## Posts tagged with: challenge

EDIT: I’m home now so this is formatted right. Suffice to say I’ll never use the Blogger Android app again.

I’m in Chicago without a computer so this will be short and sweet since I’m typing with my thumbs.

First correct and non-anonymous comment wins access to the Math Guide.

Chris bought a pumpkin last year that had a diameter of 10 inches and yielded the PERFECT amount of toasted seeds. His pumpkin this year has a diameter of 13 inches. Pumpkin skins generally have a thickness of one tenth of the radius of the pumpkin. Assume pumpkins are spherical and that each pumpkin has roughly equal amounts of seeds by volume. What will be the ratio of the amount of seeds Chris gets this year to the amount of seeds he got last year?

UPDATE: Nice work, Edward. Solution below the cut.

So the first thing you need to do is make sure you know how to find the volume of a sphere*:

Next, establish the inner volumes (the volumes that don’t include the skins) of each pumpkin. Let’s take them one at a time.

The pumpkin last year had a diameter of 10 inches, which means it had a radius of 5 inches. The skin, according to the problem, is therefore 5/10 of an inch, so the radius of the inner pumpkin (where the seeds are) is 4.5 inches. So the inner volume is:

Which works out to about 382 cubic inches. (With all the approximations and assumptions in this problem — like pumpkins as spheres — it’s silly to be any more precise.)

This year’s pumpkin has a diameter of 13 inches, so it’s got a radius of 6.5 inches and, by the same calculations as above, a skin that’s 0.65 inches thick. That means its inner radius is 5.85. Throw that into the formula to find the inner volume of this year’s pumpkin:

That works out to about 839 cubic inches.

So the new pumpkin’s volume is a lot bigger than last year’s volume.

To find the ratio of this year’s seed volume to last year’s simply divide:

839/382 = 2.2 (again, roughly).

* Please note that, like many of the Weekend Challenge question you’ll find on this site, this isn’t something you need to be able to do for the SAT, and if you ever are asked to work with the volume of a sphere, you’ll be provided the formula.

I’m going to tell you a secret: I don’t sleep well at all the night before SAT scores are released. It’s not like I even took it. I just get so excited for all the kids I’ve worked with that I can’t sleep. So last night, after a sleepless night followed by a bunch of good news, I slept like a rock. And I woke up energized, and I got some much-needed work done on my book, and now I’m posing this weekend challenge in timely manner. I’m feeling very productive today. Yay me!

By now you know the drill I’m sure. First correct (and non-anonymous) comment gets access to the Math Guide Beta (to which I just added some more hard questions this very morning).

Adina keeps track of the number of rats she sees while she waits for the subway every morning. She noticed that there was a roughly 46% increase in the amount of rats she spotted between Tuesday and Wednesday. If she saw 19 rats on Wednesday, what will be the percent decrease in rat sightings if she sees the same number Thursday that she did Tuesday? Round your answer to the nearest percent.

Good luck, y’all. And have a great weekend! I’ll post the solution early next week.

UPDATE: Nice work, Jessica. Enjoy the Beta. Solution is below the cut.

To solve this, you really just need to remember the basic formula for percent change:

What you’re given in this problem is the percent change (roughly 46%) and the final value (19). You can use that to find the original value (since it’s the number of rats Adina saw Tuesday, let’s call it t.

.46 = (19 – t)/t
.46t = 19 – t
1.46t = 19
t = 13.01…

A rat is a rat, there’s no partial rats, so we say Adina saw 13 rats on Tuesday, and thus will see 13 rats on Thursday. From here, it’s just plug-and-chug to find the answer. She’ll see 6 fewer rats on Thursday than she did on Wednesday, so the change is 6 rats. The original value is 19, because now we’re starting from Wednesday.

Percent Change = 6/19 × 100%
Percent Change = 31.57…%

Round, like the question said to, and you end up with 32%.

I didn’t do a Weekend Challenge last weekend. It’s not like I didn’t want to, y’all. Things just got totally cray-cray. My use of “cray-cray” in the previous sentence should be enough of an indication to you that things remain squarely thus.

I actually wrote two questions for this weekend, but I’m only going to use one of them. The other one is going into the strategic reserve. For emergencies.

If you’re the first to answer this correctly (and not anonymously), I’ll bestow upon you coveted access to the PWN the SAT Math Guide Beta.

Other ways you can gain access to my Magnum Opus:

• Buy it. It’s \$5. You get about 300 pages of useful SAT math help. I get a footlong meatball sub at Subway.
• Send me a question of your own making. Added bonus: this is a fantastic way to solidify your knowledge of the test.
On to the question:

In the figure above A is the center of the circle, A and D lie on BF and CE, respectively, and B, D, F, and G lie on the circle. If BC = 3, and DG (not shown) bisects BF, what is the total area of the shaded regions?

Good luck, and have a great weekend. I’ll post the solution early next week.

UPDATE: Commenter Katieluvgold got it first. Nice work Katie!  Solution posted below the cut.

This is a shaded region question, and the usual shaded region technique applies, but with a twist. Since only parts of the top half of the circle are shaded, let’s just look use the top semicircle as Awhole.

We know AD is perpendicular to CE because it’s a radius connecting to a tangent line. Those are always perpendicular. It’s a rule.  And since A and D are both on rectangle BCEF, and BC = 3, AD must also equal 3.

So the radius of the circle is 3, so the area of the circle is 9π. That means the area of the semicircle is half that. Awhole = 4.5π.

Point A is the center of the circle, so it’s obviously the center of diameter BF. When the question tells us that DG bisects BF, it’s telling us that DG is also a diameter because it also passes through the center of the circle. That means ΔCEG has both a base and a height of 6.

Because the rectangle’s top and bottom are parallel, that means the top triangle (which is our Aunshaded) is similar to the large triangle, because all the angles of the two triangles are the same (the top one is shared, and the bottom ones are corresponding angles across parallel lines). We don’t have a way of knowing what the base is automatically, but we DO know that the height of it is 3, because the height of the little triangle is a radius. So the small triangle has a base of 3 and a height of 3.

The area of a triangle is (1/2)bh, so Aunshaded = 4.5.

To calculate Ashaded, we just subtract Aunshaded from Awhole:

4.5π – 4.5 = Ashaded

If you’re taking the test tomorrow morning, you should ignore this and REST YOUR BRAIN. If you’re not taking it tomorrow, though, then this is just another regular weekend for you, and you should work your brain HARD.

The prize, as it has been every week for weeks and weeks, is access to the coveted PWN the SAT Math Guide Beta. If you wanna win, you have to comment in such a way that I can contact you. Preferably using Facebook, Twitter, or Gmail.

A warning: this one is pretty tough.

In a hand in his weekly poker game, Mike won money from both John and Sean. Mike’s percent gain on the hand was equal in magnitude to John’s percent loss. John bet twice as much as did Sean. Mike and John together held \$200 in chips before the hand began. If Sean bet \$20 in chips on the hand, what was Mike’s chip total after the hand was over?

Good luck!

UPDATE: Nobody got this yet, so I’m gonna leave it up for now unanswered. If you solve it, book access is yours.

SECOND UPDATE: OK, Eowyn got it. Nice. Solution below the cut.
There’s a lot going on here, and at first blush it may seem as though you have too many variables to work with. If you write down everything you know, though, and play around with it enough, you’ll see that you can actually get it down to just one variable in one equation!

$percent\:&space;change&space;=&space;\frac{change}{original\:&space;value}\times&space;100\%$
$\frac{Mike's\:gain}{Mike's\:original\:value}\times&space;100\%&space;=&space;\frac{John's\:loss}{John's\:original\:value}\times&space;100\%$

Of course, we can tidy this up a bit:

$\frac{Mike's\:gain}{Mike's\:original\:value}&space;=&space;\frac{John's\:loss}{John's\:original\:value}$

Now we need to start figuring out the numbers. We know that Sean bet 20, and that John bet twice as much as Sean. So John bet 40, which means John’s loss was 40. We also know that Mike’s gain was 60, since he won all the money that was bet in the hand (John’s 40 + Sean’s 20).

$\frac{60}{Mike's\:original\:value}&space;=&space;\frac{40}{John's\:original\:value}$

And we can be a bit clever here if we want to jam all the info into one equation and say that, since Mike’s and John’s chips added up to 200 before the hand, Mike’s prior chip total can be m, and John’s prior chip total can be 200 – m.

$\frac{60}{m}&space;=&space;\frac{40}{200-m}$

Solve that for m

$40m&space;=&space;60(200-m)$
$40m&space;=&space;12000-60m$
$100m&space;=&space;12000$
$m&space;=&space;120$

…and you’ll see that before the hand began, Mike had 120 dollars in chips. Since he won 60, he has 180 after the hand.

I turn 30 years old tomorrow. 30 f’ing years old. CLING TO EVERY SECOND OF YOUR YOUTH. SQUEEZE EVERY OUNCE OF JOY OUT OF IT LIKE JUICE FROM YOUR FAVORITE FRUIT. SOMEDAY YOU WILL BE OLD LIKE ME.

The prize for the challenge: Free access to the PWN the SAT Math Guide Beta.

Week 1: Served customers numbered a through b
Week 2: Served customers numbered c through d
Week 3: Served customers numbered e through f
Week 4: Served customers numbered g through h
Week 5: Served customers numbered i through j

The proprietor of a deli is trying to project how many customers he usually has on Mondays, so that he can order enough roast beef, but not order too much because nobody likes rotten roast beef. At the deli, customers take numbers before they are served, so he plans to collect data over the next 5 Mondays in the format of the list above, and is looking for an expression for average number of customers that he can plug numbers into as he collects them. What is the expression for the average number of customers (in terms of a through j) that are served in the deli on those 5 Mondays?

Good luck, children. I’ll post the solution Monday.

UPDATE: Congrats, Serrilius. I hope you enjoy the book.

Solution below the cut.

As Collin so deftly pointed out in the comments, a little plugging in does wonders here. If, for example, the deli serves customers numbered 35 through 39, how many customers were served? 35, 36, 37, 38, 39. 39-35 = 4, but 5 customers were served. So when you want to know how many customers passed through on a day when numbers a through b were served, it’s b – a + 1.

So then, to find the average, you find the sum, and then divide by 5.

[(b a + 1) + (dc + 1) + (fe + 1) + (hg + 1) + (ji + 1)]/5
= (-a + bc + de + fg + hi + j + 5)/5
= (-a + b – c + d – e + f – g + h – i + j)/5 + 1

Instead of a weekend challenge question last week, I posted a question writing contest that I thought would be a big hit, but which has thus far generated much enthusiastic sound and fury, and only one actual question. Ah well. I maintain my position that if you are able to synthesize good SAT questions, you’ll be in very good shape to do well on test day. But if you don’t want to try it, I won’t shed any additional* tears.

Anyway, a return to form: This weekend challenge is designed to frustrate, and then eventually satisfy. The prize for the first correct response will be Beta Access to my Math Guide.

A precocious toddler is making a pattern with his alphabet blocks, each of which displays a letter (A-Z) in one of 5 colors: red, blue, green, orange, yellow. He has a LOT of blocks to work with, so he sets about putting them in alphabetical order, and also color order. The pattern goes as follows: Red A, Blue B, Green C, Orange D, Yellow E, Red F, etc. When he gets to the end of the alphabet, he starts at the beginning again. Assuming he doesn’t screw up the pattern, the nth block will be his first Green N. What is n?

Put your answers in the comments, and I’ll post the solution Monday. Good luck!

UPDATE: Nice work, Robin. I hope you enjoy the Beta!

Solution below the cut.

This is basically a 2-for-the-price-of-one pattern question. I’m going to explain it the snappiest way I know how, but you should also see Robin’s winning comments for a tidy little solution if you don’t like mine.

1) Since there are 5 colors (and we use a base 10 counting system) it’s going to be easy to track the green blocks. The first green block will be the 3rd block, and then the 8th block, and then the 13th, and then the 18th, etc. Basically, if a term has a units digit of 3 or 8, it’s a green block.

2) So now all you need to do is figure out when is the first time the letter N will fall on one of those terms. N is the 14th letter, so no go there. There are 26 letters, so the next time N comes up will be the 14+26 = 40th term. Then the 66th, 92nd, and 118th. 118 ends in 8, SO THE 118th BLOCK IS GREEN AS HELL! Yessssss.

* I’ve been sobbing all week. You monsters.

This Labor Day weekend I’m going to the wedding of one of my oldest friends from my hometown. Combine the fact that the number of my friends who aren’t married is dwindling dangerously close to zero and the fact that I wasn’t able to attend this year’s fantasy football draft so I had to autodraft and the fact that this weekend marks the end of the summer and OMG WHAT IS HAPPENING TO MY LIFE.

This weekend’s prize: you guessed it — free access to the Math Guide Beta. Make sure you don’t comment anonymously if you want to win.

In a certain youth soccer league, each team plays each other team exactly one time per season. If, during a certain season, there are 231 games played, how many teams were in the league that season?

First correct answer in the comments wins Beta access, and the undying adoration of my throngs of loyal readers.

UPDATE: Nice work, “AP FRQ Solutions” (whoever you are). I’ve shared the Beta document with you. Use it in good health.

Solution below the cut.
If you look back at my post about matching questions, you’ll notice a neat little tidbit hidden in there: if you’re looking for the total number of matches that can be made between a certain number of teams, you can take the number of teams, subtract 1, and take the sum of that number and each number lower than it to find the number of matches. In other words, if you have 7 teams, they will need 6+5+4+3+2+1 matches to each play each other once. Neat, huh?

So you can use that to solve this one quick and dirty:

1+2 = 3
1+2+3 = 6
1+2+3+4 = 10…

1+2+3+4+…+20+21 = 231.

So there must be 22 teams in the league.

Note: even if you didn’t know this little factoid, you might have been able to solve this way by finding the number of matches needed for smaller leagues, and looking for patterns in the number of matches required as the leagues get bigger.

If you’re looking for a mathier explanation, I think AP FRQ Solutions did a decent job in the comments.

If you’re on the East Coast, I hope that you and yours are able to weather the storm OK. I’m not sure whether I’ll have power or not on Monday, but I’ll try to get the solution posted then if possible.

This week’s prize will again be access to the Math Guide Beta, which is looking radder and radder by the day. I’ve been doing a TON of work on it. Anyhoo, first correct (and not anonymous) answer in the comments gets access.

If (x2 + 5x)2 – 36 = (x + m)(x + n)(x + p)(x + q), what is the median of the set {m, n, p, q}?

Good luck, kiddos. And stay safe.

UPDATE: Nice work, Ammad. I hope you enjoy the book, which I’ve now shared with you in Google Docs.

Solution below the cut.
The trick to getting this one right (or at least, to getting it right without the aid of a powerful calculator) is recognizing that the left side of the original equation is actually the difference of two squares:

(x2 + 5x)2 – 36
(x2 + 5x + 6)(x2 + 5x – 6)
Of course, each of those can be factored as well:
(x + 3)(x + 2)(x + 6)(x – 1)
So although we don’t know which one is which, we do know that {m, n, p, q} = {-1, 2, 3, 6}, which has a median of 2.5.
As usual, this was a bit tougher than you’d see on the SAT, but if you figured out that, despite its complexity, this was really just a difference of two squares question, then you deserve a cookie. No…two cookies.

You know the drill by now. The prize this week will, once again, be beta access to the Math Guide, which is really coming along nicely, if I do say so myself. I kinda can’t believe it’s going to top 300 pages, but at this rate we’re definitely heading in that direction.

Anyway, to get a look at it now and come along for the ride while I finish it up, be the first to answer the following question correctly in the comments, or send me \$5 using the link above.

A stereo equipment store owner notices that all his customers spend between \$110 and \$298, inclusive, in his store when they come in. For no discernible reason other than that I need a difficult math problem, he decides to express the range of dollar amounts his customers spend, s, in an inequality of the form |s – j| ≤ k, where j and k are constants. What is jk?

I’ll post the solution Monday. Good luck!UPDATE: Nice work, Serplet. Solution below.

Questions like this aren’t super common on the SAT, but when they appear they always follow the same pattern:

|variable – middle of range| < distance from middle to ends of range

In this case, the range is 110 < s < 298, so the middle of the range is (110 + 298)/2 = 204 , and the distance from the middle to the ends of the range is (298 – 110)/2 = 94.

So j = 204, and k = 94. Their product is 19176.

For more practice with this kind of question, go here.

Something you don’t often see in SAT prep materials is a pictograph, but when you actually look at a real SAT, pictographs are all over the place!

It’s not a conspiracy by a cabal of prep writers and SAT writers; pictograph questions are usually just really easy, and so prep writers don’t pay them much mind. This morning I decided to see if I could write a pictograph question that’s worthy of being called a Weekend Challenge. I guess you guys can judge whether or not I succeeded.

The prize this week: Same as the last few weeks. First correct response in the comments gets access to the PWN the SAT Math Guide Beta Program. One note: if you comment anonymously, I have no way of contacting you, or verifying who you are if you try to claim it was you later, so you can’t win. If you want to win, don’t post anonymously.

Sam is the kind of guy who keeps track of meaningless things in unnecessarily complicated charts. The chart above depicts the average number of times Sam finds himself laughing out loud during 3 distinct parts of the day. Sam has been keeping this chart for 4 days, and calculates that he will have to be made to LOL 10 more times than usual on the 5th night when his gassy uncle comes to family dinner in order to necessitate a change of one added face on his chart. On average, how many times a day has Sam been been made to LOL in school?

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

UPDATE: Nice work, Kira! I hope you enjoy the Math Guide Beta.

Everyone else: The solution is below the cut.
So this might be a pictograph question, but it’s also a weighted average question, so the solution will involve the average table. If you’re a longtime reader, you should be nodding your head like yeah. (‘sup Miley?)

Basically, Sam averages 4n LOLs per family dinner, and it’ll take him 10 more LOLs than usual at dinner on the night in question to get his average up to 5n LOLs. Luckily, his flatulent uncle is on the way! Here’s the setup:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 4 nights 4n LOLs per night 16n LOLs + 1 night 4n + 10 LOLs + 4n + 10 LOLs = 5 nights 5n LOLs per night = 20n + 10 LOLs= 25n LOLs

There are two ways to fill in the bottom right field in the table: 1) add the right column and get 20n + 10 LOLs, or 2) multiply the bottom row to get 25n LOLs.

What that tells us is:

20n + 10 = 25n
10 = 5n
2 = n

If n = 2, then we know how many times Sam has been laughing out loud in school, on average. The pictograph tells us there are 6n LOLs per day, which means Sam’s friends must be hilarious. He averages 12 LOLs a day in school.

I’ve been thinking about symbol function questions today, and although it’s not easy to come up with a really hard symbol function, this one is probably harder than you’d see on the SAT. Not by much, though. Make it multiple choice and this could totally appear.

The prize this week: same as last week’s. Free access to the PWN the SAT Math Guide Beta Program. First correct answer in the comments doesn’t have to pay a measly \$5 to see what I’ve been working on tirelessly for weeks.

Let the ♠ symbol be defined such that b♠ equals the sum of the greatest two integer factors of b, for all integers b. If &lt; 100, what is the greatest possible value of x♠?

Good luck! I’ll post the solution Monday.

UPDATE: Nice work, Anu! Hope you enjoy the Math Guide Beta. Solution posted below the cut.

This was a bit of a tricky question. To get a handle on it, just start listing factors of the highest possible values of x:

Factors of 99: 99, 33, 11, 9, 3, 1. So 99♠ = 99 + 33 = 132.

Factors of 98: 98, 49, 14, 7, 2, 1. So 98♠ = 98 + 49 = 147.

Factors of 97: 97, 1 (97 is prime). So 97♠ = 97 + 1 = 98.

Factors of 96: 96, 48, 32, 24, 16, 12, 8, 6, 4, 3, 2, 1. So 96♠ = 96 + 48 = 144.

Are you seeing a pattern yet? x♠ will always be bigger for an even x than for an adjacent odd x. So the highest even x allowed will give us our greatest value. That’s 98♠, or 147.

I lifted a box that was too heavy this week and I screwed up my back so bad that every time I put weight on my right foot, searing pain shoots up my entire right side. Getting older is awesome!

The prize this week for the first correct response: FREE Beta Access to my book. A \$5 value! Read up on the deets, if this is the first you’re hearing of it.

How many positive integers less than 1,000 contain exactly one odd digit?

Put your answers in the comments, and I’ll post the solution and contact the winner (if there is one) on Monday. Good luck!

UPDATE: Congratulations to Chong Lee, who nailed it first. Welcome to the Beta, Chong Lee. I hope you enjoy the book.

Solution below the cut.

The best way to solve a question like this is to count possibilities for each digit. There are three places your single odd digit could appear: the hundreds digit, the tens digit, or the units digit. For each of those possibilities, you have to count all the possible positive outcomes.

If your odd digit is the hundreds digit, for example, then you count like this:

##### Hundreds digit is odd
• There are 5 odd digits (1, 3, 5, 7, 9) that could occur as the hundreds digit.
• There are 5 even digits (0, 2, 4, 6, 8) that could occur as the tens digit.
• There are 5 even digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the hundreds place.

What adds an extra degree of difficulty to this question is that the number of possible odd digits and the number of possible even digits is always the same. Sorry about that. That’s why it’s a challenge question.

Moving on, we still have to account for the odd digit coming in the tens or ones places.

##### Tens digit is odd
• There are 5 even digits that could occur as the hundreds digit. (Remember, zero still counts. 010, or just plain old 10, has one odd digit.)
• There are 5 odd digits that could occur as the tens digit.
• There are 5 even digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the tens place.

##### Ones digit is odd
• There are 5 even digits that could occur as the hundreds digit.
• There are 5 even digits that could occur as the tens digit.
• There are 5 odd digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the ones place.

##### How many positive integers less than 1000 contain exactly one odd digit?

125 + 125 + 125 = 375.

Magic, right? I know. If you want to see it done in a way that it could never be done in a test environment, I’ve confirmed the results in a spreadsheet for all the doubters. Click here.

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.

$\frac{central\:&space;angle}{360^{\circ}}=\frac{arc\:length}{circumference}$
$\frac{x^{\circ}}{360^{\circ}}=\frac{8}{18}$
$x=160$

I found this awesome graphic at Amazin’ Avenue, and it inspired me to write a baseball-themed challenge question this week. I’m a huge Mets fan, and my favorite player on the team right now is R.A. Dickey. He faces the hated Phillies tonight. I am so pumped to watch.

The prize this week: A free (imaginary) R.A. Dickey bobblehead doll! Best prize ever!

In the 2011 season, R.A. Dickey has thrown 1758 pitches and batters have completed 494 plate appearances against him as of July 15. What is the least number of batters he would need to face to have a pitch per plate appearance average under 3.50?

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

 Let’s go Mets!

UPDATE: solution below the cut.

Don’t get hung up on the terminology! Even for baseball fans, this isn’t a stat that comes up in conversation. The SAT won’t ask about something so culturally anchored as baseball, but it very well might throw a question like this at you in another context.

We know that Dickey has thrown 1758 pitches through 494 plate appearances. So right now he has a pitch per plate appearance average of 1758/494, about 3.56.

If we’re looking for the minimum number of batters he must face to get that average under 3.5, we need to figure out the minimum number of pitches that can occur in a plate appearance. That number is 1. A batter can see one pitch and, say, get a hit. He could also ground out on that pitch. Pitch per plate appearance, as you see, doesn’t tell much of a story on its own.

So the minimum number of batters he must face must be the SAME as the number of additional pitches he must throw, since it’s one pitch per batter.

Here’s your setup (using x for additional number of pitches AND additional number of plate appearances):

$\textrm{Pitch\:Per\:Plate\:Appearance\:Average}=\frac{1758+x\:\textrm{Pitches}}{494+x\:\textrm{Plate\:Appearances}}$

From here, you can either throw the equation into your trusty graphing calculator and use the table to scroll down until you see the average drop under 3.5, or you can solve algebraically. Either way, you should see that after 11 additional plate appearances and pitches, his average is just over 3.5, and after 12, it’s just under.

So the answer I’m looking for here is 12.

I’m having a lot of fun playing around with some geometry drawing software this week (nyeeerd!), so I figured I’d use it again to make another “fun” 3-D problem for the weekend challenge. This is a bit tougher than you’d find on the SAT, but the underlying concepts, as always, are important for the SAT.

The prize this week inspired by my current situation: If there should ever come a time in your life where you’re trying to move from Brooklyn to the Bronx, you’ll be smart enough not to bother trying to move yourself in your Toyota Yaris and instead just hire movers. And when you do, they won’t break your stuff.

Point B is in the center of the top face of the cube in the figure above, and point A is one of the cube’s vertices. If the distance between points A and B is d, then what is the cube’s volume in terms of d?

Put your answers in the comments, and I’ll post the solution Monday. Good luck!

UPDATE: Nice work, JD. You were wrong at first, but you corrected yourself before I did. Solution below the cut.

The most difficult 3-D problems you’ll come across on the SAT will require you to work with right triangles, usually on the same plane as one of the solid’s sides, and then on a plane that cuts through the solid. In this question, you’re going to want to make a right triangle whose hypotenuse is AB (A.K.A. d).

To do so, we’re going to have to first figure out the distance from B to the cube’s corner. Let’s assume the cube has sides of length c:

Since we’re so good with right triangles, we know that the diagonal of the square on the right is c2, and since B is the midpoint of that segment, the distance from the corner to B must be c2/2. Now we know the lengths of both legs of the right triangle we actually care about, the one with hypotenuse d. We can use it to solve for d in terms of c. (“But wait,” you say, “aren’t we looking for the volume in terms of d?” Yes, we are. We’ll get there. Promise.)

$c^2&space;+&space;\left&space;(\frac{c\sqrt{2}}{2}\right&space;)^2&space;=&space;d^2$
$c^2&space;+&space;\frac{2c^2}{4}&space;=&space;d^2$
$c^2&space;+&space;\frac{c^2}{2}&space;=&space;d^2$
$\frac{3c^2}{2}&space;=&space;d^2$
$\frac{c\sqrt3}{\sqrt2}&space;=&space;d$

Now that we’ve got d in terms of c, we can put c in terms of d, and cube it to get the volume. Wahoo!

$c&space;=&space;\frac{d\sqrt&space;2}{\sqrt&space;3}$
$c^3&space;=&space;\left&space;(&space;\frac{d\sqrt&space;2}{\sqrt&space;3}\right&space;)^3$
$c^3=\frac{2d^3\sqrt&space;2}{3\sqrt&space;3}$

Some people like to stop here, and others like to rationalize the denominator (as JD did in his response). The SAT has been known to swing both ways.

$c^3=\frac{2d^3\sqrt&space;2}{3\sqrt&space;3}\times\frac{\sqrt3}{\sqrt3}$
$c^3=\frac{2d^3\sqrt6}{9}$

Donezo.

Note: you can check your work here by plugging in. Say c = 2, so the volume of the cube is 8. That would make d = 2√3/√2. Plug that in for d in our solution. Does it give you 8? Yup.