utility and instead used as a framework in which to ask tricky reasoning questions. Also, the SAT doesn’t really test either of these concepts very often — doubly so for mode. So I’ll leave it to your college stats class to elucidate the myriad ways median and mode are useful in real life, and just show you what you need to know to PWN the rare median and/or mode question on the SAT.

**The median is the middle value in an ordered list of numbers.** If the list of numbers you’re given isn’t in numerical order, you can’t find the median until you put it in numerical order.

- {4, 6, 7, 9, 12, 16, 30}
- {9, 30, 16, 4, 7, 6, 12}
- {2, 200, 300, 700}
- {17, 22, 6, 110, 68, 52, 29, 8456}

- Which of the following CANNOT change the value of a median in a set of five numbers?

(A) Adding 0 to the set

(B) Multiplying each value by -1

(C) Increasing the least value only

(D) Increasing the greatest value only

(E) Squaring each value

This isn’t the hardest question in the world, and if you’ve seen a similar one before you probably know the answer instantly. If you haven’t, well, now you have and you’ll nail a similar one if you see it on the SAT. You’re welcome.

Let’s plug in a set to make this easier to comprehend. Say our set is {2, 3, 4, 5, 6}. The original value of the median is 4.

If we add 0 to the set like it says to do in (A), the median becomes the average of 3 and 4, or 3.5. Cross off (A).

If we multiply each value by -1, the median becomes -4. That’s not the same as 4. Cross off (B).

If we increase the least value (by more than 2) we change the median as well. Say that 2 became a 10. Now the median is 5 instead of 4. Cross off (C). Note that since the questions says “CANNOT” it doesn’t matter that we wouldn’t change the median if we only increased the least value by 1. If we can come up with a way to change the median by increasing the least value, we can cross off (C).

**Increase the greatest value as much as you like, you won’t change the order of the values at all. If we change 6 to 6,000,000, the median is still 4. (D) is our answer.**

Obviously, squaring each value changes each value, and thus changes the median value. Cross off (E).

Not so bad, right?

**The mode of a list of numbers is the number that appears most in that list.** Be aware that it’s possible for a list to have multiple modes, but all modes will appear the same number of times, and no other number will appear more often. For example: 5 and 6 are modes, in {4, 4, 5, 5, 5, 6, 6, 6, 7, 8}. Mouse over the following example sets to see their modes:

- {2, 4, 2, 7, 9, 7, 3, 2}
- {2, 4, 2, 7, 9, 7, 3, 2, 7}

{2, 3, 9, 4, 11, 4x– 8, 3y– 4}

- The modes of the set above are 2 and 11. What is one possible value of
x+y?(This is a grid-in.)

OK. In order for 2 and 11 to be the modes of the set above, each need to appear an equal number of times, and more often than any other value in the list. Which means one of two things must be true:

- 4
*x*– 8 = 2 and 3*y*– 4 = 11 - 4
*x*– 8 = 11 and 3*y*– 4 = 2

Let’s deal with possibility 1 first:

4*x* – 8 = 2

4*x* = 10

*x* = 2.5

3*y* – 4 = 11

3*y* = 15

*y* = 5

So one possible value of *x* + *y* is 2.5 + 5 = **7.5.**

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