Posts tagged with: challenge

I’ve been going to the gym lately which is torture and I hate it. I am sore everywhere. I don’t understand why anyone finds it pleasurable to go there, but I do understand not wanting the body of a 70 year old when I’m 35, so I force myself. Anyway, while I was wheezing and sputtering on the treadmill yesterday I was inspired to write a challenge question, which I humbly present to you now. Winner gets a Math Guide.

Note: Figure not drawn to scale.

The flat running surface on a certain treadmill is 5 feet long (delineated by the dashed lines in the figure). Two wheels, each with radius 4 inches, keep the belt moving smoothly as the user runs. Someone painted a yellow stripe across the belt because reasons. Mike counts the number of times he sees the yellow stripe while he runs to try to distract himself from the agony inherent in the activity. If he crosses the yellow stripe 1111 times in 20 minutes, what is his average speed over that time, in miles per hour? Round your answer to the nearest tenth.

Put your answers in the comments. First correct, non-anonymous answer wins. Full contest rules here.

UPDATE: Jayce got it first, and a book is merrily making its way to his abode. The solution is now posted below the cut. 

The first thing we need to do is figure out how long the belt is. It’s a loop, basically, that has 2 straight sides, and two semicircular sides.

The straight sides are 5 feet long. That’s easy. The radius of each semicircular part is 4 inches, which means we need to convert to feet or we’re going to be in big trouble. 

4 inches = 1/3 foot. The semicircular arc on each side, then, will be half the circumference of a circle with radius 1/3 foot. C = 2πr = (2/3)π. Since each side is only half the circle, each side is (1/3)π. Like so:

So the length of the treadmill belt is 10 + (2/3)π feet.



If we want to know how far I ran in 20 minutes, we now have the information necessary to figure it out. 

If I saw the stripe 1111 times, that means I travelled 1111(10 + (2/3)π) feet. That’s roughly 13436.9 feet, but of course we should keep our numbers precise until the final step.

To convert to miles, divide the above result by the number of feet in a mile, 5280:

So that’s how far I ran in 20 minutes. At this point, I should come clean and admit that when I go to the gym, I don’t usually make it that far in 20 minutes. But I’m getting closer.

To get a speed in miles per hour, we need to multiply that by 3, since 20 minutes is only 1/3 of an hour:

There you have it. Rounded to the nearest tenth, that’s a speed of 7.6 miles per hour.

AX - Street Fighter - Zangief
Source

I’ve got a bunch of friends at San Diego Comic-Con right now, so I figured I’d use that for the theme for this weekend’s challenge. Of course, I made most of these numbers up, but I didn’t completely pull them out of thin air.

58% of the attendees at Comic-Con this year are male. 10% of the attendees are cosplayers. 15% of the cosplayers are dressed as characters from the Marvel universe. 63% of the cosplayers are female, 20% of whom are dressed as Marvel characters. If 130,000 people attend Comic-Con, what percent of male attendees are cosplayers that are not dressed as Marvel characters? (Round your answer to the nearest tenth of a percent.)

Put your non-anonymous responses in the comments! First correct response wins a Math Guide. Full contest rules here.

UPDATE: A bunch of people got this, but Diane got it first. Solution below the cut.

The key to a question like this is just to figure out everything you can. Since each condition is an either/or condiiton (male or female, cosplay or not cosplay, Marvel or not Marvel) each statement gives you twice as much information as it initially appears to give.

Start by figuring out the total number of male attendees. The question says that 58% of the 13,000 attendees were male, so that’s 75,400. Put that number aside now— we won’t need it for a while.  (Of course, to the point above, that also tells us there are 54,600 female attendees. We don’t need that number, but we know it.)

If 10% of the attendees are cosplayers, that means 13,000 folks showed up in costume. We’re given a bunch of information about them.

First, we’re told how many of them are dressed as Marvel characters. That’s 15% of 13,000, or 1,950.

We’re also told how many of the 13,000 cosplayers are female: 63%, or 8,190 of the 13,000. Which also tells us how many of them are male: 4,810.

Of the 8,190 female cosplayers, we know 20% of them, or 1,638 are dressed as Marvel characters.

If 1,638 of the 1,950 Marvel cosplayers are female, that means the other 312 of them must be male.

Remember what we want now. We want the percentage of male attendees that came in cosplay that was NOT Marvel-inspired.

There are 4,810 male cosplayers, and 312 of them are in Marvel getup. That means 4,498 of them are NOT in Marvel costumes.

Now just find the percentage:

4,498/75,400 × 100% = 5.965517…%

Round that to the nearest tenth, and you’ve got 6.0%. 

I’m heading out this morning to go camping for the weekend (Woohoo! Bug bites!) and I won’t have any way to access the Internet reliably, so if you enter this contest please be patient. I probably won’t be able to declare a winner or post a solution until Monday evening.

I’m told the place we’ll be camping is in a wireless dead zone, which inspired this question. Note that I’m pretty sure this isn’t really how cell carriers set up their networks, but it’s a fun problem. Because it’s a bit more involved than normal, I’m going to say this challenge is worth TWO Math Guides—one for you and one for a friend. They’ll both get shipped to you, and then if you don’t have any friends you can sell it on eBay or something. I dunno. I’m sure you have friends.

Annnnd here we go!

A wireless phone company’s cell towers are arranged in regular hexagonal structure, as shown in the figure above. If each tower’s coverage radius is exactly half the length of a side of the regular hexagon, what percentage of the hexagon’s area will not be covered by the towers?

Put your answers in the comments! The usual rules apply: you must not be anonymous to win, and although you may win if you’re not in the US, you will have to help me out with shipping costs via PayPal.

Good luck, and have a great weekend.

UPDATE: Nice work, Rex T95! Two books are on their way to you! Solution below the cut.


So you’ve got a couple of tricky things going on here, but this question is not NEARLY as gnarly as it looks at first.

We’re going to treat it like a shaded region question (because that’s what it is). First, we’ll find the area of the hexagon. Then, we’ll find the areas that are covered in the hexagon by the towers. Do a little subtraction, and voila! We’ll have our answer.

Step 1: Find the area of the hexagon
There are a lot of ways to do this, but the way I find most intuitive is to remember that all regular polygons are made up of congruent isosceles triangles that meet in the center, like so:

Since the center is 360º, you can figure out the measures of all the center-touching angles by dividing 360º by the number of triangles (which, of course, is the same as the number of sides):

360º ÷ 6 = 60º

So in a regular hexagon, the central angles of the isosceles triangles are all 60º, which means these triangles are not only isosceles, but equilateral. See how I did nice to you? You’re welcome.

So we need to find the area of an equilateral triangle, and multiply it by 6 to get the area of our hexagon.

Step 1a: Find the area of an equilateral triangle
At this point, I’m going to make an executive decision and plug in 2 for the length of the sides of my hexagon. We could do this without plugging in, of course, but it’d be more of a pain and I don’t subscribe to the school of thought that math should be difficult if it doesn’t have to be. So we’re dealing with an equilateral triangle with sides of length 2:

Surely, you know by heart that the area of a triangle is (½)bh. When we drop an altitude to find the height of an equilateral triangle, we create 2 30º-60º-90º triangles, so we don’t even need to Pythagorize since we know our special rights so well

The base of the above triangle is 2, and its height is √3.

A = (½)2√3 = √3
Ahexagon = 6×A
Ahexagon = 6√3

Step 2: Find the area that is covered by the towers
This is actually a bit easier. We’ve already decided to make the sides of the hexagon equal 2, so the radii of all the circles in the diagram are 1, and therefore each has an area of π.

Furthermore, the 6 circles that are only partially included in the hexagon are all sliced the exact same way: in 120º chunks.

So let’s do a little part-whole ratio calculating to figure out the sector areas.

Of course, there are 6 of those sectors. So the total area covered by the cell towers is 2π (for the 6 sectors) + π (for the center circle) = .

Step 3: Find the uncovered area
This is the easy part! The area that’s not covered is simply AhexagonAcovered.

Auncovered = 6√3 – 3π

Step 4: Find the percentage of the hexagon that’s uncovered
This last step, where we convert to a percentage, is why it doesn’t matter that I chose to plug in a value instead of working through this whole problem a bit more algebraically.

Throw that in your calculator, and you’ll get 9.31%.

Phew!

Robin Hill Park (Jon Forster) / CC BY-SA 2.0

What an annoying weekend I had. It’s all my own poor planning, but the Math Guide went temporarily out of stock this weekend. Hopefully, everything will be back to normal by tomorrow, which means I can get right back to giving books away!

This weekend challenge is about those crazy pirate ship rides they have at amusement parks. If you’ve never seen one of these in action, here’s a video. They basically swing you around in a big circle.

Since there are so many ways to win a Math Guide these days, I’m not gonna go easy on you guys with the challenge questions. Just remember: Challenge questions are not SAT questions, they’re just for me to have fun writing and you to suffer through solving. 🙂

Anyway, here we go.

Note: Figure not drawn to scale.

Larry is seated at point L on a Pirate Ship ride that has not started moving yet, as shown. The measure of ∠ABC is 60º, and the distance between points A and C is 35 feet. With the ride at rest, the lowest point of the pirate ship is 5 feet above the ground, and Larry is 6 feet off the ground. Once the ride is in motion, how long is the arc Larry travels between the two points at which he is 40 feet off the ground? (Assume that the ship is a 60º arc.)

Put your answers in the comments. First correct (non-anonymous) answer gets a Math Guide!

Good luck!

UPDATE: wgreens got it! Nice work.

Solution below the cut
The first (and probably most important thing) to recognize is that AB and BC are both radii of the circle on which the ship travels, which means △ABC is isosceles and angles A and C are congruent. Since we also know that the measure of ∠ABC is 60º, it turns out the triangle is equilateral, and thus the circle has a radius of 35. Boom.

This, as wgreens pointed out in his answer, has some convenient implications. Specifically, it means that the center of the circle is 40 feet off the ground, so the two points at which Larry is 40 feet off the ground lie on a diameter of the circle.

That means to find the arc length, we don’t need to do any more work with angles. We just need to find the circumference, and cut it in half.

C = 2πr
C = 2π(35)
C = 70π

Since Larry only travels half the length of the circumference, he travels 35π feet.

In honor of Father’s Day, this weekend challenge is inspired by my dad’s hobby business. He sells wooden flags, like the one above, on Etsy. Basically, he finds discarded pallets (this is a pallet), chops ’em up into pieces of the size he needs, and mounts them.

As usual, these challenge questions are not really SAT questions, they just require you to work the same muscles that you need to exercise to do well on the SAT. Consider them cross-training. Here we go.

The creation of one of these flags requires 13 solid strips of wood that are 1 inch thick, 2 inches wide, and 36 inches long. If my dad makes as many flags as possible out of a pallet with 10 usable slats that are one inch thick, 7 inches wide, and 40 inches long, what is the volume of the leftover scrap in cubic inches? Disregard sawdust loss and any part of the pallet that was not originally deemed usable.

Put your answers in the comments. First correct (non-anonymous) answer gets a Math Guide. Good luck, and don’t forget to wish the fathers in your life a happy Father’s Day!

UPDATE: Congratulations to Shahriar, who got it first. I’ll be in touch shortly to get your shipping information, Shahriar, so I can send you your Math Guide.

Solution below the cut.
I realized after I posted this that my measurements weren’t very realistic at all, but whatever. It’s the math that’s important, not whether my fictional pallet measures up to industry standards.

The first step to solving this problem is to recognize that by making the thickness of the slats 1 inch, I went easy on you: the value of the volume of wood left over will be the same as the value of the area of its front-facing face. In other words, you can basically treat this as a two-dimensional problem even though I’m asking for volume. You’re welcome.

How many usable strips of wood can be made from one pallet slat? 3 of them. And once those 3 usable strips are cut away, each slat produces 36+28=64 in3 of scrap.

Of course, there are 10 such slats, so you’re looking at something like this:
Oh but wait. That produces 30 usable slats. Which is only enough to make 2 flags, since each flag requires 13 strips. I guess 4 of those strips become unusable scrap, too. 🙁

So really, the situation is that each slat produces at least 64 in3 of scrap. We have to add the volume of the 4 unnecessary strips (32×2=72 in3 each) to our final volume, since they’re not actually used in the making of flags.

(10 slats × 64 in3 scrap) + (4 strips × 72 in3 scrap) = 928 in3 of scrap

I went back to my old high school last night to attend the final concert of the choir director that presided over so many of my formative moments. I got to hang out with some of the same people with whom I used to have the kind of deep, meaning-of-life conversations that only happen in movies or in real life when you’re between the ages of 15 and 17. A good number of them are teachers now, as it turns out. Their students are lucky to have them.

I went back with a few of the people who helped make me into me, to honor a man who helped shape all of us. I saw a fantastic concert. I had a lot of hugs and handshakes. I found a plaque that still has my name on it. And then I drove home listening to a record I listened to a lot back then. Now I’m sitting in my apartment back in New York, wistful and weary. HOLD ON TO THESE MOMENTS WITH BOTH HANDS. DRINK FULLY AND RICHLY FROM THE CUP OF YOUTH. AND RETURN FOR SECONDS.

Anyway, here’s a challenge question! First correct response gets a free copy of the Math Guide.

Louis and Rebecca each had x dollars on Monday morning. On Monday afternoon Louis paid Rebecca 20% of his money for a computer that she was selling. On Tuesday morning, Rebecca paid Louis 20% of her money for a lawnmower that he was selling. On Wednesday morning, both Louis and Rebecca paid 20% of their money to Steven for…something. I don’t know. Steven was selling something. Assuming they did not spend or receive any other money in that period, how much money did Steven receive, in total, from Louis and Rebecca in terms of x?

Leave your answers in the comments. I’ll post a solution on Monday (assuming someone gets it by then).

In order to win a book, you must not comment anonymously. I need to be able to get in touch with you to get your shipping info, etc. Also, while you can still win a free book if you don’t live in the US, you’re going to have to pay for shipping. So the book is only sorta free if you’re international.

Good luck!

UPDATE: Congratulations to Jason, who nailed it with alarming speed, and to whom a Math Guide will be en route. Solution below the cut.

There’s a fast way to do this problem, and then there’s a very fast way to do this problem. I’m gonna give you both.

First, the fast way. Plug in. Say x = 100. Here’s how the money flows:

Day and timeLouisRebeccaSteven
Monday morning$100$100
Monday afternoon$80$120
Tuesday morning$104$96
Wednesday morning$83.2$76.8$20.80+$19.20=$40

$40 is 40% of x (which, as you’ll recall, we said was 100), so the answer is 40% of x, or 0.4x.

As you look at the table, you might notice that each row adds up to $200. Interesting, no? Which brings us to the very fast way to do the question. Regardless of the money that changes hands between Louis and Rebecca, the total amount of money they have together never changes. So we can just, from the get-go, say that the amount of money that gets handed to Steven is 20% of the total money Louis and Rebecca begin with.

They start with, in total, 2x dollars. 20% of 2x is 0.4x.

Have you seen this movie? I liked it.

Nobody’s answered the last challenge question yet, but whatever, here’s another one. And I’ll tell you what: First person who gets this one right gets a free Math Guide.

Norville has four different colored hula hoops around his waist, each with a diameter of 3 feet. He is standing on a flat surface. He drops a red hula hoop so that he is standing exactly in its center. He then walks 5 feet, stops, and drops a blue hula hoop (again, so that he is standing in its center). Norville then walks 10 feet in a different direction, and drops a green hula hoop in the same way. Finally, he turns and walks 8 feet in another direction before dropping his last hula hoop, a purple one, in the same manner he dropped the others. What is the greatest possible area of overlap the purple hoop can have with another hoop, and with which hoop can it have that overlap?

You can’t answer anonymously to win, but don’t like leave your address in the comments or anything—we will take care of the shipping arrangements once I declare a winner. If you live outside the US, you can still win, but you are going to have to help me cover shipping costs. Sucks, I know.

Also, I want answers in terms of π. Don’t give me any decimal approximations.

Good luck!

UPDATE: Congratulations to “Sjfour4,” whose book is en route! Solution below the cut.

I love this question because like many SAT questions, it’s not nearly as complicated as it seems at first glance. The first thing you should be asking yourself is whether Norville is able to complete a triangle with his 3 short trips. In other words, is it possible to have a triangle with sides of length 5, 10, and 8?

To evaluate the possibility, check against the triangle inequality theorem:

Is 8 + 5 > 10? YES.
Is 10 + 5 > 8? YES.
Is 10 + 8 > 5? YES.

So that triangle is possible. Here’s an artist’s rendition of Norville’s trip (using cm instead of feet because that’s how my software measures things):

The diameter of each hula hoop is 3, so the radius of each is 1.5. That means the area surrounded by each hoop is π(1.5)2 = 2.25π.
Since the triangle can be completed, the purple hoop can be dropped right on top of the red hoop, and the overlap can be the entire area covered by the hoop: 2.25π.

You guys. Crazy story. I was in a coffee shop messing around with Geometer’s Sketchpad, just slapping together some perpendicular line segments because that’s how I roll, and this guy who could see my screen thought I was an artist or something. It was kinda weird, but he wanted to buy my “art.” Naturally, I went along with it. So now I have to frame this stupid thing and deliver it to him. Thing is, he was really particular about how it should be framed. He wants both points A and J to be on one of the rectangular frame’s diagonals, and a distance of exactly 4 cm along those diagonals from A and J to the inner corners of the frame. He wants the frame to be 3 cm thick, too. Anyway, he’s busy clearing space for this thing in his house and he’s emailing me asking for dimensions and basically driving me crazy. Can you help me out? What will be the outer dimensions of the framed piece of art? I’ve included measurements of all the lengths of the segments, but don’t count those as part of my image—I’m going to delete them before I print this out.

Post your answers—rounded to 2 decimal places—in the comments. The prize for the first correct answer: I will name my next pet goldfish after you. Scout’s honor.

UPDATE: Here’s where I sheepishly admit that this is not one of my best challenges. Sorry. But Rob finally did get it, and even though I didn’t plan to when I posted this, I’m gonna send him a Math Guide for his troubles. 

Solution below the cut.


The first thing you’re gonna want to do is construct a right triangle out of this bad boy, which means you’re going to need to find the total vertical length and the total horizontal length. What a pain!

Vertical length (AK) = AB + CDEFGH + IJ = 12.87
Horizontal length (JK) = BC + DE + FG + HI = 12.88

So that’s super annoying and why this is not one of my best challenges—Geometer’s Sketchpad is an awesome program that I love and it makes making diagrams really easy and fun, but this was drawn up to be a 45º-45º-90º triangle and because of rounding now it’s not quite so pretty. We’ll be OK, but still. Annoying.

Anyway, now we can find AJ, the hypotenuse.

12.872 + 12.882 = AJ2
331.5313 = AJ2
18.2080… = AJ


OK, good enough. We’re told in the problem that the inner corners of the frame are to be 4 cm from A and J. How the heck do we do that?

Easy, friend. Similar triangles. Picture it with me. You’re going to be creating another, larger, right triangle when you extend that hypotenuse by a total of 8cm. So with a few quick ratios, we can determine the inner dimensions of the frame!

The hypotenuse of our large triangle is going to be 18.2080… + 8 = 26.2080…

Do some ratios to figure out the height and width:

Again, how annoyed am I at this rounding? Super annoyed.

But we’re almost done. Now all we need to do is account for the width of the frame, which is supposed to be 3 cm. Because the frame’s thickness will add to top and bottom, left and right, we add 6 cm to each side to arrive (finally) at our outer dimensions: 24.52 cm by 24.54 cm.

I have never seen a school bus like this. Source.

I’m going to try to ease myself back into posting math questions fairly regularly on here. No prizes for now, just the satisfaction of knowing you were the first person in the world besides me to figure this particular question out. This probably wouldn’t appear on the SAT, but it exercises the same muscles you’ll need to flex to do well.

The wheel diameter on a certain school bus is 40 inches. A sensor on one wheel indicates that over a 20 minute period, the wheel made 9250 revolutions. Assuming the wheel did not slip at all, what was the average speed of the bus, in miles per hour, for those 20 minutes? Round your answer to the nearest whole number. (12 inches = 1 foot; 5280 feet = 1 mile)

Have at it!

UPDATE: John got it. Well done, sir.

The first important thing that you need to remember about wheels is that when they roll one revolution, they travel one circumference—as long as they don’t slip. Here’s an awesome illustration of that for a wheel with a diameter of 1 (and therefore a circumference of π):

Pi-unrolled slow

So, since the wheels on our bus have a diameter of 40 inches, each revolution of one of our wheels moves the bus 40π inches.

The wheel makes 9250 revolutions in 20 minutes, so the bus travels 9250•40π ≈ 1,162,389 inches in 20 minutes.

Convert that to feet: 1,162,389 inches ÷ 12 inches/foot ≈ 96,866 feet

…then to miles: 96,866 feet ÷ 5,280 feet/mile ≈ 18.35 miles

The last step, of course, is remembering that you need to get miles per hour, and this bus has only been traveling for 20 minutes. Speed = distance ÷ time, so our bus’s speed in miles per hour is:

Of course, since you’re probably working on your calculator, you should try not to round at intermediate steps like I did. Here’s a super-precise answer from Wolfram Alpha.


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Ask me questions at qa.pwnthesat.com.
The SAT isn’t the only
nerdy thing I like.

I’m absolutely swamped with work at school. Just completely swamped. Which is why I’m sitting here trying to come up with a good weekend challenge, of course. I like procrastination almost as much as I like frozen cheese ravioli. And I like frozen ravioli a lot.

Everyone hates work problems like the one I’m about to throw at you. Even I kinda hate them, honestly. You probably wouldn’t see this on the SAT (where work problems are rare and basically limited to the easier special case of distance = rate × time problems), but of course, you’d never see most of the stuff I throw at you as Weekend Challenges on the SAT. These Weekend Challenges are like cross-training. You’re working the same muscles, but doing slightly different exercises.

Enough rambling. You know the drill by now. First correct response in the comments wins everyone else’s undying adoration. Here’s your challenge:

It would take Jerry 13 hours to paint the fence in his yard by himself. If his friend Susannah helped, it would take the two of them 6 hours to paint the fence. If their friend Lori also helped, the three could paint the whole fence in just 2 hours. How fast could Lori paint the fence by herself?

Good luck, folks. I’ll post the solution Monday (or maybe Tuesday).

UPDATE: Congrats to gottagetmealife for getting the right answer first, and thanks to JD for saving me the trouble of having to write up my own explanation. See his comment below to see how simple this question could have been. I’ll only add (to say implicitly what JD implied) that you don’t need to know at all how fast Susannah is. She’s just in there to mess with you. I hope you can find it in your hearts to forgive me for that. 🙂

I’m in Los Angeles this weekend visiting friends. It’s sunny here, and even though it’s actually not that all that warm, the palm trees make it feel that way. I know that many of you, however, are not on vacation, and that actually some of you are taking the December SAT tomorrow. If that’s you, good luck!

There’s no prize this week–just bragging rights. When the book is done, I’ll figure out creative ways to give prizes again.

John is putting his favorite movie posters in frames to hang in his house. It costs John $75 for a frame that measures 30 inches by 50 inches. It costs him $125 for a frame of similar construction but with dimensions of 40 inches by 55 inches. If 30% of the price increase is the cost of the additional wood needed for the outside of the frame, what is the cost of one foot of the wood?

Have fun, folks. I’ll post the solution early next week (might actually not be Monday or even Tuesday due to travel). Bonus bragging rights for the most coherent explanation in the comments before then.

(Side note: I know a few of you are waiting on me to answer Tumblr questions. I will probably not get to them until Monday or Tuesday since my computer time is limited.)

UPDATE: Nice work, emolano. Solution below.

This one is deceptive. The price difference between the two frames is $50. 30% of that is $15. Since the dimensions increase from 30×50 to 40×55, it’s tempting to say that you’re looking at a 15 inch increase in the amount of wood needed, and that the wood costs $1/inch. But if you did that…you’d be wrong.

If that frame is going to be rectangular, you have to add wood to all four sides of the frame! That means you’re really adding 30 inches of wood. So the price increase is $15 for 30 inches of wood, or $0.50 per inch. Since there are 12 inches in a foot, the wood costs $6 per foot.

Source: Nature’s Graffiti.

Do you guys do the Black Friday thing? One time, about 7 years ago, I got up at 3 in the morning and met some friends to wait outside Best Buy. It was complete pandemonium, and I was not fast or ruthless enough to get the TV I wanted. Then, I got the same TV for the same price a week later online. I still have it. I’m staring at it right now, actually. And that’s the story of why I hate Black Friday. The end.

As you may or may not know, the Beta for the Math Guide is closed since the book is almost done. I haven’t really figured out what to give away as a prize for these challenges now that I don’t have the Beta. I might come up with something good in the future, but this one’s just for good old fashioned bragging rights. So, I dunno, go on College Confidential and tell everyone you’re super smart. You’ll fit right in! (I kid, I kid. That site is a fantastic resource.)

x is directly proportional to y
x is inversely proportional to z

If the two statements above are true, and y = 10 and z = 5 when x = 2, what are x, y, and z when x + y + z = 32 and
x < y < z?

Good luck, troops. I’ll post the solution Monday (probably late).

UPDATE: Jeffery nailed it. Bragging rights awarded. Go lord it over everyone you know. 🙂

Solution below.

Jeffery did a nice job of laying out the pure algebra solution, so I’m basically just going to add some commentary. Let it be known that this question doesn’t really lend itself to much trickery–you’re going to have to do some equation solvin’. Remember, these Weekend Challenges are supposed to be harder than regular SAT questions!

If x is directly proportional to y, then y will always be the same quotient, no matter what. We know y = 10 when x = 2, so we know that, conveniently, y will always be 5 times bigger than x. Write the equation: y = 5x.

You can figure out the relationship between x and z in a similar way, although when we’re talking about inverse proportionality, we know that the product xz will always equal the same thing. In this case, z = 5 when x = 2, so xz = 10, always. z = 10/x.

We can now rewrite the equation from the question:

x + y + z = 32
x + 5x + 10/x = 32
6x + 10/x = 32
Now we’re going to have to solve. First, multiply everything by x to get that one out of the denominator:
6x2 + 10 = 32x
6x2 – 32+ 10 = 0
It’s not trivial to factor this, so from here you can either put it in the quadratic equation to find your roots, or you can graph it on your calculator and (depending on your calculator) figure it out that way.
The roots are x = 1/3 and x = 5.
When x = 5, y = 25, and z = 2. That adds up to 32, but breaks the x < y < z rule.
When x = 1/3, y = 5/3, and z = 10/(1/3) = 30. Yep, that adds up to 32, AND x < y < z. Nice.
Image found at this…awesome site.

This will be the last Weekend Challenge question that has access to the Math Guide Beta Program as its prize. I’m going to be closing the Beta to new users soon, as the book nears completion. If, by the way, you’re interested in the day-to-day progress I’m making on the book, you can check in on me at Google+.

When I write questions (and I have to imagine this is the same way every question writer does it) I just put in placeholder numbers while I write and then I go back and solve. If the numbers don’t work out nicely (say, a fractional child, or something) I’ll change them around. But when I wrote this one, the numbers worked out perfectly the first time. That feels awesome. Anyhoo, Beta access to the first non-anonymous commenter to PWN the following question:

At Masuk High School, 200 people are in the chorus, 130 people are in the band, and 45 people take AP calculus. If, in total, 92 people take two of the three classes, and 80 take just one of the three, how many students take all three classes?

I’ll post the solution Monday. Good luck, and have a great weekend.

UPDATE: Nice work, Jeffery. You’re officially the last person into the Beta. I hope you enjoy it.

Solution below (although Jeffery and JD both posted rather nice ones in the comments).

If you were thinking Venn diagram here, then you’re eVenn more clever than I thought!

chorus: 200 = c + p + n + x
band: 130 = b + p + m + x
AP calc: 45 = a + m + n + x
Add those all up, and you get:
375 = a + b + c + 2m + 2n + 2p + 3x
375 = a + b + c + 2(m + n + p) + 3x

From the question you know that 92 people take 2 of the three classes, so:

m + n + p = 92

You also know 80 people take only one of the three, so:

a + b + c = 80
Substitute:
375 = 80 + 2(92) + 3x
111 = 3x
37 = x
 
Awesome, right?

I’ve been having some minor technical issues with the blog. I think it has something to do with the code I wrote to make vocabulary words turn red when the page loads interacting badly with the code for DISQUS, which is the commenting system. But since I’m not a computer programmer, I have no idea how to fix it. All of which is to say: I’m sorry that this weekend challenge includes an additional element of frustration in that you may have to reload the page a few times if the comment system doesn’t properly load the first time. I’d be happy to pay a programmer to fix it if any of you want to pay me to pay the programmer. 🙂

First correct comment wins access to the coveted PWN the SAT Math Guide. Make sure your friends are sitting down when you tell them you won. We don’t want anyone to fall and sustain head trauma.

Seven years ago, Thom was half as old as Phil and three times as old as Melanie. If Thom is 19 years old now, how many years will it be until Melanie is half as old as Phil?

I’ll post the solution Monday. Have a great weekend, y’all.

Nice work Cindy and Stephanie. Solution below.

Start in the present day. Thom is 19. Which means 7 years ago, he was 12. At that time, he was half as old as Phil, so Phil was 24. Thom was 3 times as old as Melanie, so she was 4.

In the present day, Phil must be 24 + 7 = 31 years old, and Melanie is 4 + 7 = 11. You could solve algebraically:

Or you could just count:

First of all, if you’re taking the November test, good luck. If you’re prepared well, you have nothing to fear. Get in there and beast it.

Of course, many people aren’t taking the November SAT, and they deserve some love, too. If you’re not taking the test tomorrow, here’s one tough question to noodle on while your contemporaries are out achieving glory.
Prize for the first correct response: access to the PWN the SAT Math Guide. You must not be anonymous to win.
In the figure above, B is the midpoint of AC and the center of the green circle. All other labeled points are also the centers of circles. If the green area is 9π, what is the area of the circle with center E?

I actually cackled sinisterly when I made this diagram. Good luck! 🙂

UPDATE: Congrats to “emolano82” for getting it first. Solution below!

There’s no “shortcut” here, other than to rub your hands together and say “OH YEAH!” like Macho Man Randy Savage. If you’re not afraid of a bunch of circles, you’ll be OK.

The key here is to take it one circle at a time. Start with the green one. If its area is 9π, then it has to have a radius of 3, right? And a diameter of 6. Which is important because…

It’s the radius of the red circle. So…so far we know BC = 3, and that the radius of the red circle above is 6. So EC = 6, because E is on the circle, and C is its center.

Looky looky! That’s a right triangle. The short leg is 3, and the hypotenuse is 6. So you can pythagorize, or you can remember your special rights and know that this is a 30°-60°-90° right off the bat, so the long leg must be 3√3. EB = 3√3.

You can do the same thing below; BD = 3√3. So ED, which is the radius of the circle with center E, must be 6√3.

To find the area:

Area = π(6√3)2
Area = 108π