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°thus arcCE=40° and CAE = 20°

BAC = 60° thus DAE = 60° - 20° - 20° = 20°

Problem 07: -2+4i

Problem 08: 2008

$f(2002)=f(3*\dfrac{2002}{3})=\dfrac{1}{1+\dfrac{2002}{3}}=\dfrac{3}{2005}$

Problem 09: 36
4 years * 3 days * 3 months

Problem 10: 25%
$\dfrac{1}{\dfrac{4}{3}a}+\dfrac{1}{\dfrac{4}{3}b}=\dfrac{3}{4}(\dfrac{1}{a}+\dfrac{1}{b})$

Problem 11: -2 or 17
Cancel x+2, then multiply out both sides and simplify. Remember the x+2 when done.

Problem 12: $\frac{11}{21}$
The number of 3-element subsets with 3 evens: C(4,3) = 4

The number of 3-element subsets with 2 odd, 1 evens: C(5,2)*4 = 40

Problem 13: -1The number of 3-element subsets with 2 odd, 1 evens: C(5,2)*4 = 40

The terms pattern out as 1, 2, -1, 2

Problem 14: $\frac{25}{6}$
(10-x)(0.4)+x(1.0)=0.65(10)

Problem 15: $\sqrt{6}$
Square the second expression.

Problem 16: $\frac{2}{3}$
Half-angle formula for sin divided by the half-angle formula for cos.

Problem 17: $\sqrt{3}$ or $\frac{1}{9}$
OMG, this was a pain. I'll post if requested.

Problem 18: 44Problem 19: 12

Problem 20: -1

Problem 21: 26

Problem 22: $\frac{9}{2}$; 4.5

Problem 23: 11

Problem 24: 100°; length = $\dfrac{10\pi}{9}$

Problem 25: 9

Problem 26: 34

Problem 27: $6\sqrt{3}$

Problem 28: $\frac{29}{2}$; 14.5

Problem 29: $\frac{4}{9}$

Problem 30: $(\frac{3}{2}, \frac{5}{2})$

Problem 31: 264

Problem 32: 6π

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.

Update: I'll post the "trick" (which I got from the test creators; my method was hairy):

Make the substitution $x=3^t-9$ and $y=9^t-3$

Then the student should notice that the equation becomes $x^3 + y^3 = (x+y)^3$

This is only possible if the middle terms of the expansion cancel each other.

So, $3x^2y-3xy^2 = 0$, and factoring gives us $3xy(x+y)=0$ and three corresponding results.

$x = 3^t-9 = 0$ which means that t = 2

$y = 9^t-3 = 0$ which means that t = 0.5

and $9^t+3^t-12 = 0$ and t = 1

Final answer is the sum of the three values is $\dfrac{7}{2}$

Problem 38: 22/3Update: I'll post the "trick" (which I got from the test creators; my method was hairy):

Make the substitution $x=3^t-9$ and $y=9^t-3$

Then the student should notice that the equation becomes $x^3 + y^3 = (x+y)^3$

This is only possible if the middle terms of the expansion cancel each other.

So, $3x^2y-3xy^2 = 0$, and factoring gives us $3xy(x+y)=0$ and three corresponding results.

$x = 3^t-9 = 0$ which means that t = 2

$y = 9^t-3 = 0$ which means that t = 0.5

and $9^t+3^t-12 = 0$ and t = 1

Final answer is the sum of the three values is $\dfrac{7}{2}$

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= f(31) + 2 = f(15) + 1 = f(7) = f(3) - 1 = f(1) - 2 = -2

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: 74By 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.

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.

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.

Interesting. I used the sine rule for the second part of problem 38, thinking that it would at least avoid me having to solve a quadratic, but of course triangle ABD is isosceles, so everything cancels very nicely in the second application of the cosine rule. A nice question.

ReplyDeleteIn problem 40, where, exactly is H?

ReplyDeleteFH is the altitude of AFE, FH is perpendicular to AE.

ReplyDelete