%% 7.tex
\documentstyle[multicol,pictex,contest]{article}
\begin{document}
\setcounter{page}{0}

\begin{multicols}{2}
\begin{center}
\Large\bf
1996 Utah State Math Contest\\
University of Utah \\
March 15, 1996\\[3pt]
Draft Solution Set \\[3pt]
{\em Please send corrections to} \\
{\em Grant B.\ Gustafson} \\
{\em Dept Math 233 JWB} \\
{\em Univ of Utah 84112}
\end{center}

\long\def\STAT#1#2{}
\long\def\STATS#1#2#3{}

%% 7.tex
\begin{center}\Large\bf
1996 Junior Exam Grade 7
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 7} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 7
\setanswer{1.}{a} & \setanswer{11.}{a} & \setanswer{21.}{a} & \setanswer{31.}{a} \\
\setanswer{2.}{d} & \setanswer{12.}{a} & \setanswer{22.}{a} & \setanswer{32.}{b} \\
\setanswer{3.}{e} & \setanswer{13.}{d} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{a} & \setanswer{14.}{a} & \setanswer{24.}{a} & \setanswer{34.}{d} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{c} & \setanswer{35.}{c} \\
\setanswer{6.}{b} & \setanswer{16.}{c} & \setanswer{26.}{d} & \setanswer{36.}{d} \\
\setanswer{7.}{b} & \setanswer{17.}{c} & \setanswer{27.}{b} & \setanswer{37.}{e} \\
\setanswer{8.}{c} & \setanswer{18.}{b} & \setanswer{28.}{a} & \setanswer{38.}{b} \\
\setanswer{9.}{b} & \setanswer{19.}{c} & \setanswer{29.}{c} & \setanswer{39.}{d} \\
\setanswer{10.}{c} & \setanswer{20.}{b} & \setanswer{30.}{c} & \setanswer{40.}{a} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

\begin{exercises}

% 7.tex 1996

%5.95
\prob{1.}{(Defects)}% #4, 1994
A bolt has probability $3/100$ of being defective. How many defective
bolts are expected in a package of $1200$?

\rightans{a}
\ans
{$36$}
{$360$}
{$120$}
{$40$}
{$12$}
\SOL{1. \STATS{99}{95}{94.05}}{Three out of $100$ are expected to be
defective. From $1200$,
the expected number is $(1200/100)(3)=36$.}

%7.93
\prob{2.}{(Prices)}% #17, 9th 1994
Computer $X$ cost \${}$6000$ in $1987$ but prices halved every $3$
years. Find the cost of $X$ in $1996$.

\rightans{d}
\ans
{\${}$1500$}
{\${}$1250$}
{\${}$1000$}
{\${}$750$\PAR}
{\${}$500$}
\SOL{2. \STATS{9th=99}{93}{92.07}}{The price is
$6000(1/2)^{(n-1987)/3}$. For $n=1996$ the price is
$6000(1/2)^3=6000/8=750$.}


%11.8
\prob{3.}{(Doubling)}% #9,  1995
A positive number is multiplied by itself and then the new number is
doubled to obtain $32$. What is the original number?

\rightans{e}
\ans
{$16$}
{$12$}
{$10$}
{$8$}
{None}
\SOL{3. \STATS{98}{90}{88.2}}{
Let $x$ be the number. Then $2x^2=32$. To solve $x^2=16$, ask ``which
number $x$ when squared equals $16$'', to obtain $x=4$.
}

% 7.tex

%16.48
\prob{4.}{(Sequence)}% #9,  1995
What is the eighth term divided by the ninth term of the
sequence
$\dd\frac{1}{2^2}$,
$\dd\frac{2}{3^2}$,
$\dd\frac{3}{4^2}$,
$\dd\frac{4}{5^2}$, \ldots?


\rightans{a}
\ans
{$\dd\frac{800}{9^3}$}
{$\dd\frac{8}{900}$}
{$\dd\frac{700}{8^3}$}
{$\dd\frac{7}{800}$}
{None}
\SOL{4. \STATS{96}{87}{83.52}}{
The eighth term is $\dd\frac{8}{9^2}$ and the ninth term is
$\dd\frac{9}{10^2}$. Their quotient is $\frac{8}{9^2}\frac{10^2}{9}
=\frac{800}{9^3} $.
}

%17.28
\prob{5.}{(Cooks)}% #17, 1994
One Cook makes $50$ hamburgers per hour and the other $42$.
How
many
can they make together in $3$ hours?

\rightans{d}
\ans
{$92$}
{$46$}
{$138$}
{$276$}
{None}
\SOL{5. \STATS{94}{88}{82.72}}{In one hour they can make $50+42=92$. In
three hours,
$(3)(92)=276$.}

%20.79
\prob{6.}{(Sound)}% #3,  1995
Sound travels in air about $344$ meters per second. Find the
distance in meters to a lightning flash, given the
thunder was heard $4\frac{1}{3}$ seconds later, to the nearest integer.

\rightans{b}
\ans
{$1460$}
{$1491$}
{$1500$}
{$1502$}
{$1516$}
\SOL{6. \STATS{89}{89}{79.21}}{
The formula $D=RT$ is applied with $R=344$ and $T=4\frac{1}{3}$ to give
$D=344\frac{13}{3}=1490.67$, or about $1491$ kilometers.
}

%25.24
\prob{7.}{(Inequalities)}% #8, 1994
Which is a solution to the inequality $-5 < x < -\frac{11}{9}$?

\rightans{b}
\ans
{$-5$}
{$-\frac{7}{2}$}
{$-1$}
{$-6$}
{$-\frac{11}{2}$}
\SOL{7. \STATS{89}{84}{74.76}}{The inequality is strict, which
eliminates $-5$. Both $-6$ and $-1$ are out of the interval. So is
$-11/2=-5.5$.
On the number line, $-7/2$ is between $-5$ and $-11/9$, therefore
it is the answer.}

%28.75
\prob{8.}{(Coins)}% #15,  1995
A glass contains $3$ nickels, $6$ dimes, $2$ quarters and $1$ penny.
Find the probability of randomly removing a penny or quarter.

\rightans{c}
\ans
{$\frac{1}{12}$}
{$\frac{1}{6}$}
{$\frac{1}{4}$}
{$\frac{1}{3}$}
{$\frac{3}{4}$}
\SOL{8. \STATS{95}{75}{71.25}}{
The sample set consists of $12$ items and $3$ of these represent
success, so the probability is $3/12$ or $1/4$.
}

%35.14
\prob{9.}{(Shopping)}% #5, 1994
A shopper in ZCMI Center bought $7$ items at \$1.59, $5$ at \$1.89 and
$8$
at
\$1.66. Find the average dollar cost per item to the nearest cent.

\rightans{b}
\ans
{$1.68$}
{$1.69$}
{$1.70$}
{$1.71$}
{$1.72$}
\SOL{9. \STATS{94}{69}{64.86}}{The cost is
$(7)(1.59)+(5)(1.89)+(8)(1.66)=33.86$ for
$7+5+8=20$ items. The average is $33.86/20=1.693$. }

%36.8
\def\AA{
\prob{10.}{(Area)}% #4,  1995
Find the area of the quadrilateral in the figure. All angles are right
angles.
\par
\rightans{c}
\ans
{$28$}
{$29$}
{$30$\PAR}
{$31$}
{$32$}
}
\def\BB{\input {rect7a.box}}
\WRAPFIG{\AA}{\BB}{1.3in}

\SOL{10. \STATS{80}{79}{63.2}}{
Break the figure into rectangles and apply the area formula:
$area=length\times width$.
}

%39.28
\prob{11.}{(Fractions)}% #9, 1994
If $5/7$ of a number is $1200$, then what is $7/5$ of that number?

\rightans{a}
\ans
{$2352$}
{$1200$}
{$1680$}
{$857$}
{None}
\SOL{11. \STATS{92}{66}{60.72}}{Let $x$ be the number. Then
$(5/7)x=1200$,
giving $x=1680$. The
answer requested is $(7/5)x=(7/5)(1680)=2352$.}

%41.54
\prob{12.}{(Divisor)}% #11,  1995
Let $X=1337456631234376$. Which number below divides $X$ with zero
remainder?

\rightans{a}
\ans
{$8$}
{$41$}
{$251$}
{$1973$}
{None}
\SOL{12. \STATS{79}{74}{58.46}}{
Let $x=1337456631234376=1000(1337456631234)+376$. Since $1000$ is
divisible by $8$ and $376$ is divisible by $8$, then so is $x$.
Most graphing calculators in 1996 supply no real help on the
problem, due to display limitations and automatic conversions into
scientific notation.
}

%44.9
\prob{13.}{(Oranges)}% #10,  1995
Find the percentage of good oranges in a case of $85$ oranges
containing seventeen rotten oranges.

\rightans{d}
\ans
{$20$\%}
{$52$\%}
{$72$\%}
{$80$\%}
{None}
\SOL{13. \STATS{95}{58}{55.1}}{
The percentage of good oranges corresponds to $(85-17)/85=0.8$ or
$80$\%.
}

%47.86
\prob{14.}{(Inequalities)}% #21, 1994
If $x$ satisfies $x < 2$ and $2x > -6$, then also

\rightans{a}
\ans
{$-3 < x < 2$}
{$-2 < x < 3$}
{$-2 < x < 2$}
{$-12 < 2x < 3$}
{None}
\SOL{14. \STATS{79}{66}{52.14}}{The second inequality can be written as $x > -6/2$ by
dividing both sides by $2$. Together, the inequalities $x<2$ and $x>
-3$ mean $-3 < x < 2$.}

%51.16
\prob{15.}{(Systems)} % #26, 1994 modified% 1995
Find the value of $y$ in the linear system
$x+2y=4$, $\,y-3x=9$.

\rightans{d}
\ans
{$-2$}
{$1$}
{$2$}
{$3$}
{$4$}
\SOL{15. \STATS{74}{66}{48.84}}{
Solve the first equation for $x$ in terms of $y$: $x=4-2y$. Put the
answer into the second equation $y-3x=8$ to get $y-3(4-2y)=9$. Then
$7y=12+9$ and $y=3$. The other answer is $x=-2$, using the equation
$x+2y=4$ with value $y=3$.
}

%51.55
\prob{16.}{(Art Barn)}% #4, 1994
A gallery $18$ by $15$ feet by $8$ feet high will be redecorated. Find
the total square footage of the walls and ceiling.

\rightans{c}
\ans
{$399$}
{$528$}
{$798$}
{$1056$}
{$1068$}
\SOL{16. \STATS{85}{57}{48.45}}{The four walls add $(18)(8) + (18)(8) +
(15)(8)+(15)(8)$
square feet. The ceiling is $(18)(15)$ square feet. The total is
$528+270=798$ square feet.}

%54.35
\prob{17.}{(TV)}% #34, 1994
A rectangular screen has diagonals of lengths $x+10$ and
$2x-30$. Find $x$.

\rightans{c}
\ans
{$38$}
{$39$}
{$40$}
{$41$}
{$42$}
\SOL{17. \STATS{7th=55}{83}{45.65}}{The diagonals are equal:
$x+10=2x-30$. Solve for $x=40$.}


%55.28
\prob{18.}{(Quotient)}% #1,  1995
Evaluate
$\dd \frac{-(|-13|-|-3|)}{\frac{81}{7} - \frac{22}{14}}$.

\rightans{b}
\ans
{$-1.6$}
{$-1$}
{$1$}
{$1.6$}
{None}
\SOL{18. \STATS{86}{52}{44.72}}{The top of the fraction is
$|-13|-|-3|=13-3=10$.
The bottom of
the fraction is $81/7-22/14=81/7-11/7=10$. Because of the leading minus
sign, the answer is $-10/10=-1$.}

%57.76
\prob{19.}{(Bagelry)}% #19,  1995
Bagels priced at \${}$0.40$ were increased in price by $37\frac12$\% and
later sold at a $25$\% off sale. Find the sale price to the nearest
cent.

\rightans{c}
\ans
{$39$}
{$40$}
{$41$}
{$42$}
{$45$}
\SOL{19. \STATS{96}{44}{42.24}}{
Use units of cents.
The price increase was $(0.375)(40)=15$ making a new price of
$40+15=55$. The $25$\% off sale uses a sale price of
$(0.75)(55)=41.25$.
}

%64.45
\prob{20.}{(Altitude)}% #6, 1994
A triangle has base $40$ and area $200$. Find the altitude.

\rightans{b}
\ans
{$9.5$}
{$10$}
{$10.5$}
{$11$}
{$11.5$}
\SOL{20. \STATS{79}{45}{35.55}}{The area is $\frac12bh$ for base $b=40$
and unknown altitude
$h$. Given area $200$, the equation $300=\frac12(40)h$ is solved for
$h$, giving $h=10$.}

%65.96
\def\AA{
\prob{21.}{(Trapezoid)}% #14, 1994
Find the area of the trapezoid.

\rightans{a}
\ans
{$104$}
{$105$}
{$106$\PAR}
{$112$}
{$90$}
}
\def\BB{\input {trap7a.box}}
\WRAPFIG{\AA}{\BB}{1.2in}

\SOL{21. \STATS{74}{46}{34.04}}{The rectangular part has area $(8)(10)=80$. The triangular
part has legs $x$ and $8$ The hypotenuse is $10$. By similarity to a
$3$-$4$-$5$ triangle, $x=(2)(3)=6$. The total area is
$80+\frac12(8)(6)=104$. }

%68.92
\prob{22.}{(Powers)}% #20, 1994
Simplify $(-3x^3)^2 + (2x^2)^3$

\rightans{a}
\ans
{$17x^6$}
{$17x^5$}
{$-x^6$}
{$-x^5$}
{$14x^6$}
\SOL{22. \STATS{74}{42}{31.08}}{By exponent rules $(-3x^3)^2 =
(-3)^2(x^3)^2=9x^6$ and
$(2x^2)^3=(2)^3(x^2)^3=8x^6$. The sum is $(9+8)x^6$.}

%70.42
\prob{23.}{(Scores)}% #7,  1995
Find the {\em mode} for the scores $7$, $9$, $12$,
$18$, $4$, $16$, $8$, $21$, $13$.

\rightans{e}
\ans
{$11$}
{$12$}
{$13$}
{$108$}
{None}
\SOL{23. \STATS{87}{34}{29.58}}{
There is no mode because each score occurs just once.
}

%73
\def\AA{
\prob{24.}{(Payload)}% #29,  1995
Supplies on a para\-chute hang below
the
canopy
of
diameter $24$ft on ropes $20$ft long.
Find the distance $x$ in feet from the canopy to
the payload.
}
\def\BB{\input{drop7a.box}}
\WRAPFIG{\AA}{\BB}{1.25in}

\rightans{a}
\ans
{$16$}
{$11\sqrt2$}
{$9\sqrt3$}
{$7\sqrt5$}
{$4\sqrt{11}$}
\SOL{24. \STATS{45}{60}{27}}{
Let $x$ be the altitude. Then $20^2=12^2+x^2$ by
the Pythagorean theorem. The answer is $x=16$.
}

%76.8
\prob{25.}{(Remainder)}% #25, 9th 1995
What is the remainder when $7^{1025}$ is divided by $5$?

\rightans{c}
\ans
{$4$}
{$3$}
{$2$}
{$1$}
{$0$}

\SOL{25. \STATS{58}{40}{23.2}}{
The first few powers of $7$ are $7$, $49$, $343$, $2401$,
$16807$.
The remainders after division by $5$ are $2$, $4$, $3$, $1$, $2$.
The
pattern repeats, each four causing the remainders to repeat. Since
$1025/4=256\frac{1}{4}$, the remainder is the first answer in
the list of four repeated values, namely, $2$.
}




%77.5
\prob{26.}{(Vertex Angle)}% #16, 1994
Find the vertex angle of an isosceles triangle with one base angle
$40^\circ$.

\rightans{d}
\ans
{$50^\circ$}
{$80^\circ$}
{$90^\circ$}
{$100^\circ$}
{None}
\SOL{26. \STATS{50}{45}{22.5}}{The base angles are equal. So both base
angles are $40\circ$.
The sum of the angles is $180\circ$ in a triangle, therefore the vertex
angle is $180-40-40=100$.}

%88.48
\prob{27.}{(Blood Cell)}% #17,  1995
About how many DNA strands can be placed end--to--end around the equator
of a spherical human blood cell of volume $\frac{4}{3}\pi(2^{-17})^3$
cubic meters?
Use $2^{-18}$ meters for the length of a strand.

\rightans{b}
\ans
{$11$}
{$13$}
{$15$}
{$17$}
{$19$}
\SOL{27. \STATS{36}{32}{11.52}}{
The volume of a sphere of radius $R$ is $\frac{4}{3}\pi R^3$.
The radius $R$ is determined from looking at the expression for the
volume: $R=2^{-17}$. The equator has circumference $C=2\pi R$.
Divide $C$ by $S=2^{-18}$ to
find the number of strands around the equator:
$C/S=2\pi 2^{-17}/2^{-18}=4\pi=12.57$.
}

%89
\prob{28.}{(Complement)}% #23,  1995
The measure of an angle is $26^\circ$ less than four times the measure
of its complement. Find the measure of the complement.

\rightans{a}
\ans
{$23\frac{1}{5}^\circ$}
{$22\frac{4}{5}^\circ$}
{$67\frac{1}{5}^\circ$}
{$66\frac{4}{5}^\circ$}
{None}
\SOL{28. \STATS{50}{22}{11}}{
Let $x$ be the complement. Then $90-x$ is the angle and the text
implies $90-x=4x-26$. Solving gives $x=116/5=23.2$.
}

%90.82
\prob{29.}{(Spinners)}% #16,  1995
Two twelve-hour clock faces are equipped with spinners that always land
on a whole number $1$ through $12$. After spinning each, what is the
probability that one or both of the numbers is greater than $8$?

\rightans{c}
\ans
{$\frac{2}{3}$}
{$\frac{4}{9}$}
{$\frac{5}{9}$}
{$\frac{29}{36}$}
{None}
\SOL{29. \STATS{51}{18}{9.18}}{
The spins result in possible events $(a,b)$ where $a$ and $b$ are
integers $1$--$12$. Altogether there are $12^2=144$ events. An event
is successful if $a>8$ or $b>8$. These events are enumerated
in a neat table and
counted to give the success count.
The numbers $a=1$ to $a=8$ require $b$ to be selected from $9$, $10$,
$11$ and $12$. Hence there are $(8)(4)$ successful events from this
range. For $a=9$ to $a=12$, any value of $b$ is allowed, resulting in
$(4)(12)$ successful events. The total successful events number
$32+48=80$.
The probability is $80/144=5/9$.
}

%91.55
\prob{30.}{(Polynomial operations)}% #27, 1994
Simplify the expression\newline
 $5(x + 2 y)^2 - 2(x^2 - 5 y^2)$

\rightans{c}
\ans
{$3x^2+10y^2$}
{$3x^2+30y^2$}
{$3x^2+20xy+30y^2$}
{$3x^2+20xy+10y^2$}
{None}
\SOL{30. \STATS{65}{13}{8.45}}{The details: $5(x + 2 y)^2 - 2(x^2 - 5
y^2) =
5(x^2+4xy+4y^2)-2x^2+10y^2 = 5x^2+20xy+20y^2-2x^2+10y^2 =
3x^2+20xy+30y^2$.}

%% ====== Separator problems for 7-8-9 competition

\smallskip

See {\em infra} for problems 31-40.

\end{exercises}



%% 8.tex
\begin{center}\Large\bf
1996 Junior Exam Grade 8
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 8} \\ \hline
\setanswer{1.}{a} & \setanswer{11.}{e} & \setanswer{21.}{b} & \setanswer{31.}{a} \\
\setanswer{2.}{d} & \setanswer{12.}{e} & \setanswer{22.}{b} & \setanswer{32.}{b} \\
\setanswer{3.}{a} & \setanswer{13.}{b} & \setanswer{23.}{d} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{c} & \setanswer{34.}{d} \\
\setanswer{5.}{c} & \setanswer{15.}{c} & \setanswer{25.}{c} & \setanswer{35.}{c} \\
\setanswer{6.}{d} & \setanswer{16.}{c} & \setanswer{26.}{c} & \setanswer{36.}{d} \\
\setanswer{7.}{a} & \setanswer{17.}{a} & \setanswer{27.}{d} & \setanswer{37.}{e} \\
\setanswer{8.}{b} & \setanswer{18.}{a} & \setanswer{28.}{d} & \setanswer{38.}{b} \\
\setanswer{9.}{e} & \setanswer{19.}{c} & \setanswer{29.}{d} & \setanswer{39.}{d} \\
\setanswer{10.}{b} & \setanswer{20.}{e} & \setanswer{30.}{d} & \setanswer{40.}{a} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

{\bf Instructions}: Only one answer is correct. Scoring: 5(\#
right)+1(\# blank)+0(\# wrong). Time: 120 minutes.

\begin{exercises}
%% 8.tex

%16.48
\prob{1.}{(Probability)}% #11, 1994
One piece of fruit is drawn randomly from a basket with $2$ oranges, $3$
bananas and $4$ apples. Find the probability that the fruit is not an
apple and not a banana.

\rightans{a}
\ans
{$2/9$}
{$1/3$}
{$4/9$}
{$5/9$}
{$2/3$}
\SOL{1. \STATS{96}{87}{83.52}}{There are $2$ chances out of $9$ to pick
an orange.}

%20.8
\prob{2.}{(Inequalities)}% #1, 1994
Represent by an inequality: {\em Cynthia is not more than 16 years old}.

\rightans{d}
\ans
{$C<16$}
{$C>16$}
{$16\le C$}
{$C\le 16$}
{None}
\SOL{2. \STATS{99}{80}{79.2}}{It means Cynthia is possibly age $1$
through
$16$,
i.e., $C\le 16$.}

%24
\prob{3.}{(Subtraction)}% #2, 1994
Subtract the sum $[8+(-7)]$ from the sum $[12+(-10)]$.

\rightans{a}
\ans
{$1$}
{$-1$}
{$-21$}
{$-13$}
{$7$}
\SOL{3. \STATS{100}{76}{76}}{Its $[12+(-10)]-[8+(-7)]=12-10-8+7=1$.}


%24.95
\prob{4.}{(Reading)}% #26, 1994
Determine the fastest reader from the table below.

\rightans{d}
\ans
{Jack}
{Jose}
{Judy}
{Jan}
{Jill}

\begin{tabular}{|l|l|l|l|l|l|}
\hline
Name           & Jack  & Jose & Judy  & Jan  & Jill
\\ \hline
Words          & $1222$   & $1345$ & $1100$ & $734$ & $610$
%122.2
%122.272727
%122.222222
%122.333333
%122
\\ \hline
Minutes        & $10$     & $11$   & $9$    & $6$   & $5$
\\ \hline
\end{tabular}

\SOL{4. \STATS{95}{79}{75.05}}{The question reduces to finding the largest fraction among
$1222/10$, \ldots, $610/5$. With a calculator, the decision is made
after computing all decimals. Without a calculator, the final contest is
between $1100/9$ and $734/6$. Multiply by $18$ and compare
$(2)(1100)=2200$ with $(3)(734)=2202$. So Jan is fastest.}


%25.74
\prob{5.}{(Months)}% #8, 1994
February is one month before March. What month is $616$ months before
March?

\rightans{c}
\ans
{Sep}
{Oct}
{Nov}
{Dec}
{Jan}
\SOL{5. \STATS{94}{79}{74.26}}{The number $616$ represents $51$ years
and $4$ months. The question is the same as: {\em What month is $4$
months before March}?}


%27.62
\prob{6.}{(Buttons)}% #4,  9th  1995
A box with $75$ buttons has $24$\% red ones, $22\frac23$\% blue ones and
the rest are green. What fraction are green?

\rightans{d}
\ans
{$\frac{1}{3}$}
{$\frac{2}{5}$}
{$\frac{7}{15}$}
{$\frac{8}{15}$}
{None}

\SOL{6. \STATS{94}{77}{72.38}}{
The number of red ones is $0.24(75)=18$, and the number of blue ones is
$\frac{68}{3}\frac{75}{100}= 17$. The fraction of green ones is
$(75-18-17)/75=8/15$.
}

%30.34
\prob{7.}{(Midpoint)}% #8,  1995
Let $P$ be
the midpoint of the segment from $A(8,4)$ to $B(12,12)$.
Find the coordinates of $P$.

\rightans{a}
\ans
{$(10,8)$}
{$(8,10)$}
{$(20,16)$}
{$(16,20)$\PAR}
{$(4,8)$}

\SOL{7. \STATS{86}{81}{69.66}}{
The midpoint $(x,y)$ is given by $x=(8+12)/2$, $y=(4+12)/2$.
}

%37.52
\def\AA{
\prob{8.}{(Triangles)}% #20, 1994
Find the area of the right triangle, given $a=15$ and $c=17$.
\par
\rightans{b}
\ans
{$59$}
{$60$}
{$237/4$\PAR}
{$289/4$}
{$239/4$}
}
\def\BB{\input{right8a.box}}% Right triangle with a, b, c labels.
\WRAPFIG{\AA}{\BB}{1.2in}

\SOL{8. \STATS{71}{88}{62.48}}{The Pythagorean Theorem gives the
missing leg $b$ in the
right triangle from the equation $15^2+b^2=17^2$ or $b=8$. The area is
half the base times the altitude, $(1/2)(15)(8)=60$.}

%39.94
\prob{9.}{(Analytic geometry)}% #10, 1994
Determine the slope of the line through $(-1,8)$ and $(3,-2)$.

\rightans{e}
\ans
{$3$}
{$-3$}
{$\frac{3}{2}$}
{$\frac{5}{2}$}
{$-\frac{5}{2}$}
\SOL{9. \STATS{77}{78}{60.06}}{The slope is
$m=(-2-8)/(3-(-1))=-10/4=-5/2$.}

%42.73
\prob{10.}{(Trip)}% #20,  1995
Viktor leaves in his Volvo at $10$:$00$am, driving $55$mph. At $11$:$30$am,
Jana follows him in her Justy going $50$mph. When are they $100$ miles
apart?

\rightans{b}
\ans
{$2$:$45$pm}
{$3$:$00$pm}
{$3$:$15$pm}
{$3$:$30$pm\PAR}
{$3$:$45$pm}
\SOL{10. \STATS{83}{69}{57.27}}{
Jack has traveled $(1.5)(55)=82.5$ miles when Jill starts her trip.
Their
speed
difference is $5$mph. The distance between them grows by $5$ miles per
hour. When the distance grows by $17.5$ miles, then they will be $100$
miles apart. This takes $17.5/5 = 3.5$ hours. The time then will be
$3\frac12$ hours after $11$:$30$am, which is $3$:$00$pm.
}

%47.56
\prob{11.}{(Radicals)}% #12,  1995
Solve $\dd 4+\sqrt{56-15x}=3x$ for $x$.

\rightans{e}
\ans
{$-\frac53$}
{$-\frac73$}
{$\frac73$}
{$3$}
{None}

\SOL{11. \STATS{76}{69}{52.44}}{
Put $4$ across the equal sign and square both sides to
obtain $56-15x=9x^2-24x+16$. Place all terms on the right and factor
with
the FOIL method into $0=(3x+5)(3x-8)$. The answer $8/3$ is correct. The
number $-5/3$ is an {\em extraneous solution}, it does not satisfy the
original equation (the left side is always positive, hence $x>0$).
}

%48.65
\prob{12.}{(Functions)}% #14, 1994
Given $f(x)=(x+1)/(x+3)$ and $g(x)=1/(x-1)$, then find $f(g(1/3))$.

\rightans{e}
\ans
{$\frac{1}{3}$}
{$\frac{1}{6}$}
{$\frac{2}{3}$}
{$-\frac{2}{3}$}
{$-\frac{1}{3}$}
\SOL{12. \STATS{65}{79}{51.35}}{This is the value of $y=(x+1)/(x+3)$
when $x=g(1/3)= -3/2$. So
$y=(-1.5+1)/(-1.5+3)=(-0.5)/(1.5)=-1/3$.}


%49.12
\prob{13.}{(LCM)}% #13,  1995
Find the least common multiple of $45$, $30$, $25$.

\rightans{b}
\ans
{$11250$}
{$450$}
{$\frac{450}{3}$}
{$100$}
{$\frac{100}{3}$}
\SOL{13. \STATS{96}{53}{50.88}}{
The LCM of three numbers is the smallest number $N$ divisible by each of
the three numbers. Write $45=(3)(3)(5)$, $30=(2)(3)(5)$,
$25=(5)(5)$. Then $LCM=(3)(3)(5)(5)(2)=450$.
}

%51.72
\prob{14.}{(Jackpot)}% #1,  1995
A jackpot of \${}$26.50$ is divided with $60$\cents{} to each man,
$95$\cents{} to each woman and $70$\cents{} left over. A division of
$75$\cents{} per person would be short $50$\cents{}. Find the number of
women.


\rightans{c}
\ans
{$10$}
{$11$}
{$12$}
{$13$}
{$14$}
\SOL{14. \STATS{71}{68}{48.28}}{
Let $M$ be the number of men, $W$ the number of women. Then
$75(M+W)=2650+50$, $60M+95W=2650-70$. Solving gives $W=12$ and $M=24$.
}

%53.08
\prob{15.}{(Radicals)}% #16, 1994
Simplify $\dd \sqrt{98}\sqrt{75} - \sqrt{14}\sqrt{21} -
4\sqrt{108}\sqrt{2}$.

\rightans{c}
\ans
{$9.79$}
{$3\sqrt6$}
{$4\sqrt6$}
{$6\sqrt2$}
{$\pi\sqrt{10}$}
\SOL{15. \STATS{68}{69}{46.92}}{
Factor the integers as follows: $98=(2)(7^2)$, $75=(3)(5^2)$,
$14=(2)(7)$, $21=(3)(7)$, $108=(3)(2^2)(3^2)$. Then
$\sqrt{6}$ is a common factor and the sum is $(35-7-24)\sqrt{6}$.}

%55.27
\prob{16.}{(Rectangle)}% #19, 1994
A rectangle has diagonal $34$. Twice the shorter side $x$ is two more
than the longer side. Find $x$.


\rightans{c}
\ans
{$14$}
{$15$}
{$16$}
{$17$}
{None}
\SOL{16. \STATS{71}{63}{44.73}}{
Let $x$ and $2x-2$ be the sides of the rectangle.
The Pythagorean Theorem applies to give equation
$x^2+(2x-2)^2=34^2$. Simplifying, it becomes $5x^2-8x-1152=0$ which
factors by the FOIL method
into $(x-16)(5x+72)=0$. The answer is $x=16$.}

%58.56
\prob{17.}{(Rational expression)}% #29,  1995
Simplify
$$\frac{x^2-4}{x^2-7x+12}\,\frac{x^2-9}{x^2-2x-8}
\div \frac{x^2+x-6}{x^2-6x+8}.$$

\rightans{a}
\ans
{$\frac{x-2}{x-4}$}
{$\frac{x-4}{x-2}$}
{$\frac{x}{x-4}$}
{$\frac{x-2}{x+4}$}
{None}
\SOL{17. \STATS{56}{74}{41.44}}{
The factorizations are
$x^2-4=(x-2)(x+2)$,
$x^2-7x+12=(x-4)(x-3)$,
$x^2-9=(x-3)(x+3)$,
$x^2-2x-8=(x-4)(x+2)$,
$x^2+x-6=(x-2)(x+3)$,
$x^2-6x+8=(x-4)(x-2)$.
After inverting and
multiplying, the product reduces to $(x-2)/(x-4)$.
}

%59.53
\prob{18.}{(Quotients)}% #17,  1995
Find the sum of the solutions to \newline
$\dd \frac{1}{2x} - \frac{9}{x^2+6x} = \frac{2-x}{2x+12}$.

\rightans{a}
\ans
{$1$}
{$4$}
{$-1$}
{$-3$}
{None}

\SOL{18. \STATS{71}{57}{40.47}}{
Clear the fractions to obtain the quadratic $x^2-x-12=0$ which factors
into $(x-4)(x+3)=0$. The sum of the roots $4$ and $-3$ is $1$.
}

%61.46
\prob{19.}{(Auto)}% #22, 1994
The annual rate of simple interest on a one-year \${}$7200$ loan
is $12$\%. What amount in dollars must be repaid?

\rightans{c}
\ans
{$864$}
{$6336$}
{$8064$}
{$8117.98$\PAR}
{$8113.14$}
\SOL{19. \STATS{82}{47}{38.54}}{
The simple interest is \${}$864$, computed as $(0.12)(7200)$.
The amount is the interest plus the principal, $7200+864=8064$.}

%63.15
\prob{20.}{(Recording Artists)}% #15,  1995
In a survey of $900$ people concerning two vocalists, $415$
liked Paul, $269$ liked Ringo and among these, $124$ liked both.
Estimate the probability that a person dislikes both Paul and Ringo.

\rightans{e}
\ans
{$0.21$}
{$0.24$}
{$0.27$}
{$0.30$}
{$0.38$}

\SOL{20. \STATS{67}{55}{36.85}}{
The total who like one or the other or both is $415+269$ less the number
$124$ counted twice, or $560$. The total disliking both is not $216$,
but $900-560=340$. The probability of disliking both is
$340/900=17/45=0.38$.
}

%67.44
\prob{21.}{(Powers)}% #9, 1994
\def\AA{\left( x^3y^{-4} \right)^{-2}}
\def\BB{\left( x^{-2}y^{3} \right)^{3}}
Simplify $\dd\frac{\AA}{\BB}$.

\rightans{b}
\ans
{$\frac{1}{x}$}
{$\frac{1}{y}$}
{$\frac{1}{xy}$}
{$\frac{1}{y^{12}}$}
{None}
\SOL{21. \STATS{74}{44}{32.56}}{
The numerator is $x^{-6}y^{8}$ and the denominator is $x^{-6}y^9$. After
cancellations of like terms the result is $1/y$.}

%70.32
\prob{22.}{(Trapezoid)}% #24,  1995
A trapezoid of area $342$ square units has parallel sides of $x-1$ and
$2x+7$ units. Given these parallel sides are $12$ units apart, find $x$.

\rightans{b}
\ans
{$16$}
{$17$}
{$21$}
{$51$}
{None}

\SOL{22. \STATS{56}{53}{29.68}}{
The average of the parallel sides $x-1$ and $2x+7$ times the altitude
equals the area, which gives the equation $(x-1+2x+7)(12)/2=342$.
Solving
for $x$ in $(3x+6)(6)=342$ gives $x=51/3=17$.
}

%78
\prob{23.}{(Variation)}% #36, 1994
Let $x$ vary directly as $y$. Find $y$ at $x=5$, given $y=30$ at $x=3$.

\rightans{d}
\ans
{$300$}
{$150$}
{$71$}
{$50$}
{$25$}
\SOL{23. \STATS{7th=50}{44}{22}}{Let $x=ay$. The given $y=30$ at $x=3$
determines constant $a$
by the equation $3=30a$, ie, $a=1/10$. Then $y=x/a$ at $x=5$ gives
$y=5/a=5/(1/10)=50$.}

%72.86
\prob{24.}{(Color--Blind)}% #10,  9th  1995
Find the probability that two black socks are drawn randomly
without replacement from a drawer having $5$ black and $8$ blue socks.

\rightans{c}
\ans
{$\frac{3}{26}$}
{$\frac{5}{26}$}
{$\frac{5}{39}$}
{$\frac{1}{39}$}
{$\frac{14}{39}$}
\SOL{24. \STATS{59}{46}{27.14}}{
The events possible are $(a,b)$ where $a$ and $b$ are colors, black or
blue. The event size is $13(12)=156$. There are $5(4)$ events which
choose $a$ and $b$ both black. The probability is $20/156=5/39$.
}

%76.8
\prob{25.}{(Remainder)}% #25, 9th 1995
What is the remainder when $7^{1025}$ is divided by $5$?

\rightans{c}
\ans
{$4$}
{$3$}
{$2$}
{$1$}
{$0$}

\SOL{25. \STATS{58}{40}{23.2}}{
The first few powers of $7$ are $7$, $49$, $343$, $2401$,
$16807$.
The remainders after division by $5$ are $2$, $4$, $3$, $1$, $2$.
The
pattern repeats, each four causing the remainders to repeat. Since
$1025/4=256\frac{1}{4}$, the remainder is the first answer in
the list of four repeated values, namely, $2$.
}

%79.3
\prob{26.}{(Proportions)}% #20,  1995
Suppose $9$ gallons cost \${}$12.15$. Find $x$, given $x$ pints cost
\${}$2.25$.

\rightans{c}
\ans
{$ 12$}
{$\frac{38}{3}$}
{$\frac{40}{3}$}
{$ 14$}
{$\frac{20}{3}$}
\SOL{26. \STATS{69}{30}{20.7}}{
A gallon is $4$ quarts or $8$ pints. So $(8)(9)=72$ quarts cost $12.15$.
The
cost per pint is $12.15/72$. The
proportion is $\dd\frac{12.15}{72}=\frac{2.25}{x}$. The answer is
$x=\dd\frac{40}{3}$.
}

%82.95
\prob{27.}{(Distance)}% #13,  1995
The distance between points $(-a,10)$ and $(-3,5)$ is $13$. Find
$a>0$.

\rightans{d}
\ans
{$-9$}
{$9$}
{$12$}
{$15$}
{None}

\SOL{27. \STATS{55}{31}{17.05}}{
Solve $(-a+3)^2+(10-5)^2=13^2$ for $a>0$. The possible answers for $a$
are $15$ and $-9$, but only one is positive.
}

%84.18
\prob{28.}{(Polynomials)}% #26,  1995
Find the coefficient of the term $x^3$ in the polynomial
$(x+3)^2(2x-11)(5x-4)$.

\rightans{d}
\ans
{$1$}
{$3$}
{$-5$}
{$-3$}
{$-1$}
\SOL{28. \STATS{39}{38}{14.82}}{
Write out the two quadratics $(x+3)^2 = x^2 + 6x + 9$ and
$(2x-11)(5x-4)=10x^2-63x+44$. The two terms in the product of these
quadratics, which contain the term $x^3$, are $60x^3$ and $-63x^3$.
The desired term is $-3x^3$ with coefficient $-3$. Alternatively,
the sum of the roots of the quartic equals $-3 -3 + \frac{11}{2} +
\frac{4}{5}=0.3$, hence the coefficient of $x^3$ is $(10)(-1)(0.3)=-3$.
}

%89.66
\prob{29.}{(Quotient)}% #18,  1995
\def\AA{\left( 1- \frac{1}{1+x} \right)}
\def\BB{\left( \frac{1}{1-x} - 1 \right)}
Simplify:
$$\dd {\AA}^{-1} + {\BB}^{-1}$$

\rightans{d}
\ans
{$-2$}
{$2$}
{$-2x$}
{$\frac{2}{x}$}
{None}

\SOL{29. \STATS{47}{22}{10.34}}{
The first reciprocal is $\frac{1+x}{x}$. The second is
$\frac{1-x}{x}$.
Their sum is $2/x$.
}

%93.2
\prob{30.}{(Area)}% #22, 1994
Find the diameter of a circle of area $\frac{\pi}{25}$.

\rightans{d}
\ans
{$2\pi$}
{$\frac{\sqrt{\pi}}{5}$}
{$\frac{1}{5\sqrt{\pi}}$}
{$0.4$}
{$0.5$}
\SOL{30. \STATS{40}{17}{6.8}}{
The area is $\pi R^2$ in terms of the radius $R$. The
equation $\pi R^2=\pi/25$ is solved for $R=1/5$ and inserted into the
diameter
formula $D=2 R = 2/5$.}



See {\em infra} for problems 31-40.


\end{exercises}



%% 9.tex
\begin{center}\Large\bf
1996 Junior Exam Grade 9
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 9} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 9
\setanswer{1.}{b} & \setanswer{11.}{c} & \setanswer{21.}{c} & \setanswer{31.}{a} \\
\setanswer{2.}{c} & \setanswer{12.}{a} & \setanswer{22.}{c} & \setanswer{32.}{b} \\
\setanswer{3.}{b} & \setanswer{13.}{c} & \setanswer{23.}{d} & \setanswer{33.}{a} \\
\setanswer{4.}{e} & \setanswer{14.}{a} & \setanswer{24.}{c} & \setanswer{34.}{d} \\
\setanswer{5.}{d} & \setanswer{15.}{b} & \setanswer{25.}{d} & \setanswer{35.}{c} \\
\setanswer{6.}{d} & \setanswer{16.}{d} & \setanswer{26.}{c} & \setanswer{36.}{d} \\
\setanswer{7.}{c} & \setanswer{17.}{b} & \setanswer{27.}{b} & \setanswer{37.}{e} \\
\setanswer{8.}{b} & \setanswer{18.}{d} & \setanswer{28.}{d} & \setanswer{38.}{b} \\
\setanswer{9.}{a} & \setanswer{19.}{c} & \setanswer{29.}{b} & \setanswer{39.}{d} \\
\setanswer{10.}{e} & \setanswer{20.}{a} & \setanswer{30.}{a} & \setanswer{40.}{a} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

{\bf Instructions}: Only one answer is correct. Scoring: 5(\#
right)+1(\# blank)+0(\# wrong). Time: 120 minutes.

\begin{exercises}

%% 9.tex
%22.65
\prob{1.}{(Pricing)}% #3, 8th 1994
Two parkas are originally priced at \${}$150$ and \${}$200$.
Sales tax is $6$\%. Find the difference between their
register prices at a $30$\% off sale.

\rightans{b}
\ans
{\${}$15.90$}
{\${}$37.10$}
{\${}$53$}
{\${}$47.10$\PAR}
{\${}$27.10$}
\SOL{1. \STATS{91}{85}{77.35}}{The $150$-dollar parka costs
$(0.7)(150)(1.06)$ and the
$200$-dollar parka costs $(0.7)(200)(1.06)$. The difference is
$(0.7)(50)(1.06)=37.1$.}

%19.25
\prob{2.}{(Triangle)}% #7,  1995
A triangle has angles of $x$, $x+5$ and $2x+3$ degrees. Find the sum of
the two smaller angles.

\rightans{c}
\ans
{$89$}
{$90$}
{$91$}
{$132$}
{$137$}

\SOL{2. \STATS{95}{85}{80.75}}{
The angles add to $180$, giving the equation $x+x+5+2x+3=180$, having
solution $x=43$. The other angles are $(43)+5=48$ and $2(43)+3=89$.
The sum of the two smaller angles is $43+48=91$.
}

%20.1
\prob{3.}{(Equations)}% #12,  1995
The system $ax-by=6$, $ax+by=10$ has the solution $x=-1$, $y=2$. Find
$b$.

\rightans{b}
\ans
{$0$}
{$1$}
{$-8$}
{$-1$}
{$9$}

\SOL{3. \STATS{94}{85}{79.9}}{
Substitute $x=-1$ and $y=2$ to obtain the system $-a-2b=6$, $-a+2b=10$.
Add to obtain $a=-8$. Substitute $a=-8$ into $-a+2b=10$ to get $2b=2$
or
$b=1$.
}

%25.48
\prob{4.}{(Bacteria)}% #24,  1995
A population of bacteria triples each day, reaching a
size of $722925$ in $120$ hours. Find the initial population size.

\rightans{e}
\ans
{$2995$}
{$2990$}
{$2985$}
{$2980$}
{$2975$}

\SOL{4. \STATS{92}{81}{74.52}}{
Let $x$ be the initial population size. Then after $n$ hours the
population is $3^{n/24}(x)$. After $120$ hours, $n/24=120/24=5$ and
$3^5x=722925$. Dividing by $3$ five times into $722925$ gives $2975$.
}

%26.96
\prob{5.}{(Similarity)}% #13, 1994
A triangle has longest side $x+30$ and shortest side $8$. It is similar
to a triangle of longest side $x$ and shortest side $3$. Find $x$.

\rightans{d}
\ans
{$48$}
{$30$}
{$20$}
{$18$}
{None}
\SOL{5. \STATS{83}{88}{73.04}}{By similarity, $(x+30)/x=8/3$. Then
$x=18$.}

%27.62
\prob{6.}{(Buttons)}% #4,  1995
A box with $75$ buttons has $24$\% red ones, $22\frac23$\% blue ones and
the rest are green. What fraction are green?

\rightans{d}
\ans
{$\frac{1}{3}$}
{$\frac{2}{5}$}
{$\frac{7}{15}$}
{$\frac{8}{15}$}
{None}

\SOL{6. \STATS{94}{77}{72.38}}{
The number of red ones is $0.24(75)=18$, and the number of blue ones is
$\frac{68}{3}\frac{75}{100}= 17$. The fraction of green ones is
$(75-18-17)/75=8/15$.
}

%34.88
\def\AA{
\prob{7.}{(Altitude)}% #16,  1995
In the figure, $MN$ is both a median and an altitude. Given $a=x+7$ and
$2b=3x-15$, find the unknown $x$.
\par
\rightans{c}
\ans
{$18$}
{$21$}
{$29$}
{$32$\PAR}
{None}
}
\def\BB{\input{isos9a.box}}
\WRAPFIG{\AA}{\BB}{1.2in}

\SOL{7. \STATS{74}{88}{65.12}}{
The triangle must be isosceles, therefore $2a=2b$ and $3x-15=2(x+7)$.
Solving
determines $x=29$.
}

%36.25
\prob{8.}{(Inequalities)}% #5, 1994
Describe the $x$'s such that $3(2x-5)+5 > -5(1-x)-6$ and $2(x+1)+1 >
5x-9$.

\rightans{b}
\ans
{$-2 < x < 4$}
{$-1 < x < 4$}
{$0 < x < 4$}
{$1 < x < 4$}
{$2 < x < 4$}
\SOL{8. \STATS{85}{75}{63.75}}{The first reduces to $6x-14+1>5x-5-9$ or $x>-1$. The second
becomes $2x+2+1>5x-9$ or $12/3>x$. Together, $4>x>-1$.}

%41.52
\def\AA{
\prob{9.}{(Trapezoid)}% #10, 1994
Find $x$ in the trapezoid, given $A$ and $B$ are midpoints.

\rightans{a}
\ans
{$\frac56$}
{$\frac67$}
{$\sqrt{\frac34}$\PAR}
{$\sqrt{\frac45}$}
{$1$}
}
\def\BB{\input{trap9a.box}}
\WRAPFIG{\AA}{\BB}{1.4in}


\SOL{9. \STATS{68}{86}{58.48}}{The average of opposite sides $3x+4$ and
$9x+8$ is $11$. Then
$6x+6=11$ or $x=5/6$.}

%43
\prob{10.}{(Rectangle)}% #30,  1995
A rectangle has a perimeter of $42$cm. Which of the following is {\em
definitely not} its area in cm${}^2$?

\rightans{e}
\ans
{$107$}
{$108$}
{$109$}
{$110$}
{$111$}

\SOL{10. \STATS{75}{76}{57}}{
Let $a$ and $b$ be its dimensions. Then $2a+2b=42$, making $a+b=21$. The
area is $ab=a(21-a)$. The largest positive value of the quadratic
$y=x(21-x)$ occurs at its vertex, at $x=21/2$, $y=441/4$. No rectangle
can have area larger than $441/4$ and perimeter $42$. This eliminates
all but one of the answers.
}

%46.06
\prob{11.}{(Complement)}% #3,  1995
The measure of an angle is $2/3$ the measure of its supplement.
Find the measure of its complement.

\rightans{c}
\ans
{$12^\circ$}
{$16^\circ$}
{$18^\circ$}
{$36^\circ$}
{None}

\SOL{11. \STATS{87}{62}{53.94}}{
Let $x$ be the angle. The supplement is $180-x$. Then $x=2(180-x)/3$
gives $5x=360$ and $x=72$. The complement is $90-x$ or $18$.
}

%54.12
\prob{12.}{(Trapezoid)}% #26,  1995
Find the length of the median of a trapezoid with vertices $(0,0)$,
$(0,3)$, $(4,5)$ and $(4,-2)$.

\rightans{a}
\ans
{$5$}
{$4$}
{$3$}
{$2$}
{None}

\SOL{12. \STATS{62}{74}{45.88}}{
The parallel bases have lengths $3$ and $7$, by the distance formula.
The median length is half of their sum, $(3+7)/2=5$. It can also be
worked using the midpoint formula and a suitable diagram.
}

%59.3
\prob{13.}{(Line)}% #21,  1995
Determine the value of $r$ such that the line through the points
$(4,r)$ and $(r,2)$ is perpendicular to $5y-3x=1$.

\rightans{c}
\ans
{$-7$}
{$1$}
{$7$}
{$-1$}
{None}

\SOL{13. \STATS{74}{55}{40.7}}{
The slope of $5y-3x=1$ is $3/5$. A perpendicular line has slope equal to
the negative reciprocal of $3/5$, that is, $m=-5/3$.
The slope formula $m=(y_2-y_1)/(x_2-x_1)$ applies to give the equation
$-5/3=(2-r)/(r-4)$. Clearing fractions gives $-5(r-4)=3(2-r)$ which can
be solved for $r=7$.
}

%61.6
\def\AA{
\prob{14.}{(Vertex)}% #14,  1995
Triangle $ABC$ is isosceles with perimeter between $23$ and $33$. Locate
the vertex angle.
\par
\rightans{a}
\ans
{$\angle A$}
{$\angle B$}
{$\angle C$\PAR}
{undecidable}
{None}
}
\def\BB{\input{isos9b.box}}
\WRAPFIG{\AA}{\BB}{1.2in}

\SOL{14. \STATS{80}{48}{38.4}}{
The inequality $23 < x+3+2x-1+12 < 33$ means $3 < x <
\frac{19}{3}$. To be isosceles, one of these must be true: (1)
$x+3=12$, (2) $2x-1=12$ or (3) $2x-1=x+3$. The only one that can work is
(3), which reduces to $x=4$. So $AB=AC=7$ and the angle at $A$ is the
vertex.
}

%63.08
\prob{15.}{(Inequality)}% #20,  1995
Solve $|x-2|+|x+2|^2\le 4$.

\rightans{b}
\ans
{$|2x+3|\le 1$\PAR}
{$|2x+5|\le \sqrt{33}$}
{$|x-2|\le 2$}
{$|x+2|\le 2$}
{$|x+2|\le 1$}
\SOL{15. \STATS{71}{52}{36.92}}{
Since $|x-2|= \pm (x-2)$, the left side of the
inequality stands for $(x-2)+(x-2)^2$ or $-(x-2)+(x-2)^2$.
Expanding results in two quadratic inequalities, $x^2+5x-2 \le 0$ and
$x^2+3x+2 \le 0$. Add $(5/2)^2$ and $(3/2)^2$, respectively, across
these inequalities in order to complete the square on each. Taking roots
gives $|x+5/2| \le \sqrt{33}/2$ and $|x+3/2| \le 1/2$. The latter is $-2
\le x \le -1$, which happens to be included in the first, which is about
$-5.37 \le x \le .37$.
Therefore, the two inequalities can be combined into $|2x+5|\le
\sqrt{33}$.
}

%65.2
\prob{16.}{(Functions)}% #11,  1995
Given $f(x)=\dd\frac{x^2-x+2}{8x^2-5x+16}$, find\newline
$f(2+\sqrt{2})$.

\rightans{d}
\ans
{$\frac14$}
{$9$}
{$\frac94$}
{$3^{-2}$}
{$5^{-2}$}
\SOL{16. \STATS{58}{60}{34.8}}{
This is the value of $y=(x^2-x+2)/(8x^2-5x+16)$ when $x=2+\sqrt2$.  Then
$x^2-x+2=4+4\sqrt2 + 2 - 2 - \sqrt2 + 2 =6+3\sqrt2$ and
$8x^2-5x+16=8(4+4\sqrt2 + 2)-10-5\sqrt2 + 16=54+27\sqrt2$, therefore
$y=1/9$ which is $3^{-2}$.
}

%67.55
\prob{17.}{(Volume)}% #3, 1994
A cube has surface area $25\sqrt2$ cm$^2$.
Find its volume in cm$^3$.

\rightans{b}
\ans
{$\frac{12\sqrt5}{6\sqrt2}$ }
{$\frac{125\sqrt[4]{2}}{6\sqrt3}$ }
{$\frac{43}{3}$ }
{$14.3$ }
{None}
\SOL{17. \STATS{59}{55}{32.45}}{The surface area of a cube of side $x$ is $6x^2$. The equation
$6x^2=25\sqrt2$ determines $x=\frac{5\sqrt[4]{2}}{\sqrt6}$. The
volume
is
$V=x^3=\frac{5^3\sqrt{2}\sqrt[4]{2}}{6\sqrt6}$.}

%68.8
\prob{18.}{(Speedboat)}% #24, 1994
A powerboat goes upriver $30$ miles in $300$ minutes. It returns
downriver
in
$3$
hours. Find the river's speed in mi/hr.

\rightans{d}
\ans
{$5$}
{$4.95$}
{$3$}
{$2$}
{$1$}
\SOL{18. \STATS{78}{40}{31.2}}{Let $r$ be the speed of the river and $d$
the speed of the
boat in still water. Then $(d-r)(5)=30$ and $(d+r)(3)=30$. These
equations reduce to $d-r=6$ and $d+r=10$. Adding gives $2d=16$ or
$d=8$. Then $r=10-d=2$.}

%72.86
\prob{19.}{(Color--Blind)}% #10,  1995
Find the probability that two black socks are drawn randomly
without replacement from a drawer having $5$ black and $8$ blue socks.

\rightans{c}
\ans
{$\frac{3}{26}$}
{$\frac{5}{26}$}
{$\frac{5}{39}$}
{$\frac{1}{39}$}
{$\frac{14}{39}$}
\SOL{19. \STATS{59}{46}{27.14}}{
The events possible are $(a,b)$ where $a$ and $b$ are colors, black or
blue. The event size is $13(12)=156$. There are $5(4)$ events which
choose $a$ and $b$ both black. The probability is $20/156=5/39$.
}

%73.48
\prob{20.}{(Quadratics)}% #19, 1994
Find the smaller of the
Given the $T$-shaped area $A$ and the trapezoidal area $B$
below, find the smaller area, assuming $0<x<2$.

\rightans{a}
\ans
{$5x^2+4x$}
{$4x^2+6x$}
{$(2x+2)^2$\PAR}
{$4x^2+4x$}
{$-x^2+6x+6$}

\input{quad9a.box}

\SOL{20. \STATS{51}{52}{26.52}}{Let $a$ be the sum of the areas of the
two rectangles in area
$A$. Then $a=5x^2+4x$. Let $b$ be the area of the triangle plus the
square in area $qB$. Then $b=(2x+2)^2+(2x+2)$ and $b-a=-x^2+6x+6 >
-4+6x+6 > 0$ (because $0<x<2$ implies $-x^2 > -4$).
The inequality $b-a>0$ means $b>a$ and $a$ is the
smallest. }

%74.4
\prob{21.}{(Radicals)}% #23, 1994
Find the sum of the roots to $\dd \sqrt{5x^2+10x-11}=x+1$.

\rightans{c}
\ans
{$-3$}
{$-2$}
{$1$}
{$2$}
{$3$}
\SOL{21. \STATS{64}{40}{25.6}}{The equation squared is
$5x^2+10x-11=x^2+2x+1$, which simplifies into $4x^2+8x-12=0$, having
roots $1$ and $-3$. But $-3$ is not a root of the original equation. So
the sum of the roots is $1$.}

%76.8
\prob{22.}{(Remainder)}% #25,  1995
What is the remainder when $7^{1025}$ is divided by $5$?

\rightans{c}
\ans
{$4$}
{$3$}
{$2$}
{$1$}
{$0$}

\SOL{22. \STATS{58}{40}{23.2}}{
The first few powers of $7$ are $7$, $49$, $343$, $2401$,
$16807$.
The remainders after division by $5$ are $2$, $4$, $3$, $1$, $2$.
The
pattern repeats, each four causing the remainders to repeat. Since
$1025/4=256\frac{1}{4}$, the remainder is the first answer in
the list of four repeated values, namely, $2$.
}

%77.56
\prob{23.}{(Functions)}% #28, 1994
Given $f(t)=19|t+9| - |9-t|$ and $g(t)=10+t$,
evaluate $f(g(-19))$.

\rightans{d}
\ans
{$18$}
{$19$}
{$-19$}
{$-18$}
{$0$}
\SOL{23. \STATS{34}{66}{22.44}}{First, $g(-19)=-9$. Then
$f(g(-19))=f(-9)=19|-9+9| - |9-(-9)|= -18$.}

%78.4
\prob{24.}{(Short Side)}% #26, 1994
Two opposite sides of a parallelogram are $4x+5y$ and $10x+14$. The
other
sides are $x+5y$ and $y+17$. Find the shortest side.

\rightans{c}
\ans
{$4$}
{$20$}
{$21$}
{$24$}
{None}
\SOL{24. \STATS{45}{48}{21.6}}{Opposite sides of a parallelogram are
equal. The equations are
$4x+5y=10x+14$ and $x+5y=y+17$. Simplifying, they become $-6x+5y=14$ and
$x+4y=17$. The solution is $x=1$, $y=4$.
Then the sides of the parallelogram are $21$ and $24$.}

%78.92
\def\AA{
\prob{25.}{(Tangents)}% #29, 9th (altered) 1994
Given segment $\overline{AB}$ tangent at $A$, $AB=9x-22$, $BC=7x-18$ and
$DC=2x$, find the sum of all possible solutions $x$.
\par
\rightans{d}
\ans
{$\frac{8}{3}$}
{$4$}
{$\frac{16}{3}$}
{$6$\PAR}
{None}
}
\def\BB{\input{tan9a.box}}
\WRAPFIG{\AA}{\BB}{1.2in}

\SOL{25. \STATS{9th=34}{62}{21.08}}{{\bf Theorem}: {\em If a tangent
segment and a secant segment
are drawn to a circle from an exterior point, then the square of the
measure of the tangent segment is equal to the product of the measures
of the secant segment and its external secant segment}. See Merrill
Geometry (1993), p 449. Applying the theorem gives
$(9x-22)^2=(7x-18)(9x-18)$. Taking roots gives $x=10/3$ and $x=8/3$.
Neither root is extraneous. Therefore the sum of the
roots is $6$.}

%79.84
\prob{26.}{(Median)}% #17,  1995
Let $\triangle ABC$ have sides $12$, $16$ and $20$.
Find the length of the longest median to the nearest tenth.

\rightans{c}
\ans
{$14.1$}
{$14.4$}
{$17.1$}
{$17.3$}
{$10.0$}
\SOL{26. \STATS{56}{36}{20.16}}{
Assume the triangle has vertices $(0,0)$, $(12,0)$ and $(12,16)$. Using
the midpoint formula, the missing endpoint of each median can be
computed. By geometry, the shortest one connects $(0,0)$ to the midpoint
of the hypotenuse, which is at $(6,8)$. The length of that median is
$10$, by the distance formula. The other medians have lengths
$\sqrt{208}=14.42$ and $\sqrt{292}=17.09$, by application of the same
method.
}

%80.2
\prob{27.}{(Fractions)}% #20, 1994
Find the perimeter of a quadrilateral with sides
$\frac{1}{x-2}$, $\frac{x}{x-2}$,
$\frac{1}{x+2}$ and $\frac{x}{x+2}$,
given $x=1+\sqrt3$.

\rightans{b}
\ans
{$\frac{7}{2}\sqrt3$}
{$3+\frac{5}{\sqrt3}$}
{$4+\sqrt3$}
{$5+\frac{1}{\sqrt3}$\PAR}
{None}
\SOL{27. \STATS{36}{55}{19.8}}{Add the four fractions to get
$\frac{2x(1+x)}{x^2-4}$.
Set $x=1+\sqrt3$ to get $3+\frac{5}{\sqrt3}$.}

%83.6
\prob{28.}{(Area)}% #7, 1994
A circle and a square of the same area have perimeters $A$ and $B$,
respectively. Find $A/B$.

\rightans{d}
\ans
{$\pi$}
{$\frac{\pi}{2}$}
{$\sqrt{\pi}$}
{$\frac{\sqrt{\pi}}{2}$}
{None}
\SOL{28. \STATS{40}{41}{16.4}}{Let the circle have radius $R$. Let the
square have side $x$.
Then $A=2\pi R$ and $B=4x$. Given equal areas, $\pi R^2 = x^2$. Then
$x=\sqrt{\pi}R$ and the ratio is
$A/B=2\pi R/(4x)=\pi R/(2\sqrt{\pi}R) = \sqrt{\pi}/2$.}

%85.96
\prob{29.}{(Time)}% #39,  1995
Find the {\em exact} time between $7$ and $8$ when the minute hand of a
watch is precisely $10$ minutes in advance of the hour hand.

\rightans{b}
\ans
{$7$:$49$}
{$7$:$49\frac{1}{11}$}
{$7$:$49\frac{2}{11}$}
{$7$:$49\frac{3}{11}$\\[3pt]\MBL}
{$7$:$50\frac{1}{11}$}
\SOL{29.}{
%old \STATS{7th=52}{27}{14.04}},
% \STATS{8th=46}{29}{13.34},
% \STATS{9th=39}{37}{14.43}
Let $x$ be in minutes. The hour hand is at angle $H=210+30(x/60)$, the
minute hand is at angle $M=360(x/60)$ and the clock position advance for
$10$ minutes on the dial is $60$. The equation to determine $x$ is
$M-H=60$, which becomes $360(x/60) - 210 - 30(x/60)= 60$. Then
$x=540/11$, giving the answer $49\frac{1}{11}$ minutes after $7$:$00$.
}

%80.2
\prob{30.}{(Variation)}% #2, 1994
Let $z$ vary directly with $y^2$ and inversely with $x$, taking on the
value $z=a^2+1$ when $y=2/3$ and $x=4/(a^2+1)$. Find $x$ when
$z=3(a^2+1)$ and $y=a^2+1$.

\rightans{a}
\ans
{$3a^2+3$}
{$a^2+1$}
{$8a^2+8$}
{$9a^2+9$\PAR}
{None}
\SOL{30. \STATS{30}{38}{11.4}}{
Let $z=cy^2/x$ or equivalently $xz=cy^2$.
Then $y=2/3$ and $x=4/(a^2+1)$ at $z=a^2+1$ gives a value for $c$, due
to the equation
$(4/(a^2+1))a^2+1=c(2/3)^2$. Solving for $c$ produces $c=9$.
To
find
$x$, solve $3(a^2+1)x=9(a^2+1)^2$ for $x=3a^2+3$.
}

\end{exercises}

%% Top 10!
\fbox{
\begin{minipage}{3.35in}
\large\sf
Problems 31--40 have a higher weight in the
Junior competition between grades 7, 8 and 9.
\end{minipage}
}

\begin{exercises}
%37.02
\prob{31.}{(Counting)}% #2,  1994
How many integers between $100$ and $200$ contain either the digit $5$
or the digit $7$?

\rightans{a}
\ans
{$36$}
{$37$}
{$38$}
{$39$}
{$40$}
\SOL{31. \STATS{94}{67}{62.98}}{The numbers of the form $1d5$ and $1d7$
for $d$ equal to $0$,
$1$, $2$, $3$, $4$, $6$, $8$, $9$ account for $16$ numbers. The numbers
of the form $15d$ and $17d$ for $0 \le d \le 9$ account for $20$
numbers. The total is $16+20=36$.}

\prob{32.}{(Cards)}% #28,   1996, Lajos
Four cards are face down on the table. Two are Aces and two are Jacks.
Pick two and turn them over. Find the probability of getting an Ace and
a Jack.

\rightans{b}
\ans
{$\frac{3}{4}$}
{$\frac{2}{3}$}
{$\frac{1}{2}$}
{$\frac{3}{8}$}
{None}

\SOL{32. \STATS{0}{0}{0}}{
There are $12$ outcomes and $4\cdot 2$ are favorable. The probability is
$8/12$.
}



%40.06
\prob{33.}{(Jackpot)}% #34,   1995
A jackpot of \${}$47$ is divided into piles $A$, $B$, $C$, $D$ so that
$A$ and $B$ together have \${}$27$, $A$ and $C$ have \${}$25$ and
finally $A$ and $D$ have \${}$23$. Find the number of dollars in $D$.


\rightans{a}
\ans
{$9$}
{$10$}
{$14$}
{$16$}
{$17$}
\SOL{33.}{
%\STATS{7th=81}{74}{59.94},
%\STATS{8th=80}{73}{58.4},
%\STATS{9th=81}{85}{68.85}.
There are four equations in four unknowns, $A+B+C+D=47$, $A+B=27$,
$A+C=25$, $A+D=23$. Add the last three to obtain $3A+B+C+D=75$. Subtract
the first from it to get $2A=28$ and $A=14$. Then $D=23-A=9$.
}

%70.93
\prob{34.}{(Composites)}% #35,  1994
Let $X=300,100,200,517,201,275$. One number below is a factor of $X$.
Find it.

\rightans{d}
\ans
{$21$}
{$33$}
{$69$}
{$75$}
{$100$}
\SOL{34.} %\STATS{7th=51}{57}{29.07},\newline \STATS{8th=46}{60}{27.6}, \STATS{9th=46}{54}{24.84}}
{Write $X=100Y+75$. Then $X=25(4Y+3)$, so $25$
divides $X$.
The sum of the digits of $X$ is $3+1+2+5+1+7+2+1+2+7+5=36$ which is
divisible by $3$, therefore $3$ divides $X$. This shows $(3)(25)$
divides $X$. }

%73.4
\def\di#1{{\large\rm ${#1}$}}%
\prob{35.}{(Game Wheel)}% #32,  1994
A Wheel is labeled, clockwise along its circumference, with equally
spaced numerals \di{1}, \di{2}, \ldots, \di{21}{}.
A stationary marker points to
\di{1}. After $62$ seconds of counter--clockwise rotation, the marker
is at \di{8}. If numeral \di{21}{} passed the marker three times, then
find the marker numeral after only $31$ seconds.

\rightans{c}
\ans
{\di{13}}
{\di{14}}
{\di{15}}
{\di{16}}
{\di{17}}
\SOL{35.}{
% \STATS{7th=76}{35}{26.6}, \STATS{8th=79}{42}{33.18},
% \STATS{9th=83}{37}{30.71}
The wheel completes $3\frac{7}{21}$ rotations in $62$ seconds, hence
half that many in $31$ seconds, which is $1\frac{14}{21}$.
The marker points at numeral \di{15}.}

%80.4
\prob{36.}{(Fractions)}% #14,  1994
Simplify $\frac{1+3x}{1-2x}$ for $x=\frac{a-2}{2a+1}$ where $a>3$.

\rightans{d}
\ans
{$\frac{a-2}{a+2}$}
{$\frac{a-1}{a+1}$}
{$2a-1$}
{$a-1$\PAR}
{$a^2-1$}
\SOL{36. \STATS{49}{40}{19.6}}{
Given $x=(a-2)/(2a+1)$, then
the fraction $\frac{1+3x}{1-2x}$ becomes a complex fraction, in which
multiplication by $\frac{2a+1}{2a+1}$ simplifies the result to
$\frac{2a+1+3a-6}{2a+1-2a+4}$. The latter reduces to $a-1$.}

%84.09
\prob{37.}{(Odometer)}% #38,   1995
An auto odometer displays miles from $0$ to $99999$. In the first
$1000$ miles, how many displays have sum of the digits greater than
$20$?

\rightans{e}
\ans
{$42$}
{$80$}
{$82$}
{$83$}
{$84$}
\SOL{37.}{
%\STATS{7th=43}{37}{15.91}},\newline
%\STATS{8th=45}{26}{11.7},\STATS{9th=38}{32}{12.16}
The first display is $399$. The next ones are in groups starting at
$489$, $579$, $669$, $759$, $849$ and $939$. The group sizes are
$n(n+1)/2$ for $n=1,\ldots,7$. The total is $1+3+6+10+15+21+28=84$.
}

%94.6
\prob{38.}{(Pentagon)}% #39,  1994
A pentagon with sides $3$, $2$, $3$, $2$ and $5/2$ happens to have an
inscribed circle of diameter $13/4$. Find the area of the pentagon.


\rightans{b}
\ans
{$\frac{325}{16}$}
{$\frac{325}{32}$}
{$\frac{325}{64}$}
{$\frac{163}{16}$}
{$\frac{81}{8}$}

\SOL{38. \STATS{9th=26}{37}{9.62}}{
The pentagon is the sum of five triangles of altitude
$h=13/8$. The area is $3\cdot h/2 + 2\cdot h/2 + 3\cdot h/2 + 2 \cdot
h/2 + (5/2) \cdot h/2 = (25/2)(h/2) = 325/32$.}


%99.48
\prob{39.}{(Ramp)}% #28,  7th 1995
Find the slope of a ramp which raises an object
$1$m vertically for each $5$m of travel on the ramp
surface.

\rightans{d}
\ans
{$\dd\frac{1}{5}$}
{$\dd\frac{4}{\sqrt6}$}
{$\dd\frac{5}{\sqrt6}$}
{$\dd\frac{\sqrt6}{12}$}
{None}

\SOL{39. \STATS{52}{1}{0.52}}{
Form a right triangle with hypotenuse $5$ and legs $1$, $x$. Then
$x^2+1^2=5^2$, giving $x=2\sqrt6$. The slope is {\em rise/run}, which is
$1/x=1/(2\sqrt6)=\sqrt{6}/12$.
}

\prob{40.}{(Missile)}% #28,   1996
A missile hits its target with probability $0.3$. Find the least number
of missiles to be launched to insure an $80$\% probability of hitting a
target.


\rightans{a}
\ans
{$5$}
{$6$}
{$8$}
{$9$}
{None}

\SOL{40. \STATS{0}{0}{0}}{
The missile misses with probability $0.7$, so $n$ missiles miss the
target with probability $0.7^n$. To find the least $n$, solve the
inequality $1-0.7^n > 0.8$ by trial--and--error. The least one is $n=5$.
}

\end{exercises}


%% 10.tex
\begin{center}\Large\bf
1996 Senior Exam Grade 10
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 10} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 10
\setanswer{1.}{b} & \setanswer{11.}{e} & \setanswer{21.}{d} & \setanswer{31.}{c} \\
\setanswer{2.}{c} & \setanswer{12.}{c} & \setanswer{22.}{c} & \setanswer{32.}{a} \\
\setanswer{3.}{e} & \setanswer{13.}{d} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{a} & \setanswer{34.}{e} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{a} & \setanswer{35.}{a} \\
\setanswer{6.}{d} & \setanswer{16.}{a} & \setanswer{26.}{c} & \setanswer{36.}{a} \\
\setanswer{7.}{c} & \setanswer{17.}{a} & \setanswer{27.}{a} & \setanswer{37.}{b} \\
\setanswer{8.}{b} & \setanswer{18.}{c} & \setanswer{28.}{a} & \setanswer{38.}{d} \\
\setanswer{9.}{b} & \setanswer{19.}{d} & \setanswer{29.}{d} & \setanswer{39.}{c} \\
\setanswer{10.}{d} & \setanswer{20.}{e} & \setanswer{30.}{e} & \setanswer{40.}{d} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

{\bf Instructions}: Only one answer is correct. Scoring: 5(\#
right)+1(\# blank)+0(\# wrong). Time: 120 minutes.


\begin{exercises}

%% 10.tex

%%% 20.8
\prob{1.}{(Ratio)}% #6, 1995
The perimeter of a triangle is $66$ units and the sides have measures in
the ratios $2$:$4$:$5$. Find the number of units in the shortest side.

\rightans{b}
\ans
{$6$}
{$12$}
{$18$}
{$24$}
{$30$}

\SOL{1. \STATS{90}{88}{79.2}}{
The measures of the sides can be written
as $2x$, $4x$ and $5x$. The perimeter is $2x+4x+5x=66$. Then $x=6$,
but $x$ is not a side; the sides are $2x=12$, $4x=24$ and $5x=30$.
}

%%% 24.35
\def\AA{
\prob{2.}{(Geometry)}% #11, 1995
In the diagram, let segments $\bar{AB}$ and $\bar{DE}$ be parallel and
let $AC=2$, $BC=4$ and $CE=3$. Find $CD$.
\par
\rightans{c}
\ans
{$4$}
{$5$}
{$6$}
{$7$\PAR}
{None}
}
\def\BB{\input{geom10b.box}}
\WRAPFIG{\AA}{\BB}{1.1in}

\SOL{2. \STATS{85}{89}{75.65}}{
Let $x=CD$. The triangles $ABC$ and $EDC$ are similar, therefore
$\dd\frac{3}{2}=\frac{x}{4}$, which can be solved for $x=6$.
}

%%% 27.1
\prob{3.}{(Midway)}% #10, 1995
The line $8x+5y=c$ is midway between and parallel to the parallel
lines $8x+5y=3$ and $8x+5y=5$. Which is a possible value of
$c$?

\rightans{e}
\ans
{$5$}
{$6$}
{$7$}
{$8$}
{None}

\SOL{3. \STATS{81}{90}{72.9}}{
Graph the lines. The ``midway'' requirement means
$\frac{1}{2}(3+5)=c$, giving $c=4$.
}

%%% 35.62
\prob{4.}{(Midpoint)}% #19, 1995
Given $A(x,-2x+1)$ and $B(-x,2x+1)$, find the midpoint of
segment $\bar{AB}$.

\rightans{d}
\ans
{$(x,x)$}
{$(0,0)$}
{$(-1,0)$}
{$(0,1)$\PAR}
{None}
\SOL{4. \STATS{74}{87}{64.38}}{
The midpoint is $((x-x)/2,(-2x+1+2x+1)/2)=(0,1)$.
}

%%% 37.75
\prob{5.}{(Quadrilateral)}% #9, 1995
Find $x$, given a quadrilateral has exterior angles $65$, $75$,
$x$ and
$3x$ (in degrees).

\rightans{d}
\ans
{$49^\circ$}
{$51^\circ$}
{$53^\circ$}
{$55^\circ$}
{None}
\SOL{5. \STATS{75}{83}{62.25}}{
The sum of the exterior angles must be $360^\circ$. This produces the
equation $360=65+75+x+3x$ for the unknown $x$. Solving, $x=55$.
}

%%% 44.24
\prob{6.}{(Coolant)}% #1, 1995
Mix $2$, $5$ and $7$ liters of coolant at temperatures $50^\circ$,
$67^\circ$ and $73^\circ$ , respectively, to create a mixture at
$x$ degrees. Find $x$ to the nearest degree.

\rightans{d}
\ans
{$65$}
{$66$}
{$67$}
{$68$}
{$70$}
\SOL{6. \STATS{82}{68}{55.76}}{
The mixture equation is $(2+5+7)x=2(50)+5(67)+7(73)$. The answer is
$x=946/14=67.57$. The closest answer is $68$.
}

%%% 49.6
\prob{7.}{(Composition)}%#10, 1994
Simplify $(a^2+5ab+6b^2)/(a+3b)$ and evaluate at $a=3x^2-1$,
$b=1-x^2$.

\rightans{c}
\ans
{$2x^2$}
{$3x^2-1$}
{$1+x^2$}
{$x^2-1$\PAR}
{$x^2$}
\SOL{7. \STATS{72}{70}{50.4}}{ Factor $a^2+5ab+6b^2=(a+3b)(a+2b)$,
cancel $(a+3b)$ to get
$(a+2b)$. Substitute for $a$ and $b$ to obtain $1+x^2$.}

%%% 52.48
\def\AA{
\prob{8.}{(Tables)}%#19, 1994
Find an entry $x>0$ that completes the table, given
$\dd y=\frac{x}{x+4}$.
}
\def\BB{
\begin{tabular}{|l|c|c|c|}
\hline
$\EXM x$ & $0$ & $1$  &   \\[1ex] \hline
$\EXM 3xy$ & $0$ & $\frac{3}{5}$  & $\frac{1}{13}$ \\[1ex] \hline
\end{tabular}
}
\WRAPFIG{\AA}{\BB}{1.25in}

\rightans{b}
\ans
{$1/5$}
{$1/3$}
{$2/5$}
{$3/5$}
{$2/3$}

\SOL{8. \STATS{72}{66}{47.52}}{ Solve $3x^2/(x+4)=1/13$ for $x$ and
compare to the table
to select $1/3$. Alternatively, try each answer in the formula
$3xy$.}

%%% 53.08
\prob{9.}{(Interest)}%#5, 1994
An investment at $10$\%  interest compounded annually earns
\${}$60.44$ in $5$ years. Approximate the interest in dollars at six
years.

\rightans{b}
\ans
{$74$}
{$76$}
{$79$}
{$81$}
{$83$}
\SOL{9. \STATS{68}{69}{46.92}}{ The equation is $x(1.1)^5 - x = 60.44$.
Then $x(1.1)^6 - x = 76.38$.}

%%% 58
\prob{10.}{(Mixture)}%#4, 1994
Add $50$ mL of water to $125$ mL of a $20$\% salt solution. What
percentage of the new solution is salt?

\rightans{d}
\ans
{$1/7$}
{$14$}
{$14\frac17$}
{$14\frac27$}
{$14\frac37$}
\SOL{10. \STATS{75}{56}{42}}{The amount of salt is $(0.2)(125)$. The
amount of solution is $50+125=150$. The fraction of salt is
$(0.2)(125)/175$ which is $14\frac27$ percent.}

%%% 58.69
\def\AA{
\prob{11.}{(Squares)}% #17, 1995
In the big square figure, assume shapes $A$ through $G$ are also
squares. If $A$ has area $64$ cm${}^2$ and $B$ has diagonal $7\sqrt2$
cm,
then find the side of $C$.
\par
\rightans{e}
\ans
{$8$}
{$9$}
{$11$}
{$12$\PAR}
{$13$}
}
\def\BB{\input{squar10.box}}
\WRAPFIG{\AA}{\BB}{1.3in}

\SOL{11. \STATS{81}{51}{41.31}}{
Squares $A$ and $B$ have sides $8$ and $7$, respectively. Square $D$ has
side $1$. Then square $E$ has side $9$ and to its left square $F$
has side $10$. Square $G$ has side $7+8=15$, due to squares $A$
and $B$ filling a side. Then the main square has side $15+8+9=32$.
Finally, square $C$ has side $32-10-9=13$.
}

%%% 60
\prob{12.}{(Population)}% #13, 1995
In $1990$, the population of a city was $14344$, a decrease of
$8\frac{1}{3}$\% from its $1980$ population. The $1980$ population
was an increase of $8\frac{2}{3}$\% from its $1970$ population. Find
the $1970$ population.

\rightans{c}
\ans
{$14298$}
{$14392$}
{$14400$}
{$14448$\PAR}
{None}

\SOL{12. \STATS{80}{50}{40}}{
Let $P$ and $Q$ be the populations in 1970 and 1980, respectively. Then
$8\frac{1}{3}$\% is $1/12$ and $8\frac{2}{3}$\% is $13/150$, giving
$14344=Q-Q/12$ and $Q=P+13P/150$. Then
$P=\frac{150}{163}Q=\frac{150}{163}\frac{12}{11}14344=14400$.
}

%%% 61.28
\prob{13.}{(Perfect Cubes)}%#6, 1994
What fraction of the positive odd integers less than $2199$ are
perfect cubes?

\rightans{d}
\ans
{$1/100$}
{$11/2200$}
{$13/1100$}
{$7/1100$\PAR}
{$7/2200$}
\SOL{13. \STATS{88}{44}{38.72}}{ The odd perfect cubes are $1^3$, $3^3$,
\ldots, $13^3$. The
odd numbers are $1$, $3$, $5$, \ldots, $2199$. The counts are $7$ and
$1100$ for fraction $7/1100$.}

%%% 62.56
\prob{14.}{(Symmetry)}% #15, 1995
How many lines of symmetry are there for the region inside an ellipse
of eccentricity $1/2$?

\rightans{c}
\ans
{$4$}
{$3$}
{$2$}
{$1$}
{$0$}

\SOL{14. \STATS{78}{48}{37.44}}{
A line of symmetry is a line that can be drawn through the figure so
that it can be folded along the line and the two halves match exactly.
The lines of symmetry of an ellipse are just the major and minor axes.
}

%%% 65.21
\prob{15.}{(Polygon)}% #20, 1995
The sum of the measures of the interior angles in a polygon is $25020$.
Find the number of sides.

\rightans{d}
\ans
{$138$}
{$139$}
{$140$}
{$141$}
{$142$}
\SOL{15. \STATS{71}{49}{34.79}}{
The equation for the number $n$ of sides is $180(n-2)=25020$. Solving
gives $n=141$.
}

%72.86
\prob{16.}{(Socks)}% 9th grade 1995
Find the probability that two red socks are drawn randomly
without replacement from a drawer having $8$ red and $5$ blue socks.

\rightans{a}
\ans
{$\frac{14}{39}$}
{$\frac{10}{39}$}
{$\frac{5}{39}$}
{$\frac{1}{39}$}
{$\frac{56}{169}$}
\SOL{16. \STATS{59}{46}{27.14}}{
The events possible are $(a,b)$ where $a$ and $b$ are colors, red or
blue. The event size is $13(12)=156$. There are $8(7)$ events which
choose $a$ and $b$ both red. The probability is $56/156=14/39$.
}

%%% 66.52
\prob{17.}{(Tangent)}% #2, 1995
Find the equation of the line through $(3,4)$ and tangent to the circle
of radius $5$ and center $(0,0)$.

\rightans{a}
\ans
{$3x+4y=25$}
{$4x+3y=24$}
{$4y-3x=7$}
{$4x-3y=0$}
{None}
\SOL{17. \STATS{62}{54}{33.48}}{
The equation is $y-4=m(x-3)$. The point $(3,4)$ is on the circle,
therefore $m$ is the negative reciprocal of the slope of the radial line
through $(3,4)$. That line has slope $4/3$, which implies $m=-3/4$. The
equation simplifies to $3x+4y=25$.
}

%%% 68.2
\prob{18.}{(Water Tank)}%#22, 1994
A tank has a drain rate of $10$\% of capacity per hour. Its fill rate is
$15$\% of capacity per hour. When $2/3$ full and draining, about how
many hours does it take to fill?

\rightans{c}
\ans
{$6\frac13$}
{$6\frac49$}
{$6\frac23$}
{$5\frac23$}
{$5\frac13$}
\SOL{18. \STATS{53}{60}{31.8}}{ The fill time $t$ satisfies $t(0.15x -
0.1x) +2 x/3 = x$
where $x$ is the tank capacity. Cancel $x$ and solve for $t=20/3$.}

%%% 68.89
\prob{19.}{(Monopoly)}%#18, 1994
Player $A$ tosses the die and gets number $a$ ($1$ to $6$ possible).
Player $B$ tosses and gets number $b$. Find the probability that $2b>a$.

\rightans{d}
\ans
{$23/36$}
{$5/8$}
{$9/10$}
{$3/4$}
{$7/8$}
\SOL{19. \STATS{61}{51}{31.11}}{ There are $36$ states $[b,a]$.
Possibilities for success are
the states
$[1,1]$, $[2,1]$, $[2,2]$, $[2,3]$,
$[3,1]$, $[3,2]$, $[3,3]$, $[3,4]$, $[3,5]$,
$[4,1]$, $[4,2]$, $[4,3]$, $[4,4]$, $[4,5]$, $[4,6]$,
$[5,1]$,
$[5,2]$,
$[5,3]$,
$[5,4]$,
$[5,5]$,
$[5,6]$,
$[6,1]$, $[6,2]$, $[6,3]$, $[6,4]$, $[6,5]$, $[6,6]$.
The success count is $27$. The probability
is
$27/36$.}

\prob{20.}{(Inequalities)}
Find the maximum of $2x+4y$ subject to the inequalities $x\ge0$,
$y\ge0$, $x+y\le 3$ and $3x+y\le 6$.

\rightans{e}
\ans
{$0$}
{$4$}
{$14$}
{$9$}
{$12$}
\SOL{20.}{The function $z=2x+4y$ is a plane in $R^3$. The inequalities
describe a polygon in the plane. The maximum occurs at a vertex of
the polygon: $(0,0)$, $(2,0)$, $(3/2,3/2)$, $(0,3)$. The values of $z$
are $0$, $4$, $9$, $12$, respectively. The maximum of $2x+4y$ is
$12$. }

%%% 72.64
\prob{21.}{(Statistics)}%#13, 1994
Find the average of the mode and the mean for the data set
$49$, $54$, $61$, $49$, $54$, $49$, $64$ and $60$.

\rightans{d}
\ans
{$49$}
{$50$}
{$51$}
{$52$}
{$55$}
\SOL{21. \STATS{48}{57}{27.36}}{
The mode is the most frequently listed value: $\bar{m}=49$.
 The mean is the average $\bar{x}=55$.
Then $\frac12(\bar{m}+\bar{x})= \frac12(49+55)=52$}

%%% 77.5
\prob{22.}{(Magnitude)}%#7, 1994
Given $a^3 > b > 1$, which number is largest?

\rightans{c}
\ans
{$(a^2b^2)^{101}$}
{$a^{600}$}
{$(a^2)^{408}$}
{$(a^4b^4)^{51}$\PAR}
{$(ab)^{203}$}
\SOL{22. \STATS{75}{30}{22.5}}{ Change each to powers of $a$ and $b$.
Cancel the common factor of $a^{202}$ and apply comparisons to reduce
the contest to $(a^2)^{408}$ and $(a^4b^4)^{51}$.
Dividing these two gives $(a^3/b)^{204}$.
Then $a^3 > b$ implies the quotient is greater than one, hence
$(a^2)^{408}$ is largest.
}

%%% 78.94
\prob{23.}{(Disk)}% #28, 1995
Select from the list below a point which is inside the circle of
center $(-1,2)$ and radius 3 and outside the circle of center $(-1,1)$
and radius $2$.

\rightans{e}
\ans
{$(-2,1)$}
{$(-1-\sqrt2,1)$}
{$(-1-\sqrt3,1)$\PAR}
{$(-3,1)$}
{$(-1-\sqrt5,1)$}
\SOL{23. \STATS{54}{39}{21.06}}{
The inequalities are $(x+1)^2+(y-2)^2<9$
and $(x+1)^2+(y-1)^2>4$. There is little to do except try the points.
}

%%% 79.48
\prob{24.}{(Sector)}%#17, 1994
Find the area of a sector of arc length $\pi/6$ and radius $1/\pi$.

\rightans{a}
\ans
{$\frac{1}{12}$}
{$\frac{\pi}{36}$}
{$\frac{11}{126}$}
{$\frac{\pi}{48}$}
{None}
\SOL{24. \STATS{54}{38}{20.52}}{The area is a fraction of the area of
the whole circle. The fraction is the arc length $s$ of the sector
divided by the arc length of the circle: $s/(2\pi r)$. The whole circle
area is $\pi r^2$. The sector area is $\pi r^2 \frac{s}{2\pi
r}=\frac{sr}{2}$. The answer is therefore
$\frac12\frac{\pi}{6}\frac{1}{\pi}=\frac{1}{12}$.}

%%% 82.66
\prob{25.}{(Tiles)}%#8, 1994
Rectangular tiles of width $1$ and lengths $1$, $2$, $3$ and $5$ touch
each other to form a figure. Let $x$ and $y$ be the maximum and minimum
perimeters among all figures. Find $x/y$.

\rightans{a}
\ans
{$\frac{15}{8}$}
{$2$}
{$\frac{17}{8}$}
{$\frac{9}{4}$}
{$\frac{19}{9}$}
\SOL{25. \STATS{51}{34}{17.34}}{ The maximum is $30$ obtained by the
tiles touching only at a
vertex. The minimum is $16$ obtained by packing the tiles into a
rectangle (various geometries are possible). }

%%% 83.6
\prob{26.}{(Polynomials)}%#14, 1994
Let $f$ and $g$ be polynomials of degrees $2$ and $6$, respectively.
Find the degree of the quotient $g(x^2)/f(x^3)$, ignoring any
remainder.

\rightans{c}
\ans
{$2$}
{$4$}
{$6$}
{$7$}
{$12$}
\SOL{26. \STATS{40}{41}{16.4}}{ The degree of $f(x^3)$ is $6$ and the
degree
of $g(x^2)$ is $12$. The quotient has degree $6$.}

%%% 84.4
\prob{27.}{(Trigonometry)}%#27, 1994
A triangle has sides of length $4$, $5$ and $8.5$ units. Find the cosine
of the largest angle.


\rightans{a}
\ans
{$-\frac{25}{32}$}
{$-\frac34$}
{$-\frac{13}{16}$}
{$-\frac{27}{32}$}
{None}
\SOL{27. \STATS{40}{39}{15.6}}{
The largest angle is opposite the longest side. Apply the Law of
Cosines, $c^2=a^2+b^2 - 2ab\cos C$ to obtain $\cos C =
(a^2+b^2-c^2)/(2ab)$. Then $\cos C = -25/32$.
}

%%% 86.36
\prob{28.}{(Guessing)}% #4, 1995
Elena guesses on $10$ true-false questions. Find the probability that
she gets exactly $2$ wrong.

\rightans{a}
\ans
{$45/1024$}
{$4/5$}
{$1/8$}
{$33/1024$\PAR}
{$45/512$}
\SOL{28. \STATS{44}{31}{13.64}}{ There are $2^{10}$ possible answer
sets. The $10$ questions can be marked with $8$ correct and $2$
incorrect in $1+2+\cdots+9=9\cdot10/2$
ways. The probability is $45/1024$.}

%%% 88.78
\prob{29.}{(Models)}%#30, 1994
Let $f(a)$ be the circumference in feet for a circle of area $a$ square
feet. Which answer below is the circumference of a circle of diameter
$4$?

\rightans{d}
\ans
{$f(2\pi)$}
{$f(4\pi)/\pi$}
{$f(\pi)$}
{$\pi f(4/\pi)$\PAR}
{$4f(\pi/2)$}
\SOL{29. \STATS{34}{33}{11.22}}{Use $a=\pi r^2$ and $c=2\pi r$. By the
area formula, $f(a)=c=2\pi r=2\sqrt{\pi}\sqrt{\pi r^2}=2\sqrt{a\pi}$.
The area of a circle of diameter $4$ (radius $2$)
is $4\pi$. The answers given are (a) $2\pi$, (b) $2\sqrt2$, (c) $4$, (d)
$4\pi$ and (e) $4\sqrt{2}\pi$. The match is (d).}

\prob{30.}{(Logarithms)}
Assume $y>0$ and $x>0$. Solve for $y$: $\log_4 x^2 + 5\log_4 y^2 = 4
\log_{16} y^{6}$

\rightans{e}
\ans
{$5x^3$}
{$2^x$}
{$4^x$}
{$x^2$}
{$x$}
\SOL{30.}{ Use  $\log_b u = \log_c u /\log_c b$ to write the equation
as $\log_4(x^2y^{10})=4\log_4 y^6/\log_4 16$. Then write as
$\log_4(x^2y^{10})=\log_4 y^{12}$ to get $x^2y^{10}=y^{12}$. Cancel to
obtain $y=x$.}

\smallskip

See {\em infra} for problems 31-40.

\smallskip

\end{exercises}

%% 11.tex
\begin{center}\Large\bf
1996 Senior Exam Grade 11
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 11} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 11
\setanswer{1.}{d} & \setanswer{11.}{d} & \setanswer{21.}{c} & \setanswer{31.}{c} \\
\setanswer{2.}{b} & \setanswer{12.}{d} & \setanswer{22.}{d} & \setanswer{32.}{a} \\
\setanswer{3.}{d} & \setanswer{13.}{e} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{d} & \setanswer{34.}{e} \\
\setanswer{5.}{a} & \setanswer{15.}{b} & \setanswer{25.}{d} & \setanswer{35.}{a} \\
\setanswer{6.}{e} & \setanswer{16.}{e} & \setanswer{26.}{c} & \setanswer{36.}{a} \\
\setanswer{7.}{a} & \setanswer{17.}{c} & \setanswer{27.}{c} & \setanswer{37.}{b} \\
\setanswer{8.}{a} & \setanswer{18.}{e} & \setanswer{28.}{a} & \setanswer{38.}{d} \\
\setanswer{9.}{e} & \setanswer{19.}{a} & \setanswer{29.}{c} & \setanswer{39.}{c} \\
\setanswer{10.}{a} & \setanswer{20.}{b} & \setanswer{30.}{b} & \setanswer{40.}{d} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

{\bf Instructions}: Only one answer is correct. Scoring: 5(\#
right)+1(\# blank)+0(\# wrong). Time: 120 minutes.


\begin{exercises}

% 11th
%% The following are from the 1996 Grade 10 exam.
%% It is normal to put in problems from the Spring of the 10th grade.
%% Here we are in catchup-mode, because the 10th grade was too hard
%% for the years 1994 and 1995 (over-zealous exam writer (GBG)).

%%% 44.24
\prob{1.}{(Coolant)}% #6 grade 10 #1, 1995, #6 1996
Mix $2$, $5$ and $7$ liters of coolant at temperatures $50^\circ$,
$67^\circ$ and $73^\circ$ , respectively, to create a mixture at
$x$ degrees. Find $x$ to the nearest degree.

\rightans{d}
\ans
{$946$}
{$66$}
{$67$}
{$68$}
{$70$}
\SOL{1. \STATS{82}{68}{55.76}}{
The mixture equation is $(2+5+7)x=2(50)+5(67)+7(73)$. The answer is
$x=946/14=67.57$. The closest answer is $68$.
}


%%% 52.48
\def\AA{
\prob{2.}{(Tables)}% #8,  Grade 10, #19, 1994, #8 1996
Find an entry $x>0$ that completes the table, given
$\dd y=\frac{x}{x+4}$.
}
\def\BB{
\begin{tabular}{|l|c|c|c|}
\hline
$\EXM x$ & $0$ & $1$  &   \\[1ex] \hline
$\EXM 3xy$ & $0$ & $\frac{3}{5}$  & $\frac{1}{13}$ \\[1ex] \hline
\end{tabular}
}
\WRAPFIG{\AA}{\BB}{1.25in}

\rightans{b}
\ans
{$1/5$}
{$1/3$}
{$2/5$}
{$3/5$}
{$2/3$}

\SOL{2. \STATS{72}{66}{47.52}}{ Solve $3x^2/(x+4)=1/13$ for $x$ and
compare to the table
to select $1/3$. Alternatively, try each answer in the formula
$3xy$.}

%%% 61.28
\prob{3.}{(Perfect Cubes)}% #13,  Grade 10, #6, 1994, #13, 1996
What fraction of the positive odd integers less than $2199$ are
perfect cubes?

\rightans{d}
\ans
{$1/100$}
{$11/2200$}
{$13/1100$}
{$7/1100$\PAR}
{$7/2200$}
\SOL{3. \STATS{88}{44}{38.72}}{ The odd perfect cubes are $1^3$, $3^3$,
\ldots, $13^3$. The
odd numbers are $1$, $3$, $5$, \ldots, $2199$. The counts are $7$ and
$1100$ for fraction $7/1100$.}

%%% 72.64
\prob{4.}{(Statistics)}% #21,  Grade 10, #13, 1994, #21, 1996
Find the average of the mode and the mean for the data set
$49$, $54$, $61$, $49$, $54$, $49$, $64$ and $60$.

\rightans{d}
\ans
{$49$}
{$50$}
{$51$}
{$52$}
{$55$}
\SOL{4. \STATS{48}{57}{27.36}}{
The mode is the most frequently listed value: $\bar{m}=49$.
 The mean is the average $\bar{x}=55$.
Then $\frac12(\bar{m}+\bar{x})= \frac12(49+55)=52$}

%%% 84.4
\prob{5.}{(Trigonometry)}% #27,  Grade 10, #27, 1994, #27, 1996
A triangle has sides of length $4$, $5$ and $8.5$ units. Find the cosine
of the largest angle.


\rightans{a}
\ans
{$-\frac{25}{32}$}
{$-\frac34$}
{$-\frac{13}{16}$}
{$-\frac{27}{32}$}
{None}
\SOL{5. \STATS{40}{39}{15.6}}{
The largest angle is opposite the longest side. Apply the Law of
Cosines, $c^2=a^2+b^2 - 2ab\cos C$ to obtain $\cos C =
(a^2+b^2-c^2)/(2ab)$. Then $\cos C = -25/32$.
}






%59.85
\def\AA{
\prob{6.}{(Triangle)}% #22,  9th 1994
Find side $w$ in the triangle.
\par
\rightans{e}
\ans
{$6$}
{$16/3$}
{$11/2$\PAR}
{$14/3$}
{$5$}
}
\def\BB{\input{tri9b.box}}
\WRAPFIG{\AA}{\BB}{1.35in}

\SOL{6. \STATS{55}{73}{40.15}}{The small right triangle is similar to the large right
triangle. Therefore, $w/8=(7+8)/24$ and $w=5$.}

%78.44
\prob{7.}{(Triangle)}% #23,  9th 1995
Classify the triangle $ABC$ with vertices $A(2,-4)$, $B(-3,1)$ and
$C(1,6)$.


\rightans{a}
\ans
{obtuse scalene}
{acute scalene\PAR}
{right scalene}
{isosceles}
{None}

\SOL{7. \STATS{77}{28}{21.56}}{
By the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ the sides
have lengths $\sqrt{50}$, $\sqrt{41}$ and $\sqrt{101}$, so it's scalene.
The angle at $B$ is not a right angle because the Pythagorean relation
fails: $\sqrt{50}^2+\sqrt{41}^2 \ne \sqrt{101}^2$. In fact, this
relation says that the angle at $B$ is obtuse.
}

%92.8
\def\AA{
\prob{8.}{(Photo)}% #8, 9th 1995
A square picture frame is to be designed as in the figure so that
$\frac{2}{3}$ of the total area is used by the inner frame. Find
the area of the inner frame in square units.
}
\def\BB{\input{photo8.box}}
\WRAPFIG{\AA}{\BB}{1.1in}

\rightans{a}
\ans
{$40+16\sqrt{6}$}
{$40+23\sqrt{3}$}
{$40+28\sqrt{2}$\PAR}
{$33+19\sqrt{6}$}
{None}

\SOL{8. \STATS{30}{24}{7.2}}{
The inner square has area $x^2$ and the outer one has area $(x+2)^2$,
therefore $\frac23(x+2)^2=x^2$. Solving for $x=4+2\sqrt{6}$ results in
inner area $x^2=40+16\sqrt{6}$.
}

%97
\prob{9.}{(Triangle)}% #25, 9th 1994
Sides $a$ and $b$ of a triangle satisfy the equation $6a^2+7ab+6b^2=30$. Side $c$ is
divided by the bisector of the opposite angle in the ratio $3$:$2$.
Find the smaller of $a$ and $b$.

\rightans{e}
\ans
{$\sqrt{7/6}$}
{$\sqrt{7/5}$}
{$3/2$}
{$2/3$}
{$1$}
\SOL{9. \STATS{20}{15}{3}}{The relation $a/b=3/2$ is given by the bisector hypothesis.
Solve for $a=3b/2$ and insert into the quadratic equation to get
$30b^2=30$. Then $b=1$ and $a=3/2$.}



%% 11.tex

%17.28
% 11th
\prob{10.}{(Rectangle)}% #7,  1995
Given a rectangle of area $1248$ square units and diagonal $52$ units,
determine the ratio of its sides.

\rightans{a}
\ans
{$\frac32$}
{$\frac65$}
{$\frac43$}
{$\frac75$}
{$\frac97$}

\SOL{10. \STATS{88}{94}{82.72}}{
Let $x$ and $y$ be the two sides. Then
$x^2+y^2=52^2$ and $xy=1248$. Hence $(x+y)^2 = x^2+2xy+y^2=5200$ and
$x+y=20\sqrt{13}$. Then $1248=xy=x(20\sqrt{13}-x)$ is a quadratic
equation for $x$ with solutions determined by the FOIL factorization
$(x-12\sqrt{13})(x-8\sqrt{13})=0$. The ratio is $12/8=3/2$.
}

%28.6
% 11th
\prob{11.}{(Navigation)}% #3,  1995
A lighthouse and a ship are $x$ {\em feet} apart, both at sea level, with
the light located $400$ feet above the sea. The {\em angle of
depression} is $15^\circ$, from the light to the ship. Approximate $x$.

\rightans{d}
\ans
{$1100$}
{$1451$}
{$1472$}
{$1493$}
{None}
\SOL{11. \STATS{85}{84}{71.4}}{
Using right triangle geometry, $400/x = \tan(15)$, giving
$x=400\cot(15)=1493$, approximately.
The value of $\sin(\theta/2)$ can be obtained from the half-angle
identity. Similarly for $\cos(\theta/2)$. Therefore, the value of
$\tan(15)$ can be deduced from $\sin(30)$ and $\cos(30)$.
}

%43.91
\def\AA{
\prob{12.}{(Skid)}% #13,  1995
Jane measured the skid marks when the brakes of her Justy were applied
at
different speeds. Use a simple data fitting model to estimate how far in
feet the Justy would skid if the brakes were applied at $70$ mph.
}
\def\BB{
{\small\rm
\begin{tabular}{|c|c|}
\hline {\bf ft} & {\bf mph} \\ \hline
 $44$ & $30$ \\ \hline
 $78$ & $40$ \\ \hline
$123$ & $50$ \\ \hline
$176$ & $60$ \\ \hline
\end{tabular}
}
}
\WRAPFIG{\AA}{\BB}{1in}

\rightans{d}
\ans
{$210$}
{$220$}
{$230$}
{$240$}
{$250$}
%72./40^2;
%p:=[[20,20],[30,44],[40,78],[50,123],[60,176]];
%for i from 1 to 5 do print(0.05*(p[i][1])^2); od;
\SOL{12. \STATS{71}{79}{56.09}}{
The graph of the data looks quadratic.
The best model seems to be $y=kx^2$ where $y$ is the skid length in feet
and $x$ is the speed in mph. The first data point gives
$k=44/30^2=0.049$. Testing $y=0.049x^2$ reproduces the table. The
prediction is $y$ at $x=70$, which is $y=0.049(70)^2=240.1$ feet.
Other values of $k$ can be obtained from the other data items, but there
is not much change and the logical answer is $240$.
}

%46.99
\prob{13.}{(Intercept)}% #2, 1994
Find the $y$-intercept of the line through $(5,2)$ and perpendicular to
$5x-7y=17$.

\rightans{e}
\ans
{$-7$}
{$-8$}
{$7$}
{$8$}
{$9$}
\SOL{13. \STATS{93}{57}{53.01}}{
The line is $y-(2)=m(x-5)$ where $-1/m$ is the slope of
$5x-7y=17$. Then $-1/m=5/7$ giving $m=-7/5$. Set $x=0$ in the equation
to get the $y$-intercept $y=2+m(0-5)=9$.}

%%% 68.2
\prob{14.}{(Water Tank)}% #18, Grade 10, #22, 1994, #18, 1996
A tank has a drain rate of $10$\% of capacity per hour. Its fill rate is
$15$\% of capacity per hour. When $2/3$ full and draining, about how
many hours does it take to fill?

\rightans{c}
\ans
{$6\frac13$}
{$6\frac49$}
{$6\frac23$}
{$5\frac23$}
{$5\frac13$}
\SOL{14. \STATS{53}{60}{31.8}}{ The fill time $t$ satisfies $t(0.15x -
0.1x) +2 x/3 = x$
where $x$ is the tank capacity. Cancel $x$ and solve for $t=20/3$.}

%48.9
\prob{15.}{(Complex)}% #1, 1994
Simplify the number $8(1+i)(1-i)^{-6}$.

\rightans{b}
\ans
{$1+i$}
{$1-i$}
{$2+2i)$}
{$8i$}
{$i$}
\SOL{15. \STATS{73}{70}{51.1}}{
First, $(1-i)^2 = 1-2i+i^2=-2i$, so
$(1-i)^6 = (-2i)^3 = -8 i^3=8i$. The answer is $8(1+i)/(8i) =
(i+i^2)/i^2 = 1-i$.}

%50.3
\prob{16.}{(Monotone)}% #12, 1994
On which interval below is $y=x^2-5x+6$ strictly increasing?

\rightans{e}
\ans
{$1 < x < 5$}
{$0 < x < 5$}
{$2 < x < 3$\PAR}
{$2 < x < 4$}
{$3 < x < 5$}
\SOL{16. \STATS{71}{70}{49.7}}{
Factor into the form $y=(x-\frac52)^2 +6 - \frac{25}{4}$. Then $y$
increases
to the right of the vertex point $x=2.5$. }

%51.36
\prob{17.}{(Linear System)}% #10, 1994
Compute the value of $80x$:
$$
\left\{
\begin{array}{rcr} x - 2y - 3z &=& 2\\ x - 4y + 3z&=&a \\ - 3x + 5y
+ 4z &=& 0
\end{array}
\right.
$$

\rightans{c}
\ans
{$5a-70$}
{$-130-25a$}
{$-310-35a$\PAR}
{$1+a$}
{None}
\SOL{17. \STATS{76}{64}{48.64}}{
Cramer's Rule applies. The elimination method applies. The
solution is $80x=-310-35a$, $80y=-130-25a$, $80z=5a-70$.}

%51.76
% 11th
\prob{18.}{(Distance)}% #1,  1995
Find the distance between the points $(-1,4,2)$ and $(1,2,-2)$.

\rightans{e}
\ans
{$2\sqrt{2}$}
{$2\sqrt{3}$}
{$4$}
{$2\sqrt{5}$}
{$2\sqrt{6}$}
\SOL{18. \STATS{67}{72}{48.24}}{
Use the distance formula $\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$ to
obtain $\sqrt{(1-(-1))^2 +(2-4)^2+(-2-2)^2}=2\sqrt6$.
}

%51.81
\prob{19.}{(Sequence)}% #12,  1995
Find the
value of $\dd a_{12}/b_{9}$,
given\newline
$a_n=3200(0.8)^{2n+1}$ and $b_n=1600(0.2)^{3n-2}$.

\rightans{a}
\ans
{$2^{51}$}
{$4^{25}$}
{$4^{26}$}
{$2^{26}$}
{None}
\SOL{19. \STATS{61}{79}{48.19}}{
The values are $a_{12}=3200(0.8)^{25}$ and $b_{9}=1600(0.2)^{25}$.
Then $a_{12}/b_{9}=2(0.8)^{25}/(0.2)^{25}=2(0.8/0.2)^{25}=
2(4^{25})=2^{51}$.
}

%62.56
\prob{20.}{(Lines)}% #21, 1994
Find the distance from the point $(3,5)$ to the line $12y=-5x-3$.

\rightans{b}
\ans
{$5\frac{12}{13}$}
{$6$}
{$6\frac{1}{13}$}
{$6\frac{2}{3}$}
{$6\frac{2}{13}$}
\SOL{20. \STATS{78}{48}{37.44}}{
The standard form of the  line is $5x+12y+3=0$. Using a classical
formula $\sqrt{A^2+B^2} d = Ax_1+By_1 + C$ for the distance $d$ gives
the answer $\sqrt{5^2+12^2}d=5(3)+12(5)+3=78$ or $13d=78$, and finally
$d=6$.
}

%64.6
\prob{21.}{(Asymptote)}% #7, 1994
Find the slant or oblique asymptote for $\dd
y=\frac{2x^3+x^2+x+1}{x^2+1}$.

\rightans{c}
\ans
{$y=x+1$}
{$y=x-1$}
{$y=2x+1$\PAR}
{$y=2x-1$}
{$y=x-2$}
\SOL{21. \STATS{59}{60}{35.4}}{Divide $2x^3+x^2+x+1$ by $x^2+1$ using
long
division. The quotient
is $2x+1$ and the remainder is $-x/(x^2+1)$. The slant asymptote is
$y=2x+1$.}

%68
\prob{22.}{(Roots)}% #6, 1994
Apply the {\em Rational Root Theorem} to select from the list below a
{\em candidate} for a rational root of $7x^9-5x^7+2x^3+210=0$.
%$7x^9-5x^7+2x^3+210=0$.

\rightans{d}
\ans
{$9$}
{$\frac87$}
{$\frac{5}{14}$}
{$-\frac{6}{7}$}
{$11$}
\SOL{22. \STATS{50}{64}{32}}{A possible rational root must be a factor of
$210=(2)(3)(5)(7)$
divided by a factor of $7$. The only one that qualifies from the given
list is $-6$. A candidate is not necessarily a root!}

%%% 78.94
\prob{23.}{(Disk)}% #23,  #28, 1995
Select from the list below a point which is inside the circle of
center $(-1,2)$ and radius 3 and outside the circle of center $(-1,1)$
and radius $2$.

\rightans{e}
\ans
{$(-2,1)$}
{$(-1-\sqrt2,1)$}
{$(-1-\sqrt3,1)$\PAR}
{$(-3,1)$}
{$(-1-\sqrt5,1)$}
\SOL{23. \STATS{54}{39}{21.06}}{
The inequalities are $(x+1)^2+(y-2)^2<9$
and $(x+1)^2+(y-1)^2>4$.
Try the points, one at a time.
}

%77.56
\prob{24.}{(Symmetry)}% #3, 1994
Find the axis of symmetry for the parabola that passes through the
points $(-3,47)$, $(2,7)$ and $(3,23)$.

% y=4x^2-4x-1
\rightans{d}
\ans
{$x=-\frac14$}
{$x=-\frac12$}
{$x=\frac14$}
{$x=\frac12$\PAR}
{$x=\frac34$}
\SOL{24. \STATS{51}{44}{22.44}}{The axis of symmetry is the vertical line through the vertex.
The equation of the parabola is $y=a+b(x+3))+c(x+3)(x-2)$ (any quadratic
will do). Stuff the data points into this equation. Then $a=47$,
$b=-8$ and $c=4$. The equation is
$y=47-8(x+3)+4(x+3)(x-2)$. The vertex is at $x=1/2$.}

%78.54
\prob{25.}{(Partial fractions)}% #15, 1994
Find $A+B+C$:\\ $\dd \frac{x-9}{(x^2-4x+4)(x+3)} =
\frac{A}{(x-2)^2}+\frac{B}{x-2}+\frac{C}{x+3}$.

\rightans{d}
\ans
{$\frac{7}{5}$}
{$\frac{12}{25}$}
{$-\frac{12}{25}$}
{$-\frac{7}{5}$}
{$0$}
\SOL{25. \STATS{37}{58}{21.46}}{
Multiply by $(x-2)^2$ and cancel like terms to get
$(x-9)/(x+3)=A+ B(x-2)+C(x-2)^2/(x+3)$. Let $x=2$. Then $-7/5=A+0+0$
giving $A=-7/5$. Similarly, $C=-12/25$. Substitute $x=0$ to get an
equation for $B$, obtaining $B=12/25$. Then $A+B+C=-7/5$.}

%80.92
\prob{26.}{(Intersect)}% #28, 1994
The graphs $y=4x^3+9x^2+2x+5$ and $y=3x+8$
intersect at three $x$-values
$x=x_1$, $x=x_2$ and $x=x_3$.   Find $x_1+x_2+x_3$ to the nearest
hundredth.

\rightans{c}
\ans
{$-2.27$}
{$-2.26$}
{$-2.25$}
{$-2.24$\PAR}
{$-2.23$}
\SOL{26. \STATS{36}{53}{19.08}}{
The fastest method to solve this problem is to apply the theorem on the
sum of the roots of a polynomial. The polynomial is $4x^3+9x^2+2x+5 -
(3x+8) = 4x^3+9x^2-x-3$. The sum of the roots is $\frac{1}{4}(-1)(9)=
-\frac{9}{4} = -2.25$.}

%84.36
\prob{27.}{(Inverse)}% #8, 1994
Let $f(x)=(x+1)/(x+2)$. Compute the value of the inverse of $f$ at
$y=\sqrt2/(1+\sqrt2)$.

\rightans{c}
\ans
{$\frac{\sqrt2+1}{\sqrt2-1}$}
{$\sqrt2+1$}
{$\sqrt2-1$}
{$\sqrt2$\PAR}
{$-\sqrt2$}
\SOL{27. \STATS{34}{46}{15.64}}{
Let $y=(x+1)/(x+2)$. Solve for $x$ in terms of $y$. Then
$x=(1-2y)/(y-1)$. Let $y=\sqrt2/(1+\sqrt2)$. Then $x=\sqrt{2} - 1$.}

%85.43
\prob{28.}{(Maximum)}% #25, 1994
Find the maximum area for a rectangle of sides $x$ and $y$
subject to the restraint
$y+2x=5$.

\rightans{a}
\ans
{$\frac{25}{8}$}
{$\frac{25}{4}$}
{$\frac{75}{8}$}
{$\frac{25}{2}$}
{$\frac{5}{4}$}
\SOL{28. \STATS{47}{31}{14.57}}{
Solve for $y=5-2x$, insert into the area formula $xy$ to get the
function $f(x)=x(5-2x)$, to be maximized.
The
maximum is at the vertex of the parabola, which is at $x=5/4$. Then
$f(x)=x(5-2x)=(5/4)(5-5/2)=25/8$.}

%92.17
\prob{29.}{(Radius)}% #22, 1994
Find the radius of the inscribed circle to the nearest tenth
for a triangle of sides $12$,
$15$ and $25$.

\rightans{c}
\ans
{$2.0$}
{$2.2$}
{$2.4$}
{$2.6$}
{$2.8$}
\SOL{29. \STATS{29}{27}{7.83}}{
This problem is solved using Heron's formula
$r=\frac{1}{s}\sqrt{s(s-a)s-b)(s-c)}$ where $s=(a+b+c)/2$ is the
half--perimeter and $a$, $b$, $c$ are the sides of the triangle.
In this problem, $r=\frac{1}{26}\sqrt{(26)(14)(11)(1)}=2.43373723$.}

%94.6
\prob{30.}{(Planes)}% #27, 1994
Find the distance to the nearest tenth
from $(24,-5,2)$ to the plane $4x-12y-3z+6=0$.

\rightans{b}
\ans
{$14.4$}
{$12$}
{$10.8$}
{$9$}
{None}
\SOL{30. \STATS{18}{30}{5.4}}{
Apply the standard formula $\sqrt{A^2+B^2+C^2} d = Ax_1+By_1+Cz_1+D$ for
the distance $d$ from a point to a plane.
The result is $\sqrt{4^2+12^2+3^2} \, d = 4(24)-12(-5)-3(2)+6$, which
reduces to $d=156/13=12$.}

\smallskip

See {\em infra} for problems 31-40.

\smallskip

\end{exercises}

%% 12.tex
\begin{center}\Large\bf
1996 Senior Exam Grade 12
\ifKEY with Key \ifsolutions and Solutions \fi \fi
\ifsolutions
\\[1pc]\normalsize
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 12} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 12
\setanswer{1.}{a} & \setanswer{11.}{b} & \setanswer{21.}{c} & \setanswer{31.}{c} \\
\setanswer{2.}{a} & \setanswer{12.}{a} & \setanswer{22.}{b} & \setanswer{32.}{a} \\
\setanswer{3.}{c} & \setanswer{13.}{c} & \setanswer{23.}{a} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{c} & \setanswer{34.}{e} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{b} & \setanswer{35.}{a} \\
\setanswer{6.}{b} & \setanswer{16.}{d} & \setanswer{26.}{d} & \setanswer{36.}{a} \\
\setanswer{7.}{a} & \setanswer{17.}{d} & \setanswer{27.}{c} & \setanswer{37.}{b} \\
\setanswer{8.}{e} & \setanswer{18.}{c} & \setanswer{28.}{e} & \setanswer{38.}{d} \\
\setanswer{9.}{a} & \setanswer{19.}{c} & \setanswer{29.}{d} & \setanswer{39.}{c} \\
\setanswer{10.}{d} & \setanswer{20.}{a} & \setanswer{30.}{a} & \setanswer{40.}{d} \\
\hline
\end{tabular}
\medskip
\fi
\end{center}

{\bf Instructions}: Only one answer is correct. Scoring: 5(\#
right)+1(\# blank)+0(\# wrong). Time: 120 minutes.


\begin{exercises}

%92.8
\def\AA{
\prob{1.}{(Photo)}% #8,  9th 1995
A square picture frame is to be designed as in the figure so that
$\frac{2}{3}$ of the total area is used by the inner frame. Find
the area of the inner frame in square units.
}
\def\BB{\input{photo8.box}}
\WRAPFIG{\AA}{\BB}{1.1in}

\rightans{a}
\ans
{$40+16\sqrt{6}$}
{$40+23\sqrt{3}$}
{$40+28\sqrt{2}$\PAR}
{$33+19\sqrt{6}$}
{None}

\SOL{1. \STATS{30}{24}{7.2}}{
The inner square has area $x^2$ and the outer one has area $(x+2)^2$,
therefore $\frac23(x+2)^2=x^2$. Solving for $x=4+2\sqrt{6}$ results in
inner area $x^2=40+16\sqrt{6}$.
}

%% 12.tex


%17.2
\prob{2.}{(Series)}% #2,   1995
Find the sum the infinite series\newline
\centerline{$0.002+0.0002+0.00002+\cdots$}

\rightans{a}
\ans
{$\frac{2}{900}$}
{$\frac{1}{300}$}
{$\frac{5}{900}$}
{$\frac{4}{900}$}
{None}
\SOL{2. \STATS{90}{92}{82.8}}{
The series is geometric with common ratio $r$.
Write it as $0.002(1+r+r^2+\cdots)$ where $r=0.1$. Then the
sum is $0.002/(1-r)=0.002(10/9)=2/900$.
}

%28.8
\prob{3.}{(Max-Min)}% #25,  1994
Find the maximum value of $y=16x^2-x^3$ on $0 \le x <
\infty$ to the nearest integer.

\rightans{c}
\ans
{$598$}
{$605$}
{$607$}
{$609$}
{$611$}
\SOL{3. \STATS{80}{89}{71.2}}{The function $y$ takes on positive values and is negative for
large values of $x$. Therefore, its maximum is assumed on some closed
interval $[0,R]$. If $R$ is large, then $y$ is negative at $x=R$ and
$y$ is zero at $x=0$, so the maximum must be in $(0,R)$. A
maximum occurs at a critical point of $y$. The derivative is
$y'=32x-3x^2=x(32-3x)$. The interior critical point is $x=32/3$. The
maximum value at $x=32/3$ is $y=(32/3)^2(16-32/3)=\frac{2^{14}}{3^3}=606.81$.
A graphing calculator, properly used, can zero on the value without
using calculus. This kind of problem was nearly the most successful 12th
grade problem in 1995!
}

%30.65
\prob{4.}{(Lines)}% #1,  1994
Along the line through $(3,4)$ and parallel to $5x+4y=5$, find the value
of $y$ when $x=-3$.

\rightans{d}
\ans
{$9$}
{$10$}
{$11$}
{$23/2$}
{None}
\SOL{4. \STATS{95}{73}{69.35}}{A parallel line is $5x+4y=c$ for any $c$.
To insure $(3,4)$ on
this line, take $c=5(3)+4(4)=31$. Then at $x=-3$, $4y=c-5x=31-5(-3)=46$
and finally $y=23/2$.}

%32
\prob{5.}{(Conics)}% #2,  1994
Find the center of the circle whose equation is $x^2+y^2 +2x + 8y = 8$.

\rightans{d}
\ans
{$(1,-4)$}
{$(1,4)$}
{$(-1,4)$}
{$(-1,-4)$\PAR}
{$(2,-4)$}
\SOL{5. \STATS{85}{80}{68}}{Complete the square and write as
$(x+1)^2+(y+4)^2=8+1+16$.
Then the center is $(-1,-4)$. Alternatively, using calculus, the center
is where the partial derivatives in $x$ and $y$ vanish.}

%32.5
\prob{6.}{(Area)}% #26,  1994
Find the area bounded by $x=0$, $x=1$, $y=0$ and $y=x^2(1+x^2)$.

\rightans{b}
\ans
{$\frac{7}{15}$}
{$\frac{8}{15}$}
{$\frac{3}{5}$}
{$\frac{2}{3}$}
{$\frac{11}{15}$}
\SOL{6. \STATS{75}{90}{67.5}}{The function $x^2(1+x^2)$ is positive on
$[0,1]$. So the area
is $\int_0^1 x^2(1+x^2)dx = [x^3/3 + x^5/5] |_0^1 =
\frac13+\frac15=8/15$.

}

%51.81
\prob{7.}{(Sequence)}% #6,   #12,  1995
Find the
value of $\dd a_{12}/b_{9}$,
given\newline
$a_n=3200(0.8)^{2n+1}$ and $b_n=1600(0.2)^{3n-2}$.

\rightans{a}
\ans
{$2^{51}$}
{$4^{25}$}
{$4^{26}$}
{$2^{26}$}
{None}
\SOL{7. \STATS{83}{80}{66.4}}{
The values are $a_{12}=3200(0.8)^{25}$ and $b_{9}=1600(0.2)^{25}$.
Then $a_{12}/b_{9}=2(0.8)^{25}/(0.2)^{25}=2(0.8/0.2)^{25}=
2(4^{25})=2^{51}$.
}

%36.86
\prob{8.}{(Minima)}% #25,   1995
Determine a value of $x$ for which the cubic $y-1=5x^3-x^2$ has a
local minimum.

\rightans{e}
\ans
{$\frac{1}{15}$}
{$\frac{4}{15}$}
{$\frac{1}{3}$}
{$0$}
{None}
\SOL{8. \STATS{82}{77}{63.14}}{
Calculus method: the critical points are roots of the quadratic
$15x^2-2x$ ($x=0$ and $x=2/15$ are critical points).
\par
A graphical method, without calculus: the graph suggests a relative
minimum near $2/15$.
}

%46.72
\prob{9.}{(Sequences)}% #10,  1994
The first four terms of an arithmetic sequence are $1$, $a$, $b$,
$16$. Find term $47$.

\rightans{a}
\ans
{$231$}
{$235$}
{$239$}
{$243$}
{$247$}
\SOL{9. \STATS{74}{72}{53.28}}{
The common difference $x$ satisfies $x=a-1=b-a=16-b$. Adding
these gives $3x=(a-1)+(b-a)+(16-b)=15$ or $x=5$. The terms are $1$,
$6$, $11$, $16$, \ldots. The $n^{\mbox{th}}$ term is $x_n = 1+5(n-1)$ so
$x_{47} = 1+5(46)=231$.}

%47.56
\prob{10.}{(Tangent)}% #27,  1994
Find the slope of the line tangent at $x=e^{-3}$ to the curve $y=x\ln(x^2)$.

\rightans{d}
\ans
{$-1$}
{$-2$}
{$-3$}
{$-4$}
{None}
\SOL{10. \STATS{69}{76}{52.44}}{
The derivative is $y'=2(1+\ln(x))=2+2\ln(e^{-3})=2+2(-3)\ln(e)=-4$.}

%% 11 so far...

%62.56
\prob{11.}{(Normal)}% #24,   1995
A function $f(x)$ has slope $-1/5$ at $(1,2)$. Find the $y$-intercept
of the normal line at $(1,2)$.

\rightans{b}
\ans
{$3$}
{$-3$}
{$2$}
{$-2$}
{$1$}
\SOL{11. \STATS{72}{52}{37.44}}{
The tangent line is given by $y=2-(x-1)/5$.  The normal line is given by
$y=2+5(x-1)$, because the slope of the normal line is the negative
reciprocal of the slope of the tangent line. The $y$-intercept is the
solution of the equation $y=2+5(0-1)$. Then $y=-3$.
}

%66.4
\prob{12.}{(Quadratics)}% #22,  1994
Given real $a$ and $b$, which inequality implies $x^2+bx+4a=0$ has a
complex root?

\rightans{a}
\ans
{$b^2<16a$}
{$b^2< 4a$}
{$a>4b^2$}
{$b^2>4a$\PAR}
{$a^2<4b$}
\SOL{12. \STATS{56}{60}{33.6}}{Let $A=1$, $B=b$, $C=4a$ in
$Ax^2+Bx+C=0$.
The roots are
complex if and only if the discriminant $B^2-4AC = b^2-4(1)(2a)$ is
negative. Then $b^2 < 16a$.}

%68.65
\prob{13.}{(Ellipse)}% #19,   1995
Find an axis of symmetry of the ellipse\newline
 $4x^2+25y^2+250y+525=0$.

\rightans{c}
\ans
{$y=5$}
{$x=5$}
{$y=-5$}
{$x=-5$\PAR}
{$y=0$}
\SOL{13. \STATS{57}{55}{31.35}}{
Complete the square to write the equation in the equivalent form
$4x^2+25(y^2+10y+25)-25^2+525$ and finally as $4x^2+25(y+5)^2=100$.
The center is $(0,-5)$. Axes of symmetry are $x=0$ and $y=-5$.
}

%69.75
\prob{14.}{(Arcsine)}% #17,  1994
Simplify $\cos({\rm Sin}^{-1} (x/2))$ for $|x| \le 2$.

\rightans{c}
\ans
{$x/2$}
{$1-x^2/4$}
{$\sqrt{1-x^2/4}$\PAR}
{$4/(4+x^2)$}
{None}
\SOL{14. \STATS{55}{55}{30.25}}{
The principal arcsine is defined on $[-1,1]$ and takes
values in $[-\pi/2,\pi/2]$. Let $\theta={\rm Sin}^{-1}(x/2)$. Then
$\sin\theta
=
x/2$ and $\theta$ between $-\pi/2$ and $\pi/2$ implies the cosine is
positive
and
$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-(x/2)^2}$.
}

%70
\prob{15.}{(Derivative)}% #29,  1994
Let $f$ be defined on $(-\infty,\infty)$ and differentiable with
$f'(3)=\pi$ and $f(3)=1$. Find
$\frac{d}{dx}[f(x)f(x^2-6)]$ at $x=3$.

\rightans{d}
\ans
{$4\pi$}
{$5\pi$}
{$6\pi$}
{$7\pi$}
{None}
\SOL{15. \STATS{60}{50}{30}}{Let $z=f(x)f(x^2-6)$. By the product rule
it follows that
$z' = f'(x)f(x^2-6) + f(x)[f(x^2-6)]'$. By the chain rule,
$[f(x^2-6)]' = f'(x^2-6)(2x)$. At
$x=3$, the value is $z' = f'(3)f(3) + f(3)f'(3)(2)(3) =
\pi + 6\pi = 7\pi$.
}

%74.56
\prob{16.}{(Inequalities)}% #38,  1994
Find the maximum of $4x+5y+3$ subject to the inequalities $x\ge0$,
$y\ge0$, $x+y\le 3$ and $3x+y\le 6$.

\rightans{d}
\ans
{$11$}
{$12$}
{$16$}
{$18$}
{$19$}
\SOL{16. \STATS{12th=54}{50}{27}}{
The function $z=4x+5y+3$ is a plane in $R^3$. The inequalities
describe a polygon in the plane. The maximum occurs at a vertex of
the polygon: $(0,0)$, $(2,0)$, $(3/2,3/2)$, $(0,3)$. The values of $z$
are $3$, $11$, $16.5$, $18$, respectively. The maximum of $z$ is
$18$. }


%78.54
\prob{17.}{(Partial fractions)}% #15, 1994
Find $A+B+C$:\\ $\dd \frac{x-9}{(x^2-4x+4)(x+3)} =
\frac{A}{(x-2)^2}+\frac{B}{x-2}+\frac{C}{x+3}$.

\rightans{d}
\ans
{$\frac{7}{5}$}
{$\frac{12}{25}$}
{$-\frac{12}{25}$}
{$-\frac{7}{5}$}
{$0$}
\SOL{17. \STATS{37}{58}{21.46}}{
Multiply by $(x-2)^2$ and cancel like terms to get
$(x-9)/(x+3)=A+ B(x-2)+C(x-2)^2/(x+3)$. Let $x=2$. Then $-7/5=A+0+0$
giving $A=-7/5$. Similarly, $C=-12/25$. Substitute $x=0$ to get an
equation for $B$, obtaining $B=12/25$. Then $A+B+C=-7/5$.}

%80.92
\prob{18.}{(Intersect)}% #28, 1994
The graphs $y=4x^3+9x^2+2x+5$ and $y=3x+8$
intersect at three $x$-values
$x=x_1$, $x=x_2$ and $x=x_3$.   Find $x_1+x_2+x_3$ to the nearest
hundredth.

\rightans{c}
\ans
{$-2.27$}
{$-2.26$}
{$-2.25$}
{$-2.24$\PAR}
{$-2.23$}
\SOL{18. \STATS{36}{53}{19.08}}{
The fastest method to solve this problem is to apply the theorem on the
sum of the roots of a polynomial. The polynomial is $4x^3+9x^2+2x+5 -
(3x+8) = 4x^3+9x^2-x-3$. The sum of the roots is $\frac{1}{4}(-1)(9)=
-\frac{9}{4} = -2.25$.}

%84.36
\prob{19.}{(Inverse)}% #8, 1994
Let $f(x)=(x+1)/(x+2)$. Compute the value of the inverse of $f$ at
$y=\sqrt2/(1+\sqrt2)$.

\rightans{c}
\ans
{$\frac{\sqrt2+1}{\sqrt2-1}$}
{$\sqrt2+1$}
{$\sqrt2-1$}
{$\sqrt2$\PAR}
{$-\sqrt2$}
\SOL{19. \STATS{34}{46}{15.64}}{
Let $y=(x+1)/(x+2)$. Solve for $x$ in terms of $y$. Then
$x=(1-2y)/(y-1)$. Let $y=\sqrt2/(1+\sqrt2)$. Then $x=\sqrt{2} - 1$.}

%85.43
\prob{20.}{(Maximum)}% #25, 1994
Find the maximum area for a rectangle of sides $x$ and $y$
subject to the restraint
$y+2x=5$.

\rightans{a}
\ans
{$\frac{25}{8}$}
{$\frac{25}{4}$}
{$\frac{75}{8}$}
{$\frac{25}{2}$}
{$\frac{5}{4}$}
\SOL{20. \STATS{47}{31}{14.57}}{
Solve for $y=5-2x$, insert into the area formula $xy$ to get the
function $f(x)=x(5-2x)$, to be maximized.
The
maximum is at the vertex of the parabola, which is at $x=5/4$. Then
$f(x)=x(5-2x)=(5/4)(5-5/2)=25/8$.}

%92.17
\prob{21.}{(Radius)}% #22, 1994
Find the radius to the nearest tenth
of the inscribed circle for a triangle of sides $12$,
$15$ and $25$.

\rightans{c}
\ans
{$2.0$}
{$2.2$}
{$2.4$}
{$2.6$}
{$2.8$}
\SOL{21. \STATS{29}{27}{7.83}}{
This problem is solved using Heron's formula
$r=\frac{1}{s}\sqrt{s(s-a)s-b)(s-c)}$ where $s=(a+b+c)/2$ is the
half--perimeter and $a$, $b$, $c$ are the sides of the triangle.
In this problem, $r=\frac{1}{26}\sqrt{(26)(14)(11)(1)}=2.43373723$.}

%94.6
\prob{22.}{(Planes)}% #27, 1994
Find the distance from $(24,-5,2)$ to the plane $4x-12y-3z+6=0$, to the
nearest tenth.

\rightans{b}
\ans
{$14.4$}
{$12$}
{$10.8$}
{$9$}
{None}
\SOL{22. \STATS{18}{30}{5.4}}{
Apply the standard formula $\sqrt{A^2+B^2+C^2} d = Ax_1+By_1+Cz_1+D$ for
the distance $d$ from a point to a plane.
The result is $\sqrt{4^2+12^2+3^2} \, d = 4(24)-12(-5)-3(2)+6$, which
reduces to $d=156/13=12$.}


%82.36
\prob{23.}{(Asymptotes)}% #5,  1994
The curve $2xy=x^2-6x+4y+10$ has slant (oblique) and vertical
asymptotes. One is $y=x/2-2$. Find the other.

\rightans{a}
\ans
{$x=2$}
{$y=2$}
{$y=x/2+1$\PAR}
{$y=2x$}
{None}
\SOL{23. \STATS{42}{42}{17.64}}{In normal form, $y=(x^2-6x+10)/(2x-4) =
x/2 - 2 + 2/(2x-4)$.
The vertical asymptote is $x=2$ and the slant asymptote is $y=x/2-2$.}

%86.5
\prob{24.}{(Cofactor)}% #7,  1994
Find the cofactor (signed minor) of row $3$ and column $2$ in the
determinant $\left|\begin{array}{rrr}
a & 0 & b\\
b & a & 0\\
0 & b & a
\end{array} \right|$.

\rightans{c}
\ans
{$ab$}
{$-b^2$}
{$b^2$}
{$a^2$}
{$a^3+b^3$}
\SOL{24. \STATS{54}{25}{13.5}}{The minor is
$\left|\begin{array}{rr}
a & b\\
b & 0 \end{array} \right| = -b^2$. The
sign is $(-1)^{3+2} = -1$. The cofactor is $(-1)(-b^2)$.}

%86.95
\prob{25.}{(Integral)}% #28,   1995
Let $f$ be the unique quadratic polynomial which passes through $(0,1)$,
$(1,-1)$ and $(2,3)$. Find $\int_0^2 |f(x)|^2\,dx$.

\rightans{b}
\ans
{$\frac{31}{15}$}
{$\frac{34}{15}$}
{$\frac{7}{3}$}
{$\frac{8}{3}$}
{None}
\SOL{25. \STATS{29}{45}{13.05}}{
The polynomial is $y=3x^2-5x+1$.
To find this polynomial, let $f(x)=a+bx+cx(x-1)$ and set up three
equations in three unknowns $a$, $b$, $c$ using the given data points.
Solve for $a$, $b$, $c$ by elimination or Cramer's rule.
The integral is $\int_0^2 (3x^2-5x+1)^2 dx =
(9/5)x^5-(15/2)x^4+(31/3)x^3-5x^2+x |_0^2 = 34/15$.
}

%97.28
\prob{26.}{(Series)}% #43,   #21,   1996
Sum the series
 $\dd\sum_{n=1}^\infty 5^{-2n+2} \, 4^{2n-3}$.

\rightans{d}
\ans
{$\frac{25}{16}$}
{$\frac{25}{18}$}
{$\frac{25}{24}$}
{$\frac{25}{36}$}
{None}
\SOL{26. \STATS{0}{0}{0}}{
Write the term as $(5^2)(4^{-3})5^{-2n}\, 4^{2n}
=c(16/25)^n$ where $c=5^2/4^3$. Then the summation, by geometric series
formulas, is $c(16/25)/(1-(16/25)=16c/9=25/36$.
}


%91.6
\prob{27.}{(Inverse)}% #10,   1995
Find the determinant of $2A^{-2}+A^{-3}$, given\par
\centerline{$
A=\left (\begin {array}{rrr}
1  & 2 & 1 \\\noalign{\medskip}
1  & 1 & 1 \\\noalign{\medskip}
2  &-x &-1 \end {array}
\right )
$}


% x:='x':A:=matrix([[1,2,1],[1,1,1],[2,-x,-1]]);
\rightans{c}
\ans
{$3$}
{$x$}
{$\frac{4x+7}{27}$}
{$\frac{4x+7}{9}$}
{$\frac{4x+7}{3}$}

\SOL{27. \STATS{28}{30}{8.4}}{
The determinant rule $\det(AB)=\det(A)\det(B)$ can be applied to
$2A^{-2}+A^{-3}=A^{-3}(I+2A)$ to give the answer as the product of
$\det(A^{-3})$ and $\det(I+2A)$. The first factor is
$1/(\det(A))^3=1/27$ because of $\det(A)=3$ and determinant rules.
The second factor is a direct evaluation,
$$
I+2A=\left (\begin {array}{rrr}
3  & 4 & 2 \\\noalign{\medskip}
2  & 3 & 2 \\\noalign{\medskip}         %3(-3+4x) -2(-4+4x) + 4(8-6)
4  &-2x &-1 \end {array}                %-9+12x+ 8-8x+ 8 = 4x+7
\right )
$$
which has value $4x+7$.
The product of the factors is $(4x+7)/27$.
}

%97
\prob{28.}{(Triangle)}% #25,  9th 1994
Sides $a$ and $b$ of a triangle satisfy the equation $6a^2+7ab+6b^2=30$. Side $c$ is
divided by the bisector of the opposite angle in the ratio $3$:$2$.
Find the smaller of $a$ and $b$.

\rightans{e}
\ans
{$\sqrt{7/6}$}
{$\sqrt{7/5}$}
{$3/2$}
{$2/3$}
{$1$}
\SOL{28. \STATS{20}{15}{3}}{The relation $a/b=3/2$ is given by the
bisector hypothesis.
Solve for $a=3b/2$ and insert into the quadratic equation to get
$30b^2=30$. Then $b=1$ and $a=3/2$.}





%94.06
\prob{29.}{(Probability)}% #21,  1994
A coin is tossed three times. Find the probability that at most two
heads show given that at least one head shows.

\rightans{d}
\ans
{$3/8$}
{$5/8$}
{$5/7$}
{$6/7$}
{$3/4$}
\SOL{29. \STATS{54}{11}{5.94}}{This can be solved using conditional probability
$P(A/B)=P(A\mbox{~and~}B)/P(B)$. A simpler method is to use the reduced
sample space of events $htt$, $tht$, $tth$, $hht$, $hth$, $thh$, $hhh$
consisting of all events with at least one head. The probability is
then the number of events with one or two $h$'s divided by the total
number of events: $6/7$.}

%97.04
\prob{30.}{(Skeet)}% #32,  1994
Jack normally hits $4$ skeets out of $6$ attempts. Find the
probability that he hits exactly $2$ in $4$ attempts.

\rightans{a}
\ans
{$24/81$}
{$(2/3)^4$}
{$32/81$}
{$\frac{6!}{4!2!}\frac{4}{81}$\PAR}
{None}
\SOL{30. \STATS{12th=35}{21}{7.35}}{
By advanced textbook methods the probability is $Ca^kb^{n-k}$
where $n=4$, $k=2$, $C=\choose{n}{k}=6$, $a=4/6$ and $b=2/6$. See
Merrill Algebra 2 (1992), p 720. The answer is
$(6)(4/6)^2(2/6)^2=24/81$. An elementary solution starts with the
probability of a successful single trial of two hits and two misses. The
probability of success is $2/3$ and of failure $1/3$ ($4$ hits out $6$).
Two hits and two misses has probability
$\frac23\frac23\frac13\frac13=4/81$. There are $6$ ways for two hits and
two misses: $HHMM$, $HMHM$, $HMMH$, $MHHM$, $MHMH$, $MMHH$. Adding the
probabilities of each single event gives $24/81$. }

% Top 10!
\end{exercises}

\fbox{
\begin{minipage}{3.35in}
\large\sf
Problems 31--40 have a higher weight in the
Senior competition between grades 10, 11 and 12.
\end{minipage}
}

\begin{exercises}

%88.39
\prob{31.}{(Polynomials)}% #29,  9th 1995
Let $p$ be a cubic polynomial in $x$ with roots $-2$, $2$ and $5$.
Assume $p=20$ when $x=0$. Find $p$ when $x=1$.

\rightans{c}
\ans
{$10$}
{$11$}
{$12$}
{$13$}
{$14$}
\SOL{31. \STATS{27}{43}{11.61}}{
By the factor theorem, $(x+2)$, $(x-2)$ and $(x-5)$ are factors of $p$.
The polynomial $p$ must be a constant multiple of $(x+2)(x-2)(x-5)$.
Briefly, this means $p=c(x+2)(x-2)(x-5)$ for some constant $c$. In
expanded form, $p=c(x^3-5x^2-4x+20)$. Since $p=20$ at $x=0$, then $c=1$
and $p=x^3-5x^2-4x+20$. The value of $p$ at $x=1$ is $1-5-4+20=12$.
}

\prob{32.}{(Missile)}% #28,   1996
A missile hits its target with probability $0.3$. Find the least number
of missiles to be launched to insure an $80$\% probability of hitting the
target.

\rightans{a}
\ans
{$5$}
{$6$}
{$8$}
{$9$}
{None}

\SOL{32. \STATS{0}{0}{0}}{
The missile misses with probability $0.7$, so $n$ missiles miss the
target with probability $0.7^n$. To find the least $n$, solve the
inequality $1-0.7^n > 0.8$ by trial--and--error. The least one is $n=5$.
}


%97.28
\prob{33.}{(Perimeter)}% #21,   1996
Find the perimeter of a right triangle having an inscribed circle of
diameter $1$ and one leg of $7/5$ units.

\rightans{a}
\ans
{$6.3$}
{$\frac{19}{3}$}
{$\frac{59}{9}$}
{$6$}
{None}
\SOL{33. \STATS{0}{0}{0}}{
Let $r=1/2$ be the radius of the circle. Let $a=7/5$ be the known leg.
The circle's contact points divide the triangle's sides into segments of
lengths $r$, $x$; $x$, $a-r$; $a-r$, $r$, where $x$ is unknown. An
equation for $x$ is obtained by finding the area in two ways: half the
base times the altitude, and half the perimeter times $r$. This equality
is $\frac12a(r+x)=(a+x)r$, hence $x=ar/(a-2r)$. The triangle perimeter
is $2r+2x+2(a-r)=2a(a-r)/(a-2r)=63/10$.
}


%97.28
\prob{34.}{(Shells)}% #21,   1996
Two spherical surfaces of radii $13$ and $15$ have centers $14$ apart.
Find the radius of the circle of intersection.

\rightans{e}
\ans
{$10\frac37$}
{$11\frac47$}
{$12\frac17$}
{$12\frac27$}
{None}
\SOL{34. \STATS{0}{0}{0}}{
Let $r$ be the unknown radius. The circle is in a plane at distance $x$
from the center of the smaller sphere, and at distance $14-x$ from the
center of the larger sphere. The Pythagorean Theorem implies that
$x^2+r^2=13^2$ and also $(14-x)^2 + r^2 = 15^2$. Subtract these two
equations to isolate $x=140/28=5$. Then $r^2=13^2-x^2=144$ and $r=12$.
}

%97.28
\prob{35.}{(Cubic)}% #21,   1996
Find the sum of the reciprocals of the roots of $3x^3+2x^2-5x+2=0$.

\rightans{a}
\ans
{$\frac{5}{2}$}
{$\frac{5}{3}$}
{$-\frac{5}{3}$}
{$\frac{2}{3}$}
{$-\frac{2}{3}$}

\SOL{35. \STATS{0}{0}{0}}{
Replace $x$ by $1/x$ to get $2x^3-5x^2+2x+3=0$. Then $-5/2$ equals the
negative of the sum of the roots, giving answer $5/2$.
}

%97.28
\prob{36.}{(Ratio)}% #21,   1996
A parallelogram of altitude $30$ and with diagonals $34$ and $78$ has
area $A$. There is another parallelogram with the same altitude and
diagonals having area $B>A$. Find $A/B$.

\rightans{a}
\ans
{$\frac{7}{11}$}
{$\frac{2}{3}$}
{$\frac{9}{13}$}
{$\frac{5}{7}$}
{$\frac{11}{15}$}
\SOL{36. \STATS{0}{0}{0}}{
Draw a figure for each. One has base $44$, the other has base $28$.
The areas are then $base\times altitude$, $A=(28)(30)$ and $B=(44)(30)$.
The ratio $A/B$ is $28/44$ or $7/11$.
}

%97.28
\def\AA{
\prob{37.}{(Yardstick)}% #21,   1996
A yardstick contacts a large tomato tin and a small tomato paste tin,
as in the figure. Find the distance between the contact points, given
the tins have diameters $20$ and $4$, with centers $17$ apart.
}
\def\BB{\input{yard.box}}
\WRAPFIG{\AA}{\BB}{1.3in}

\rightans{b}
\ans
{$\frac{29}{8}\sqrt{17}$}
{$15$}
{$\frac{15}{4}\sqrt{17}$}
{$16$}
{None}


\SOL{37. \STATS{0}{0}{0}}{
The contact points are along an external tangent to the two circles.
Draw a line parallel to this tangent and passing through the center of
the small circle. Complete the sides of a rectangle of dimensions $2$
and $x$, with parallel sides along radii of the two circles. Isolate
from this figure a right triangle of sides $x$, $8$, $17$. By the
Pythagorean Theorem, $x=15$.

}

%78.92
\def\AA{
\prob{38.}{(Tangents)}% #25,   #29, 9th (altered) 1994
Given segment $\overline{AB}$ tangent at $A$, $AB=9x-22$, $BC=7x-18$ and
$DC=2x$, find the sum of all possible solutions $x$.

\rightans{d}
\ans
{$\frac{8}{3}$}
{$4$}
{$\frac{16}{3}$}
{$6$\PAR}
{None}
}
\def\BB{\input{tan9a.box}}
\WRAPFIG{\AA}{\BB}{1.25in}

\SOL{38. \STATS{9th=34}{62}{21.08}}{{\bf Theorem}: {\em If a tangent
segment and a secant segment
are drawn to a circle from an exterior point, then the square of the
measure of the tangent segment is equal to the product of the measures
of the secant segment and its external secant segment}. See Merrill
Geometry (1993), p 449. Applying the theorem gives
$(9x-22)^2=(7x-18)(9x-18)$. Taking roots gives $x=10/3$ and $x=8/3$.
Both are solutions (neither is extraneous).
Therefore the sum of the roots is $6$.}

%97.28
\prob{39.}{(Urn)}% #21,   1996
Urn A has $2$ white and $4$ red balls; Urn B has $5$ white and $3$ red.
A ball is randomly chosen from urn A, deposited in urn B, then a ball is
selected randomly from urn B. Given that a white ball was selected from
B, find the probability that the ball transferred from A was white.

\rightans{c}
\ans
{$\frac12$}
{$\frac13$}
{$\frac38$}
{$\frac49$}
{None}
\SOL{39. \STATS{68}{04}{2.72}}{
The scheme uses conditional probability calculations, as follows:
$P(B_1=W|B_2=W) = r/(s+t)$, where
$r=P(B_1=W)P(B_2=W|B_1=W)$,
$s=P(B_1=R)P(B_2=W|B_1=R)$,
$t=P(B_1=W)P(B_2=W|B_1=W)$.
Then $r=t=(1/3)(2/3)$ and $s=(2/3)(5/9)$. The probability is
$r/(s+t)=3/8$.
}

%97.28
\prob{40.}{(Median)}% #21,   1996
A scalene triangle of area $15\sqrt{11}$ has medians $9$ and $15$.
Determine the unknown median.

\rightans{d}
\ans
{$9$}
{$24$}
{$6\sqrt{15}$}
{$3\sqrt{59}$}
{$3\sqrt{57}$}
\SOL{40. \STATS{0}{0}{0}}{
Let $M$ be the intersection point of the medians in triangle $ABC$.
Construct a second triangle $ACD$. Label sides $AD=6$, $DC=10$ and draw
the medians through $M$, with segment lengths $3$, $6$, $5$, $10$, $y$,
$2y$, with $y$ unknown. Then $DM$ is also divided into segments of
lengths $y$ and $y$. The area of triangle $AMC$ is $1/3$ of the area of
triangle $ABC$, making $5\sqrt{11}$. Apply Heron's area formula for
triangles to obtain the second equation
$5\sqrt{11}=\sqrt{(y-2)(y+2)(8-y)(8+y)}$. This produces the clever
equation $(y^2-59)(y^2-9)=0$ for unknown $y$. The unknown median has
length $3y$, so $9$ and $3\sqrt{59}$ are the candidates. The value $9$
is impossible because the triangle is scalene.
}

\end{exercises}
\end{multicols}
\eject

\begin{center}\LARGE\bf
1996 Utah State Math Contest Answer Keys
\end{center}

\begin{multicols}{2}
\large
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 7} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 7
%\setanswer{1.}{a}
%\setanswer{2.}{d}
%\setanswer{3.}{e}
%\setanswer{4.}{a}
%\setanswer{5.}{d}
%\setanswer{6.}{b}
%\setanswer{7.}{b}
%\setanswer{8.}{c}
%\setanswer{9.}{b}
%\setanswer{10.}{c}
%\setanswer{11.}{a}
%\setanswer{12.}{a}
%\setanswer{13.}{d}
%\setanswer{14.}{a}
%\setanswer{15.}{d}
%\setanswer{16.}{c}
%\setanswer{17.}{c}
%\setanswer{18.}{b}
%\setanswer{19.}{c}
%\setanswer{20.}{b}
%\setanswer{21.}{a}
%\setanswer{22.}{a}
%\setanswer{23.}{e}
%\setanswer{24.}{a}
%\setanswer{25.}{c}
%\setanswer{26.}{d}
%\setanswer{27.}{b}
%\setanswer{28.}{a}
%\setanswer{29.}{c}
%\setanswer{30.}{c}
%\setanswer{31.}{a}
%\setanswer{32.}{b}
%\setanswer{33.}{a}
%\setanswer{34.}{d}
%\setanswer{35.}{c}
%\setanswer{36.}{d}
%\setanswer{37.}{e}
%\setanswer{38.}{b}
%\setanswer{39.}{d}
%\setanswer{40.}{a}
\setanswer{1.}{a} & \setanswer{11.}{a} & \setanswer{21.}{a} & \setanswer{31.}{a} \\
\setanswer{2.}{d} & \setanswer{12.}{a} & \setanswer{22.}{a} & \setanswer{32.}{b} \\
\setanswer{3.}{e} & \setanswer{13.}{d} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{a} & \setanswer{14.}{a} & \setanswer{24.}{a} & \setanswer{34.}{d} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{c} & \setanswer{35.}{c} \\
\setanswer{6.}{b} & \setanswer{16.}{c} & \setanswer{26.}{d} & \setanswer{36.}{d} \\
\setanswer{7.}{b} & \setanswer{17.}{c} & \setanswer{27.}{b} & \setanswer{37.}{e} \\
\setanswer{8.}{c} & \setanswer{18.}{b} & \setanswer{28.}{a} & \setanswer{38.}{b} \\
\setanswer{9.}{b} & \setanswer{19.}{c} & \setanswer{29.}{c} & \setanswer{39.}{d} \\
\setanswer{10.}{c} & \setanswer{20.}{b} & \setanswer{30.}{c} & \setanswer{40.}{a} \\
\hline
\end{tabular}

\medskip

\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 8} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 8
%\setanswer{1.}{a}
%\setanswer{2.}{d}
%\setanswer{3.}{a}
%\setanswer{4.}{d}
%\setanswer{5.}{c}
%\setanswer{6.}{d}
%\setanswer{7.}{a}
%\setanswer{8.}{b}
%\setanswer{9.}{e}
%\setanswer{10.}{b}
%\setanswer{11.}{e}
%\setanswer{12.}{e}
%\setanswer{13.}{b}
%\setanswer{14.}{c}
%\setanswer{15.}{c}
%\setanswer{16.}{c}
%\setanswer{17.}{a}
%\setanswer{18.}{a}
%\setanswer{19.}{c}
%\setanswer{20.}{e}
%\setanswer{21.}{b}
%\setanswer{22.}{b}
%\setanswer{23.}{d}
%\setanswer{24.}{c}
%\setanswer{25.}{c}
%\setanswer{26.}{c}
%\setanswer{27.}{d}
%\setanswer{28.}{d}
%\setanswer{29.}{d}
%\setanswer{30.}{d}
%\setanswer{31.}{a}
%\setanswer{32.}{b}
%\setanswer{33.}{a}
%\setanswer{34.}{d}
%\setanswer{35.}{c}
%\setanswer{36.}{d}
%\setanswer{37.}{e}
%\setanswer{38.}{b}
%\setanswer{39.}{d}
%\setanswer{40.}{a}
\setanswer{1.}{a} & \setanswer{11.}{e} & \setanswer{21.}{b} & \setanswer{31.}{a} \\
\setanswer{2.}{d} & \setanswer{12.}{e} & \setanswer{22.}{b} & \setanswer{32.}{b} \\
\setanswer{3.}{a} & \setanswer{13.}{b} & \setanswer{23.}{d} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{c} & \setanswer{34.}{d} \\
\setanswer{5.}{c} & \setanswer{15.}{c} & \setanswer{25.}{c} & \setanswer{35.}{c} \\
\setanswer{6.}{d} & \setanswer{16.}{c} & \setanswer{26.}{c} & \setanswer{36.}{d} \\
\setanswer{7.}{a} & \setanswer{17.}{a} & \setanswer{27.}{d} & \setanswer{37.}{e} \\
\setanswer{8.}{b} & \setanswer{18.}{a} & \setanswer{28.}{d} & \setanswer{38.}{b} \\
\setanswer{9.}{e} & \setanswer{19.}{c} & \setanswer{29.}{d} & \setanswer{39.}{d} \\
\setanswer{10.}{b} & \setanswer{20.}{e} & \setanswer{30.}{d} & \setanswer{40.}{a} \\
\hline
\end{tabular}

\medskip

\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 9} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 9
%\setanswer{1.}{b}
%\setanswer{2.}{c}
%\setanswer{3.}{b}
%\setanswer{4.}{e}
%\setanswer{5.}{d}
%\setanswer{6.}{d}
%\setanswer{7.}{c}
%\setanswer{8.}{b}
%\setanswer{9.}{a}
%\setanswer{10.}{e}
%\setanswer{11.}{c}
%\setanswer{12.}{a}
%\setanswer{13.}{c}
%\setanswer{14.}{a}
%\setanswer{15.}{b}
%\setanswer{16.}{d}
%\setanswer{17.}{b}
%\setanswer{18.}{d}
%\setanswer{19.}{c}
%\setanswer{20.}{a}
%\setanswer{21.}{c}
%\setanswer{22.}{c}
%\setanswer{23.}{d}
%\setanswer{24.}{c}
%\setanswer{25.}{d}
%\setanswer{26.}{c}
%\setanswer{27.}{b}
%\setanswer{28.}{d}
%\setanswer{29.}{b}
%\setanswer{30.}{a}
%\setanswer{31.}{a}
%\setanswer{32.}{b}
%\setanswer{33.}{a}
%\setanswer{34.}{d}
%\setanswer{35.}{c}
%\setanswer{36.}{d}
%\setanswer{37.}{e}
%\setanswer{38.}{b}
%\setanswer{39.}{d}
%\setanswer{40.}{a}
\setanswer{1.}{b} & \setanswer{11.}{c} & \setanswer{21.}{c} & \setanswer{31.}{a} \\
\setanswer{2.}{c} & \setanswer{12.}{a} & \setanswer{22.}{c} & \setanswer{32.}{b} \\
\setanswer{3.}{b} & \setanswer{13.}{c} & \setanswer{23.}{d} & \setanswer{33.}{a} \\
\setanswer{4.}{e} & \setanswer{14.}{a} & \setanswer{24.}{c} & \setanswer{34.}{d} \\
\setanswer{5.}{d} & \setanswer{15.}{b} & \setanswer{25.}{d} & \setanswer{35.}{c} \\
\setanswer{6.}{d} & \setanswer{16.}{d} & \setanswer{26.}{c} & \setanswer{36.}{d} \\
\setanswer{7.}{c} & \setanswer{17.}{b} & \setanswer{27.}{b} & \setanswer{37.}{e} \\
\setanswer{8.}{b} & \setanswer{18.}{d} & \setanswer{28.}{d} & \setanswer{38.}{b} \\
\setanswer{9.}{a} & \setanswer{19.}{c} & \setanswer{29.}{b} & \setanswer{39.}{d} \\
\setanswer{10.}{e} & \setanswer{20.}{a} & \setanswer{30.}{a} & \setanswer{40.}{a} \\
\hline
\end{tabular}

\medskip

\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 10} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 10
%\setanswer{1.}{b}
%\setanswer{2.}{c}
%\setanswer{3.}{e}
%\setanswer{4.}{d}
%\setanswer{5.}{d}
%\setanswer{6.}{d}
%\setanswer{7.}{c}
%\setanswer{8.}{b}
%\setanswer{9.}{b}
%\setanswer{10.}{d}
%\setanswer{11.}{e}
%\setanswer{12.}{c}
%\setanswer{13.}{d}
%\setanswer{14.}{c}
%\setanswer{15.}{d}
%\setanswer{16.}{a}
%\setanswer{17.}{a}
%\setanswer{18.}{c}
%\setanswer{19.}{d}
%\setanswer{20.}{e}
%\setanswer{21.}{d}
%\setanswer{22.}{c}
%\setanswer{23.}{e}
%\setanswer{24.}{a}
%\setanswer{25.}{a}
%\setanswer{26.}{c}
%\setanswer{27.}{a}
%\setanswer{28.}{a}
%\setanswer{29.}{d}
%\setanswer{30.}{e}
%\setanswer{31.}{c}
%\setanswer{32.}{a}
%\setanswer{33.}{a}
%\setanswer{34.}{e}
%\setanswer{35.}{a}
%\setanswer{36.}{a}
%\setanswer{37.}{b}
%\setanswer{38.}{d}
%\setanswer{39.}{c}
%\setanswer{40.}{d}
\setanswer{1.}{b} & \setanswer{11.}{e} & \setanswer{21.}{d} & \setanswer{31.}{c} \\
\setanswer{2.}{c} & \setanswer{12.}{c} & \setanswer{22.}{c} & \setanswer{32.}{a} \\
\setanswer{3.}{e} & \setanswer{13.}{d} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{a} & \setanswer{34.}{e} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{a} & \setanswer{35.}{a} \\
\setanswer{6.}{d} & \setanswer{16.}{a} & \setanswer{26.}{c} & \setanswer{36.}{a} \\
\setanswer{7.}{c} & \setanswer{17.}{a} & \setanswer{27.}{a} & \setanswer{37.}{b} \\
\setanswer{8.}{b} & \setanswer{18.}{c} & \setanswer{28.}{a} & \setanswer{38.}{d} \\
\setanswer{9.}{b} & \setanswer{19.}{d} & \setanswer{29.}{d} & \setanswer{39.}{c} \\
\setanswer{10.}{d} & \setanswer{20.}{e} & \setanswer{30.}{e} & \setanswer{40.}{d} \\
\hline
\end{tabular}

\medskip
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 11} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 11
%\setanswer{1.}{d}
%\setanswer{2.}{b}
%\setanswer{3.}{d}
%\setanswer{4.}{d}
%\setanswer{5.}{a}
%\setanswer{6.}{e}
%\setanswer{7.}{a}
%\setanswer{8.}{a}
%\setanswer{9.}{e}
%\setanswer{10.}{a}
%\setanswer{11.}{d}
%\setanswer{12.}{d}
%\setanswer{13.}{e}
%\setanswer{14.}{c}
%\setanswer{15.}{b}
%\setanswer{16.}{e}
%\setanswer{17.}{c}
%\setanswer{18.}{e}
%\setanswer{19.}{a}
%\setanswer{20.}{b}
%\setanswer{21.}{c}
%\setanswer{22.}{d}
%\setanswer{23.}{e}
%\setanswer{24.}{d}
%\setanswer{25.}{d}
%\setanswer{26.}{c}
%\setanswer{27.}{c}
%\setanswer{28.}{a}
%\setanswer{29.}{c}
%\setanswer{30.}{b}
%\setanswer{31.}{c}
%\setanswer{32.}{a}
%\setanswer{33.}{a}
%\setanswer{34.}{e}
%\setanswer{35.}{a}
%\setanswer{36.}{a}
%\setanswer{37.}{b}
%\setanswer{38.}{d}
%\setanswer{39.}{c}
%\setanswer{40.}{d}
\setanswer{1.}{d} & \setanswer{11.}{d} & \setanswer{21.}{c} & \setanswer{31.}{c} \\
\setanswer{2.}{b} & \setanswer{12.}{d} & \setanswer{22.}{d} & \setanswer{32.}{a} \\
\setanswer{3.}{d} & \setanswer{13.}{e} & \setanswer{23.}{e} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{d} & \setanswer{34.}{e} \\
\setanswer{5.}{a} & \setanswer{15.}{b} & \setanswer{25.}{d} & \setanswer{35.}{a} \\
\setanswer{6.}{e} & \setanswer{16.}{e} & \setanswer{26.}{c} & \setanswer{36.}{a} \\
\setanswer{7.}{a} & \setanswer{17.}{c} & \setanswer{27.}{c} & \setanswer{37.}{b} \\
\setanswer{8.}{a} & \setanswer{18.}{e} & \setanswer{28.}{a} & \setanswer{38.}{d} \\
\setanswer{9.}{e} & \setanswer{19.}{a} & \setanswer{29.}{c} & \setanswer{39.}{c} \\
\setanswer{10.}{a} & \setanswer{20.}{b} & \setanswer{30.}{b} & \setanswer{40.}{d} \\
\hline
\end{tabular}

\medskip
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{\bf Answers to}\\
\multicolumn{4}{|c|}{\bf 1996 Exam Grade 12} \\ \hline
%\setanswer{1.}{c}   &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\
%grade 12
%\setanswer{1.}{a}
%\setanswer{2.}{a}
%\setanswer{3.}{c}
%\setanswer{4.}{d}
%\setanswer{5.}{d}
%\setanswer{6.}{b}
%\setanswer{7.}{a}
%\setanswer{8.}{e}
%\setanswer{9.}{a}
%\setanswer{10.}{d}
%\setanswer{11.}{b}
%\setanswer{12.}{a}
%\setanswer{13.}{c}
%\setanswer{14.}{c}
%\setanswer{15.}{d}
%\setanswer{16.}{d}
%\setanswer{17.}{d}
%\setanswer{18.}{c}
%\setanswer{19.}{c}
%\setanswer{20.}{a}
%\setanswer{21.}{c}
%\setanswer{22.}{b}
%\setanswer{23.}{a}
%\setanswer{24.}{c}
%\setanswer{25.}{b}
%\setanswer{26.}{d}
%\setanswer{27.}{c}
%\setanswer{28.}{e}
%\setanswer{29.}{d}
%\setanswer{30.}{a}
%\setanswer{31.}{c}
%\setanswer{32.}{a}
%\setanswer{33.}{a}
%\setanswer{34.}{e}
%\setanswer{35.}{a}
%\setanswer{36.}{a}
%\setanswer{37.}{b}
%\setanswer{38.}{d}
%\setanswer{39.}{c}
%\setanswer{40.}{d}
\setanswer{1.}{a} & \setanswer{11.}{b} & \setanswer{21.}{c} & \setanswer{31.}{c} \\
\setanswer{2.}{a} & \setanswer{12.}{a} & \setanswer{22.}{b} & \setanswer{32.}{a} \\
\setanswer{3.}{c} & \setanswer{13.}{c} & \setanswer{23.}{a} & \setanswer{33.}{a} \\
\setanswer{4.}{d} & \setanswer{14.}{c} & \setanswer{24.}{c} & \setanswer{34.}{e} \\
\setanswer{5.}{d} & \setanswer{15.}{d} & \setanswer{25.}{b} & \setanswer{35.}{a} \\
\setanswer{6.}{b} & \setanswer{16.}{d} & \setanswer{26.}{d} & \setanswer{36.}{a} \\
\setanswer{7.}{a} & \setanswer{17.}{d} & \setanswer{27.}{c} & \setanswer{37.}{b} \\
\setanswer{8.}{e} & \setanswer{18.}{c} & \setanswer{28.}{e} & \setanswer{38.}{d} \\
\setanswer{9.}{a} & \setanswer{19.}{c} & \setanswer{29.}{d} & \setanswer{39.}{c} \\
\setanswer{10.}{d} & \setanswer{20.}{a} & \setanswer{30.}{a} & \setanswer{40.}{d} \\
\hline
\end{tabular}
\end{center}
\end{multicols}
\end{document}