You guys stumped or something?

Anyone?

Oops - didn't check this thread yesterday. Glad the next round hasn't started yet.

No longer very easy math I see.. Or, on the other hand, I managed without using any forms for aids (apart from the brain). I think I might have a bit unortodox explenation on some of them, that perhaps quite a few would want to see.

1 Correct (D)
2 Correct (C)
3 Correct (C)
4 Not quite good answer. The correct answer is D. You are able to take out a factor 2 from each of your suggested factors.
5 Correct (A)
6 Not answered. The correct answer is B. (solve it by multiplying out the alternatives, or trugh the following observations:
q^3-q^2=q^2(q-1), 2q-2=2(q-1). You now have divided it into two terms with q-1 as common factor, so that you may further factorize it to (q-1)(q^2+2)
7 Correct (D)
8 Not answered. The correct answer is C. Only B and C has a constant term that is a factor in the constant term of the product, and the -16 term from B would never be outweighted by the positive 4 term in order to get a positive first order term koeficient.
9 Correct (C)
10 Correct (A)
11 Not answered. Correct answer is D. The others may be elliminated by siple insertion. Use of the second degree formula is the most stright-forward way of solving this without alternatives that I know of.
12 Not answered. Correct answer is D. This can be factorized to (r-8)(r+1).
13 Not answered. Correct answer is A. This is a well known square (z-2)^2
14 Not answered. Correct answer is B. ((11-1)^1-4*(121-96).. (Was a bit though to do in the head)
15 Not answered. Correct answer is A. The discriminant is 3^2-4*4*28, witch is clearly less than 0, thus the equation has no solutions.
16 Not answered. Correct answer is A. Easy to see by elimination. Otherwise solve the equation x(x-3)=340.
17 Not answered. Correct answer is A. An hypotenuse of 26 and whole number sides tells me that this is probably the pythagorean tripple 5, 12, 13 times 2. And this matches the required difference between the legs. ( what you were suposed to do was solve the equation (x^2+(x+14)^2=26^2 )
18 Not answered. Correct answer is A. Same method as 16.
19 Not answered. Correct answer is C. Move terms to left side and you get the well known square (x-3)^2=0.
20 Not answered. correct answer is D. Again you'll probably have to use the second degree formula. (A is easily eliminated. You also can eliminate C and D by the middle of the solutions has to be -4/2)

1D correct
2C correct
3C correct
4D correct
5A correct
6B correct
7D correct
8C - the other root is 2s+16
9C correct
10A correct
11D - working backwards, (2+2sqrt2)(-2+2sqrt2)=4 and (-2y-2y sqrt2)+(-2y+2y sqrt2)= -4y. So the roots work.
12D - factorize to (r-8)(r+1)
13A - factorize to (z-2)(z-2)
14B - from x^2 + 10x + 25, use b^2 - 4ac formula to get 100 - 4(25)=0
15A - negative discriminant, no real roots
16A - x^2 -3x - 340 = (x+17)(x-20)
17A - (x+24)(2x-20)
18A - (x-8)(x+10)
19C - (x-3)(x-3)
20D - working backwards, (2+sqrt7)(2-sqrt7)= -3, so it works

I must say it is amusing how 3 persons managed to post answers within 5 minutes of each other :D

[Reserved for posting round 5 answers when i feel less lazy]

Damn almost missed this round.

1. ok
2. ok
3. ok
4. wrong (D)
5. ok
6. wrong (B)
7. ok
8. wrong (C)
9. ok
10. ok
11. wrong (D)
12. wrong (D)
13. wrong (A)
14. wrong (B)
15. wrong (A)
16. wrong (A)
17. wrong (A)
18. wrong (A)
19. wrong (C)
20. wrong (D)

Round 6
Difficulty level 6

*Incorrect/Blank*

3. C. (-5 v^2) / (12 t^2 u^2) | Divide by 4, cancel out the variables | (-20 t^5 u^2 v^3) / (48 t^7 u^4 v) = (-5 v^3-1) / (12 t^7-5 u^4-2)

4. B. (w + 2) / (w - 4) | Factor the equations: [(w + 3) (w + 2)] / [(w - 4) (w + 3)] | Cancel out (w + 3)

5. C. b = -2 and 4 | Forget the numerator and just look at the denominator, factor it: (x - 4) (x - 2) | The illegal values would make the denominator equal to 0, so x - 4 = 0 and x + 2 = 0, x = 4 and x = -2

6. A. 1 / (y^2 z ^2) | Multiply across: (6 x y^2 z^2) / (6 x y^4 z^4) | Cancel out: 1 / (y^4-2 z^4-2)

7. C. r + 2 | Factor and Cancel: [(r + 5) (r + 2) / 3] * [(3r - 30) / (r + 5) (r - 10)] --> [(r + 2) / 1] * [(r - 10) / (r - 10)] --> (r + 2) * 1 = r + 2

8. C. c = -7, c = -3, c = 2, c = 5 | Factor the denominators: (c + 7) (c - 2), (c - 5) (c + 3) | Set the c's to 0: c + 7 = 0, c - 2 = 0, c - 5 = 0, c + 3 = 0 | Solve for each: c = -7, c = 2, c = 5, c = -3

9. D. 1 | Set it up: (40x / 64y) * (8y / 5x) | Cancel out: (8 / 8) = 1

10. A. (m n^5) / p^4 | Set it up: [(m^2 n^3) / p^3] * (n^2 / mp) | Cancel and multiply across: (m n^5) / p^4

11. D. (r + 5) / (r + 7)(r + 7)(r -3) | Set it up: [(r + 5) / (r +7) (r -2)] * [(r - 2) / (r + 7) (r -3)] | Factor and multiply across: (r + 5) / (r + 7)(r + 7)(r -3)

12. C. 8 | Find the smallest multiple both denominators (4, 8) fit in: 8

13. B. x^3 y^4 | Take the highest power of each variable: x^3, y^4

14. D. (p + 5)(p + 2)(p + 3) | Take the factors of each denominator and eliminate any duplicates, then put them together

15. C. w^2 - 7w + 12 | Break down the denominator to find what to multiply it by: (w + 5) (w - 4) | Multiply the top by (w - 4): w^2 - 7w + 12

16. D. 5/36 | Find the LCD: 36 | Solve: (32/36) - (27/36) = (5/36)

17. A. (q^2 + 2q + 3) / [(q + 1)(q + 2)(q + 3) | Find the LCD: (q +1)(q +2)(q +3) | [q (q + 1) / (q + 2)(q + 3)(q + 1)] + [1 (q + 3) / (q + 1)(q + 2)(q + 3) | (q^2 + q + q + 3) / (q + 1)(q + 2)(q + 3)

18.B. 3(t + 2) | Factor and set it up: [4(t - 2)(t + 2) / 8] * [6 / (t -2)] | Cancel: (t + 2) * 3

19. B. y = 2+3/8 | Multiply both sides by (y - 2): 3 = 8y - 16 | Set to y = form: 8y = 19 | y = 19/8 = 2+3/8

20. C. a = 9+1/2 | Multiply each side by (a +3) and (a - 2): 5a - 10 = 3a + 9 | Set to a = form: 2a = 19 --> a = 19/2 = 9 1/2

*Correct*

1,2

So... many... blank questions...

1. ok
2. ok
3. wrong (C)
4. wrong (B)
5. wrong (C)
6. wrong (A)
7. wrong (C)
8. wrong (C)
9. wrong (D)
10. wrong (A)
11. wrong (D)
12. wrong (C)
13. wrong (B)
14. wrong (D)
15. wrong (C)
16. wrong (D)
17. wrong (A)
18. wrong (B)
19. wrong (B)
20. wrong (C)

Since you usally reply asap, just letting you know if you don't know already,
That Round 6 is up.

Thanks. I appreciate that.

1. right (B)
2. right (C) - correction: I initially wrote D because of misreading the question. pointed out here
3. C - by inspection
4. B - (w + 2)(w + 3) / (w - 4)(w + 3)
5. C - denominator = (b-4)(b+2)
6. A - by inspection
7. C - (r+5)(r+2)(r-10)3 / 3(r-10)(r+5)
8. C - denominator = (c+7)(c-2)(c-5)(c+3)
9. D - by inspection
10. A - by inspection
11. D - (r+5)(r-2) / (r+7)(r-2)(r+7)(r-3)
12. C - (3/4) = (6/8)
13. B - by inspection
14. D - denominators are (p+2)(p+5) and (p+2)(p+3)
15. C - w^2 + w - 20 = (w-4)(w+5), so multiply by (w-4)
16. D - by inspection
17. A - denominators are (q+2)(q+3) and (q+2)(q+1), so numerator q(q+1)+(q+3)
18. B - 3(t - 2)(t+2) / (t-2)
19. B - 0 = 8y - 16 - 3
20. C - 5a -10 = 3a + 9 --> 2a = 19

I'll participate when the contest starts demanding some real math.

Come on, elevate it to freshman calculus, at least!