Posts tagged with: challenge

It is my hope that, as soon as you’re done figuring this problem out, you’ll be on your way to a wonderful (and professionally arranged and totally safe) fireworks display. What follows is not an SAT question, but if you enjoy math enough to visit my blog regularly and try these challenge questions, my hope is that you’ll think it’s fun anyway.

The prize this week: you’ll be close enough to the fireworks to feel the explosions in your gut. I love that. If you don’t love that, your prize can be different. I’m easy!

The height, in feet, of a pyrotechnic rocket t seconds after the fuse is lit is given by the function above, where r is a constant. If the rocket explodes 6.5 seconds after the fuse is lit at a height of 324 feet, for how many seconds does the fuse burn before the rocket takes off?

Put your answers in the comments. I’ll post the solution on Tuesday. Happy 4th of July, if you’re in the States. 🙂

UPDATE: Solution below the cut.

I think the fastest way to get to a solution here is to recognize that every time t appears in the function, it’s accompanied by ” – r“. If you’re really kicking butt on function translation, you might even recognize that the constant r is applying a rightward shift to the parabolic function.

Quick and dirty: substitute a single variable (I’m going to use x) in for (t – r):

y = 144x – 16x2
And graph it. Use your fancy calculator to calculate the maximum, which is at x = 4.5, y = 324.
If t = 6.5 when the rocket hits 324 feet, then 4.5 must equal 6.5 – r, and therefore r = 2.
So the answer I’m looking for, the time it takes for the fuse to burn down, is 2 seconds. That’s the rightward shift of the graph, which causes the rocket not to take off right at t = 0. Take a look at the graph as an illustration of the position of the rocket above the ground as a function of time (remembering that in our piece-wise function, all of the values less than 2 would actually just be zero — the rocket is just sitting on the ground):
Of course, there’s also an algebraic solution: substitute 324 in for h(t), substitute 6.5 in for t, and solve for r. It’s a bit more labor intensive, but it’s a fine way to go, especially if you don’t use a graphing calculator. You still, of course, need to recognize that r is the fuse time.

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.

1. After m days, 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 is m?

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

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

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

 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 87m. 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:

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

 m 87 87m 1 99 99 m + 1 89 87m + 99 or 89(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!

87m + 99 = 89(m + 1)
87m + 99 = 89m + 89
87m + 10 = 89m
10 = 2m
5 = m

Oh hell yes.
 Source: Married to the Sea.

This started out as an Old MacDonald’s farm question. No wait, I thought to myself, not depressing enough.

The prize this week: You’ll get the satisfaction of knowing that you probably solved this problem in less time than I spent staring at my computer screen trying to come up with a clever prize for this week. I swear I used to be more creative.

In Wendell’s house, the ratio of unopened credit card offers to out-of-date phone books is 9 to 5. The ratio of magazines to crushed loose cigarettes is 25 to 7, and the ratio of McDonald’s Happy Meal toys to rotting, half-eaten pizzas is 3 to 2. There are 6 used-up batteries lying around for each broken VCR. The ratio of crushed loose cigarettes to McDonald’s Happy Meal toys is 5 to 8, and the ratio of used-up batteries to out-of-date phone books is 5 to 7. There are 30 magazines for each 4 broken VCRs. If there were 108 unopened credit card offers, how many rotting, half-eaten pizzas would Wendell have in his house?

Please note that, as usual for the weekend challenge, I’m taking a concept that SAT has been known to test, and extending it to an extreme to which the SAT would not go (not only in subject matter, but in scope as well). All of this is to say that these weekend challenges are meant to be fun for you, not to freak you out if you can’t get them. They’re usually a degree or two harder than what the SAT will throw your way.
Put your answers in the comments. I’ll post a solution Monday (probably late in the day). Good luck!UPDATE: Nice work, Chris. And special thanks to Catherine from kitchen table math, who pointed out that my original wording (“If there are…how many does…”) made non-integer quantities (fractional batteries??) quite unsavory. I’ve changed the wording of the last sentence a bit now for the benefit of people who find this in the archives and want to torture themselves with this question later.

Solution below the cut.

The SAT would never throw such a complex question at you, but the solution I advocate is one that might help you on the harder ratio questions the SAT will toss your way. Remember that on ratio questions, units are paramount. When you’re presented with ratios of more than two things and asked to suss out the relationship between just two of those things, the best and most elegant solution is to line up all the fractions and multiply them together, eliminating unwanted units along the way. Let me show you what I mean:

What we’re given: a bunch of ratios relating together the following things (in order of appearance):

credit card offers (CCO)
phone books (PBK)
magazines (MGZ)
cigarettes (CIG)
Happy Meal toys (HMT)
half-eaten pizzas (HEP)
batteries (BAT)
VCRs (same, duh.)

What we want: the ratio of unopened credit card offers (CCO) to rotting, half-eaten pizzas (HEP). Once we have that ratio, then we’ll deal with the 108 CCO.

How we get there: Start by listing the ratios that contain the units you want. Make sure to put CCO on top, and HEP on bottom.

Simplify…

We need to get rid of PBK and HMT, so let’s find some ratios we can use to do so, one at a time.

Simplify…

And then we just keep going. This might get monotonous (it’s a challenge question), but it’s really just the same procedure over and over again until the desired result.

Simplify…

Repeat:

Simplify…

Repeat:

Simplify…

Repeat:

Gasp…Deep breath…Simplify…

That wasn’t so bad now, was it?

Now set up a proportion to see how many HEP will exist if there are 108 CCO:
Note: Although it will get gnarly with things like fractional batteries, if you’re good with your calculator you can also do this by plugging in 108 for CCO and solving consecutive ratios until you end at 16 HEP. I’m not going to spell that way out here. For some more realistic SAT-style questions that can be solved similarly, look here.

I’ve spruced it up a tad, but an extremely similar question was #17 (not even #20!) on an SAT in 2006. The prize this week (pardon my proselytizing): Someday you will be as good at something as Mike Miller is at songwriting.

UPDATE: A couple folks nailed this one, and doing so represents, in my opinion, a promising level of nimbleness. Color me impressed. Solution below the cut. The first thing you need to recognize here is that the pink shaded region is a quadrilateral.

The angles in a quadrilateral will always add up to 360°, just like the angles in a triangle will always add up to 180°. Since v + = 72, the two unlabeled angles must add up to 360° – 72° = 288°. Since the question states that all the angles in the polygon are the same, we can conclude than each of the unlabeled angles is 144°.

From here, there are a few ways to get the solution. Since this is a rare bird of a question, I’m going to solve it for you first in a way that I think might actually have some utility when applied to other questions, and then I’ll give you the quick solution.

Triangles, friend-o

The SAT friggin’ loves triangles. So even though it’s highly unlikely that you’ll come across a question like this on the SAT since I’ve only seen it once, practice with breaking confusing diagrams into triangles might pay off for you on game day. Since this is a regular polygon (that is, it has equal sides and equal angles), it can be broken down into a bunch of triangles that meet at its center. Those triangles will all be isosceles, and their legs will bisect the interior angles of the polygon. This is easier shown than said:

So the question becomes “how many 36° angles fit in the center of this shape before it goes all the way around (or, how many times does 36° go into 360°? And your answer, of course, is 10. 10 of these triangles will form a 10-sided polygon.

The other way

I’m sure some of you are scratching your head, wondering why I haven’t used the (n – 2)×180° formula. Quite simply, not everyone knows it and it’s extremely rare that you’d actually need that formula on the SAT. I don’t want to fill anybody’s head with unnecessary formulas, and I think the triangle solution is elegant. But I’d be remiss not to mention that the sum of the degree measure of all the angles in a polygon can be calculated using (n – 2)×180°, where n is the number of sides of the polygon. For example, the sum of the degree measures of the angles in a 6-sided figure is (6 – 2)×180° = 720°. If you’re clever, you can use this formula once you know the degree measure of one angle in a regular polygon to calculate the number of sides. If the polygon is regular, that means all the angles will be the same, so the total measure divided by the number of sides will give the measure of one of the angles. In other words:

We can plug 144° in for the measure of the single angle, and solve for n:

144n = (n – 2)×180
144n = 180n – 360
-36n = -360
n = 10

Bangarang.

Good luck to all you warriors out there giving it your all one last time before the summer. May your June SAT scores be well worth all your hard work.

This weekend’s challenge is a bit of a logic question. The prize if you get it: nobody in your testing room will assault your senses with unbridled body odor tomorrow morning. Awesome, right? I know.

If g, h, j, m, n, p, q, r, and s are all positive integer constants, and (g + h + j)(m + n + p + q)(r + s) is odd, what is the difference between the maximum number of even integers and the minimum number of even integers that could be in {ghjmnpqrs}?

UPDATE: We had two correct answers again this week! I’m going to have to start making these harder.

Solution below the cut.

The thing you need to recognize here is that there are certain properties of even and odd numbers that are true no matter what the actual numbers are. (Yes, this is fair game for the SAT to throw your way.)

1. even × even = even
2. odd × odd = odd
3. even × odd = even
4. even + even = even
5. odd + odd = even
6. even + odd = odd

So first: if (g + h + j)(m + n + p + q)(r + s) is odd, then (g + h + j), (m + n + p + q), and (r + s) all have to be odd (Rule 2). In order for that to happen, we’re going to need to use rules 4-6 on the sums within the parentheses.

We’re looking for the minimum and the maximum number of even integers that could be used, so let’s examine each separately. (I’m going to opt for clarity over conciseness here, but see the SECOND comment below from StaceyHL for the quick and dirty.)

MINIMUM evens

You know this won’t work, but try plugging in nothing but odds (I’ll use 3):

(g + h + j)(m + n + p + q)(r + s)
(3 + 3 + 3)(3 + 3 + 3 + 3)(3 + 3)
odd × even × even = even

The (g + h + j) is already odd! (m + n + p + q) and (r + s) are even, though, so we need to find a way to make them odd so that our product will be odd. Note that if I change one value from (m + n + p + q) to even (say, from 3 to 2) then I’ll have an odd sum. Same goes for (r + s). By starting with absolutely no evens, I’m able to see the minimum number of evens by dealing with one sum at a time. Once each sum is odd, the whole expression will be odd.

(g + h + j)(m + n + p + q)(r + s)
(3 + 3 + 3)(3 + 3 + 3 + 2)(3 + 2)
odd × odd × odd = odd

So the minimum number of even integers in the set is 2.

MAXIMUM evens

Just like we did above, I’m going to start by assuming something I know can’t be true: that all the values are even. Then I’m going to see how few I can change to make my product odd.

(g + h + j)(m + n + p + q)(r + s)
(2 + 2 + 2)(2 + 2 + 2 + 2)(2 + 2)
even × even × even = even

Oh, that won’t do at all! What is the smallest number of values I’d need to change to make all those sums odd? Looks like I’m going to have to change one in each set of parentheses, right?

(g + h + j)(m + n + p + q)(r + s)
(2 + 2 + 3)(2 + 2 + 2 + 3)(2 + 3)
odd × odd × odd = odd

So the maximum number of even integers in the set is 6.

The big finish

If the maximum number of even integers in the set is 6, and the minimum number of even integers in the set is 2, then the answer to this weekend’s challenge is:

6 – 2 = 4
 Krispy Pizza, one of my local joints. Source.

This is a bit off-topic, but I saw Arcadia last night on Broadway, and it blew me away. I can’t stop thinking about it. If theater is your thing and tickets are within your means, then you should run and not walk. I still have chills. Even if theater is not your thing, it’s worth a read.

And now back to your regularly scheduled frustration. Prize this week: when you graduate and move to New York, the guy at your local pizza place will come to know you by first name and give you the biggest slice available whenever you come in. You will know, then, that you have made it.

In the figure above, the red lines divide the circle into 8 equal parts. What is z?

UPDATE: Nice work to John and Debbie, who both got this and who both will hopefully know the pleasure of being a regular at a local pizza joint one day, if they don’t already. Solution below the cut.

As is true in most geometry questions, it’s possible to solve for z in a number of different ways here. If your way was different than mine but you still got the same answer, that’s just fine. You say potayto, I say potahto.

First, recognize that if a circle is divided up evenly into 8 sections, then each of the central angles is going to be 360° ÷ 8 = 45°. Mark the ones that will probably matter for us:

Note that the triangle formed by the pink segments and the black one is isosceles, because the pink segments are both radii. That means, since the vertex angle is 135°, the other two must each be 22.5°.

We now have two of the three angles in the green triangle below, so we can solve for the third.

180° – 45° – 22.5° = 112.5°. Since the angle we’re looking for is vertical to that angle, z = 112.5 also.

I actually wasn’t going to do a weekend challenge this week because it was a super busy week, culminating in today’s flight to Minnesota with my whole family for a wedding we’re attending tomorrow, but this question came to me on the plane and I think it’s good enough to post from the hotel.

Prize this week: It won’t raaaeeaaaaiiiiin on your wedding day.
Note: figure not drawn to scale.

If the legs of a right triangle are x + m and x + n and the hypotenuse is √ax² + 16x + c as shown above for all non-negative values of x, and a, c, m and n are positive integer constants, what is the greatest possible value of ac?

The triangle is green because weddings are festive.

UPDATE: nobody got this one. As usual, this challenge question is a level or two harder than you’d see on the SAT, so don’t sweat it too hard. However, I did draw my inspiration here from a real question from a real SAT, and you can see it in the Blue Book: test 3 section 5 #8.

Also, here’s a mea culpa: because I initially posted this from a hotel in Minnesota between wedding festivities, I didn’t originally word the question as carefully as I should have. If you saw this right when it was posted, I apologize. Careful wording is important in math questions, and I hold myself to a higher standard than that.

Solution below the cut.

Alright, so let’s take the obvious first step and pythagorize:

(x + m)2 + (x + n)2 = (√ax² + 16x + c)2
x2 + 2mx + m2 + x2 + 2nx + n2ax² + 16x + c
Now do a little rearranging:

2x2 + 2mx + 2nx + m2 + n2ax² + 16x + c
2x2 + (2m + 2n)x + m2 + n2 = ax² + 16x + c

The important thing here is that this needs to be true for ALL non-negative values of x, which means that the coefficients on the left and right of this equation have to be equal to each other*. In other words, 2 = a2m + 2n = 16, and m2 + n2 = c.

Now you start to see where this is going. Simplify 2m + 2n = 16 to m + n = 8, and start looking at positive integer possibilities for m and n that will add up to 8, and then sum their squares to find values:

7 + 1 = 8; 72 + 12 = 50
6 + 2 = 8; 62 + 22 = 40
5 + 3 = 8; 52 + 32 = 34
4 + 4 = 8; 42 + 42 = 32
Greatest value for c is 50. Since we saw before that the value of a is 2, the greatest possible value of ac is 100. Phew.
__________
* I’m not going to do it rigorously here, but I know this is a sticking point for a lot of people, so you can prove this to yourself by plugging in different values for x and seeing what happens. For example: if you say x = 1, then your left side will be 2 + 2m + 2n + m2 + n2 and your right side will be a + 16 + c. Then you could say m = 2 and n = 3, and your left side would be 2 + 4 + 6 + 4 + 9 = 25. You could then set a = 5 and c = 4 to make the right side = 25 as well. Those same numbers will NOT work when you set x = 2. If the equation has to work for all values of x, then the coefficients of the terms containing x have to be equal on either side of the equation.

Very sorry for the disappearing and reappearing blog the past 24 hours or so. Blogger has had some difficulties, but I’ve used it for the better part of a decade without incident, so this one won’t send me running for the hills. I’m assured that the super long post I wrote and posted yesterday about obsessive vocabulary studying will be back shortly. Or I will seriously freak out.

Prize for answering this weekend’s challenge question: I’ll use an image of your choice in a future post. Image can’t be copyrighted, profane, or have anything to do with the Philadelphia Phillies. I reserve the right not to post anything I think sucks.

If p and q are positive integers, what is p + q?

UPDATE: Congrats to JD for getting it the long way, and then the short way. Solution below the cut.

It’s funny: I write these questions in a vacuum and sometimes I don’t realize exactly how hard they’ll be until I discuss them with a few people. This one, as these challenge questions go, was especially difficult. I knew it was devious to use 243 in the exponent and the denominator, since the exponent literally could have been anything, but I just couldn’t help myself. So I’m sorry if this one drove you nuts.

There are two insights required to solve this:

1. 15243 = 32435243. This is easier to see when you’re dealing with variables (ex: (xy)2 = x2y2), but you can always factor numbers that are raised to exponents, if that’ll help you towards a solution. In this case we’re trying to get to 3p5q; that’s a clue that you’re going to want to factor the 15 out.
2. 243 = 35. Yeah. That’s gonna be important.
Given those things, let’s attack the problem (remember, every step you take should bring you closer, somehow, to 3p5q.

Now you can cancel the 35 out of the denominator by subtracting 5 from the exponent in 3243 in the numerator. (For a review of this and other exponent rules, click here.)

And there you have it. If p and q are positive integers, then they have to be 238 and 243, respectively. So p + q = 481.

First of all, if you’re taking the May SAT tomorrow then I wish you the best of luck. You might want to sit this one out, at least until after your test. The prize this week for the first correct answer: domesticated animals will be able to understand what you say to them for 24 hours. Use this opportunity to tell dogs how awesome they are, cats what ingrates they are, and goldfish how sad and lonely their lives are as if they didn’t already know. I have no idea what to say to your ferret. Right, onwards:

If the difference between the areas of two circles is 10π and the sum of their diameters is 4√5, the radius of the larger circle is how much bigger than the radius of the smaller circle?

UPDATE: Commenter John Gutsch nailed it. Solution below the cut.

At first glance this looks like a circle question, but as you’ll see in a moment, it’s really only a circle question in a cosmetic sense. Really, it’s a solve for expressions question. Let’s begin.

Say the radius of the larger circle is m, and the radius of the smaller circle is n. Write an equation to represent the first thing the question tells you: the difference between the areas of the circles is 10π:

πm2 – πn2 = 10π
Oh snap! Looks like there’s a π in each term, so let’s divide that right out of there:
m2 – n2 = 10
Oh double-snap! Looks like we’ve got a difference of two squares! Factor it:
(n)(– n) = 10

Now what? Well, remember what else the question told us, that “the sum of their diameters is 4√5.” In other words:

2m + 2n = 4√5
2(m + n) = 4√5
m + n = 2√5

Wow. Substitute 2√5 for m + n, and you get:

(n)(– n) = 10
2√5(– n) = 10
– n = 10/(2√5)
m – n = √5

Or, as commenter John Gutsch put it, about 2.236. Great job!

 NOT A REAL PLANE ON FIRE. Source.

Appropriately for May Day, this question will cause you to send distress signals. Prize this week: you get to strut around all weekend telling people how much smarter you are than they, and they have to humbly agree. You may force up to three strangers to “kiss the ring.”

Note: Figure not drawn to scale.

In the figure above, P is the center of a circle, and Q and R lie on the circle. If the area of the sector PQR is 25π3 and the length of arc QR is 5π2, what is the measure of ∠RPQ?

UPDATE: props and ring kisses to Elias, who answered on Facebook, and to JD, who did his thing right here on the site (and spelled out the solution quite elegantly). Side note: I’d love to be able to integrate those two commenting systems — is that possible?

Solution below the cut:

Solution

This difficult circle question its, at its core, a ratio question. Remember that, with circles, part/whole = part/whole, or:

(For a more rigorous explanation and practice with this, see this post.)

We can set set up ratios using this framework for both the area of the sector and the length of the arc:

Sector Area:
Arc Length:

At this point, you could just use Wolfram Alpha, but surely you’d rather do it the long way. I know your pulse quickened a bit when you noticed that both equations share the same left side, and can therefore be combined:

And neatly solved for r:

$\dpi{150}&space;\fn_cm&space;\frac{25\pi&space;^2}{r^2}=\frac{5\pi}{2&space;r}$
$\dpi{150}&space;\fn_cm&space;25\pi&space;^2(2r)=5\pi&space;(r^2)$
$\dpi{150}&space;\fn_cm&space;50\pi&space;^2r=5\pi&space;r^2$
$\dpi{150}&space;\fn_cm&space;10\pi=r$

Sweet. If r = 10π, we can now solve for x using one of our original ratios:

$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{5\pi^2}{2\pi(10\pi)}$
$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{5\pi^2}{20\pi^2}$
$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{1}{4}$

$\dpi{150}&space;\fn_cm&space;x=90$

Oh heck yes. Laser show.

Prize this week: you get to decide, once and for all, whether Marshmallow Peeps or Cadbury Creme Eggs are the better candy. No longer will disagreements be chalked up to “difference of opinion.” Once you issue your decree, it will be a matter of record. Minor dissent will be tantamount to outright prevarication. Think you can handle all that power?

Right, the question. Here we go.

Four adjacent offices are to be assigned to four employees at random. What is the probability that Scooter and The Big Man (two of the employees) will be placed next to each other?

UPDATE: Props to Elias, who nailed it on Facebook. He’s going with whichever candy has the least packaging (a man after my own sustainable heart). I’m thinking that’s the Creme Egg.

Solution below the cut…

The biggest mistake kids make with probability questions is to overcomplicate them. If you’ve done a unit on probability in school, you know things can become very complicated very quickly, and that it’s pretty easy to make a mistake under pressure. On the SAT, of course, that’s doubly true. However, on the SAT you’ll always be dealing with small numbers (remember: it’s a calculator optional test), so you can sidestep all the necessary calculations if you just get fast at LISTING COMBINATIONS!

Let’s just number the employees 1-4, and list all the possible ways they could be distributed to the four offices:

1234   2134   3124   4123
1243   2143   3142   4132
1324   2314   3214   4213
1342   2341   3241   4231
1423   2413   3412   4312
1432   2431   3421   4321
Take a minute to understand the order in which I listed those, because that’s important. I listed them from least to greatest, taking care to list all the ones that started with “12,” then all the ones that started with “13,” etc. I broke them into columns every time I had a new digit in the first position. I can check my work to make sure I’ve listed them all by making sure all the columns are the same length.
Now that I’ve listed every single possibility, I only need to count the ones that fit the question. How many have the same two people adjacent to each other? Let’s say Scooter is 1 and The Big Man is 2:
1234 2134 3124 4123
1243 2143 3142   4132
1324   2314 3214 4213
1342   2341   3241   4231
1423   2413 3412 4312
1432   2431 3421 4321
Count ’em up! Scooter and The Big Man are next to each other 12 of the 24 times! That’s a probability of 12/24, which simplifies to 1/2, or 0.5.
Remember that the numbers will stay small on the SAT. If you get fast at listing possibilities, this is probably the biggest matrix of this kind you’ll ever have to make. If you have to make one with 5 possibilities (which I’ve seen once), it’s still doable. It’s a much faster and more sure-fire way to get a question like this correct than trying to work through all the crazy probability calculations.
Ok, but what about the math? What makes this question so difficult to solve from a math perspective (and it is difficult — hundreds of you viewed this post for an average time of 3:23 without answering) is that you have a combination of ORs and ANDs:
• Scooter could get the first slot AND The Big Man could get the second slot,
• OR Scooter could get the second slot AND The Big Man could get the first slot OR the third slot,
• OR Scooter could get the third slot AND The Big Man could get the second slot OR the fourth slot,
• OR Scooter could get the fourth slot AND The Big Man could get the third slot.
Whoa. Break that down into a big ol’ probability expression (more on how to do this in my massive probability post):
So there you go. Of course, the math works out nicely as well, once you know how to set it up. I stress again: for this kind of question it is easier and less time consuming to do it the “long” way, unless you are an absolute prodigy at probabilities. When I take the test, I who PWNs the SAT and gets 2400s, I list all the possibilities and count. But hey man, it’s your life.

Same prize as last weekend for whoever gets this first (either on the blog, or on Facebook; you’re competing against each other here). Any \$5 album from Amazon.com. Ready?

A small town has 3 theaters in it. Last Saturday night, the average age in the first theater was 29, and the average age in the second theater was 24. If the overall average theatergoer on that night was 36 years old and the ratio of attendees for the three theaters was 2:3:10, respectively, then what was the average age of the attendees in the third theater?

Good luck, folks. I’ll post the answer and contact the winner (if there is one) on Monday. Remember: you can’t win if I can’t contact you.

UPDATE: Congrats to Codi, who nailed it on Facebook. Answer below the cut.

This is an average question, so let’s use the average table! Start by filling in what the questions tells us (and plugging in obvious values for numbers of people in the theaters based on the given ratio):

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people (1st theater) 29 years per person 3 people (2nd theater) 24 years per person 10 people (3rd theater) 15 people (all theaters) 36 years per person

Now fill in the sums:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people(1st theater) 29 years per person 58 years 3 people(2nd theater) 24 years per person 72 years 10 people(3rd theater) 15 people(all theaters) 36 years per person 540 years

Hopefully you see that the sum of the ages in the first and second theaters is 58+72=130. Therefore, in order for the sum of everyone’s ages to be 540, the sum of the ages of the people in the third theater has to be 540-130=410. If there are 10 people in a room whose ages add up to 410, the average age in the room has to be 41! To finish the table:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people(1st theater) 29 years per person 58 years 3 people(2nd theater) 24 years per person 72 years 10 people(3rd theater) 41 years per person 410 years 15 people(all theaters) 36 years per person 540 years

Note that the average table allows you to check your work to make sure you didn’t make any mistakes. In the left-hand column 2+3+10=15, and in the right-hand column 58+72+410=540. That’s a beautiful thing. I literally weep when I see it.

Hey all. Here’s your weekend challenge question. The prize this weekend: any \$5 album from the Amazon mp3 store. I can only give you the prize if I can get in touch with you (using your email or Disqus/Google/Yahoo/Twitter/Facebook account), so please don’t be completely anonymous if you want the prize.

(m + n + p + 180)(qbd – 2eac + r) = 3x + y

Based on the figure and equation above, what is x in terms of y?

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

UPDATE: solution below the cut.

Nobody got this, although I did get a few emails with valiant attempts. The answer is: x = –y/3.

Why? Well, it’s a geometry plug-in. You’re given a bunch of shapes and not a single angle to call your own, so MAKE THEM UP. You just need to make sure you don’t break anything in the process — all your triangles need to add up to 180°, all your straight lines need to add up to 180° and in this case your pentagon needs to add up to 540° (because that’s how a pentagon rolls).

Easiest way to go is just to pretend your pentagon is regular (even though it’s clearly not) and plug in 108° for m, n, p, q, and r. That’s going to make all the base angles of the triangles come out to 72°, and a, b, c, d, and e all equal 36°.

The (m + n + p + 180) part of the question is horse-hockey. I only put it in there because I didn’t use those variables elsewhere in the problem, but I named them in the diagram.

The (q – b – d – 2e – a – c + r) part, using the numbers I just plugged in above (or any numbers that don’t break triangles, straight lines, or pentagons) comes out to ZERO. (Since nobody won the prize for this question, I’ll award it to anyone who feels like writing a general proof of this. I won’t hold my breath.)

So you’re left with:

0 = 3x + y

Which is why x = –y/3.

 Come at me, bro. (Source.)

Pretty easy question here, so I’ll attach a prize of fairly small value, but that might still be fun: the first person to comment with the answer gets to name a character in a future word problem on this site, and what they do/sell/wear/eat. For example: “Rita is the diaper changer at a daycare center that feeds the kids nothing but corn.” Of course, I have no way of contacting an anonymous commenter, so you’re going to have to identify yourself. Cool? Let’s goooo.

Four kids are in a room; their average age is 8 years old. Then an adult enters the room, and the average age becomes 16. A tense conversation quickly escalates, culminating in one of the children screaming “You’re not even my real dad!” and leaving the room, but the average age in the room stays the same. A few minutes later, another kid leaves the room in search of a sandwich. If the last kid to leave was 4 years old, then the adult is how much older than the average age of the people remaining in the room, awkwardly staring at each other?

MOAR UPDATEZ and SOLUTION: Congratulations to “Amy” who answered correctly, albeit with some degree of uncertainty. Good enough, kid. Email me to claim your prize.

Solution after the jump.

Solution

We’re going to solve this bad boy with the average table. Because it is awesome and if you don’t love it then you probably are incapable of love. Let’s start by finding the age of the adult. Here’s what the problem tells us:

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

 4 people 8 years/person +1 person 5 people 16 years/person

In order to find the adult’s age, just fill in the sums of the ages in the first and third rows, and figure out the difference. That is to say: if we know the sum of the ages when it was just the 4 kids in the room was 32, and then the sum of ages became 80 when the adult walked in, how old must the adult have been?

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

 4 people 8 years/person 32 years +1 person 48 years/person +48 years 5 people 16 years/person 80 years

OK, so if the adult is 48 years old, we have half of what we need. Now let’s figure out what happened after the adult walked in. We know two people left the room, and we know enough about each to figure out the average age in the room when the dust eventually settles. Picking up from where we left off, let’s deal with the first kid having a little hissy fit and leaving:

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

 5 people 16 years/person 80 years -1 person 4 people 16 years/person

This looks a lot like the table above did, so we’ll fill it in the same way. Note that I filled in the age of the kid who left, although it doesn’t actually matter at all to the solution of the problem.

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

 5 people 16 years/person 80 years -1 person 16 years/person -16 years 4 people 16 years/person 64 years

Ok, so since the average age stayed the same after that kid left, the sum of the ages in the room is now 64. When one more kid (age 4) leaves, we’re left with 3 people in the room, aged a total of 60 years. What’s the average age in the room? 20 years:

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

 4 people 16 years/person 64 years -1 person 4 years/person -4 years 3 people 20 years/person 60 years

The adult is how much older than the average age in the room?

48 – 20 = 28.
Sweet.

This is a little harder than a typical SAT question, and obviously formatted with a bit more flair and color than the College Board would use, but it deals with the same concepts you’ll need to master to kick the SAT where it hurts most, so have a go at it:

I looked out my window the other day and realized that the shadow that my apartment building was casting was a perfect square. If the distance from the edge of the shadow to the top of the building (the dotted line on the diagram) was 100 feet, and my building (which has a square base) is 7 times as tall as it is wide, what was the area of the shadow?

I’ll refrain from answering this myself just yet. Put your answer in the comments, and check back in a few days to see if you were right!

Update: This post had LOTS of views and very few folks attempting an answer, so either people were finding this post by mistake when they Googled for something else, people are shy, or this question was harder than I thought. Answer and explanation below the cut.

Basically, this problem requires three insights to solve:

1. If the shadow is a perfect square, then it has to be exactly as long as it is wide. That means it has to be as long as the building is wide. Once you’ve got that figured out…
2. This is a right triangle problem. The dotted line in the diagram is the hypotenuse of the triangle formed by the height of the building and the length of the shadow (which is the same as the width of the building).
3. The area of the shadow will be the width of the building squared.
So let’s call the width of the building x and get to work:

Pythagorize it:

x2 + (7x)2 = 1002
50x2 = 10000
x2 = 200

There you go. The area of the shadow is 200 ft2. Nice job to the one kid who got it right, and all the others who got it right and didn’t have the confidence to post it. Maybe next time I’ll offer a prize or something.