Diagnostic Math Drill #2 - MUSCLE UP.

Here are twenty more moderate-to-high difficulty questions for you to wrestle with. This drill is a bit harder than my first one, and is probably not the first thing you should try on this site unless you consider yourself quite advanced. It's all doable, of course, but it's meant to push you, and expose any weaknesses you still have. Do NOT try to limit yourself to 25 minutes here.

It's also, of course, not exhaustive. I've tried to write questions that cover a broad range of possible SAT questions, but in no way can 20 questions (or 40, if you count the first drill too) encompass every kind of question that will be thrown your way. I'm trying to supplement, not supplant, the work you should be doing in the Blue Book to prepare for the SAT.

The answer key, as always, is linked from the end of the drill so that you can work through the drill without peeking, but you can also access it directly here.

Feel free to print, share, etc.

--OR--

Good luck!

1. These problems look awesome. Gonna tackle them for fun if I get out of work on time.

Thanks

Alexis Avila, Private Tutor and Founder of Prepped & Polished, LLC

2. Thanks! I'd love your feedback if you get a chance to try the drill.

3. great drill-it's definitely a medium-and difficult problem only drill but very helpful for students trying to get into the mid 600's and higher. I like that you used a bit of each type of math problem throughout. Really well thought out problems. I'm going to use this with my student tonight. thanks.

Alexis

4. Thank you, that's very kind. It's definitely aimed at kids who aspire to the highest scores.

5. Catherine JohnsonMay 21, 2011 at 7:41 PM

Hi!

Catherine here (Debbie's friend).

C. and I just took the test and I can't figure out the solution to number 12 - do you have solutions posted anywhere?

Thanks!

6. Catherine JohnsonMay 21, 2011 at 9:43 PM

I had an 'inflexible knowledge' episode with number 12 - I've got it now.

7. Sorry you got back to it before I got back to you; I was at a wedding last night. I'm glad you figured it out! There actually aren't solutions to these questions posted anywhere yet, just the answer keys. I'll get around to the detailed solutions one of these days. :)

8. These are great!  And to Catherine and Debbie: try plugging in numbers for x and y, say 10 for each.  The big triangle would then have an area of 50.  The base and height of the little triangle would be 4/5ths of 10, or 8 each.  So the little triangle's area would be 32.  Therefore, by subtraction,  the shaded region would have an area of 18.  Since x times y = 100 (using our plugged in #s), the answer is D.

9. yes I got a 19/20!  But I didnt time my self so that isnt very accurate. Thanks for the problems bro!

10. Glad you liked the drill, and nice work!

11. here is the more algebraic way. The little triangle has a base of .8y because 1-.2 =.8. Since the triangles are similar the same goes for x. It is 4/5 of 1. So the height of the small triangle is .8x. To the find the shaded region its big triangle - small triangle area. XY/2 - (.8x.8y)/2.

xy - .64xy/2

.36xy/2 = .18xy or D

12. Could you explain numbers 4, 7, and 18? Thanks.

13. Sure. I'm gonna do them in 3 different responses to make sure I don't type them all into one and then close the browser by mistake, since this comment form doesn't autosave. :\

For #4, the slickest way to look at this is to realize that if you put those two right triangles together in a different way, you'd create a square with sides equal to the circle's radius. The area of the whole circle is Ï€r^2, and the area of the square is r^2. So the fraction of the circle that's shaded is (r^2)/(Ï€r^2) = 1/Ï€.

You can also plug in, which I think makes things a bit clearer. Say the radius is 2. Then the area of the circle is 4Ï€, and you've got 2 isosceles right triangles with legs of length 2 that are shaded. Each one of those has an area of ½(2)(2) = 2, so the total shaded area is 4. Again, the fraction of the circle that's shaded is 4/(4Ï€), or just 1/Ï€.

14. For #7, picture a basketball in a box, the way they're sold at sporting goods stores sometimes.

Plug in might help again, although it's not necessary. Say the radius of the sphere is 2. Then a cube that contains it must be, at the very least, as high as the sphere stands, or 4. The length and width, likewise, would have to be 4. So the volume of the cube would be 4×4×4 = 64. Go to the answer choices to see which one gives you 64 when r.
If you don't want to plug in, of course, you could do the same thought experiment and just leave the radius as r. The height, length, and width of the cube have to be 2r, so its volume is 2r×2r×2r = 8r^3

15. For #18, the basic idea is this:

If the area of the square is 1, then each of its sides are equal to 1. If those four circles are congruent and tangent to each other like the question says, then they all must have a radius of half the side of the square, or 1/2.

The circumference of each circle, then, is 2Ï€(1/2), or just Ï€. But for our purposes, we only care about 3/4 of each circle (270Âº/360Âº, since the square slices out 90Âº from each). So each circle contributes (3/4)Ï€ to the perimeter, and there are 4 of them.

4×(3/4)Ï€ = 3Ï€

16. Thank you so much for taking the time to answer each one in depth. I love this site!

17. You're welcome :)

18. An explanation for #15 would greatly be appreciated. I'm pretty lost right now -_-.

19. If a circle touches a line at exactly one point, that means it's tangent to it. Since this circle does so for both the x- and y-axes, its center must be equidistant from both. So, for example, it could be at (5, 5). If it was at (5, 5), though, that'd only eliminate choice (A). To eliminate the other incorrect choices, you need to consider where else the center could be. It could also be at (5, -5), (-5, 5), or (-5, -5). If you evaluate each of those points, you'll eliminate every choice but the correct one. :)

See the image below for a chart of Sharon and Michelle's water balloon game.

20. 19/20 !! For, QN 20, i thought it was A, Can you please explain? And for QN18, It says the shaded part...So, (3/4)(4)(2.pi.) (1/2) is for those four squares..What about the perimeter of square? square is shaded as well..So, don't we need to add 4(the perimeter of square) as well?

21. Nice work! You can eliminate (A) on #20 with a quick plug in. Pick a score that should NOT work (like 10) and see if the inequality in (A) is still true: |10 - 39| ≤ 75. That's true, which means the inequality in choice (A) doesn't represent the same range as the scores in the class. See my post on absolute values for a quick way to solve a question like this.

The square in #18 is important, as you figured out, but it doesn't count towards the perimeter because it's surrounded by the larger shape. When I adapted this drill for my book, I made the square a dotted line, to make that more clear. :)

22. How do you do 9? :(

23. Solve for expressions! If you add the two equations together, you get 8x - 8y = 48. Since you want x - y, divide everything by 8.

24. Hi, can you please explain number 5? When I first did it, I somehow didn't have issues with it. A month later I'm reviewing it and I don't understand it. I made the table just like how you make it. My issue is where did you get -10 from in the sum column. I understand where the -36 is coming from but if the sum of the four numbers is 10, then why is it put as negative? If it was positive the answer would be -26. Sorry, but this really confused me >.<

25. I know--tricky, right? :)

The negative 10 comes from the fact that we are TAKING OUT the 4 numbers from the 6 that add up to -36. Think about it: if 6 numbers add up to -36, but 4 of them add up to positive 10, the other 2 must add up to -46.

The average table can get tricky when you have negative sums and averages. Just remember that you have to add a negative in front of a # of numbers if you're removing it from the rest of the numbers, and when you do that, you have to add a negative in front of the sum as well. If it's already negative, it becomes positive. But in this case, it's a positive sum that we take away from a number that's already negative.

26. Can you explain #6? I know that the slope is p-2/2 so you have to do the negative of the reciprocal which is -(2/p-2) but how does that give you -2/p-2? I end up with -2/p+2 :(

27. Also can you explain #12?

28. Careful! You're applying the negative to both the top and bottom of the fraction. You wouldn't change -(2/3) to (-2)/(-3). For proof, plug in for p. You'll see more clearly what's going on .

29. See this scan from my book...

30. Thank you so much!

31. Could you please explain Q3 and Q14? I couldn't do the 3rd one, and my answer on the 14th one is wrong.

32. Well, you can disqualify the first condition because the first term is 1, but 4^1 = 4. So clearly the terms are not 4^n,

By the same logic, you can confirm that the second condition is true:
2^(2(1-1) = 2^0 = 1
2^(2(2-1) = 2^2 = 4
2^(2(3-1) = 2^4 = 16
etc.

Note that you can simplify 2^(2(n-1): it's the same as saying 4^(n-1).

So the pattern IS being multiplied by 4 each time, it just started at 4^0, not 4^1.

To figure out the last one, note that positive powers of 4 always have a units digit of 4 or 6, and they alternate. So in this pattern, all even-numbered terms will have units digits of 4, and all odd-numbered terms will have units digits of 6.

33. Hi, could you explain number 17? Thanks in advance! :)

34. Sure. First, find n by referring to the table to see what f(1) is. f(1) = 2, so n = 2.

Now all we need to do is figure out what g(n – 2) is. Since n = 2, n – 2 = 0. So really, we're looking for g(0). The table tells us that g(0) = –2, so the answer is (B).

35. Can you explain no. 2

36. You bet. Assume Lindsey picks first. The probability of her picking a green marble is 2/6. Since she doesn't put the marble back, now the probability of Jordan picking the only red marble is 1/5. Since we want the probability of these two events BOTH happening, we have to multiply those probabilities: (2/6)(1/5) = 2/30 = 1/15.

Now here's the cool part: If Jordan picked first instead, we'd have the same probability! He'd have a 1/6 chance of picking red, then Lindsey would have a 2/5 chance of picking green. Still 1/15.

37. Can you please explain 19 ? :)

38. I recommend plugging in for this one. Say the legs of the original right triangle are 10 and 20. That way, the area of the original triangle is ½(10)(20) = 100.

Now grow the 10 leg by 15% and shrink the 20 leg by 20%. Your new legs are 11.5 and 16.

Your new area is ½(11.5)(16) = 92.

Since we were total smarty-pantses and set up the original area to be 100, we're done without any additional calculations. The new triangle's area is 92% of the original one's area.

39. AGH that was so easy! I don't know how I didn't understand -_- Ii always manage to solve the supposedly harder ones and then get confused by THESE?? :'(

40. If you're not familiar with function notation, this isn't easy. Don't get mad, get even. :) http://blog.pwnthesat.com/2011/03/plain-old-non-symbol-function-questions.html

41. Perfect :) Thank you! I LOVE THIS SITE. Oh, and one question, how do i view an answer i've posted on your ask any question section, after i posted as an anon?

42. I have a couple math questions to discuss from SAT papers, where could i contact you? Would email be alright? Please answer ASAP

43. Can you submit them to qa.pwnthesat.com/ask ?

44. Haha, thanks! If you ask as Anon, you're just going to have to keep your eye on the site (qa.pwnthesat.com). I answer almost everything within 24 hours. I sometimes answer a lot of questions at once, so you might have to scroll down a bit to find your question.

45. You're very welcome! Alright, thanks! I'll see if i can get the hang of it...

46. Haha finished in 24 minutes! Didn't read the warning above. Got a 19/20 :( stupid mistake on number 5, forgot to divide by two. Do you have any tips to minimize stupid mistakes like this?

47. LOL nice work. First tip: SLOW DOWN. If you can do this drill in 24 mins, you probably have all kinds of extra time on a real SAT section. Second tip: try this.

48. Thanks, I never really thought of it that way. I usually end up finishing 10 minutes early on the math section of a practice test, so I just assume that I didn't make any stupid mistakes. This is kinda off topic, but I was just wondering if you know about AoPS? I'm interested in math contests and I was curious about how you would recommend preparing for them/your thoughts on them.

49. I don't know much about math contests, unfortunately. I think they're super cool and probably a lot of fun, but I was never aware of them in high school so I never did any myself. I'd imagine preparing for them would be similar to the SAT in some ways, though. Get your hands on old problems, do them, learn anything they reveal you don't know, repeat.

50. well this was a blow to my confidence.........could you explain all of them please :( :( :( :'(

52. actually never mind. they're not that bad haha

53. Can you explain how to solve #11 both with plug-ins and intuitively? Thanks!

54. Plug in first: say m = 2 and n = 3. Then 4^(m + n) becomes 4^(2 + 3) = 4^5 = 1024. So q = 1024.

Now evaluate 4^[2(2) + 2(3)]. That simplifies to 4^10, which is 1048576. Pretty big number! All you have to do now, though, is see which answer choice gives you that number for q = 1024. Try each choice—only (C) works. 1024^2 = 1048576.

The other way to get this is with exponent rules. Note that the exponent in 4^(2m + 2n) is double the exponent in 4^(m + n). In other words, the exponent is multiplied by 2. How do you multiply an exponent by 2? You raise the whole exponential expression to the 2nd power!

if 4^(m + n) = q, then:

[4^(m + n)]^2 = q^2
4^[2(m + n)] = q^2
4^(2m + 2n) = q^2

55. Thank you so much!