This question was submitted to me the last time I did a question writing contest. It’s pretty good, right? |

Here’s something I believe to be true: if you can write a convincing SAT math question, right down to realistic wrong answer choices, you’re close to mastering the concepts in the question. That’s because question writing forces you not only to understand the rules, but also to think through the tricky bits of the rules, and all the rules those are based on, and all the rules *those* are based on, all the way down to the basic building blocks of math.

To make a long story short, I think this is a fantastic, not to mention *fun*, way to hone your SAT math skills.

To make it even *more* fun, I’m going to hold a contest. For the month of December 2013, I will give away up to a book (or up to 3 books) every Friday for the best SAT math question(s) submitted that week. The plan is to give away 1 book per week. I say “up to 3” because I want flexibility if I can’t choose a favorite from a bunch of good submissions.

##### Here’s how it’s going to work

- Submit your question in the comments on this post. If your question includes a figure, attach an image of the figure to the comment. You’re all teenagers, so I trust you can figure out how to attach an image with DISQUS.
**You may submit 1 question per week.**- It must be multiple choice. Part of the feel of a real SAT question is that its incorrect answer choices are not completely random. They often anticipate likely mistakes a test taker might make.

*a lot*of SAT math questions, so I have a good sense of how a question should feel. You need to match that. Other than feel…

- Your question should be challenging (level 4 or 5 difficulty), but not impossible. Remember: real SAT questions usually look harder than they are.
- Your question can’t require knowledge that the SAT doesn’t require.
- Your question should not require lengthy calculations to solve.
- If other people like your question, and upvote it in the comments, I will take that into consideration. But not
*much*consideration.

##### UPDATE 1

I *love* the discussion that’s been going on in the comments of this post. You guys have really embraced this contest and there were a bunch of great submissions for week 1. The winner this week is Martin, who submitted this gem:

What I love about it is that it puts one good twist on a concept everyone knows. It’s the perfect level of difficulty, it’s formatted well, and it’s worded carefully and precisely. Excellent work!

Let’s keep the conversation going strong for the rest of the month! There are more books to be won!

##### UPDATE 2

Two winners this week! Here they are, for your PWNing pleasure:

Congratulations Damon and…Guest!

Remember, this contest continues throughout December, so if you haven’t won yet, there’s still time! Keep these great questions coming!

##### UPDATE 3

After a week 3 that yielded some interesting questions but no winners, week 4 has two winners! Congratulations are in order to Jeff and Trippy, whose questions are below. They both hit all the marks—they test only concepts the SAT tests, they require one small flash of insight, and very little work beyond that. They’re both probably *slightly* harder than questions you’ll see on your SAT, but they’re great practice.

And with that, this contest is over. Thanks to everyone who wrote questions and made this so much fun!

## Comments (121)

Really cool idea,Mike! I will think of some challenging questions 🙂

Is the answer to the picture E? Either way, can you explain the process please?

The radius of the largest circle is 9 so the area of the largest circle is 81pi. Now get the area of the second largest circle. Subtract the area of the smallest circle from the area of the second largest circle to get “X.” Subtract X from 81pi. That is the concept, but how do you get the radii of the two smaller circles? Never mind I read the question incorrectly.

Hint: focus on the diameters of the circles first, and then figure out the radii from there.

I figure out the radii. The biggest circle is 9, then 7.5 and then 5.5. Now what? I figured out the areas but how do you know how much that small sliver is? I logically guessed E based on what I could calculate and it didn’t say anywhere that it’s not drawn to scale;)

Do biggest area – medium area + small area 🙂

BUT—make sure you’ve got your radii right! I’d recheck that 5.5 one…

How much does creativity come into play? Is our job to make a question that “nobody has seen anything like this before” or can we skillfully renumber a challenging question?

There’s not much skill involved in renumbering a question. I doubt anyone will come up with anything that’s unlike anything ever seen before, but creativity will be looked upon favorably. So will combining multiple commonly tested concepts.

#creativity (I have to make an impression, being one of the first ones)

If 10 pokéballs in a bag weigh 4 pounds and 12 pokéballs in a bag weigh 4.6 pounds, they occupy 1/5 and 6/25 of the bag’s total volume respectively, how much does a bag full of pokéballs weigh?

a) 15 pounds

b) 16 pounds

c) 17 pounds

d) 18 pounds

e) It cannot be determined from the given information.

Is it A?

My math: 10 balls is a 1/5 of the bag so 50 balls is a full bag. Now 2 balls = .6 lbs. (I took the diff between the weights of 12 balls and 10 balls) and now multiply .6×25 because .6 is 2 balls and you get 15 lbs!

I’m going to go with E because the pokéballs have different densities (mass/ball) despite occupying the same volume. If you were to fill the bag entirely with first pokéball (4lbs/10 balls= density 2/5 or .4), your weight would be .4*50 (10 balls occupy 1/5 of volume; 50 balls of original density poké is a full bag), or 20 lbs. Things change if you were to fill the bag with 40 dense pokeballs (16 lbs), which occupies 4/5ths of the volume, and 10 less-dence ones (3 lbs), which occupies the remaining 5/25ths of the bag. Things get weird because you will have 50 pokéballs in a full bag no matter what; you could revise the problem such that each pokéball has distinct individual (lbs/ball) and volumetric (balls/volume) densities and then give the number of pokéballs in a full bag. This would require figuring out everything used in this problem and coming up with a system of equations. Yay! Math!

Unless…the weights you give include the weight of the bag. In which case the answer is 16. ‘Cuz the 50 balls that make up the bag weigh .3 pounds each, and the bag is 1 pound.

Mike- can you help us with this problem? I don’t think the problem is worded correctly- or it might just be me? I personally have no idea. Do the weight of the balls and the amount correspond to the volume?

I think what this discussion shows is how important the wording is on an SAT question, and how carefully a real question will be worded. Leah’s two comments are right on—it really depends whether the author intended the weight of the bag to count or not. Perhaps he can chime in and clarify?

also, I have never learned about density or anything in school yet. Will that be on the SAT?

No…you won’t need density on the SAT

Functions:

If f(x) = 3x-4

and g(x) = (4^3) + 3 – f(x^2)

What is g(2radical 2)?

Sorry, I don’t know how to insert radical symbol.

A) (6 radical 2) – 4

B) 47

C) 59

D) 69

E) 173

the answer should be 47… But I can’t find that answer choice… 2root2 squared is 8 and then you can plug that in into f(x) to get 20. 4^3+3-20 is 47

Perfect! @kylezhu:disqus.

This is an awesome question! I guess the answer is probably E (81pie+36pie-(56.25pie)=60.75pie) 😀

The area of each face of a cube is x. What is the volume of the cube?

A. 6x

B. x^2

C. x^3

D. x^(3/2)

E. x^6

Answer is D

yep. answer should be D for koko’s question

Come on everyone! Post questions and let us challenge each other ! 🙂

I really like this competition.

I smell a plug-in.

Is the answer A?

ka-ching!

Oh yeah!!!

Hello,

This is a great problem! I was able to solve the problem without using h or j… I found out that the length of a was supposed to be 2 root 2, using the 45 45 90 triangle. But I don’t understand what h and j have to with the problem… Any advice or tips please?

Honestly, I have no clue. I guessed that part and i thought maybe it is Delta y divided by Delta X type of thing so that’s why i chose an answer that had y/x. But, @Justin, maybe you want to explain?

Hey Trippy, I responded to the question you posted- care to check if my answer was right? 🙂 thanks!

Yo,

In order to solve for this in terms of h and j, you have to understand that the line alpha is perpendicular to the line drawn from point C to point P. You can then imply that the slope of CP is j/h, and then slope of the line alpha is (m + r) / (r + x). Here is an attached worked out solution.

Looking back on this, I believe that I should have also made it clear about point C being in the origin.

Yeah, that would have helped!

Makes sense, that’s what i figured.

See my response for trippy

Wait, why did you have -h/j? Shouldn’t it be h/j? Sorry to be bothering you with these naive questions…

The slope of alpha is -(m + r / x + r), since the line is going down. I just decided to negate h/j to make the other slope positive.

sorry, I don’t get what you are saying. m+r and x+r are the y values right? Slope is y2-y1/x2-x1.

m + r is just the magnitude of the *y delta* of line AB, and x + r is the magnitude of the *x delta*. Since from the image you see the line goes down (m +r) units and (x + r) units right, you can conclude the slope of the line AB is -(m+r)/(x+r)

Would your type of question appear on the SAT? It seems way harder than other questions i have seen on this subject…

Pete the Polite Dragon pillages 2 villages per month, but only eats 20% of the people from each one. The original population of each village is 50% greater than that of the previous one. After 2 months of pillaging, 540 people remain in the last pillaged village. What is Pete’s average meal size, in people, rounded to the nearest whole number?

(A) 51

(B) 61

(C) 65

(D) 78

(E) 81

How many days are in the month? 30 or 31? How many meals does he eat a day?

You don’t actually need to know either of those! Each month, he pillages 2 villages (eats 2 meals)

Is the answer B? @kylekuzman:disqus

My math:

In order to find original population, take last village of 540 people and multiply by 1.2 = 648. Now we can find all the original village populations.

First village: 81 ppl

second: 162 ppl

third: 324 ppl

fourth village: 648 ppl.

then you said he only eats 20% of the people from each village.

so that gives us 16.2 ppl for the first village.

second village: 32.4 ppl eaten

third village: 64.8 ppl eaten

fourth village: 129.6 ppl eaten

add up all the people eaten and divide by 4 since there are 4 villages and that gives us 60.75 average which rounds to 61!

No actually, it’s not B. Your idea for how to solve it is good, but you have a couple mistakes in your calculations

Where did i make a mistake? @kylekuzman:disqus

You have one in your very first calculation, 540 x 1.2. The math is correct, but the logic behind it is not. The mistake was repeated later as well

Sorry to be so vague, but I’m trying to give you hints and not just give the answer away!

OMG! I just realized what i did! I can’t believe i did that! Okay, hold on, I am re-doing it!:)

Haha cool, I’m glad you see it!

Is it A?

Nope 😉

Okay, now I am confused. Can you please explain it to me?

Sure. You need to figure out the percent change correctly first (http://blog.pwnthesat.com/2011/03/percent-change.html)

Sure. Your mistakes mainly had to do with percent change.

To find the original population of the last village, it’s not 540 x 1.2. That would be 540 + 20%, which does not equal the original population – 20%. If we call the original population y, 540 = 0.8y, so y = 675.

Likewise, population of village 4 divided by 2 does not equal the population of village 3. To find that, set up 675 = 1.5z, where z = population of village 3. z = 450, and so on until village 1

Can you understand my bad explanation?

Yes, thank you

Great one! Is it E?

It was a tricky one, you really have to be careful dealing with percentage.

Yep, nice job solving it! I tried to make it very easy to get confused, just like the SAT does. Out of curiosity, did any of the other answer choices seem right to you while you were solving it?

Yeah, you did a nice job with those answers. The goal is to get the original population of the last village right. Once you got that, it gets easier. But yeah, the 61 one nearly got me!

This question gets high marks for creativity, and concept tested, but it would be much clearer (but not much easier) if you just said in the problem that he ate four meals! One thing I hope you guys will come away from as a result of this contest is a respect (maybe even an awe) for the precision with which real questions are written. 🙂

Thanks Mike! I was actually really proud of it, so am kinda disappointed that it didn’t win, but I do see what you mean with the clarity. For next week’s competition, would you accept a revised version of it or should I write a new question entirely?

May I post not only my final problem, but my shortlist and my longlist?

Probably my question would be a combinatorial one… Would that be okay?

To keep things fair, you can only submit one question per week.

If you’re going to make a question using combinations, make sure it’s like SAT questions that use combinations, which is to say make sure nCr notation is not necessary to solve it!

So… here’s my question! Probably Level 5 because of the students’ unfamiliarity with these kind of questions, but could pass for a level 3/4, because if you look at it from a certain perspective, the problem becomes quite easy.

Alan, Bob, and Charlie are at the same Math class, which has 16 students. The teacher randomly grouped them into to 2 groups of equal size. After grouping the students, the teacher announced that Bob and Charlie are in the same group. What is the probability (chance) that Alan and Bob are in the same group?

This is supposed to be a grid-in question but if you want choices, here they are:

(A) 3/8

(B) 3/7

(C) 7/16

(D) 1/2

(E) 4/7

If I win, may I ask for a kindle copy of your essay guide instead? (you may send an amazon gift of your book to my email.

Thanks a lot for this competition!

Let’s generalize this problem. Let the math class have xy students that are to be grouped into x groups. Everything else is the same.

You may post in-thread or email me at my email (please ask Mike) if you have / want a solution.

The answer to my question is (B) 3/7.

How is my question? Why no responses?

May I ask what do you guys, especially Mike, think about my question?

Thanks!

The narrative of this problem makes it very difficult to work out; the timing is unclear, and timing is very important in probability problems.

The probability of flipping a coin and getting 3 heads in a row is (1/2)(1/2)(1/2) = 1/8. But if I told you I just flipped 2 coins and got heads, and then asked you the probability of flipping a 3rd coin and getting heads, you’d be wrong to say 1/8. The first two coins were already flipped, so the probability of THOSE being heads is now 1. Do you see what I’m getting at?

When you write a probability question, you need to be VERY clear about what happens when.

EDIT: Let me dial back my previous critique a little bit. I was referring to the question as you originally posted it, which is what came into my email. I didn’t read the version that’s up there as carefully and you addressed some of my concerns. I think the main issue with this question is that it’s way beyond the difficulty of SAT probability questions.

Oh, I didn’t know that edits would not come up on the email…

Are you sure that it is THAT hard? It is actually simple. Consider that B and C are in one group. Then there are 14 spaces left, 6 which of are in the group where B and C are. Since the groupings are done randomly, the probability that A is in the group that B is in is basically 6/14, which is 3/7.

I thought that that is an example of a good SAT question: a question that seems hard at a first glance but is really simple if you think of it in the right way.

EDIT: sorry for being too cocky / rude, if ever you got offended…

Here’s my question, hope you’ll like it.

Nice one! Answer C?

Thanks, I’m glad you liked it. Yes, C is the right answer.

Sorry for the late answer, I’m from Europe, so my timing is a little off:)

No not at all, I understand, I was recently studying in Germany for 4 months 🙂 Would you like to try my challenge question? No one’s gotten the answer yet!

Is it C?

That’s correct, the answer is C.

This question is awesome. It’s my choice for week 1 winner!

IS this even a level 4/5 question. IT seems like a level 2/3 to me, as it is really straightforward.

Hey congrats on the win, you deserve it for that great question 🙂 I was wondering, what software do you use to create the diagram?

Hey thanks, hope you’ll be the winner this week, as your question was pretty good too:) Well I have to admit I haven’t had the time (and didn’t now better) so I drew it in Word. But you could use GeoGebra, it’s free online and you can also download it for free. But I don’t know, I’m not an expert. Hope this helps though:)

Here is my 2nd question:

2 circles, 1 square, and 1 triangle are on the same plane. What is the maximum possible intersection points that they have?

HINT: Don’t even attempt drawing it out.

Could be grid-in, but if you really want, here are the choices.

(A) 28

(B) 30

(C) 32

(D) 36

(E) 40

Let x$ be defined as x$= x^2 – x + 1 for all values of x. If c$= (c-3)$ what is the value of c?

A. 1

B. 1/2

C. 9/4

D. 2

E. 8/5

Love the blog..

Here is a question I made up (it may be a little more difficult than an actual SAT question):

wait, but the sides of the triangle don’t have to be distinct right? for instance, we can have 1, 1, 1

Yes, that’s right.

In the figure, A square ABCD is inscribed in a circle with center O. Let E be a point on the minor arc CD such that EAC=12. Find the measure of EBD.

(A) 24

(B) 27

(C) 30

(D) 33

(E) 36

I meant that angle EAC’s measure is 12 degrees.

This is a nice effort, but it’s too hard to be an SAT question. The solution requires, as best as I can tell, a rule that’s not tested on the SAT. Do you have a solution that doesn’t require knowing that angle AEB is 45º?

Wait… so this week has no winner?

here is it

Did you even read my solution?

Of course I did.

What is that orange thing beside the comment? I think that this question does not need any advanced knowledge…

Nooo… I have to think of several SAT-level questions. 🙁

The sad thing is, that I refrain from boring problems that look like regular school problems…

I like critical thinking problems…

The point of this contest is to help students better understand how SAT questions are constructed so that they’re better prepared to do well on the SAT. I’m sorry it’s frustrating you, but if at the end of this you have a better sense of what SAT questions don’t look like (in this case, yours doesn’t look like an SAT question because there are too many steps required to solve it) then that’s a tiny victory.

Thanks for telling me that, because this question’s solution came out to me naturally… (maybe that’s not the case for most students).

Let me think of another question…

That’s right. The contest is to make a question that could actually appear on the SAT, and no submissions met that requirement this week.

Here is my question! Let’s take a look at it

The underlying concept here is great, but there are a few issues that prevent it from being a winner:

– Grid-ins will never involve pi (or force you to make decimal approximations of it)

– I said in the rules you can’t submit grid-ins

– You don’t actually say in the question that we’re dealing with semicircles. Remember, SAT questions need to tell you

exactlywhat’s going on in the figure in very clear language, especially when the figure’s not drawn to scale.If you address these concerns and resubmit, I will reconsider for the final week of the contest!

I have seen this question before in the BB…

If 2(2^12) + 3(2^13) + 4(2^14) + 5(2^15) = 2^x, what is the value of x?

A. 15

B. 16

C. 17

D. 18

E. 19

is the answer d?

Awesome question.

A degree 4 polynomial p(x) that has 4 distinct real roots satisfy the condition p(2013-x)=p(x). Find the sum of its 4 roots.

(A) 0

(B) 2013

(C) 4026

(D) 8052

(E) 16104

So… this week has no winner again?

🙁

Apologies–I got tied up last night and wasn’t able to post. There is a winner this week.

This is not an SAT-type question…you will never see the SAT say “degree 4 polynomial.”

🙁 It seems like in the weeks 2 and 4 all questions except mine were chosen…

I am back in the PWNing game since I didn’t get my goal score…

Here is my question. If….

-4z -2y + 2x = 48

-6y – 3z + 4x = -57

4z + 5x – 1y = 63

What is x+y+z?

A. 27

B. 43

C. 56

D. 79

E. 168

Great question!

Wooohooo!

on the math portion of the SAT, the raw score is calculated as; 1 point is awarded for each answer, and 1/4 point is deducted for each wrong answer. If jack answered all w questions on the test and earned a raw score of 15, how many questions did he answer correctly?

A, w – 15

b, w/5

c (w/5) – 10

D, (W-10)/5

E. 8 + (W/5)

Thanks, but this contest is over now. 🙁

Ok thak you though 🙂

Can you tell me how to do the first question from update 2? I’m really struggling.

Here’s a hint: you can find the intersection point by setting the two functions equal to each other.

Hey Mike,

Can you post the answers to the questions please?

The answer to the question in Update 1 is (C).

The answers to the questions in Update 2 are (C) for the parabolas question and (D) for Damon’s question.

The answers to the questions in Update 3 are (C) for Trippy’s question and (D) for Jeff’s question.