Answers to 2002.

Problem 01: 4/5
Problem 02: 8 pounds

3/4 of the gas weighs 18 pounds. Gas = 24 pounds. Thus tank is 8

Problem 03: 30

$\frac{1}{(\frac{1}{5}-\frac{1}{6})} = \frac{1}{\frac{1}{30}} = 30$

Problem 04: 99%
Problem 05: 20°

BAD = 20° arcAC = 140° arcACE = 180°,
thus arcCE=40° and CAE = 20°
BAC = 60° thus DAE = 60° - 20° - 20° = 20°

Problem 06: 30°
Problem 07: -2+4i
Problem 08: 2008
Problem 09: 36
Problem 10: 25%

Problem 33: d = 9/5 miles
Problem 34: n = 1, 2
Problem 35: 105 ways
Problem 36: 1001/334
Problem 37: 7/2

Pretty hefty polynomial expansion. I'll post it if requested.

Problem 38: 22/3

Law of Cosines, solve for cos A: $cosA = \frac{11}{21}$, then plug back in for side AD.

Problem 39: -2

f(2002) = f(1001) + 1 = f(500) = f(250) + 1 = f(125) + 2 = f(62) + 1
= f(31) + 2 = f(15) + 1 = f(7) = f(3) - 1 = f(1) - 2 = -2

Problem 40: 4/3

FH is the altitude.
By similar triangles, AH/FH = 4/1 and HE/FH = AE/1.
Add the equations: AH/FH + HE/FH = 4 + AE; AE/FH = 4 +AE
Swap: FH = AE /(4 + AE)
1/6 = 1/2*AE*(FH) = 1/2 AE*[AE/(4+AE)]
Algebraic manipulation and x = AE: 3x² - x - 4 = (3x - 4)(x + 1)
AE is either 4/3 or -1.

Problem 41: 74

3 squares = 50. Original triangle + other easy triangle = 12.
Angle α in the upper left so the obtuse angle is 180°-α. Sin(α) = 4/5 = sin(180°-α), making the area of the upper obtuse triangle A = ½*3*5*4/5 = 6. Similar logic gives us the area of the lower triangle A = ½*4*5*3/5 = 6.
Total 62 + 6 + 6 = 74.