%% solutions2.tex \documentstyle[multicol,pictex,contest]{article} \begin{document} \setcounter{page}{0} \begin{multicols}{2} \begin{center} \Large\bf 1995 Utah State Math Contest\\ University of Utah \\ March 17, 1995\\[3pt] Teacher's Solution Set \end{center} The statistics of each problem is encoded below using the notation {\bf [60\%:19\%]}, which means $60$\% of the contestants attempted the problem and $19$\% of those attempts resulted in the correct answer. \def\STAT#1#2{[#1\%:#2\%]} %% 7.tex \begin{center}\Large\bf 1995 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 1995 Exam Grade 7} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 7 \setanswer{1.}{d} &\setanswer{11.}{a} &\setanswer{21.}{b} &\setanswer{31.}{a} \\ \setanswer{2.}{b} &\setanswer{12.}{b} &\setanswer{22.}{a} &\setanswer{32.}{a} \\ \setanswer{3.}{b} &\setanswer{13.}{c} &\setanswer{23.}{a} &\setanswer{33.}{d} \\ \setanswer{4.}{a} &\setanswer{14.}{d} &\setanswer{24.}{c} &\setanswer{34.}{b} \\ \setanswer{5.}{d} &\setanswer{15.}{c} &\setanswer{25.}{d} &\setanswer{35.}{d} \\ \setanswer{6.}{b} &\setanswer{16.}{c} &\setanswer{26.}{a} &\setanswer{36.}{d} \\ \setanswer{7.}{e} &\setanswer{17.}{d} &\setanswer{27.}{c} &\setanswer{37.}{e} \\ \setanswer{8.}{b} &\setanswer{18.}{c} &\setanswer{28.}{d} &\setanswer{38.}{c} \\ \setanswer{9.}{a} &\setanswer{19.}{b} &\setanswer{29.}{b} &\setanswer{39.}{d} \\ \setanswer{10.}{d} &\setanswer{20.}{c} &\setanswer{30.}{d} &\setanswer{40.}{c} \\ \hline \end{tabular} \medskip \fi \end{center} \smallskip {\bf Instructions}: Only one answer is correct. Scoring: 5(\# right)+1(\# blank)+0(\# wrong). Time: 120 minutes. \begin{exercises} \prob{1.}{(Quotient)} Evaluate $\dd \frac{-(|-13|-|-3|)}{\frac{81}{7} - \frac{22}{14}}$. \rightans{d} \ans {$1.6$} {$-1.6$} {$1$} {$-1$} {None} \SOL{1. \STAT{86}{52}}{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 $-10/10=-1$.} \prob{2.}{(Tank)} A gas tank becomes half full after adding $2.5$ gallons to a tank $1/3$ full. Find the tank capacity in gallons. \rightans{b} \ans {$16$} {$15$} {$14$} {$12$} {$11$} \SOL{2. \STAT{77}{80}}{ Let $x$ be the capacity. Then $x/3 + 2.5=x/2$. Solving for $x$ gives $x=15$. } \prob{3.}{(Sound)} Sound travels in air about $344$ meters per second. Find the approximate distance in meters to a lightning flash, given the thunder was heard $4$ seconds later. \rightans{b} \ans {$1300$} {$1400$} {$1500$} {$1600$} {$1700$} \SOL{3. \STAT{89}{89}}{ The formula $D=RT$ is applied with $R=344$ and $T=4$ to give $D= 1376$, or about $1400$ kilometers. } \def\AA{ \prob{4.}{(Area)} Find the area of the quadrilateral in the figure. All angles are right angles. \par \rightans{a} \ans {$30$} {$31$} {$32$\PAR} {$33$} {$34$} } \def\BB{\input {rect7.box}} \WRAPFIG{\AA}{\BB}{1.2in} \SOL{4. \STAT{80}{79}}{ Break the figure into rectangles and apply the area formula: $area=length\times width$. } \prob{5.}{(Integers)} Find the greatest integer whose product with $-4$ is greater than $-24$. \rightans{d} \ans {$11$} {$1$} {$3$} {$5$} {$7$} \SOL{5. \STAT{92}{54}}{ Let $x$ be the integer. Then $-4x > -24$ implies $-x > -6$ or $x<6$. The greatest integer is $x=5$. } \def\AA{ \prob{6.}{(Area)} In the figure, let rectangle $C$ be six times the area of rectangle $A$, and let rectangle $B$ have area $48$ units, which is $12$ units more than the area of rectangle $A$. Find the area of rectangle $C$. } \def\BB{\input {rect7a.box}} \WRAPFIG{\AA}{\BB}{0.8in} \rightans{b} \ans {$256$} {$216$} {$204$} {$196$} {$180$} \SOL{6. \STAT{93}{95}}{ Use lower case letters for areas. Then $c=6a$, $b=48$, $b=12+a$. Substituting gives $48=12+a$ or $a=36$. Then $c=6a=6(36)=216$. } \prob{7.}{(Mode)} Find the mode for the basketball scores $7$, $9$, $12$, $17$, $4$, $16$, $8$, $22$, $13$. \rightans{e} \ans {$9$} {$12$} {$12.5$} {$13$} {None} \SOL{7. \STAT{87}{34}}{ There is no mode because each score occurs just once. } \def\AA{ \prob{8.}{(Wyoming)} Estimate the area in square miles of the state of Wyoming in the figure, assuming it has the shape of a trapezoid, with parallel north and south borders $275$ miles apart. Units are miles. } \def\BB{\input{wyoming.box}} \WRAPFIG{\AA}{\BB}{1.1in} \rightans{b} \ans {$97700$} {$97800$} {$97900$} {$98000$\PAR} {$98100$} \SOL{8. \STAT{63}{44}}{ Multiply the sum of the bases by the altitude and divide by two. The area is $(349+362)(275)(1/2)=97762.5$. The actual area of Wyoming is $97809$. } \prob{9.}{(Sequence)} What is the eighth term divided by the ninth term of the sequence $\dd\frac{1}{2}$, $\dd\frac{2}{3}$, $\dd\frac{3}{4}$, $\dd\frac{4}{5}$, \ldots? \rightans{a} \ans {$\dd\frac{80}{81}$} {$\dd\frac{8}{10}$} {$\dd\frac{9}{10}$} {$\dd\frac{17}{19}$} {$\dd\frac{56}{90}$} \SOL{9. \STAT{96}{87}}{ The eighth term is $\dd\frac{8}{9}$ and the ninth term is $\dd\frac{9}{10}$. Their quotient is $\frac{80}{81}$. } \prob{10.}{(Apples)} Find the percentage of good apples in a basket of $65$ apples containing thirteen rotten apples. \rightans{d} \ans {$20$\%} {$52$\%} {$72$\%} {$80$\%} {None} \SOL{10. \STAT{95}{58}}{ The percentage of good apples corresponds to $(65-13)/65=0.8$ or $80$\%. } \prob{11.}{(Divisor)} Which number below divides\\ $1337456631234376$? \rightans{a} \ans {$8$} {$41$} {$251$} {$1973$} {None} \SOL{11. \STAT{79}{74}}{ 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 1995 supply no real help on the problem, due to display limitations. } \prob{12.}{(Fractions)} Find a fraction equivalent to $\dd \frac{40p^2q}{52pq^2}$. \rightans{b} \ans {$\dd\frac{p}{q}$} {$\dd\frac{10p}{13q}$} {$\dd\frac{10p}{q}$} {$\dd\frac{10}{13}$} {None} \SOL{12. \STAT{86}{83}}{ Write as $\dd \frac{(2)(4)(5)ppq}{(4)(13)pqq}$. Cancel like terms to obtain $\dd \frac{(2)(5)p}{13q}$ which equals $\dd \frac{10p}{13q}$. } \prob{13.}{(LCM)} Find the least common multiple of $45$, $30$, $105$. \rightans{c} \ans {$5$} {$315$} {$630$} {$15$} {None} \SOL{13. \STAT{96}{53}}{ The LCM of three numbers is the smallest number $N$ divisible by each of the three numbers. Write $45=(3)(3)(5)$, $30=(2)(3)(5)$, $105=(5)(3)(7)$. Then $LCM=(3)(3)(5)(2)(7)=630$. } \prob{14.}{(Wrench)} A bolt measures about $15/32$ inches. Which wrench size in inches below is the best choice to fit the bolt? \rightans{d} \ans {$5/8$} {$11/16$} {$7/16$} {$1/2$} {$9/16$} \SOL{14. \STAT{88}{52}}{ The various values $x$ given in the answers have to satisfy $x>15/32$ in order to qualify, however, the right answer is the smallest such $x$. The tests are $15/32<5/8$, $15/32<11/16$, $15/32<7/16$, $15/32<1/2$, $15/32<9/16$. Only $15/32<7/16$ fails. The smallest qualifying size is $x=1/2$. } \prob{15.}{(Marbles)} A bag of marbles has $3$ blue, $6$ red, $2$ green and $1$ black. Find the probability of randomly drawing a green or black marble. \rightans{c} \ans {$\frac{1}{12}$} {$\frac{1}{6}$} {$\frac{1}{4}$} {$\frac{1}{3}$} {None} \SOL{15. \STAT{95}{75}}{ The sample set consists of $12$ items and $3$ of these represent success, so the probability is $3/12$ or $1/4$. } \prob{16.}{(Spinners)} 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 the sum of the two numbers is greater than $8$? \rightans{c} \ans {$\frac{23}{36}$} {$\frac{25}{36}$} {$\frac{29}{36}$} {$\frac{31}{36}$} {None} \SOL{16. \STAT{51}{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 $a+b>8$. These events are $(1,8)$, $(1,9)$, $(1,10)$, $(1,11)$, $(1,12)$, $(2,7)$, $(2,8)$, \ldots, $(2,12)$, $(3,6)$, $(3,7)$, \ldots, $(3,12)$ and so on. Arranging these in a neat table and counting gives the success count as $5+6+7+8+9+10+11$ plus $5(12)$ which is $56+60=116$. The probability is $116/144$. } \prob{17.}{(DNA strand)} How many DNA strands can be laid end-to-end across the diameter of a spherical human blood cell of volume $\frac{4}{3}\pi(2^{-17})^3$m? Use $2^{-18}$m for the length of a strand. \rightans{d} \ans {$1$} {$2$} {$3$} {$4$} {$5$} \SOL{17. \STAT{36}{32}}{ The volume of a sphere of radius $R$ is $\frac{4}{3}\pi R^3$. The human blood cell has diameter $2R=2^{-16}$. Divide by $S=2^{-18}$ to find the number of strands across a diameter: $2R/S=2^{-16}/2^{-18}=2^2=4$. } \def\AA{ \prob{18.}{(Combinatorics)} %1994 modified How many different square $4$-panel window designs can be made using two {\em Gray} panes, one {\em Frosted} pane and one {\em Clear} pane? } \def\BB{\input{pane7.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{c} \ans {$24$} {$18$} {$12$} {$10$} {$4$} \SOL{18. \STAT{91}{60}}{ The patterns labeled the same way we read a page of text are FGGC, FGCG, FCGG, GFCG, GFGC, GGCF, GGFC, GCFG, GCGF, CFGG, CGFG, CGGF. } \prob{19.}{(Discount)} Boots priced at \${}$42$ were increased in price by $25$\% and later sold at a $25$\% off sale. What was the final sales price? \rightans{b} \ans {\${}$42$} {\${}$39.38$} {\${}$52.50$} {\${}$38$} {\${}$39$} \SOL{19. \STAT{96}{44}}{ The price increase was $(0.25)(42)=10.50$ making a new price of $42+10.50=52.50$. The $25$\% off sale uses a sale price of $(0.75)(52.50)=39.375$ or \${}$39.38$. } \prob{20.}{(Proportions)} Suppose $9$ gallons cost \${}$12.15$. Find $x$, if $x$ quarts cost \${}$2.25$. \rightans{c} \ans {$ 6$} {$\frac{19}{3}$} {$\frac{20}{3}$} {$ 7$} {$\frac{5}{3}$} {None} \SOL{20. \STAT{69}{30}}{ A gallon is $4$ quarts. So $(4)(9)=36$ quarts cost $12.15$. The proportion is $\dd\frac{36}{12.15}=\frac{x}{2.25}$. The answer is $x=\dd\frac{20}{3}$. } \prob{21.}{(Linear equation)} Solve for $x$: $2(x-1)+5=3(x-1)$. \rightans{b} \ans {$5$} {$6$} {$-5$} {$-6$} {None} \SOL{21. \STAT{90}{71}}{ Swap sides, add $-2(x-1)$ to each side. Then $3(x-1)-2(x-1)=5$. The result is $x-1=5$ which gives $x=6$. } \prob{22.}{(Squares)} The sum of the squares of two positive integers is $289$. What is their sum? \rightans{a} \ans {$23$} {$22$} {$21$} {$20$} {None} \SOL{22. \STAT{57}{43}}{ This triangle is supposed to be a familiar one of sides $8$, $15$, $17$. If so, then the problem is immediately solved as $8+15=23$. \par To decide upon the answer without such knowledge, taking into account all possible combinations, let $a$ and $b$ be the integers satisfying $a^2+b^2=289=17^2$. Then $a$ and $b$ are the legs of a right triangle, which according to the triangle inequality must each be less than $17$. There are only a finite number of possibilities to try: $1\le a\le 16$, $b=\sqrt{289-a^2}$. The possibilities are tried: $a=1$, $b=12\sqrt2$, \ldots, $a=16$, $b=\sqrt{33}$. The only integer answers are $a=8$, $b=15$ and $a=15$, $b=8$. Then $a+b=23$. } \prob{23.}{(Complement)} The measure of an angle is $26^\circ$ more than three times the measure of its complement. Find the measure of the complement. \rightans{a} \ans {$16^\circ$} {$32^\circ$} {$38.5^\circ$} {$74^\circ$} {None} \SOL{23. \STAT{50}{22}}{ Let $x$ be the complement. Then $90-x$ is the angle and the text implies $90-x=26+3x$. Solving gives $x=16$. } \def\AA{ \prob{24.}{(Transversal)} Let lines $L$ and $M$ be parallel and cut by the transversal $T$ as in the figure, forming $8$ angles. Given $a=7x$ and $b=2x+75$, in degrees, find the measure in degrees of the largest of the $8$ angles. } \def\BB{\input{trans7.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{c} \ans {$15^\circ$} {$75^\circ$} {$105^\circ$} {$120^\circ$} {$125^\circ$} \SOL{24. \STAT{50}{39}}{ A transversal cuts parallel lines in equal angles, therefore, $7x=2x+75$. Solving gives $x=15$ and the two angles are $7x=105$ and its supplement $180-105=75$. The largest is $105$. } \prob{25.}{(Area)} Which dimension below is a side of a rectangle of perimeter $16$ and area $12$? \rightans{d} \ans {$3$} {$4$} {$5$} {$6$} {$8$} \SOL{25. \STAT{86}{60}}{ Draw a figure and try each dimension in the list (a) -- (e). The only rectangle with perimeter $16$ and area $12$ has dimensions $2$ and $6$. } \prob{26.}{(Systems)} %1994 modified Find the value of $x$ in the linear system $x+y=4$, $\,y-3x=8$. \rightans{a} \ans {$-1$} {$0$} {$2$} {$4$} {$5$} \SOL{26. \STAT{74}{66}}{ Solve the first equation for $x$ in terms of $y$: $x=4-y$. Put the answer into the second equation $y-3x=8$ to get $y-3(4-y)=8$. Then $4y=12+8$ and $y=5$. Finally, $x=-1$, by using the equation $x+y=4$ with the value $y=5$. Faster method: subtract the equations to get $(x+y)-(y-3x)=4-8$. This reduces to $4x=-4$ or $x=-1$. } \def\AA{ \prob{27.}{(Triangles)} %1994 modified Find area $A$. \par \rightans{c} \ans {$4100$} {$4102$\PAR} {$4104$} {$4110$\PAR} {$4116$} } \def\BB{\input{tri7.box}} \WRAPFIG{\AA}{\BB}{1.6in} \SOL{27. \STAT{42}{58}}{ The Pythagorean theorem applied to the left triangle reveals the altitude $x$ satisfies $x^2+96^2=120^2$ or $x=72$. The base is $114$. The area is half the base times the altitude, or $(1/2)(114)(72)=4104$. } \prob{28.}{(Inclined plane)} Find the slope of an inclined plane which raises an object $1$m vertically for each $3$m of travel on the inclined surface. \rightans{d} \ans {$\dd\frac{1}{3}$} {$\dd\frac{1}{\sqrt2}$} {$\dd\frac{3}{2\sqrt2}$} {$\dd\frac{\sqrt2}{4}$} {None} \SOL{28. \STAT{52}{1}}{ Form a right triangle with hypotenuse $3$ and legs $1$, $x$. Then $x^2+1^2=3^2$, giving $x=2\sqrt2$. The slope is {\em rise/run}, which is $1/x=1/(2\sqrt2)=\sqrt{2}/4$. } \def\AA{ \prob{29.}{(Supply drop)} The load of supplies on a parachute hangs $16$ft below the canopy, which has diameter $24$ft. Find the measure $x$ of a suspension line in feet. } \def\BB{\input{drop7.box}} \WRAPFIG{\AA}{\BB}{1.25in} \rightans{b} \ans {$8\sqrt{13}$} {$20$} {$\frac{59}{3}$} {$21$} {$4\sqrt{13}$} \SOL{29. \STAT{45}{60}}{ Let $x$ be the length of the suspension line. Then $x^2=12^2+16^2$ by the Pythagorean theorem. The answer is $x=20$. } \prob{30.}{(Perimeter)} %1994 modified Find the perimeter of an equilateral triangle of area $25\sqrt{3}$. \rightans{d} \ans {$10$} {$20$} {$25$} {$30$} {None} \SOL{30. \STAT{38}{25}}{Let $x$ be a side of the triangle. Cut the triangle in half to give two equal triangles with sides in the ratio $\frac{1}{2}$:$\frac{\sqrt{3}}{2}$:$1$. The altitude is $\sqrt{3}x/2$ and the base is $x$, so the triangle's area is $(1/2)(x)(\sqrt{3}x/2)$, which also equals $25\sqrt{3}$. Solving $(1/2)(x)(\sqrt{3}x/2)=25\sqrt{3}$ gives $x=10$. The perimeter is $3x=30$. } See {\em infra} for problems 31-40. \end{exercises} %% 8.tex \begin{center}\Large\bf 1995 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 1995 Exam Grade 8} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 8 \setanswer{1.}{c} &\setanswer{11.}{c} &\setanswer{21.}{d} &\setanswer{31.}{a} \\ \setanswer{2.}{e} &\setanswer{12.}{e} &\setanswer{22.}{c} &\setanswer{32.}{a} \\ \setanswer{3.}{b} &\setanswer{13.}{d} &\setanswer{23.}{b} &\setanswer{33.}{d} \\ \setanswer{4.}{e} &\setanswer{14.}{c} &\setanswer{24.}{c} &\setanswer{34.}{b} \\ \setanswer{5.}{e} &\setanswer{15.}{a} &\setanswer{25.}{c} &\setanswer{35.}{d} \\ \setanswer{6.}{d} &\setanswer{16.}{c} &\setanswer{26.}{b} &\setanswer{36.}{d} \\ \setanswer{7.}{c} &\setanswer{17.}{c} &\setanswer{27.}{a} &\setanswer{37.}{e} \\ \setanswer{8.}{b} &\setanswer{18.}{a} &\setanswer{28.}{d} &\setanswer{38.}{c} \\ \setanswer{9.}{b} &\setanswer{19.}{b} &\setanswer{29.}{c} &\setanswer{39.}{d} \\ \setanswer{10.}{b} &\setanswer{20.}{b} &\setanswer{30.}{d} &\setanswer{40.}{c} \\ \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} \prob{1.}{(Travel)} Jack travels $7$ miles the first day and $51$ miles the last day, increasing his daily mileage $4$ miles per day. Find his total mileage. \rightans{c} \ans {$336$} {$340$} {$348$} {$352$} {$356$} \SOL{1. \STAT{93}{91}}{ The trip is $12$ days, with daily mileages $7$, $11$, $15$, \ldots, $51$. The sum of these mileages is $348$. } \prob{2.}{(Inequality)} Given $|5-3x|\ge 1$, which of the following is necessarily true? \rightans{e} \ans {$x\ge 2$} {$x=2$} {$x< 2$} {$x>4/3$\PAR} {None} \SOL{2. \STAT{89}{25}}{ The inequality means $5-3x\ge 1$ or $-(5-3x)\ge 1$. On the number line, these options describe two disjoint intervals. None of the answers supplied describe two intervals. The answer is {\em None}. } \prob{3.}{(Volume)} Find the volume in cubic units of a parallelepiped (box) of sides $yz^4$, $yz^2$ and $y^2z$ units. \rightans{b} \ans {$y^4z^9$} {$(yz)^4z^3$} {$y^4z^6$} {$\frac{4}{3}y^4z^7$\PAR} {None} \SOL{3. \STAT{77}{30}}{ The volume of a box is the product of the side dimensions, which is $yz^4yz^2y^2z=y^4z^7$. } \prob{4.}{(Powers)} Given $a>0$, simplify $\dd\frac{\left(a^{x+2}\right)^2}{\left(a^{x-3}\right)^2\left(-a\right)^5}$. \rightans{e} \ans {$a^6$} {$-a^{6}$} {$-a^{-3}$} {$a^5$} {$-a^5$} \SOL{4. \STAT{46}{33}}{ The fraction numerator is $a^{2x+4}$ and the denominator is $(-1)^5a^{2x-6+5}$. The first reduction is $(-1)^5=-1$. Powers combine to make $a^{2x+4-2x+6-5}$, giving the simplification $-a^5$. } \prob{5.}{(Fractional equation)} Solve $\dd\frac{3}{x+1}=\frac{1}{x+1}-7$. \rightans{e} \ans {$\frac{-7}{9}$} {$\frac{11}{7}$} {$\frac{-11}{7}$} {$\frac{9}{7}$} {$\frac{-9}{7}$} \SOL{5. \STAT{62}{63}}{ Combine to $2/(x+1)=-7$, then take reciprocals to get $(x+1)/2=-1/7$. Solving as a linear equation, $x=-1-2/7=-9/7$. } \prob{6.}{(Percent)} How many elevenths are equivalent to the fraction of $75$ percent? \rightans{d} \ans {$\frac{3}{4}$} {$\frac{9}{11}$} {$\frac{12}{11}$} {$8\frac{1}{4}$} {None} \SOL{6. \STAT{88}{40}}{ The equation to be solved is $x/11=0.75$, which has solution $8.25$. } \prob{7.}{(Line)} Determine the value of $r$ such that the line through the points $(4,r)$ and $(r,2)$ has slope $-5/3$. \rightans{c} \ans {$-7$} {$1$} {$7$} {$-1$} {None} \SOL{7. \STAT{67}{48}}{ 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 $r=7$. } \prob{8.}{(Midpoint)} Let $P$ be the midpoint of the segment from $A(8,4)$ to $B(12,12)$. Find the coordinates of $P$. \rightans{b} \ans {$(8,10)$} {$(10,8)$} {$(20,16)$} {$(16,20)$\PAR} {$(4,8)$} \SOL{8. \STAT{86}{81}}{ The midpoint $(x,y)$ is given by $x=(8+12)/2$, $y=(4+12)/2$. } \prob{9.}{(Region)} Find the point $(x,y)$ below which is above the curve $y=|x|$ and below the line $2y+x=3$. \rightans{b} \ans {$(1.5,1)$} {$(-1,1.5)$} {$(1.5,-1)$\PAR} {$(-1,0.75)$} {None} \SOL{9. \STAT{47}{40}}{ The region above the curve $y=|x|$ is $y > |x|$. The only candidate is $(-1,1.5)$. The region below the line $2y+x=3$ is $y<(3-x)/2$. Again, $(-1,1.5)$ satisfies $1.5 < (3-(-1))/2$. The answer is $(-1,1.5)$. } \prob{10.}{(Length)} The length of a rectangle of perimeter $50$ cm exceeds its width by $7$ cm. Find the length in centimeters. \rightans{b} \ans {$28.5$} {$16$} {$14$} {$12$} {$9$} \SOL{10. \STAT{85}{61}}{ The equations are $x=y+7$, $2x+2y=50$. Solving gives $x=16$, $y=9$. } \prob{11.}{(Radicals)} Simplify $\sqrt{(x-1)^2}$. \rightans{c} \ans {$x-1$} {$1-x$} {$|1-x|$} {$-|1-x|$\PAR} {None} \SOL{11. \STAT{85}{4}}{ The square root is always positive or zero, by convention. The identity $\sqrt{u^2}=|u|$ is applied to give the answer $|x-1|$ which is also $|1-x|$. } \prob{12.}{(Radicals)} Solve $\dd \sqrt{3x-14}+x=6$ for $x$. \rightans{e} \ans {$-5$} {$4$} {$6$} {$10$} {None} \SOL{12. \STAT{76}{69}}{ Put $x$ on the right side of the equation and square both sides to obtain $3x-14=36-12x+x^2$. Place all terms on the right and factor with the FOIL method into $0=(x-5)(x-10)$. The answer $5$ is correct. The number $10$ is an {\em extraneous solution}, it does not satisfy the equation. } \prob{13.}{(Distance)} The distance between points $(-a,11)$ and $(-3,5)$ is $10$. Find $a>0$. \rightans{d} \ans {$3$} {$5$} {$10$} {$11$} {None} \SOL{13. \STAT{55}{31}}{ Solve $(-a+3)^2+(11-5)^2=10^2$ for $a>0$. The possible answers for $a$ are $11$ and $-5$, but only one is positive. } \prob{14.}{(Cube)} A white plastic cube is painted black and cut into $64$ unit cubes. Find the number of unit cubes with exactly two black faces. \rightans{c} \ans {$12$} {$16$} {$24$} {$32$} {None} \SOL{14. \STAT{78}{45}}{ There are twelve edges on the original cube, each of which produced two unit cubes having exactly two black faces. No other unit cubes have two black faces. } \prob{15.}{(Probability)} In a survey of $700$ people concerning two ice cream flavors, $415$ liked chocolate, $269$ liked vanilla and among these, $124$ liked both flavors. Estimate the probability that a person dislikes both flavors. \rightans{a} \ans {$0.20$} {$0.22$} {$0.24$} {$0.26$} {$0.28$} \SOL{15. \STAT{67}{55}}{ 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 $16$, but $700-560=140$. The probability of disliking both is $140/700=1/5=0.2$. } \prob{16.}{(Median)} Use the set of data $12$, $24$, $18$, $22$, $17$, $10$. Find the median of all values greater than the median of the whole set. \rightans{c} \ans {$12.5$} {$18$} {$22$} {$24$} {None} \SOL{16. \STAT{74}{57}}{ The median divides the data set into two halves. From the sorted order $10$, $12$, $17$, $18$, $22$, $24$ the median of the whole set is $(17+18)/2$. The data set greater than this value is $18$, $22$, $24$, which has median $22$. } \prob{17.}{(Quotients)} Simplify $\dd\frac{x^4-16}{x^4-8x^2+16}$. \rightans{c} \ans {$-1$} {$\frac{x+2}{x-2}$} {$\frac{x^2+4}{x^2-4}$} {$\frac{x^2}{x^2-8}$} {None} \SOL{17. \STAT{71}{57}}{ The factorizations $x^4-16=(x^2-4)(x^2+4)$ and $x^4-8x^2+16=(x^2-4)^2$ are used to obtain $(x^2+4)/(x^2-4)$. } \prob{18.}{(Quotient)} Simplify $\dd \left(1-\frac{1}{1+x}\right)^{-1} - \left(\frac{1}{1-x}-1\right)^{-1}$. \rightans{a} \ans {$2$} {$-2$} {$\frac{2}{x}$} {$\frac{-2}{x}$} {None} \SOL{18. \STAT{47}{22}}{ The first reciprocal is $\frac{1+x}{x}$. The second is $\frac{1-x}{x}$. Their difference is $2x/x$ or $2$. } \prob{19.}{(Functions)} Given $f(x)=(2x+3)^2$ and $g(y)=\sqrt{y}-5$, which answer below is a solution $x$ of the equation $g(f(x))=8$? \rightans{b} \ans {$8$} {$-8$} {$13$} {$-5$} {None} \SOL{19. \STAT{36}{26}}{ The equation is $\sqrt{y}-5=8$ with $y=(2x+3)^2$. The answer is obtained from $|2x+3|=13$ as $2x+3=13$ or $-2x-3=13$. The first gives $x=5$ while the second gives $x=-8$. } \prob{20.}{(Trip)} Jack begins a trip at $10$:$00$am, driving $50$mph. Jill follows him at $11$:$30$am, going $45$mph. When are they $100$ miles apart? \rightans{b} \ans {$3$:$30$pm} {$4$:$30$pm} {$4$:$45$pm} {$5$:$00$pm\PAR} {$5$:$30$pm} \SOL{20. \STAT{83}{69}}{ Jack has traveled $75$ miles when Jill starts her trip. Their speed difference is $5$mph. The distance between them grows by $5$ miles per hour, therefore, after $5$ hours an additional $25$ miles has accumulated, making $75+25=100$ miles. Five hours after $11$:$30$am is $4$:$30$pm. } \prob{21.}{(Chemistry)} Ten liters of a $50$\% alcohol solution are replaced by $100$\% solution to make a $75$\% solution. Find the number of liters in the first solution. \rightans{d} \ans {$35$} {$30$} {$25$} {$20$} {$15$} \SOL{21. \STAT{67}{56}}{ Let $x$ be the number of liters of solution before replacement. Then $0.5(x-10)+1.0(10)=0.75x$, giving the linear equation $2x+20=3x$. Solving, $x=20$. } \def\AA{ \prob{22.}{(Rectangle)} Find the area in cm${}^2$ of the region inside the rectangle and outside the equilateral triangle of side $x$. Units are centimeters. } \def\BB{\input{tri8a.box}} \WRAPFIG{\AA}{\BB}{1.1in} \rightans{c} \ans {$20$} {$8\sqrt{6}$} {$12\sqrt{3}$} {$21$} {$24\sqrt{3}$} \SOL{22. \STAT{42}{41}}{ The area of the rectangle is $6x$ and of the triangle $\frac{1}{2}(x)(6)$, so the region of interest has area $6x-3x$ or $3x$. The value of $x$ is found from the Pythagorean theorem by using sides $6$ and $x/2$ with hypotenuse $x$: $x^2=6^2+(x/2)^2$. Solving gives $x=4\sqrt{3}$ and then the area requested is $3x=12\sqrt{3}$. } \prob{23.}{(Acute)} For which values of $d$ is $(30-d/2)$ the measure of a positive acute angle in degrees? \rightans{b} \ans {$-600$ when $y=2$ and $x=8/a$. Find $y$ when $z=a^2$ and $x=(2a)^3$. \rightans{b} \ans {$a^5$} {$2a^5$} {$4a^5$} {$a^5/2$} {None} \SOL{2. \STAT{30}{38}}{ The variation has the form $z=cy/x$ for some $c$. The value of $c$ is known from $z=a$, $x=8/a$, $y=2$ by substitution into $z=cy/x$: $a=2c/(8/a)$ or $c=4$. Then $z=4y/x$ gives $y=xz/4$ and at $z=a^2$, $x=(2a)^3$ the answer is $y=(2a)^3a^2/4=2a^5$. } \prob{3.}{(Supplement)} The measure of an angle is one-fourth the measure of its supplement. Find the measure of its complement. \rightans{c} \ans {$126^\circ$} {$36^\circ$} {$54^\circ$} {$144^\circ$} {None} \SOL{3. \STAT{87}{62}}{ Let $x$ be the angle. The supplement is $180-x$. Then $x=(180-x)/4$ gives $5x=180$ and $x=36$. The complement is $90-x$ or $54$. } \prob{4.}{(Pens)} A box with $75$ pens has $24$\% red pens, $17$ blue pens 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{4. \STAT{94}{77}}{ The number of red pens is $0.24(75)=18$, so there are $75-17-18=40$ green pens. The fraction is $40/75$ or $8/15$. } \def\AA{ \prob{5.}{(Inequality)} Decide the correct statement for lengths $a$ and $b$ in the figure. \par \rightans{c} \ans {$a=b$} {$a>b$\PAR} {$b>a$} {undecidable\PAR} {None} } \def\BB{\input{ineq9.box}} \WRAPFIG{\AA}{\BB}{1.5in} \SOL{5. \STAT{94}{84}}{ The triangle with equal base angles of $43^\circ$ is isosceles. The lengths $a$ and $b$ are the third sides of two triangles having two congruent sides. The SAS inequality (Hinge Theorem) implies that $a < b$, because the first included angle $35^\circ$ is less than the second included angle $37^\circ$. } \prob{6.}{(Algebra)} Let $x=a-b$ and $y=a^3-3a^2b+3ab^2-b^3$. Find the formula valid for all $a$ and $b$. \rightans{a} \ans {$y^2=x^6$} {$y=x^4$} {$y^3=x$} {$y=-x^3$} {None} \SOL{6. \STAT{52}{37}}{ The expansion of $(a-b)^3$ is $y$, therefore $y=x^3$ and $y^2=x^6$. } \prob{7.}{(Triangle)} A triangle has angles of $x$, $x+5$ and $2x+3$ degrees. Find the number of degrees in the largest angle. \rightans{d} \ans {$43$} {$48$} {$79$} {$89$} {$90$} \SOL{7. \STAT{95}{85}}{ 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 largest is $89$. } \def\AA{ \prob{8.}{(Photo)} 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. \STAT{30}{24}}{ 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 $x^2=40+16\sqrt{6}$. } \prob{9.}{(Doubling)} 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{9. \STAT{98}{90}}{ Let $x$ be the number. Then $2x^2=32$. To solve $x^2=16$, ask ``which number $x$ when squared equals $16$'', to obtain $x=4$. } \prob{10.}{(Socks)} Find the probability that two black socks are drawn randomly without replacement from a drawer having $8$ black and $5$ blue socks. \rightans{a} \ans {$\frac{14}{39}$} {$\frac{10}{39}$} {$\frac{5}{39}$} {$\frac{1}{39}$} {$\frac{56}{169}$} \SOL{10. \STAT{59}{46}}{ The events possible are $(a,b)$ where $a$ and $b$ are colors, black or blue. The event size is $13(12)=156$. There are $8(7)$ events which choose $a$ and $b$ both black. The probability is $56/156=14/39$. } \prob{11.}{(Functions)} Given $f(x)=\dd\frac{x^2+2x+1}{x^2+4x+4}$, find $f(-1/2)$. \rightans{e} \ans {$\frac14$} {$9$} {$\frac94$} {$5^{-2}$} {$3^{-2}$} \SOL{11. \STAT{58}{60}}{ This is the value of $y=(x^2+2x+1)/(x^2+4x+4)$ when $x=-1/2$. Then $x^2+2x+1=(x+1)^2=(1/2)^2=1/4$ and $x^2+4x+4=(x+2)^2=(3/2)^2=9/4$. So $y=(1/4)/(9/4)=1/9$. } \prob{12.}{(Equations)} The system $ax+y=6$, $ax+by=10$ has the solution $x=-1$, $y=2$. Find $b$. \rightans{c} \ans {$-4$} {$4$} {$3$} {$-3$} {$7$} \SOL{12. \STAT{94}{85}}{ Substitute $x=-1$ and $y=2$ to obtain the system $-a+2=6$, $-a+2b=10$. Then $a=-4$. Substitute $a=-4$ into $-a+2b=10$ to obtain $2b=6$ or $b=3$. } \prob{13.}{(Webbing)} Given seven points in the plane, no three on the same line, find the number of line segments needed to connect each pair of points. \rightans{d} \ans {$11$} {$17$} {$20$} {$21$} {$23$} \SOL{13. \STAT{84}{73}}{ Take a given point of the seven and connect a line segment to each of the other six points. It takes $6$ segments. Repeat with a different point, then only $6-1$ segments are needed, since the first point was already connected. The pattern: $6+5+4+3+2+1=21$ segments are needed. } \def\AA{ \prob{14.}{(Vertex)} Triangle $ABC$ is isosceles with perimeter between $23$ and $32$. Locate the vertex angle. \par \rightans{c} \ans {$\angle A$} {$\angle B$} {$\angle C$\PAR} {undecidable} {None} } \def\BB{\input{isos9.box}} \WRAPFIG{\AA}{\BB}{1.2in} \SOL{14. \STAT{80}{48}}{ The inequality $23 < 2x+2+10+x+4 < 32$ means $\frac73 < x < \frac{16}{3}$. To be isosceles, one of these must be true: (1) $2x+2=10$, (2) $2x+2=x+4$ or (3) $x+4=10$. The only one that can work is $x=4$. So $2x+2=10$ and the angle at $C$ is the vertex. } \prob{15.}{(Clock)} As the hands of a clock move from $3$am to $2$:$59$pm, how many times do they form a right angle? \rightans{b} \ans {$18$} {$22$} {$23$} {$24$} {$25$} \SOL{15. \STAT{69}{13}}{ This problem can be solved by drawing $24$ clock diagrams or by winding a standard watch with hour and minute hands. We explain below why the answer is not $24$ and not $23$. \par If the minute hand and the hour hand form a right angle, then a fixed number of minutes $a$ elapses before the hands can form another right angle. A simple estimate is $a=32.75$, approximately. The fact $a>30$ is enough to argue that $24$ is the wrong answer. As a check on the logic, $32.75(22)>720$, which accounts for the full number of minutes $30(24)=720$ from $3$am to $3$pm. \par In the first hour, $3$am to $4$am, the hands are at right angles $2$ times, once at $3$am and then after $a$ minutes. During the four hour period from $4$am to $7$:$59$am, the hands are at right angles two times per hour for $4$ hours, making $8$ times. In the time period, $8$am to $8$:$59$pm, they are at right angles only once. The total for five hours and $59$ minutes is $2+8+1=11$. The geometry from $3$am to $8$:$59$am is the same as $9$am to $2$:$59$pm. So the total is $11+11=22$. \par A precise solution can be based upon the angle increase between hands, which is $330^\circ$ per hour. Let $t$ be in hours and let $m$ stand for an integer. For a right angle to occur, $330t$ must be an integral multiple of $180$, i.e., $330t=180m$. The domain of $t$ is $0\le t \le 11\frac{59}{60}$. Therefore, $m$ satisfies $0\le m=\frac{330}{180}t\le \frac{330}{180}(11+\frac{59}{60})=\frac{7909}{360}<22$. This implies that $m$ can be any integer from $0$ to $21$, corresponding to $22$ times $t$. } \def\AA{ \prob{16.}{(Altitude)} In the figure, $MN$ is both a median and an altitude. Given $a=x+7$ and $b=2x-15$, find the unknown $x$. \par \rightans{c} \ans {$12$} {$16$} {$22$} {$44$\PAR} {None} } \def\BB{\input{isos9a.box}} \WRAPFIG{\AA}{\BB}{1.2in} \SOL{16. \STAT{74}{88}}{ The triangle must be isosceles, therefore $a=b$ and $2x-15=x+7$. Solving determines $x=22$. } \prob{17.}{(Median)} Let right $\triangle ABC$ have legs $12$ and $16$. Find the length of the shortest median to the nearest tenth. \rightans{c} \ans {$6.0$} {$8.0$} {$10.0$} {$14.4$} {$17.0$} \SOL{17. \STAT{56}{36}}{ Assume the triangle has vertices $(0,0)$, $(16,0)$ and $(0,12)$. 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 $(8,6)$. The length of that median is $10$, by the distance formula. The other medians have lengths $4\sqrt{13}$ and $2\sqrt{73}$, by application of the same method. } \prob{18.}{(Cube)} The diagonal of a cube is $96$cm. Find the volume in cm${}^3$. \rightans{d} \ans {$98294\sqrt{3}$} {$98296\sqrt{3}$} {$98300\sqrt{3}$\PAR} {$98304\sqrt{3}$} {$98306\sqrt{3}$} \SOL{18. \STAT{25}{45}}{ The diagonal $96$ is the hypotenuse of a right triangle of legs $a$ and $b$. Let $b$ equal the measure of a side of the cube. Then leg $a$ is the hypotenuse of a right triangle of sides $b$ and $b$, making $a^2=b^2+b^2$. Finally, $96^2=a^2+b^2=3b^2$ and $b=32\sqrt{3}$, giving volume $b^3=\sqrt{27}(2^{15})=98304\sqrt{3}$. } \prob{19.}{(Planes)} How many planes are determined by five points in space, no four of which are coplanar? \rightans{b} \ans {$8$} {$10$} {$11$} {$14$} {None} \SOL{19. \STAT{62}{39}}{ Each three set of points determines a plane, but taking all combinations leads to duplicate planes. The distinct ones number $10$ in all. } \prob{20.}{(Inequality)} Solve $|x-2|+|x+2|\le 4$. \rightans{b} \ans {$|x|\le 1$} {$|x|\le 2$} {$-1 \le x \le 2$\PAR} {$-2 \le x \le 1$} {None} \SOL{20. \STAT{71}{52}}{ Since $|x-2|= \pm (x-2)$ and $|x+2|=\pm (x+2)$ the left side of the inequality stands for one of $(x-2)+(x+2)$, $-(x-2)+(x+2)$, $(x-2)-(x+2)$ or $-(x-2)-(x+2)$. The cases are $2x\le 4$, $4\le 4$, $-4\le 4$ or $-2x\le 4$. The only conditions on $x$ are $-2\le x \le 2$, or $|x|\le 2$. } \prob{21.}{(Line)} Determine the value of $r$ such that the line through the points $(4,r)$ and $(r,2)$ has slope $-5/3$. \rightans{c} \ans {$-7$} {$1$} {$7$} {$-1$} {None} \SOL{21. \STAT{74}{55}}{ 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 $r=7$. } \prob{22.}{(Parallelogram)} Three vertices of a parallelogram are $(10,3)$, $(-1,2)$ and $(1,-1)$. Which point below is a possible fourth vertex? \rightans{a} \ans {$(-10,-2)$} {$(12,2)$} {$(8,5)$} {$(7,6)$\PAR} {None} \SOL{22. \STAT{77}{26}}{ Draw a diagram. Only one seems worth testing, namely $(-10,-2)$. Opposite sides have the same slope, so it's a parallelogram. } \prob{23.}{(Triangle)} 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{23. \STAT{77}{28}}{ 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. } \prob{24.}{(Bacteria)} A population of bacteria triples each day, reaching a size of $2187000$ in $144$ hours. Find the initial population. \rightans{a} \ans {$3000$} {$2500$} {$2000$} {$1000$} {None} \SOL{24. \STAT{92}{81}}{ Let $x$ be the initial population size. Then after $n$ hours the population is $3^{n/24}(x)$. After $144$ hours, $n/24=144/24=6$ and $3^6x=2187000$. Dividing by $3$ six times into $2187000$ gives $3000$. } \prob{25.}{(Remainder)} What is the remainder when $3^{251}$ is divided by $5$? \rightans{c} \ans {$4$} {$3$} {$2$} {$1$} {$0$} \SOL{25. \STAT{58}{40}}{ The first few powers are $3$, $3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$. The remainders after division by $5$ are $3$, $4$, $2$, $1$, $3$. The pattern repeats, each four causing the remainders to repeat. Since $251/4=62\frac{3}{4}$, the remainder is the third answer in the list of four repeated values, namely, $2$. } \prob{26.}{(Trapezoid)} Find the length of the median of a trapezoid with vertices $(0,0)$, $(0,3)$, $(4,4)$ and $(4,-1)$. \rightans{b} \ans {$5$} {$4$} {$3$} {$2$} {None} \SOL{26. \STAT{62}{74}}{ The parallel bases have lengths $3$ and $5$, by the distance formula. The median length is half of their sum, $(3+5)/2=4$. It can also be worked using the midpoint formula and a suitable diagram. } \prob{27.}{(Paint)} Two rectangular solids are similar with the ratios between corresponding sides at $3$:$1$. The smaller requires $x$ gallons to paint its surface. Find the number of gallons required to paint the surface of the larger. \rightans{c} \ans {$x^2$} {$3x$} {$9x$} {$27x$} {None} \SOL{27. \STAT{86}{35}}{ The dimensions $a$, $b$, $c$ of the smaller correspond to dimensions $3a$, $3b$, $3c$ of the larger. Therefore, the surface area of the larger solid is $9$ times larger than the surface area of the smaller solid. The paint requirement is $9x$ gallons. } \prob{28.}{(Series)} Sum the series $\frac12+ (\frac13+\frac23)+ (\frac14+\frac24+\frac34)+\\[3pt] (\frac15+\frac25+\frac35+\frac45)+ \cdots+ (\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+\cdots+\frac{99}{100})$. \rightans{a} \ans {$2475$} {$4950$} {$2476$} {$4952$} {None} \SOL{28. \STAT{42}{28}}{ The terms in parentheses can be simplified by using the formula $1+2+3+\cdots+n=n(n+1)/2$ (Gauss' childhood sum formula). The simplifications are $\frac12$, $\frac{2(3)/2}{3}$, $\frac{3(4)/2}{4}$, $\frac{4(5)/2}{5}$, \ldots, $\frac{99(100)/2}{100}$. This is half of $1+2+3+4+\cdots+99=99(99+1)/2=4950$. The series sums to $2475$. } \prob{29.}{(Polynomials)} 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{29. \STAT{27}{43}}{ 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{30.}{(Rectangle)} A rectangle has a perimeter of $42$cm. Which of the following could be its area in cm${}^2$? \rightans{e} \ans {$125$} {$121$} {$120$} {$116$} {$110$} \SOL{30. \STAT{75}{76}}{ 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. To obtain a rectangle of area $110$, solve $110=x(21-x)$ for $x=10$, $x=11$. Possible rectangles have dimensions $10$, $11$. } \smallskip {\large\bf Common Problems 31-40} There were 348 grade 7, 369 grade 8 and 441 grade 9 exams. Statistics are presented for each grade on the common problems 31--40. \smallskip %% ====== Separator problems for 7-8-9 competition, 31-40 \def\AA{ \prob{31.}{(Probability)} A moth lands on some square in the figure, then randomly moves just once horizontally or vertically to an adjacent square. Find the probability that the moth ends up on the square with the pentagon. } \def\BB{\input{moth8.box}} \WRAPFIG{\AA}{\BB}{0.8in} \rightans{a} \ans {$\frac{1}{4}$} {$\frac{1}{5}$} {$\frac{1}{24}$} {$\frac{1}{8}$} {None} \SOL{31.}{ {\bf \STAT{7th=76}{16}},\newline {\bf \STAT{8th=85}{16}}, {\bf \STAT{9th=78}{17}}.\newline Label the squares as $1$ through $6$, numbering $1$, $2$, $3$ across the top of the ``T'' ($2$ on the triangle) and $4$, $5$, $6$ down the column of the ``T'' ($5$ is the pentagon). An event is a pair $(a,b)$ where the moth lands on $a$ and moves to $b$. The possible outcomes are $(1,2)$, $(2,3)$, $(2,4)$, $(2,1)$, $(3,2)$, $(4,2)$, $(4,5)$, $(5,4)$, $(5,6)$ and $(6,5)$. The successful outcomes are $(4,5)$ and $(6,5)$ (they have to end up on $5$). However, the outcomes have different individual probabilities. They are not equally likely events, therefore simple counting of events does not suffice to solve the problem. The probability for event $(4,5)$ is $\frac16$ (one chance in six of landing on $4$) times $\frac12$ (one chance in two of moving to $5$), i.e., $\frac16\frac12=\frac{1}{12}$. The probability for event $(6,5)$ is $\frac16$ (one chance in six of landing on $4$) times $1$ (only one move is possible), i.e., $\frac16(1)=\frac{1}{6}$. The total probability is the sum $\frac{1}{12}+\frac{1}{6}=\frac{1}{4}$. } \def\AA{ \prob{32.}{(Triangles)} The biggest rectangle in the figure consists entirely of right triangles. Find the perimeter of this rectangle in feet, given $\tan A =\frac25$, $\tan B =\frac45$, $\tan C = \tan D =\frac12$, $\tan E = 4$, $\tan F = 1$, $\tan G =\frac13$ and $x=15$ ft. } \def\BB{\input{latrec8.box}} \WRAPFIG{\AA}{\BB}{1.45in} \rightans{a} \ans {$138$} {$139$} {$140$} {$142$} {$143$} {\footnotesize\em {\bf Def}: $\tan A$ in a right triangle is the quotient of opposite and adjacent sides.} \SOL{32.}{ {\bf \STAT{7th=22}{19}},\newline {\bf \STAT{8th=28}{32}}, {\bf \STAT{9th=26}{30}}.\newline The tangent of an angle is the quotient of the opposite and adjacent sides. Using this fact allows the sides to be determined as $33$ and $36$ feet, hence the perimeter is $138$. A difficulty arises with angle $F$, because to find the legs of that triangle it is necessary to observe that the legs are equal ($\tan F=1$). Let $x$ be the unknown length of the equal legs. By this stage the horizontal dimension of the rectangle is known to be $36$ft. So $\tan G=\frac13=\frac{x}{36-x}$, which determines $x=9$ and finally the vertical dimension of the rectangle is $33$ft. } \def\AA{ \prob{33.}{(Folding)} A rectangular sheet of paper is folded across a corner as shown in the figure, making angle $\theta$. Find $\tan\theta$. \par \rightans{d} \ans {$\frac67$} {$\frac34$} {$\frac78$} {$\frac{\sqrt{7}}{3}$\PAR} {$\frac{\sqrt{7}}{4}$} } \def\BB{\input {fold7.box}} \WRAPFIG{\AA}{\BB}{1.0in} \SOL{33.}{ {\bf \STAT{7th=28}{28}},\newline {\bf \STAT{8th=35}{37}}, {\bf \STAT{9th=53}{57}}.\newline When unfolded, the paper has a crease along the hypotenuse of a right triangle. The missing side $x$ satisfies by the Pythagorean theorem $x^2+6^2=8^2$, which can be solved for $x=2\sqrt{7}$. The tangent is the quotient of the sides of the right triangle, $\tan\theta=2\sqrt{7}/6=\sqrt{7}/3$. } \prob{34.}{(Jackpot)} 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 $A$. \rightans{b} \ans {$13$} {$14$} {$15$} {$16$} {$17$} \SOL{34.}{ {\bf \STAT{7th=81}{74}},\newline {\bf \STAT{8th=80}{73}}, {\bf \STAT{9th=81}{85}}.\newline 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 $A=14$. } \def\AA{ \prob{35.}{(Paper helix)} A paper sheet $10\times 25$ is cut by the method shown in the figure. By keeping the scissors always $1$ unit from the edge, cutting results in a ribbon of $1$-unit wide rectangles joined at right angles, which when lifted, forms a helix (a coil spring). Find the total perimeter. } \def\BB{\input {mow7.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{d} \ans {$216$} {$432$} {$430$} {$502$} {$503$} \SOL{35.}{ {\bf \STAT{7th=48}{17}},\newline {\bf \STAT{8th=45}{15}}, {\bf \STAT{9th=45}{11}}.\newline On the first path, the scissors travels $24$. Then $8$, $23$, $7$, $22$ and so on, the total being $24+8+23+7+22+\cdots+1+16=216$. The scissors creates two edges simultaneously, therefore the created perimeter is $216+216=432$. The $10\times 25$ rectangle perimeter is $70$, making the total perimeter $432+70=502$. } \def\AA{ \prob{36.}{(Pizza)} Jeff cuts unequal pieces from a round pizza using $4$ straight cuts. What is the greatest number of pieces he can cut? } \def\BB{\input{pizza7.box}} \WRAPFIG{\AA}{\BB}{0.6in} \rightans{d} \ans {$8$} {$9$} {$10$} {$11$} {$12$} \SOL{36.}{ {\bf \STAT{7th=93}{22}},\newline {\bf \STAT{8th=93}{31}}, {\bf \STAT{9th=92}{27}}.\newline By considering cases, $11$ is the maximum. This is found by making each new cut cross the maximum number of previous cuts. } \prob{37.}{(Chicken pen)} A triangular pen is constructed from $50$ feet of fencing, using sides of length $11.5$, $9.4$ and $x$ feet. The $x$ actually used appears below. Find it. \rightans{e} \ans {$30$ft} {$29.1$ft} {$22$ft} {$21$ft} {$16$ft} \SOL{37.}{ {\bf \STAT{7th=85}{2}},\newline {\bf \STAT{8th=86}{4}}, {\bf \STAT{9th=91}{6}}.\newline The two sides add to $11.5+9.4=20.9$. There is $50-20.9=29.1$ft of fencing remaining, but an allowed dimension must satisfy the triangle inequality. All given dimensions except $16$ violate the triangle inequality. } \def\AA{ \prob{38.}{(Clock)} A $12$-hour digital clock has $720$ different displays, $60$ per hour. How many displays have the sum of the digits greater than $15$? } \def\BB{\input{clock7.box}} \WRAPFIG{\AA}{\BB}{0.8in} \rightans{c} \ans {$100$} {$116$} {$120$} {$124$} {$125$} \SOL{38.}{ {\bf \STAT{7th=43}{37}},\newline {\bf \STAT{8th=45}{26}}, {\bf \STAT{9th=38}{32}}.\newline The greatest sum of digits that can result from the hours digits is $9$ and from the minutes digits $5+9=14$. The times $1$:$00$ to $1$:$59$ will therefore contribute no displays with sum of the digits greater than $15$. \par The times $2$:$00$ to $3$:$00$ produce only the display $2$:$59$pm, giving just one qualifying time. The next hour produces times $3$:$49$, $3$:$58$ and $3$:$59$. The following hour produces times $4$:$39$, $4$:$48$, $4$:$49$, $4$:$57$, $4$:$58$ and $4$:$59$. The pattern emerges as $0+1+3+6+10+15+21+27+33+0+1+3\cdots$, which is the sum of terms $n(n-1)/2$ from $n=1$ to $n=9$ less $1+3$ plus $0+1+3$. The total is $120$ displays (not $124$ as often claimed). } %count:=0: %for i from 1 to 12 do % u:=floor(i/10.): v:=i-10*u: % hourdigits:=u+v: hour:=i: % w:=0: % for j from 0 to 59 do % minute:=j: % u:=floor(minute/10.): v:=minute-10*u: % sumofdigits:=hourdigits+u+v: % if sumofdigits>15 then % count:=count+1: w:=w+1: % fi: % od: % print(w,` displays for time `,i); %od: %print(count,` displays found`); \prob{39.}{(Time)} Find the {\em exact} time between $4$ and $5$ when the minute hand of a watch is precisely $5$ minutes in advance of the hour hand. \rightans{d} \ans {$4$:$26\frac12$} {$4$:$26\frac34$} {$4$:$27\frac{1}{11}$} {$4$:$27\frac{3}{11}$\\[3pt]\MBL} {$4$:$28\frac{1}{11}$} \SOL{39.}{ {\bf \STAT{7th=52}{27}},\newline {\bf \STAT{8th=46}{29}}, {\bf \STAT{9th=39}{37}}.\newline Let $M$ be the reading of the minute hand in minutes after $4$:$00$. Let $H$ be the reading of the hour hand, read on the minutes scale. The time $4$:$00$ corresponds to $H=20$. And $H$ advances $5$ when the hour hand moves from $4$ to $5$. Then at a time $t$ minutes after $4$:$00$, $0\le t \le 60$, we have $M=t$ and $H=20+t/12$. The time sought is defined by $M=H+5$. Inserting relations gives $t=25+t/12$ which is solved for $t=300/11=27 + 3/11$ minutes after $4$. \par An alternative solution, in terms of angles in degrees, can be constructed as follows. Let $x$ be in minutes. The hour hand is at angle $120+30(x/60)$, the minute hand is at angle $360(x/60)$ and the clock position $5$ minutes on the dial after $4$:$00$pm is $30$. The equation to determine $x$ is $360(x/60) - 120 - 30(x/60)= 30$. Then $x=300/11$, giving the same answer as above. } \prob{40.}{(Intercepts)} How many lines through $(4,3)$ have intercepts $(a,0)$ and $(0,b)$ with $a$ and $b$ being positive integers? \rightans{c} \ans {$3$} {$5$} {$6$} {$7$} {Infinitely many} \SOL{40.}{ {\bf \STAT{7th=65}{9}},\newline {\bf \STAT{8th=69}{5}}, {\bf \STAT{9th=71}{6}}.\newline Draw a line through $(a,0)$, $(4,3)$ and $(0,b)$. The slopes of the two segments in the figure are $3/(4-a)$ and $(b-3)/(-4)$. They are equal, which gives the equation $(a-4)(b-3)=12$. By geometry, the values of $a$ are restrained to $5 \le a$ and the values of $b$ are restricted to $b\ge 4$. The question reduces to determining which positive integers $p$ and $q$ satisfy $pq=12$. This finite list is easily written down. Then $a=p+4$ and $b=q+3$. The possible pairs $[a,b]$ are $[5,15]$, $[6,9]$, $[7,7]$, $[8,6]$, $[10,5]$, $[16,4]$. There are six possible lines. } \end{exercises} %\end{multicols} %% 10.tex \begin{center}\Large\bf 1995 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 1995 Exam Grade 10} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 10 \setanswer{1.}{d} &\setanswer{11.}{a} &\setanswer{21.}{b} &\setanswer{31.}{e} \\ \setanswer{2.}{a} &\setanswer{12.}{a} &\setanswer{22.}{b} &\setanswer{32.}{d} \\ \setanswer{3.}{c} &\setanswer{13.}{c} &\setanswer{23.}{d} &\setanswer{33.}{a} \\ \setanswer{4.}{a} &\setanswer{14.}{c} &\setanswer{24.}{b} &\setanswer{34.}{b} \\ \setanswer{5.}{d} &\setanswer{15.}{c} &\setanswer{25.}{a} &\setanswer{35.}{d} \\ \setanswer{6.}{b} &\setanswer{16.}{b} &\setanswer{26.}{c} &\setanswer{36.}{d} \\ \setanswer{7.}{c} &\setanswer{17.}{c} &\setanswer{27.}{a} &\setanswer{37.}{d} \\ \setanswer{8.}{b} &\setanswer{18.}{c} &\setanswer{28.}{e} &\setanswer{38.}{b} \\ \setanswer{9.}{b} &\setanswer{19.}{b} &\setanswer{29.}{a} &\setanswer{39.}{c} \\ \setanswer{10.}{b} &\setanswer{20.}{b} &\setanswer{30.}{c} &\setanswer{40.}{e} \\ \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} \prob{1.}{(Coolant)} Mix $3$, $5$ and $7$ liters of coolant at temperatures $60^\circ$, $67^\circ$ 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{1. \STAT{82}{68}}{ The mixture equation is $(3+5+7)x=3(60)+5(67)+7(73)$. The answer is $x=342/5=68.4$. The closest answer is $68$. } \prob{2.}{(Tangent)} Find the equation of the line through $(4,3)$ and tangent to the circle of radius $5$ and center $(0,0)$. \rightans{a} \ans {$4x+3y=25$} {$3x+4y=24$} {$4x-3y=7$} {$3x-4y=0$} {None} \SOL{2. \STAT{62}{54}}{ The equation is $y-3=m(x-4)$. The point $(4,3)$ is on the circle, therefore $m$ is the negative reciprocal of the slope of the radial line through $(4,3)$. That line has slope $3/4$, which implies $m=-4/3$. } \prob{3.}{(Roots)} Apply the rational root theorem to the cubic $x^3-8x^2-7935=0$ and match a {\em candidate} for a root from the list below. \rightans{c} \ans {$35$} {$71$} {$115$} {$2643$} {$-343$} \SOL{3. \STAT{33}{55}}{ By the theorem, a potential root must divide $7935$. The only divisor from the list is $115$.} \prob{4.}{(Guessing)} Jack guesses on $10$ true-false questions. Find the probability that he gets exactly $8$ correct. \rightans{a} \ans {$45/1024$} {$4/5$} {$1/8$} {$33/1024$\PAR} {$45/512$} \SOL{4. \STAT{44}{31}}{ 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 $45/1024$.} \prob{5.}{(Determinant)} Evaluate $\left|\begin{array}{rrr} 1 & 1 & 1 \\ a & 1 & -2 \\ a^2 & 1 & 4 \end{array}\right|.$ \rightans{d} \ans {$2+5a-a^2$} {$(3+a)(2+a)$} {$-(1+a)(2+a)$} {$(3-3a)(2+a)$} {None} \SOL{5. \STAT{37}{39}}{ Since $a=1$ makes the determinant zero, because of equal columns, then only $(3-3a)(2+a)$ is a candidate. The determinant has value $(-2-1)(-2-a)(1-a)$ by direct expansion using Sarrus' rule or a cofactor expansion along column one. } \prob{6.}{(Ratio)} The perimeter of a triangle is $72$ units and the sides have measures in the ratios $3$:$4$:$5$. Find the number of units in the shortest side. \rightans{b} \ans {$12$} {$18$} {$24$} {$30$} {None} \SOL{6. \STAT{90}{88}}{ The measures of the sides can be written as $3x$, $4x$ and $5x$. The perimeter is $3x+4x+5x=72$. Then $x=6$, but $x$ is not a side; the sides are $3x=18$, $4x=24$ and $5x=30$. } \prob{7.}{(Straws)} Five straws of lengths $3$, $4$, $5$, $6$ and $12$ units are used to form different triangles. How many have perimeter divisible by $3$? \rightans{c} \ans {$4$} {$3$} {$2$} {$1$} {None} \SOL{7. \STAT{89}{21}}{ There are $4$ possible triangles, 3-4-5, 3-4-6, 3-5-6 and 4-5-6. Only two have perimeter divisible by $3$. } \prob{8.}{(Polygons)} Which exterior angle is impossible for a regular polygon? \rightans{b} \ans {$9^\circ$} {$16^\circ$} {$22\frac12^\circ$} {$45^\circ$} {$51\frac37^\circ$} \SOL{8. \STAT{66}{45}}{ The number $n$ of sides of a regular polygon and the number $x$ of degrees in an exterior angle satisfy the relation $n=360/x$. All of the answers $x$ produce an integer for $n$, except for $x=16$, which gives the impossibility $n=360/16=22.5$. } \prob{9.}{(Pentagon)} Find $x$, given a pentagon has exterior angles $63$, $75$, $58$, $x$, $3x$ (in degrees). \rightans{b} \ans {$39^\circ$} {$41^\circ$} {$43^\circ$} {$45^\circ$} {None} \SOL{9. \STAT{75}{83}}{ The sum of the exterior angles must be $360^\circ$. This produces the equation $360=63+75+58+x+3x$ for the unknown $x$. Solving, $x=41$. } \prob{10.}{(Midway)} The line $5x+8y=4.5$ is midway between and parallel to the parallel lines $5x+8y=3$ and $5x+8y=c$. Which is a possible value of $c$? \rightans{b} \ans {$5$} {$6$} {$7$} {$8$} {$9$} \SOL{10. \STAT{81}{90}}{ Graph the lines. The ``midway'' requirement means $\frac{1}{2}(3+c)=4.5$, giving $c=6$. } \def\AA{ \prob{11.}{(Geometry)} In the diagram, let segments $\bar{AB}$ and $\bar{DE}$ be parallel and let $AC=2$, $CD=6$ and $CE=3$. Find $BC$. \par \rightans{a} \ans {$4$} {$5$} {$6$} {$7$\PAR} {None} } \def\BB{\input{geom10a.box}} \WRAPFIG{\AA}{\BB}{1.1in} \SOL{11. \STAT{85}{89}}{ Let $x=BC$. The triangles $ABC$ and $EDC$ are similar, therefore $\dd\frac{3}{2}=\frac{6}{x}$, which can be solved for $x=4$. } \def\AA{ \prob{12.}{(Stop)} A stop sign with eight sides is to be cut from sheet metal that is stocked in square pieces. Given the stop sign edges are $1$ ft, find the square footage of an uncut stock piece which minimizes waste. } \def\BB{\input{stop.box}} \WRAPFIG{\AA}{\BB}{0.85in} \rightans{a} \ans {$3+2\sqrt2$} {$2+3\sqrt2$} {$2+2\sqrt3$\PAR} {$4+2\sqrt5$} {$8$} \SOL{12. \STAT{56}{66}}{ The minimum square will touch on four sides of the octagon. The stop sign is made by cutting off corners of the square, leaving a right triangle of hypotenuse $1$ with legs $x$ and $x$. The Pythagorean theorem implies $2x^2=1$ or $x=1/\sqrt{2}$. The side of the square is $1+2x=1+\sqrt{2}$ and the area is $(1+\sqrt2)^2=3+2\sqrt{2}$. } \prob{13.}{(Population)} In $1990$, the population of a city was $12298$, a decrease of $8\frac{1}{3}$\% from its $1980$ population. The $1980$ population was an increase of $7\frac{1}{2}$\% from its $1970$ population. Find the $1970$ population. \rightans{c} \ans {$12360$} {$12380$} {$12480$} {$12500$\PAR} {None} \SOL{13. \STAT{80}{50}}{ Let $P$ and $Q$ be the populations in 1970 and 1980, respectively. Then $8\frac{1}{3}$\% is $1/12$ and $7\frac{1}{2}$\% is $3/40$, giving $12298=Q-Q/12$ and $Q=P+3P/40$. Then $P=\frac{40}{43}Q=\frac{40}{43}\frac{12}{11}12298=12480$. } \prob{14.}{(Integers)} If the product of three positive consecutive integers is divided by each of them in turn, then the sum of the three quotients is $74$. Find the smallest of the three integers. \rightans{c} \ans {$2$} {$3$} {$4$} {$5$} {$-6$} \SOL{14. \STAT{74}{77}}{ Let $N=x(x+1)(x+2)$. Then $\dd\frac{N}{x}+\frac{N}{x+1}+\frac{N}{x+2}=74$. This gives the quadratic equation $x^2+2x-24=0$ which has roots $4$ and $-6$. The answer is $x=4$, because $x>0$. } \prob{15.}{(Symmetry)} How many lines of symmetry are there for the region inside a non-rectangular rhombus? \rightans{c} \ans {$4$} {$3$} {$2$} {$1$} {$0$} \SOL{15. \STAT{78}{48}}{ 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. Only the diagonals of a rhombus have this property, unless the rhombus is a square. } \def\AA{ \prob{16.}{(Altitude)} Given right triangle $ABC$ with hypotenuse $AB$ and measures $AD=5$, $DB=8$, then find altitude $x=CD$. } \def\BB{\input{tri10.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{b} \ans {$\sqrt{10}$} {$2\sqrt{10}$} {$2\sqrt{11}$} {$\frac53\sqrt{13}$\PAR} {$\frac74\sqrt{13}$} \SOL{16. \STAT{44}{79}}{ The hypotenuse $b$ of the small triangle can be computed by the geometric mean equation $\frac{5}{b}=\frac{b}{13}$. Then $b^2=65$. By the Pythagorean theorem, $b^2=5^2+x^2$, hence $x^2=40$ and $x=2\sqrt{10}$. This solution is equivalent to a trigonometric approach which equates the tangents of angles $CAB$ and $CBD$. } \def\AA{ \prob{17.}{(Area)} In the big square figure, assume shapes $A$ through $G$ are also squares. If $A$ has area $64$ cm${}^2$ and $B$ has area $49$ cm${}^2$, then find the area in cm${}^2$ of square $C$. \par \rightans{c} \ans {$100$} {$121$} {$169$\PAR} {$225$} {$324$} } \def\BB{\input{squar10.box}} \WRAPFIG{\AA}{\BB}{1.3in} \SOL{17. \STAT{81}{51}}{ 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 $32-10-9=13$ with area $13^2=169$. } \def\AA{ \prob{18.}{(Triangle)} The triangle in the figure has sides of $14$, $9$ and $17$ units. Point $B$ is on the inscribed circle. Find the number of units in $AB$. } \def\BB{\input{tri10a.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{c} \ans {$9$} {$10$} {$11$} {$12$} {None} \SOL{18. \STAT{28}{26}}{ Let $x=AB$. Let the sides of lengths $14$, $9$ and $17$ be divided in ratios $x$:$z$, $z$:$y$ and $y$:$x$, respectively, by the contact points with the inscribed circle. We used: {\em If two segments from the same exterior point are tangent to a circle, then they are congruent}. Then $x+z=14$, $y+z=9$ and $y+x=17$. Solving these three equations in three unknown gives $x=11$, $y=6$, $z=3$. } \prob{19.}{(Midpoint)} Given $A(x,-2x+1)$ and $B(-x,2x-1)$, find the midpoint of segment $\bar{AB}$. \rightans{b} \ans {$(x,x)$} {$(0,0)$} {$(-1,0)$} {$(0,-1)$\PAR} {None} \SOL{19. \STAT{74}{87}}{ The midpoint is $((x-x)/2,(-2x+1+2x-1)/2)=(0,0)$. } \prob{20.}{(Polygon)} The sum of the measures of the interior angles in a polygon is $24840$. Find the number of sides. \rightans{b} \ans {$138$} {$140$} {$142$} {$144$} {$145$} \SOL{20. \STAT{71}{49}}{ The equation for the number $n$ of sides is $180(n-2)=24840$. Solving gives $n=140$. } \def\AA{ \prob{21.}{(Angles)} Given angles $a=50^\circ$ and $d=65^\circ$ as in the figure, then find the sum of the measures of angles $ECA$ and $BCD$. \par \rightans{b} \ans {$220$} {$230$} {$232$} {$245$} {$250$} } \def\BB{\input{circ10.box}} \WRAPFIG{\AA}{\BB}{0.9in} \SOL{21. \STAT{36}{53}}{ The two triangles are isosceles. The base angles adjacent to the circle are $65$, $65$ and $50$, $50$, respectively. So the central angles used by the triangles are $180-65-65=50$ and $180-50-50=80$. These two central angles sum to $50+80=130$. Then $a+d+130=360$, or $a+d=230$. } \def\AA{ \prob{22.}{(Chord)} If $B$ is a point on a semicircle $XBY$ with $\bar{AB} \perp \bar{XY}$ at point $A$, $XA=2\frac{1}{12}$ and $AY=12$, then find $BY$. } \def\BB{\input{chord10a.box}} \WRAPFIG{\AA}{\BB}{1.2in} \rightans{b} \ans {$\frac{25}{2}$} {$13$} {$9\sqrt2$} {$8\sqrt3$} {$6\sqrt5$} \SOL{22. \STAT{38}{73}}{ Let $x$ be the unknown length $BY$. Let $u=XB$ and $v=AB$. The Pythagorean theorem implies $x^2=v^2+12^2$ and $u^2=(2+\frac{1}{12})^2+v^2$. Since angle $B$ is $90^\circ$, then the Pythagorean theorem implies $u^2+x^2=(12+2+\frac{1}{12})^2$. Eliminating between the two equatons for $x^2$ gives $u^2+v^2=50+(25/12)^2$. Using the other equation for $u^2$ and $v^2$ gives $v=5$. Then $x^2=v^2+12^2=13^2$ and $x=13$. } \def\AA{ \prob{23.}{(Perimeter)} The six congruent squares in the figure have total area $486$ cm${}^2$. Find the perimeter of the figure in cm. } \def\BB{\input{perim10.box}} \WRAPFIG{\AA}{\BB}{0.5in} \rightans{d} \ans {$120$} {$122$} {$124$} {$126$} {$130$} \SOL{23. \STAT{90}{93}}{ Let $x$ be the side of a square. Then $6x^2=486$, giving $x=9$. The perimeter, using the figure, is $14x=(14)(9)=126$. } \def\AA{ \prob{24.}{(Frustrum)} The figure is a right circular cone of base radius $12$ cm that has been cut off parallel to the base at height $12$ cm. Find the lateral surface area in cm${}^2$. } \def\BB{\input{frustrum.box}} \WRAPFIG{\AA}{\BB}{1in} \rightans{b} \ans {$245\pi$} {$247\pi$} {$248\pi$} {$250\pi$} {$256\pi$} \SOL{24. \STAT{18}{47}}{ The portion shown in the figure is called a {\em frustrum of a cone}. The lateral area of a frustrum is the difference of the lateral areas of the larger cone and the smaller cone. Lateral areas are computed for cones by the formula $L=\pi R\ell$, where $R$ is the radius of the base and $\ell$ is the slant height of the cone. The lateral area does not include the area of the circular base. The slant height of the frustrum is $13=\sqrt{5^2+12^2}$, using the Pythagorean theorem. The slant height $x$ of the smaller excised cone must satisfy by similar triangles the equation $\frac{x}{7}=\frac{13}{5}$. The difference of the lateral areas of the larger and smaller cones is $12\pi(13+x)-7\pi x=156\pi+5\pi x =156\pi + 91\pi=247\pi$. } \prob{25.}{(Prism)} Find the surface area in cm${}^2$ of a right prism of height $15$ cm and right triangular base with legs of $3$ cm and $4$ cm. \rightans{a} \ans {$192$} {$180$} {$240$} {$190$} {$220$} \SOL{25. \STAT{41}{60}}{ The surface area is the sum of the areas of the faces. There are two triangular faces, each of area $\frac12(3)(4)$, making $12$. There are three rectangular faces, of dimensions $15\times 3$, $15\times 4$ and $15 \times \sqrt{3^2+4^2}$. The last dimension is obtained by an application of the Pythagorean theorem. The total area is $12+(15)(3+4+5)=192$. } \prob{26.}{(Complex)} Find the conjugate inverse of the complex number $\frac{1}{z^2}$ where $z=\frac{i}{1-i}$ and $i=\sqrt{-1}$. \rightans{c} \ans {$i$} {$-i$} {$\frac{i}{2}$} {$\frac{-i}{2}$} {None} \SOL{26. \STAT{48}{28}}{ The inverse is $z^2$. The conjugate of $z^2$ is the square of the conjugate of $z$. Then $\bar{z}= \frac{-i}{1+i}$ implies $\bar{z}^2=\frac{i^2}{(1+i)^2}=\frac{-1}{2i}=\frac{i}{2}$. } \prob{27.}{(Magnitude)} Let $f(x)=\sin^2(x)+\cos(x)$, $g(x)=\frac{2x^2+100x+11}{(3x-1)(x+1)}$, $h(x)=e^{x^2}\cos(x)$. If $x$ is an odd integer multiple of $\pi/2$ and $x>10^{10}$, then which function value below is largest? \rightans{a} \ans {$f(x)$} {$g(x)$} {$h(x)$} {$f(x)g(x)$\PAR} {$f(x)h(x)$} \SOL{27. \STAT{23}{19}}{ All values of $h(x)$ are zero. The value of $g(x)$ is about $2/3$, because by the division algorithm for polynomials, $g(x)=2/3+r(x)$ where $r(x)$ is nearly zero. The value of $f(x)$ is $1$ for odd integer multiples of $\pi/2$. The largest value is given by $f(x)$. } \prob{28.}{(Disk)} Select from the list below a point on the line $2x+3y=1$ which is also inside the circle of radius $3$ and center $(-1,2)$. \rightans{e} \ans {$(1,-\frac13)$} {$(-4,3)$} {$(\frac98,-\frac{5}{12})$} {$(\frac54,-\frac12)$\PAR} {$(-\frac{15}{4},\frac{17}{6})$} \SOL{28. \STAT{54}{39}}{ The circle equation is $(x+1)^2+(y-2)^2=9$. Insert the line equation $y=(1-2x)/3$ into the circle equation to find the $x$-values where the line crosses the circle. They occur at approximately $0.9$ and $-3.9$. So it is difficult to eliminate any of the answers. The best attack is to check which ones are on the line and then test them against the inequality $(x+1)^2+(y-2)^2<9$. All five points are on the line $2x+3y=1$. The values of $(x+1)^2+(y-2)^2$ at the five points are $9.4$, $10$, $10.4$, $9.4$, $8.3$. Therefore, $(-\frac{15}{4},\frac{17}{6})$ qualifies as on the line and inside the circle. } \prob{29.}{(Series)} Sum the series $\frac12+ (\frac13+\frac23)+ (\frac14+\frac24+\frac34)+\\[3pt] (\frac15+\frac25+\frac35+\frac45)+ \cdots+ (\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+\cdots+\frac{99}{100})$. \rightans{a} \ans {$2475$} {$4950$} {$2476$} {$4952$} {None} \SOL{29. \STAT{38}{43}}{ The terms in parentheses can be simplified by using the formula $1+2+3+\cdots+n=n(n+1)/2$ (Gauss' childhood sum formula). The simplifications are $\frac12$, $\frac{2(3)/2}{3}$, $\frac{3(4)/2}{4}$, $\frac{4(5)/2}{5}$, \ldots, $\frac{99(100)/2}{100}$. This is half of $1+2+3+4+\cdots+99=99(99+1)/2=4950$. The series sum is $2475$. } \prob{30.}{(Polynomials)} Let $p$ be a quintic polynomial in $x$ with roots $1-2i$, $1+2i$, $\sqrt{2}$, $-\sqrt{2}$ and $1$. Assume $p=160$ when $x=-1$. Find $p$ when $x=\frac32$. \rightans{c} \ans {$\frac{23}{2}$} {$12$} {$\frac{25}{2}$} {$13$} {$\frac{125}{4}$} \SOL{30. \STAT{12}{29}}{ By the factor theorem, $(x-1+2i)$, $(x-1-2i)$, $(x-\sqrt2)$, $(x+\sqrt2)$ and $(x-1)$ are factors of $p$. The polynomial $p$ must be a constant multiple of $(x^2+4)(x^2-2)(x-1)$. Briefly, this means $p=c(x^2+4)(x^2-2)(x-1)$ for some constant $c$. In expanded form, $p=c({x}^{5}-{x}^{4}+2\,{x}^{3}-2\,{x}^{2}-8\,x+8)$. Since $p=160$ at $x=-1$, then $160=c(1+4)(1-2)(-1-1)$ or $c=16$. The value of $p$ at $x=\frac32$ is $16(\frac94+4)(\frac94-2)(\frac32-1)=\frac{25}{2}$. } See {\em infra} for problems 31-40. \end{exercises} %% 11.tex \begin{center}\Large\bf 1995 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 1995 Exam Grade 11} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 11 \setanswer{1.}{e} &\setanswer{11.}{b} &\setanswer{21.}{e} &\setanswer{31.}{e} \\ \setanswer{2.}{a} &\setanswer{12.}{a} &\setanswer{22.}{e} &\setanswer{32.}{d} \\ \setanswer{3.}{a} &\setanswer{13.}{c} &\setanswer{23.}{d} &\setanswer{33.}{a} \\ \setanswer{4.}{b} &\setanswer{14.}{d} &\setanswer{24.}{e} &\setanswer{34.}{b} \\ \setanswer{5.}{a} &\setanswer{15.}{e} &\setanswer{25.}{b} &\setanswer{35.}{d} \\ \setanswer{6.}{d} &\setanswer{16.}{d} &\setanswer{26.}{c} &\setanswer{36.}{d} \\ \setanswer{7.}{b} &\setanswer{17.}{a} &\setanswer{27.}{d} &\setanswer{37.}{d} \\ \setanswer{8.}{e} &\setanswer{18.}{d} &\setanswer{28.}{d} &\setanswer{38.}{b} \\ \setanswer{9.}{a} &\setanswer{19.}{e} &\setanswer{29.}{e} &\setanswer{39.}{c} \\ \setanswer{10.}{a} &\setanswer{20.}{c} &\setanswer{30.}{a} &\setanswer{40.}{e} \\ \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 \prob{1.}{(Distance)} Find the distance between the points $(-1,2,4)$ and $(1,-2,2)$. \rightans{e} \ans {$2\sqrt{2}$} {$2\sqrt{3}$} {$4$} {$2\sqrt{5}$} {$2\sqrt{6}$} \SOL{1. \STAT{67}{72}}{ Use the distance formula $\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$ to obtain $\sqrt{((2)^2 +(-4)^2+(-2)^2}=2\sqrt6$. } % 11th \prob{2.}{(Area)} Find the approximate area of a regular pentagon of perimeter $45$ cm. \rightans{a} \ans {$101.25\cot(\frac{\pi}{5})$} {$101.25\cot(\frac{2\pi}{5})$\PAR} {$101\cot(\frac{\pi}{5})$} {$101\cot(\frac{2\pi}{5})$} {$100\cot(\frac{\pi}{5})$} \SOL{2. \STAT{31}{47}}{ The length of each side is $45/5=9$. The length of an apothem is $(9/2)/\tan(36^\circ)$, because each central angle is $360/5=72$. The area is $\frac12$ the perimeter times the apothem, or $\frac{45}{2}\frac{9}{2}\cot(36^\circ)$. The radian measure of $36^\circ$ is $\frac{\pi}{5}$. } % 11th \prob{3.}{(Shadow)} When the angle of elevation of the sun is $44^\circ$, the shadow cast by a building is $35$ meters. Approximate the building height in meters. \rightans{a} \ans {$33.8$} {$35.1$} {$36.1$} {$36.5$} {$36.9$} \SOL{3. \STAT{85}{84}}{ Draw a figure using a $44$-$46$-$90$ triangle with legs $35$ and $x$. Then $x/35=\tan(44^\circ)$. The answer is $x=35\tan(44^\circ)=33.8$. If the angle was $45$, then the answer would be $35\tan(45^\circ)=35(1)=35$. The only choice is $33.8$. } % 11th \prob{4.}{(Planes)} How many planes are determined by five points in space, no four of which are coplanar? \rightans{b} \ans {$8$} {$10$} {$11$} {$14$} {None} \SOL{4. \STAT{61}{50}}{ Each three set of points determines a plane, but taking all combinations leads to duplicate planes. The distinct ones number $10$ in all. } % 11th \prob{5.}{(Traffic)} Three out of five cars arriving at a signal light have a green or yellow light. Estimate the probability that exactly $4$ of the next $7$ cars will have a red light. \rightans{a} \ans {$0.19$} {$0.20$} {$0.22$} {$0.24$} {$0.29$} \SOL{5. \STAT{38}{25}}{ Let $p=\frac{2}{5}$, $q=\frac{3}{5}$. Then $p+q=1$ and $p$ is the probability of a red light for one car. The probability of exactly four out of the next seven being red is the term of the binomial expansion $(p+q)^7$ with power $p^4q^3$. This term is $\choose{7}{4}p^4q^3=35(\frac25)^4(\frac35)^3=0.193536$. } % 11th \prob{6.}{(Rhombus)} A rhombus with each side $26$ cm has one diagonal $48$ cm. Find the number of centimeters in the other diagonal. \rightans{d} \ans {$10$} {$14$} {$16$} {$20$} {$40$} \SOL{6. \STAT{64}{64}}{ The diagonals meet at right angles and bisect each other, therefore the other diagonal $x$ must satisfy, by the Pythagorean theorem, $24^2+(x/2)^2=26^2$. Solving, $x=20$. } % 11th \prob{7.}{(Rectangle)} Find the larger dimension of a rectangle of perimeter $32$ units and area $48$ square units. \rightans{b} \ans {$11$} {$12$} {$13$} {$14$} {$4$} \SOL{7. \STAT{88}{94}}{ The sides $a$ and $b$ of the rectangle satisfy $2a+2b=32$, which implies the dimensions are $a$ and $16-a$. So the area is $a(16-a)=48$. Trying numbers for $a$ from $1$ to $15$ gives $a=4$. The larger of $4$ and $16-4$ is $12$. } % 11th \prob{8.}{(Volume)} Given a regular right prism $40$ cm high whose cross section is a regular hexagon of perimeter $30$ cm, find the volume in cm${}^3$. \rightans{e} \ans {$500\sqrt3$} {$1837\sqrt2$} {$612\sqrt2$} {$2598$\PAR} {$1500\sqrt3$} \SOL{8. \STAT{39}{44}}{ The volume is the base area $B$ times the height $40$. To find $B$, divide the hexagon into $6$ equilateral triangles of side $30/6=5$ cm. Each equilateral triangle has area $25\sqrt3/4$, so six of them gives $B=75\sqrt3/2$. The volume is $40B=(20)(75)\sqrt3=2598.1$. } \prob{9.}{(Interest)} Find the least number of full years for a deposit to double using continuous interest at $7$\% per annum. Use $\ln(2)=0.693$. \rightans{a} \ans {$10$} {$9$} {$8$} {$7$} {$6$} \SOL{9. \STAT{58}{73}}{ Let $x(t)$ be the amount and $x(0)$ the initial amount. Then $x(t)=x(0)e^{rt}$ where $r=7/100$ is the per annum interest rate. Doubling means $x(t)=2x(0)$, therefore the equation to be solved for $t$ is $2x(0)=x(0)e^{rt}$. Cancel $x(0)$ and take logarithms to obtain $\ln(2)=\ln(e^{rt})$ and finally $t=\frac{1}{r}\ln(2)=\frac{100}{7}\ln(2)$, which is about $9.9$ years. } \prob{10.}{(Distance)} The curves $\frac{x^2}{9}+\frac{y^2}{16}=1$ and $x+y=1$ meet in exactly two points $P$ and $Q$. Find the distance from $P$ to $Q$. \rightans{a} \ans {$\frac{96}{25}\sqrt3$} {$\frac{19}{5}\sqrt3$} {$\frac{48}{13}\sqrt3$} {$\frac{47}{13}\sqrt3$\PAR} {$4\sqrt3$} \SOL{10. \STAT{36}{45}}{ The points $P$ and $Q$ have the form $(x_1,1-x_1)$ and $(x_2,1-x_2)$ where $x_1$ and $x_2$ are the roots of the quadratic equation $x^2/9+(1-x)^2/16=1$. The distance is $\sqrt{(x_1-x_2)^2+(-x_2+x_1)^2}=\sqrt{2}|x_1-x_2|$. Since the roots are $(9\pm 24\sqrt6)/25$, the answer is $\sqrt2(48\sqrt6/25)=96\sqrt3/25$. } \prob{11.}{(Inverse)} Which expression below for $f(x)$ is also the expression for $f^{-1}(x)$? \rightans{b} \ans {$\frac{1}{x^2}$} {$-x$} {$\log(x)$} {$\frac{x}{1+x}$} {None} \SOL{11. \STAT{82}{47}}{ To test (a), let $y$ be the given expression, swap $x$ and $y$ to get $x=1/y^2$, then solve for $y$ in terms of $x$. This results in $y=1/\sqrt{x}$, therefore (a) does not qualify as the inverse of itself. Both (c) and (d) fail, but (b) works, because $y=-x$ swaps to $x=-y$ and then $y=-x$, reproducing the original equation. } \prob{12.}{(Sequence)} Find the value of $\dd a_{12}/b_{9}$, given\newline $a_n=32000(0.8)^{2n+1}$ and $b_n=16000(0.4)^{3n-2}$. \rightans{a} \ans {$2^{26}$} {$2^{25}$} {$2^{-26}$} {$2^{-25}$} {$2^{24}$} \SOL{12. \STAT{61}{79}}{ The values are $a_{12}=32000(0.8)^{25}$ and $b_{9}=16000(0.4)^{25}$. Then $a_{12}/b_{9}=2(0.8)^{25}/(0.4)^{25}=2(0.8/0.4)^{25}= 2^{26}$. } \def\AA{ \prob{13.}{(Skid)} Viktor measured skid marks when the brakes of his Volvo were applied at different speeds. Use a simple data fitting model to estimate how far in feet the Volvo would skid if the brakes were applied at $70$ mph. } \def\BB{ {\small\rm \begin{tabular}{|c|c|} \hline {\bf ft} & {\bf mph} \\ \hline $18$ & $20$ \\ \hline $41$ & $30$ \\ \hline $72$ & $40$ \\ \hline $113$ & $50$ \\ \hline $162$ & $60$ \\ \hline \end{tabular} } } \WRAPFIG{\AA}{\BB}{1in} \rightans{c} \ans {$201$} {$211$} {$221$} {$232$} {$241$} %72./40^2; %p:=[[20,18],[30,41],[40,72],[50,113],[60,162]]; %for i from 1 to 5 do print(0.045*(p[i][1])^2); od; \SOL{13. \STAT{71}{79}}{ 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 $k=18/20^2=0.045$. Testing $y=0.045x^2$ reproduces the table. The prediction is $y$ at $x=70$, which is $y=0.045(70)^2=220.5$ feet. } \def\AA{ \prob{14.}{(Inscribed)} The quadrilateral inscribed in the circle has angles $a=2x$, $b=2x+10$, $c=2x-10$, all in degrees. Find the measure of angle $d$ in degrees. } \def\BB{\input{inscribe.box}} \WRAPFIG{\AA}{\BB}{0.7in} \rightans{d} \ans {$85$} {$95$} {$100$} {$105$} {$110$} \SOL{14. \STAT{41}{45}}{ Opposite angles in an inscribed quadrilateral add to $180^\circ$. This produces from $a+b=180$ the equation $2x+2x+10=180$ and finally $x=85/2$. Then $c=2x-10=75$. Finally, $c+d=180$, giving $d=105$. } \prob{15.}{(Scaling)} Given rectangle $ABCD$ where $A=(-2,4)$, $B=(-2,-4)$, $C=(-6,-4)$, $D=(-6,4)$, let $T$ be the image of $ABCD$ obtained by applying the matrix product $\left(\begin{array}{cc}5 & 0\\ 0 & 5\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)$ to each point $(x,y)$ of $ABCD$. Find the area of $T$ in square units. \rightans{e} \ans {$160$} {$320$} {$400$} {$600$} {$800$} \SOL{15. \STAT{46}{52}}{ The matrix applies a scale factor of $5$ to the figure, therefore the edges of the image figure are $5$ times longer than the original. Rectangle $ABCD$ has sides $4$ and $8$, so $T$ is a rectangle with sides $20$ and $40$. The area of $T$ is $(20)(40)=800$. } \prob{16.}{(Bisector)} Find the $y$-intercept of the perpendicular bisector of the segment joining $(9,5)$ and $(-7,11)$. \rightans{d} \ans {$5$} {$\frac{31}{6}$} {$\frac{11}{2}$} {$\frac{16}{3}$} {$\frac{17}{3}$} \SOL{16. \STAT{55}{70}}{ The segment has slope $m=(11-5)/(-7-9)=-3/8$, therefore the line perpendicular to it has slope $-1/m=8/3$. The midpoint of the segment is $(\frac{9+(-7)}{2},\frac{5+11}{2})=(1,8)$. The perpendicular bisector therefore has equation $y-8=(8/3)(x-1)$. It has $y$-intercept satisfying the equation $y-8=(8/3)(0-1)$, which gives $y=16/3$. } % 11th \prob{17.}{(Systems)} % A:=matrix([[1,2,1],[1,-4,2],[2,-2,3]]);augment(A,vector([1,1,2])); % rref("); Count the number of solutions for\par \centerline{ $\left(\begin{array}{rrr} 1 & 2 & 1 \\ 1 & -4 & 2 \\ 2 & -2 & 3 \end{array}\right) \left(\begin{array}{r} x \\ y \\ z \end{array}\right) = \left(\begin{array}{r} 1 \\ 1 \\ 2 \end{array}\right).$} \rightans{a} \ans {Infinite} {$0$} {$1$} {$2$} {None} \SOL{17. \STAT{42}{22}}{ The determinant is zero, therefore it has either $0$ or infinitely many solutions. Since $x=1$, $y=0$, $z=0$ works in the equation, there are infinitely many. } \prob{18.}{(Determinant)} Given $c-a=3x$ and $c-b=2x$, then find in terms of $x$ the value of \par \centerline{ $\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right|$ } \rightans{d} \ans {$-3x^3$} {$3x^3$} {$-6x^3$} {$6x^3$} {None} \SOL{18. \STAT{22}{27}}{ The determinant has value $(c-a)(c-b)(b-a)$ by direct expansion using Sarrus' rule or a cofactor expansion along row one. The determinant is known in the literature as {\em Vandermonde's determinant}. The value of $b-a$ is obtained by subtracting $c-b$ from $c-a$. Substitution of $c-a=3x$, $c-b=2x$, $b-a=x$ gives the answer $6x^3$. } \prob{19.}{(Geodesic)} Find the distance from the point $(2,3)$ to the line $3x+4y=7$. \rightans{e} \ans {$2$} {$\frac{12}{5}$} {$\frac{9}{4}$} {$\frac{55}{24}$} {$\frac{11}{5}$} \SOL{19. \STAT{55}{51}}{ The distance from a point to a line is measured as the perpendicular distance, taken along a line normal to the given line and through the given point $(2,3)$. The slope of this line is the negative reciprocal of the slope of $3x+4y=7$. Using $y-y_0=m(x-x_0)$ as the line equation, then $x_0=2$, $y_0=3$, $m=-1/(-3/4)=4/3$ and the line is $y-3=(4/3)(x-2)$. The lines meet when $y=3+(4/3)(x-2)=(7-3x)/4$, which gives $x=17/25$ and $y=31/25$. The distance is $\sqrt{(2-17/25)^2+(3-31/25)^2}=11/5$. } \def\AA{ \prob{20.}{(Fan)} An electric fan is switched off, making $n$ turns in the first second. Each additional second, it loses $\sigma$ percent of its revolutions. Find the maximum number of fan revolutions. } \def\BB{\input{fan12.box}} \WRAPFIG{\AA}{\BB}{1in} \rightans{c} \ans {$102\frac{n}{\sigma}$} {$101\frac{n}{\sigma}$} {$100\frac{n}{\sigma}$} {$99\frac{n}{\sigma}$} {$98\frac{n}{\sigma}$} \SOL{20. \STAT{35}{63}}{ The series whose sum is the maximum number of turns is $$n\sum_{k=0}^\infty \left(1-\frac{\sigma}{100}\right)^k.$$ This is a geometric series of common ratio $r=1-\sigma/100$ with sum $n/(1-r)$. The sum simplifies to $100n/\sigma$. To derive the series formula, observe the number of revolutions during a later second is the number from the preceding second multiplied by the fraction $1-0.01\sigma$, because $0.01\sigma$ revolutions are lost. } \prob{21.}{(Zeros)} Count the number of real roots $x$ of the equation $\ln|x^2-3x+1|=0$. \rightans{e} \ans {$0$} {$1$} {$2$} {$3$} {$4$} \SOL{21. \STAT{78}{27}}{ The equation is equivalent to two quadratic equations, $x^2-3x+1=1$ and $x^2-3x+1=-1$. The first has roots $x=0$ and $x=3$. The second has roots $x=1$ and $x=2$. There are four real roots. } % 11th \prob{22.}{(Parallel)} The lines $5x+8y=9$ and $5x+8y=3$ are parallel. Find the distance between them. \rightans{e} \ans {$5$} {$6$} {$\frac{3}{\sqrt{22}}$} {$\frac{\sqrt2}{\sqrt5}$} {$\frac{6}{\sqrt{89}}$} \SOL{22. \STAT{51}{38}}{ A line perpendicular to both is $y-y_0=m(x-x_0)$ where $(x_0,y_0)$ is on one of the lines and $m$ is the negative reciprocal of the slope common to both lines. Choose $x_0=0$, $y_0=3/8$ (on $5x+8y=3$) and let $m=-1/(-5/8)=8/5$. Then $y-3/8=8x/5$ is the perpendicular line through $P=(0,3/8)$. Let $Q$ be the intersection with the other line. Then the required distance is the distance from $P$ to $Q$. Solving equations leads to $8x/5+3/8=y=9/8-5x/8$ with solution $x=30/89$, giving $Q=(30/89,651/712)$. Then $|P-Q|=\sqrt{(30/89)^2+(651/712-3/8)^2}=6/\sqrt{89}$. } \prob{23.}{(Fractions)} Find $D$ in the partial fraction expansion\par \centerline{$\dd \frac{p}{q} = \frac{A}{x+1}+\frac{B(x+1)+C}{x^2+2x+2}+ \frac{D(x+1)+E}{(x^2+2x+2)^2}$,} if $p=x^2+3x+1$ and $q=(x+1)(x^2+2x+2)^2$. \rightans{d} \ans {$-1$} {$0$} {$1$} {$2$} {None} \SOL{23. \STAT{24}{23}}{ The easiest method to apply is multiplication by $(x^2+2x+2)^2$ to clear the quadratic term. Then substitute a complex root of that factor, ie, $x=-1+i$. The result is a complex equation for the values $D$ and $E$. Solving them gives $D=2$ and $E=1$. The other values are $A=-1$, $B=1$, $C=0$. } % 10th \prob{24.}{(Disk)} Select from the list below a point on the line $2x+3y=1$ inside the circle of radius $3$ and center $(-1,2)$. \rightans{e} \ans {$(1,-\frac13)$} {$(-4,3)$} {$(\frac98,-\frac{5}{12})$} {$(\frac54,-\frac12)$\PAR} {$(-\frac{15}{4},\frac{17}{6})$} \SOL{24. \STAT{61}{39}}{ The circle equation is $(x+1)^2+(y-2)^2=9$. Insert the line equation $y=(1-2x)/3$ into the circle equation to find the $x$-values where the line crosses the circle. They occur at approximately $0.9$ and $-3.9$. So it is difficult to eliminate any of the answers. The best attack is to check which ones are on the line and then test them against the inequality $(x+1)^2+(y-2)^2<9$. All five points are on the line $2x+3y=1$. The values of $(x+1)^2+(y-2)^2$ at the five points are $9.4$, $10$, $10.4$, $9.4$, $8.3$. Therefore, $(-\frac{15}{4},\frac{17}{6})$ qualifies as on the line and inside the circle. } % 10th \def\AA{ \prob{25.}{(Chord)} If $B$ is a point on a semicircle $XBY$ with $\bar{AB} \perp \bar{XY}$ at point $A$, $XA=2\frac{1}{12}$ and $AY=12$, then find $BY$. } \def\BB{\input{chord10a.box}} \WRAPFIG{\AA}{\BB}{1.2in} \rightans{b} \ans {$\frac{25}{2}$} {$13$} {$9\sqrt2$} {$8\sqrt3$} {$6\sqrt5$} \SOL{25. \STAT{37}{44}}{ Let $x$ be the unknown length $BY$. Let $u=XB$ and $v=AB$. The Pythagorean theorem implies $x^2=v^2+12^2$ and $u^2=(2+\frac{1}{12})^2+v^2$. Since angle $B$ is $90^\circ$, then the Pythagorean theorem implies $u^2+x^2=(12+2+\frac{1}{12})^2$. Eliminating between the two equatons for $x^2$ gives $u^2+v^2=50+(25/12)^2$. Using the other equation for $u^2$ and $v^2$ gives $v=5$. Then $x^2=v^2+12^2=13^2$ and $x=13$. } \prob{26.}{(Intercepts)} How many lines through $(4,3)$ have intercepts $(a,0)$ and $(0,b)$ with $a$ and $b$ being positive integers? \rightans{c} \ans {$3$} {$5$} {$6$} {$7$} {Infinitely many} \SOL{26. \STAT{71}{6}}{ Draw a line through $(a,0)$, $(4,3)$ and $(0,b)$. The slopes of the two segments in the figure are $3/(4-a)$ and $(b-3)/(-4)$. They are equal, which gives the equation $(a-4)(b-3)=12$. By geometry, the values of $a$ are restrained to $5 \le a$ and the values of $b$ are restricted to $b\ge 4$. The question reduces to determining which positive integers $p$ and $q$ satisfy $pq=12$. This finite list is easily written down. Then $a=p+4$ and $b=q+3$. The possible pairs $[a,b]$ are $[5,15]$, $[6,9]$, $[7,7]$, $[8,6]$, $[10,5]$, $[16,4]$. There are six possible lines. } \prob{27.}{(Profit)} A plant produces $x$ calculators per week at total cost $\frac15x^2+15x+200$ dollars. Given unit price $\frac13(125-x)$ dollars, find $x$ to maximize revenue. \rightans{d} \ans {$22$} {$23$} {$24$} {$25$} {$26$} \SOL{27. \STAT{36}{42}}{ The profit $P$ is the revenue minus the cost. It is defined on $010^{10}$, which function value below is largest? \rightans{b} \ans {$f(x^2)$} {$g(\sqrt{x})$} {$h(x)$} {$g(1/x)$\PAR} {$h(x^3)$} \SOL{18. \STAT{62}{58}}{ The value of $f(x)$ is not larger than $1$. The value of $g(x)$ is about $4/3$, because by the division algorithm for polynomials, $g(x)=4/3+r(x)$ where $r(x)$ is nearly zero. The value of $g(1/x)$ is negative, about $-11$. The value of $h(x^3)$ is about $0$. The largest value is given by $g(\sqrt{x})$, which is about $4/3$. } \prob{19.}{(Ellipse)} Find below one of the vertices of the ellipse $4x^2+25y^2+250y+525=0$. \rightans{b} \ans {$(-5,0)$} {$(-5,-5)$} {$(0,-5)$} {$(0,3)$} {$(0,7)$} \SOL{19. \STAT{57}{55}}{ 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 $(0,-5)$. The vertices are found by setting $x=0$ or $y=-5$, which are the center coordinates, to obtain $(0,-7)$, $(0,-3)$, $(5,-5)$, $(-5,-5)$. The center can be found directly with calculus, eliminating the square completion step. Take the partial derivative of the equation in each variable and find the critical point: it will be the center. For example, the partial in the $y$ variable is $50y+250=0$ which gives $y=-5$. } \prob{20.}{(Comet)} Halley's comet has an elliptical path. Approximate the length of the semi-minor axis in astronomical units ($1$ AU $=9.3\times 10^7$ miles), given the eccentricity is $0.9672759$ and the semi-major axis is $17.94104$ AU. \rightans{b} \ans {$4.9$} {$4.6$} {$4.2$} {$4.0$} {$3.8$} \SOL{20. \STAT{18}{36}}{ The key knowledge required is the formula $b^2=a^2(1-e^2)$. From this formula it is clear that $b$ is the unknown and $e$, $a$ are given in the problem. Then $b=17.94104\sqrt{1-(0.9672759)^2}=4.552124232$. Calculator use on this problem is not required, because the square root can be estimated as $0.25$, leaving only $4.6$ as a reasonable answer. } \prob{21.}{(Radius)} Find the radius of the circle that goes through the three points of intersection with the ellipse $x^2+4y^2=36$ and the parabola $x^2+y=3$. \rightans{e} \ans {$\frac{223}{72}$} {$\frac{11}{4}$} {$\frac{3}{8}$} {$\frac{23}{6}$} {$\frac{27}{8}$} \SOL{21. \STAT{35}{21}}{ Solve the parabola equation for $x^2=3-y$, stuff it into the ellipse equation to get $3-y+4y^2=36$, which in normal form is the quadratic $4y^2-y-33=0$ with roots $y=3$ and $y=-11/4$. Using the parabola equation, then $P=(0,3)$, $Q=(\sqrt{23}/2,-11/4)$ and $R=(-\sqrt{23}/2,-11/4)$ are the three points of intersection. By geometry, the center of the circle through $P$, $Q$ and $R$ is on the line $x=0$ at some point $y=c$. The distance formula can be used to get a quadratic equation for $c$, whose only solution is $c=-3/8$. The center of the circle is therefore at $(0,-3/8)$. The radius is $|3-(-3/8)|=\frac{27}{8}$. } \prob{22.}{(Train)} Commuter service costs \$1.50 per ride and $600$ people ride the train daily. Each $5$ cent fare increase causes $40$ people to quit riding and each $5$ cent fare decrease causes $40$ more to ride. Approximate the fare which maximizes the revenue. \rightans{b} \ans {\${}$1.11$} {\${}$1.13$} {\${}$1.15$} {\${}$1.17$\PAR} {\${}$1.19$} \SOL{22. \STAT{60}{53}}{ Let $P=600+N$ be the number of passengers, $F=150-5N/40$ the fare per passenger in cents, $R=FP=(1200-N)(600+N)/8$ the total revenue. Using calculus, $0=8R'=600-2N$ gives critical point $N=300$ at which level the fare is $F=(1200-300)/8=225/2=112.5$. Without calculus, complete the square on the quadratic $(1200-N)(600+N)$ to write it as $810000-(N-300)^2$, which says the maximum is at the vertex $N=300$. } \def\AA{ \prob{23.}{(Telephone)} A phone line runs from $A$ to $P$ to $B$, in the figure, with $P$ selected to minimize the cost. Line $\bar{AP}$ and line $\bar{PB}$ have costs in the ratio $8$:$3$. Approximate the cost-minimizing distance $BP$ in miles. } \def\BB{\input{phone.box}} \WRAPFIG{\AA}{\BB}{0.9in} \rightans{d} \ans {$6.2$} {$6.4$} {$6.6$} {$6.8$} {$7.0$} \SOL{23. \STAT{38}{31}}{ The cost of a stretch of line is the product of its length and the cost per unit length. Let $c>0$ be chosen so that the expensive line costs $8c$ and the cheaper line costs $3c$, per unit length. Then the cost of the line is $y=8c\sqrt{(x-8)^2+9}+3cx$ where $x=BP$, $0 \le x \le 8$. The critical points satisfy $y'=0$, which reduces to $55(8-x)^2=81$ and finally $x=8-9/\sqrt{55}$ ($x$ must be positive). Checking endpoints, $y(8)=48$, $y(0)=8\sqrt{73}$. At the critical point, $y(8-9/\sqrt{55})=3\sqrt{55}+24$, which is the minimum value of $y$ on $0\le x \le 8$. Then $BP=8-9/\sqrt{55}=6.8$. \par This problem can be solved without calculus using a graphing calculator to localize the minimum. The plot should use $y=8\sqrt{(8-x)^2+9}+3x$ on $0\le x \le 8$. A replot on $6.5\le x \le 7.0$ will close in on the minimum. The minimum value of $46.25$ at $x=6.786$ is relatively simple to isolate. } \prob{24.}{(Normal)} A function $f(x)$ has slope $-1/5$ at $(-1,2)$. Find the $x$-intercept of the normal line at $(-1,2)$. \rightans{b} \ans {$7$} {$-\frac{7}{5}$} {$\frac{9}{5}$} {$\frac52$} {$-\frac{2}{5}$} \SOL{24. \STAT{72}{52}}{ 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 $x$-intercept is the solution of the equation $0=2+5(x+1)$. Then $x=-7/5$. } \prob{25.}{(Inflection)} Determine a value of $x$ for which the cubic $y-1=2x^3-x^2$ has an inflection point. \rightans{b} \ans {$\frac{1}{3}$} {$\frac{1}{6}$} {$0$} {$\frac{1}{2}$} {$1$} \SOL{25. \STAT{82}{77}}{ Calculus method: the critical points are roots of the quadratic $6x^2-2x$ ($x=0$ and $x=1/3$ are critical points). The inflection points can be found from the critical points of this quadratic, ie, from the equation $12x-2=0$. \par Graphical method, without calculus: the graph suggests a relative minimum near $0.3$, a relative maximum near $0$. A logical guess is that an inflection occurs at $1/6$. } \prob{26.}{(Derivative)} Find $dy/dx$ at $x=2$, given $y=\sqrt{z}$, $z=u/v$ and $u=2x+1$ and $v=x^2-3$. \rightans{a} \ans {$\frac {-9}{\sqrt 5}$} {$\frac {9}{\sqrt 5}$} {$\frac {-18}{\sqrt 5}$} {$\frac {18}{\sqrt 5}$} {None} \SOL{26. \STAT{80}{62}}{ The chain rule implies $y'= -\left({x}^{2}+3+x\right) \left(\frac {1+2x}{x^2-3} \right)^{-1/2} \left ({x}^{2}-3\right )^{-2}$. Substitution of $x=2$ gives $\frac {-9}{\sqrt 5}$. } \prob{27.}{(Quartic)} Select from below the equation of a tangent line to an inflection point of the graph of the quartic $y=x^4-6x^3+12x^2-3x+1$. % f:=x->x^4-6*x^3+12*x^2-3*x+1; % diff(f(x),x);g:=unapply(",x);diff(g(x),x); \rightans{a} \ans {$y=5x+1$} {$y=5x-5$} {$y=7x+2$\PAR} {$y=7x-9$} {$y=5x+13$} \SOL{27. \STAT{51}{62}}{ The slope of the tangent line at $x$ is $4x^3-18x^2+24x-3$. Inflection points are those values of $x$ such that $12x^2-36x+24$ changes sign at $x$. The latter is $12(x^2-3x+2)=12(x-2)(x-1)$, which changes sign at $x=1$ and $x=2$. Points on the curve corresponding to these two values of $x$ are $(1,5)$, $(2,11)$. The slopes there are $7$ and $5$, respectively. The tangent line equations are $y-5=7(x-1)$ and $y-11=5(x-2)$. Of these, only the second appears in the list, $y=5x+1$. } \prob{28.}{(Integral)} Let $f$ be the unique quadratic polynomial which passes through $(0,3)$, $(1,-4)$ and $(2,-15)$. Find $\int_0^2 f(x)dx$. \rightans{b} \ans {$-9$} {$-\frac{28}{3}$} {$-\frac{29}{3}$} {$-10$} {$-\frac{31}{3}$} \SOL{28. \STAT{29}{45}}{ The exact value is given by Simpson's rule, according to numerical integration theory. The answer can be reported as the interval length $2$ divided by $6$ times the sum $(3)+4(-4)+(-15)=-28$, giving $-28(2)/6=-28/3$. \par Alternatively, compute the polynomial $f(x)=(x+3)(1-2x)=-2x^2-5x+3$ from the given information and integrate to obtain $-28/3$. To find the 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: $3=a$, $-4=a+b$, $-15=a+2b+2c$. Solve for $a$, $b$, $c$ by elimination or Cramer's rule. } \prob{29.}{(Integral)} Let $f$ be nonnegative and have two continuous nonnegative derivatives. Assume $f(0)=3$, $f(1)=3.5$, $f(2)=4$ and $f(3)=5$. Select the smallest value $y$ from below with $\int_0^3 f(x)dx\le y$. \rightans{c} \ans {$10.1$} {$10.4$} {$11.5$} {$12.5$} {$13.0$} \SOL{29. \STAT{32}{44}}{ The idea is to approximate the integral from below using the Rectangular rule and from above using the Trapezoidal rule. The Rectangular rule implies $\int_0^3 f(x)dx\ge 10.5$. The approximation should use integrand values $3$, $3.5$ and $4$. The assumption $f'(x)\ge 0$ implies $f$ is increasing, therefore the rectangular approximation is from below. The Trapezoidal approximation uses three trapezoids, each of area $\frac12(b+B)$ ($b$ and $B$ are the bases), to give upper estimate $\frac{3+3.5}{2}(1) + \frac{3.5+4}{2}(1) + \frac{4+5}{2}(1)=\frac{23}{2} = 11.5$. The assumption $f''(x)\ge 0$ implies convexity of $f$, which forces the polygonal curve of the Trapezoidal approximation to stay above the curve $y=f(x)$. } \prob{30.}{(Jars)} Twelve identical straws are distributed at random into three jars. Find the approximate probability that the first jar contains three straws. \rightans{a} \ans {$18(\frac{2}{3})^{11}$} {$19(\frac{2}{3})^{11}$} {$20(\frac{2}{3})^{11}$} {$21(\frac{2}{3})^{11}$\\[3pt]\MBL} {$22(\frac{2}{3})^{11}$} \SOL{30. \STAT{24}{26}}{ The first solution is by counting favorable events. The sample space contains $3^{12}$ events. If a favorable event occurs, then the remaining $9$ straws are distributed among $2$ jars. The number of events which forms the jar with three straws is $12!/(3!9!)=(12)(11)(10)/6=(20)(11)$. The other two jars can be filled in $2^9$ ways. So the favorable events amount to $(20)(11)(2^9)=(55)(2^{11})$. The probability is $(55)2^{11}/3^{12}$. The fraction $55/3$ is $18.33$, therefore the closest answer is $18(\frac{2}{3})^{11}$. \par A simpler solution identifies the answer as arising from probabilities $p=\frac{1}{3}$ and $q=\frac{2}{3}$ and the binomial sum for $(p+q)^{12}$. The term that represents the desired probability is the one for $p^3q^9$, which is $\choose{12}{3}p^3q^9=\frac{55}{3}(\frac{2}{3})^{11}$. } \smallskip {\large\bf Common Problems 31-40} There were 357 grade 10, 364 grade 11 and 312 grade 12 exams. Statistics on problems 31--40 appear below for each grade. \smallskip %% separ10-12.tex %% Separator problems for grades 10, 11, 12 \prob{31.}{(Inequalities)} Find the maximum of $3x+2y$ subject to the inequalities $x\ge0$, $y\ge0$, $x+y\ge 5$ and $2x+y\le 8$. \rightans{e} \ans {$10$} {$13$} {$14$} {$15$} {$16$} \SOL{31.}{ {\bf \STAT{10th=62}{25}},\newline {\bf \STAT{11th=55}{26}}, {\bf \STAT{12th=56}{26}}.\newline The function $z=3x+2y$ is a plane in $R^3$. The inequalities describe a scalene triangle in the plane. The maximum occurs at one of the triangle's vertices $(0,5)$, $(0,8)$, $(3,2)$. The values of $z$ are $10$, $16$, $13$, respectively, and the maximum of $3x+2y$ is $16$. } \prob{32.}{(Range)} Select below a possible value of $\dd\sqrt{D}$, given $D=a^2b+ab$ and integers $a$ and $b$ satisfy $|a-b|=1$. \rightans{d} \ans {$68\sqrt{2}$}% 91.9 {$55\sqrt{3}$}% 95.3 {$43\sqrt{5}$}% 96.2 {$50\sqrt{6}$\PAR}% 122.5 {$55\sqrt{5}$}% 122.98 \SOL{32.}{ {\bf \STAT{10th=16}{49}},\newline {\bf \STAT{11th=18}{26}}, {\bf \STAT{12th=23}{34}}.\newline Consider first the case $a=x$, $b=x+1$. Then $D=x^2(x+1)+x(x+1)=x^3+2x^2+x=x(x+1)^2$ and $\sqrt{D}=\sqrt{x}(x+1)$. The values given range from $91$ to $123$. Solving $x(x+1)^2=91^2$ and $x(x+1)^2=123^2$ tells us to check values $x=19$ to $x=25$. Only $x=24$ gives a value in the list, which is $50\sqrt{6}$. \par The second case $a=x+1$, $b=x$ results in $D=x(x+1)^2+x(x+1)=x(x+1)(x+2)$. The relevant range for $x$ is $21$ to $24$, which produces no answer from the list. } \prob{33.}{(Rhombus)} A rhombus of area $20$ cm${}^2$ has diagonals the in ratio of $2$:$1$. Find the side in centimeters. \rightans{a} \ans {$5$} {$23/5$} {$2\sqrt{5}$} {$14/3$} {$11/2$} \SOL{33.}{ {\bf \STAT{10th=38}{38}},\newline {\bf \STAT{11th=41}{37}}, {\bf \STAT{12th=44}{42}}.\newline The diagonals of a rhombus are perpendicular, therefore the area of a subtriangle is $(1/2)(2a)(a)$ where the diagonals have lengths $2a$ and $4a$. By the Pythagorean Theorem, $4a^2+a^2=x^2$, where $x$ is the unknown side. The area hypothesis gives $4a^2=20$. Finally, $x^2=25$ and the answer is $x=5$. } \prob{34.}{(Clock)} As the hands of a clock move from $3$am to $2$:$59$pm, how many times do they form a right angle? \rightans{b} \ans {$18$} {$22$} {$23$} {$24$} {None} \SOL{34.}{ {\bf \STAT{10th=67}{15}},\newline {\bf \STAT{11th=66}{17}}, {\bf \STAT{12th=68}{15}}.\newline This problem can be solved by drawing $24$ clock diagrams or by winding a standard watch with hour and minute hands. We explain below why the answer is not $24$ and not $23$. \par If the minute hand and the hour hand form a right angle, then a fixed number of minutes $a$ elapses before the hands can form another right angle. A simple estimate is $a=32.75$, approximately. The fact $a>30$ is enough to argue that $24$ is the wrong answer. As a check on the logic, $32.75(22)>720$, which accounts for the full number of minutes $30(24)=720$ from $3$am to $3$pm. \par In the first hour, $3$am to $4$am, the hands are at right angles $2$ times, once at $3$am and then after $a$ minutes. During the four hour period from $4$am to $7$:$59$am, the hands are at right angles two times per hour for $4$ hours, making $8$ times. In the time period, $8$am to $8$:$59$pm, they are at right angles only once. The total for five hours and $59$ minutes is $2+8+1=11$. The geometry from $3$am to $8$:$59$am is the same as $9$am to $2$:$59$pm. So the total is $11+11=22$. \par A precise solution can be based upon the angle increase between hands, which is $330^\circ$ per hour. Let $t$ be in hours and let $m$ stand for an integer. For a right angle to occur, $330t$ must be an integral multiple of $180$, i.e., $330t=180m$. The domain of $t$ is $0\le t \le 11\frac{59}{60}$. Therefore, $m$ satisfies $0\le m=\frac{330}{180}t\le \frac{330}{180}(11+\frac{59}{60})=\frac{7909}{360}<22$. This implies that $m$ can be any integer from $0$ to $21$, corresponding to $22$ times $t$. } \prob{35.}{(Dice)} Two dice are tossed $100$ times. Find the probability of getting $12$ points one time or two times. \rightans{d} \ans {$\frac {422}{63}(\frac{35}{36})^{99}$} {$\frac {281}{42}(\frac{35}{36})^{99}$} {$\frac {47}{7}(\frac{35}{36})^{99}$\\[3pt]\MBL} {$\frac {845}{126}(\frac{35}{36})^{99}$} {None} \SOL{35.}{ {\bf \STAT{10th=13}{15}},\newline {\bf \STAT{11th=19}{15}}, {\bf \STAT{12th=22}{9}}.\newline Let $p=\frac{1}{36}$, $q=\frac{35}{36}$. Then $p+q=1$ and $p$ is the probability of $12$ points in one toss. The probability of one or two successful tosses out of the next $100$ is the sum of the terms of the binomial expansion $(p+q)^{100}$ with powers $pq^{99}$ and $p^2q^{98}$. These terms add to $$\choose{100}{1}p^{1}q^{99}+ \choose{100}{2}p^{2}q^{98}$$ which equals $$q^{98}(100pq + 50(99)p^2)= q^{98}\frac{4225}{648}.$$ This agrees with $\frac {845}{126}(\frac{35}{36})^{99}$. } \prob{36.}{(Factorial)} Evaluate the sum \newline \centerline{$2(4)(2!)^2+3(5)(3!)^2+4(6)(4!)^2+ \cdots +99(101)(99!)^2$.} \rightans{d} \ans {$(100!)^2$} {$(99!)^2-4$} {$(100!)^2-1$\PAR} {$(100!)^2-4$} {$(99!)^2-1$} \SOL{36.}{ {\bf \STAT{10th=19}{24}},\newline {\bf \STAT{11th=30}{19}}, {\bf \STAT{12th=35}{25}}.\newline This is $\sum_{n=2}^{99} n(n+2)(n!)^2$. Write $n(n+2)(n!)^2=(n+1)^2(n!)^2-(n!)^2=((n+1)!)^2-(n!)^2$ and observe that the summation is a telescoping sum. The answer is $(100!)^2-(2!)^2$. } \def\AA{ \prob{37.}{(Triangle)} In the figure, let triangles $ABD$ and $BCD$ be given with $AB=2$, $BC=1$, $AD=x$, $BD=4-x$ and $CD=1+x$. Find $x$. } \def\BB{\input{drills.box}}% Created from drill problem, 1965 MAA \WRAPFIG{\AA}{\BB}{1.1in} \rightans{d} \ans {$\frac{27}{14}$} {$\frac{93}{50}$} {$\frac{64}{35}$} {$\frac{13}{7}$} {None} \SOL{37.}{ {\bf \STAT{10th=16}{19}},\newline {\bf \STAT{11th=23}{17}}, {\bf \STAT{12th=25}{21}}.\newline The problem can solved with the law of cosines $c^2=a^2+b^2-2ab\cos C$ or with Heron's formula $A=\sqrt{s(s-a)s-b)s-c)}$, $s=\frac12(a+b+c)$. \par The trigonometric solution starts with defining angle $\theta$ to be angle $BCD$. Then $2(2)(x)\cos \theta=x^2+2^2-(4-x)^2=8x-12$ and $2(3)(x)\cos\theta = 3^2+x^2-(1+x)^2=8-2x$. Equating cosine terms results in the linear equation $3(8x-12)=2(8-2x)$, which has solution $x=13/7$. \par The Heron's formula solution, which uses no trigonometry, proceeds as follows. Drop an altitude $\bar{DE}$ to a point $E$ along $\bar{AB}$ which divides the segment $\bar{AB}$ as $t=AE$ and $2-t=EB$. Let $h=DE$. The Pythagorean theorem implies $h^2=x^2-t^2=(4-x)^2-(2-t)^2$, whereby $t=2x-3$ and $h^2=12x-3x^2-9$. By positivity of $t$, we have $x>3/2$. The area $\alpha$ of the largest triangle satisfies $\alpha^2=(3h/2)^2$ and also $\alpha^2=s(s-x)s-1-x)(s-3)$ where $s=(x+1+x+3)/2$ is the semiperimeter used in Heron's formula. Set the two formulas for $\alpha^2$ equal to obtain a quadratic equation for $x$, $7\,{x}^{2}+13-20\,x=0$, with roots $13/7$ and $1$. The requirement $x>3/2$ implies $x=1$ is invalid. The answer must be $x=13/7$. } \def\AA{% Modified version of 1965 MAA problem \prob{38.}{(Fold)} A rectangular piece of paper of height $6$ and width $x<5$ is folded such that two diagonally opposite vertices coincide, causing a crease in the paper of length $\sqrt{7}$. Find $x$. } \def\BB{\input{folding.box}} \WRAPFIG{\AA}{\BB}{0.8in} \rightans{b} \ans {$\sqrt{7}$} {$\sqrt{6}$} {$\frac{7}{2\sqrt{2}}$} {$\frac{6}{\sqrt{7}}$} {$\frac{2\sqrt{7}}{\sqrt{5}}$} \SOL{38.}{ {\bf \STAT{10th=17}{20}},\newline {\bf \STAT{11th=22}{16}}, {\bf \STAT{12th=22}{21}}.\newline The right figure does the problem. \par \def\AA{ Add to the given figure to obtain the figure at the right. Using triangle $T$, we have $x^2+(2u-6)^2=7$. Using triangle $S$, we have $(6-u)^2+x^2=u^2$, because the folding reproduces the dimension $u$ on the hypotenuse of triangle $S$. } \def\BB{\input{folding0.box}} \WRAPFIG{\AA}{\BB}{0.9in} \par Solving gives $4u^2-12u-7=0$ or $u=7/2$. Then $x^2+(2u-6)^2=7$ implies the result $x^2=6$. } \def\AA{ \prob{39.}{(Segment)} The triangle $ABC$ in the figure has sides of $14$, $9$ and $17$ units. A copy of segment $\bar{AB}$ is rotated from vertex $A$ until it meets the inscribed circle at the } \def\BB{\input{tanseg.box}} \WRAPFIG{\AA}{\BB}{1.2in} \par points $D$ and $E$. Given $AE=AB$, approximate $AD$. \rightans{c} \ans {$3.6$} {$3.8$} {$4.0$} {$4.2$} {$4.4$} \SOL{39.}{ {\bf \STAT{10th=24}{33}},\newline {\bf \STAT{11th=26}{24}}, {\bf \STAT{12th=27}{19}}.\newline The idea is to apply \begin{quote}\footnotesize\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 equals the product of the measures of the secant segment and its external secant segment. \end{quote} The secant segment is $AE=9$. The tangent segment can be shown to be $6$ (see below). Then $6^2=(AE)(AD)$, which gives $AD=4$. To establish the tangent segment length of $6$, argue as follows. Let the sides of lengths $14$, $9$ and $17$ be divided in ratios $x$:$z$, $z$:$y$ and $y$:$x$, respectively, by the contact points with the inscribed circle. We used: {\em If two segments from the same exterior point are tangent to a circle, then they are congruent}. Then $x+z=14$, $y+z=9$ and $y+x=17$. Solving these three equations in three unknowns gives $x=11$, $y=6$, $z=3$. } \def\AA{ \prob{40.}{(Diameter)} Let points $A$, $B$, $C$, $D$ be on the circumference of a circle of diameter $10$, as in the figure. Assume segments $\bar{AC}$ and $\bar{BD}$ meet orthogonally at $M$ with $AM=x$, $CM=6$ and $DM=3$. Find $x$. } \def\BB{\input{diam.box}} \WRAPFIG{\AA}{\BB}{1.0in} \rightans{e} \ans {$3$} {$\frac{7}{2}$} {$\frac{2}{5}\sqrt{65}$} {$\sqrt{10}$} {$\sqrt{11}$} \SOL{40.}{ {\bf \STAT{10th=22}{9}},\newline {\bf \STAT{11th=27}{5}}, {\bf \STAT{12th=30}{10}}.\newline The first step is to establish the measure $BM=2x$ from the ratio relation $(CM)(AM)=(DM)(BM)$. Locate the center $N$ of the circle by constructing perpendiculars from the midpoints of the two chords $\bar{AC}$ and $\bar{BD}$. Then $\bar{BN}$ is the hypotenuse of a right triangle of legs $a$ and $b$, calculated as $a=\frac{6+x}{2}-x=3-\frac12x$ and $b=\frac12(BD)= =x+\frac32$. Since $BN=10/2$, the Pythagorean theorem $a^2+b^2=5^2$ gives a quadratic equation $(3-x/2)^2+(x+3/2)^2=25$ that can be solved for $x=\sqrt{11}$. } \end{exercises} \end{multicols} \begin{multicols}{2} \begin{center}\Large\bf 1995 Utah State Math Contest Answer Keys \end{center} \begin{center} \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 7} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 7 \setanswer{1.}{d} &\setanswer{11.}{a} &\setanswer{21.}{b} &\setanswer{31.}{a} \\ \setanswer{2.}{b} &\setanswer{12.}{b} &\setanswer{22.}{a} &\setanswer{32.}{a} \\ \setanswer{3.}{b} &\setanswer{13.}{c} &\setanswer{23.}{a} &\setanswer{33.}{d} \\ \setanswer{4.}{a} &\setanswer{14.}{d} &\setanswer{24.}{c} &\setanswer{34.}{b} \\ \setanswer{5.}{d} &\setanswer{15.}{c} &\setanswer{25.}{d} &\setanswer{35.}{d} \\ \setanswer{6.}{b} &\setanswer{16.}{c} &\setanswer{26.}{a} &\setanswer{36.}{d} \\ \setanswer{7.}{e} &\setanswer{17.}{d} &\setanswer{27.}{c} &\setanswer{37.}{e} \\ \setanswer{8.}{b} &\setanswer{18.}{c} &\setanswer{28.}{d} &\setanswer{38.}{c} \\ \setanswer{9.}{a} &\setanswer{19.}{b} &\setanswer{29.}{b} &\setanswer{39.}{d} \\ \setanswer{10.}{d} &\setanswer{20.}{c} &\setanswer{30.}{d} &\setanswer{40.}{c} \\ \hline \end{tabular} \medskip \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 8} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 8 \setanswer{1.}{c} &\setanswer{11.}{c} &\setanswer{21.}{d} &\setanswer{31.}{a} \\ \setanswer{2.}{e} &\setanswer{12.}{e} &\setanswer{22.}{c} &\setanswer{32.}{a} \\ \setanswer{3.}{b} &\setanswer{13.}{d} &\setanswer{23.}{b} &\setanswer{33.}{d} \\ \setanswer{4.}{e} &\setanswer{14.}{c} &\setanswer{24.}{c} &\setanswer{34.}{b} \\ \setanswer{5.}{e} &\setanswer{15.}{a} &\setanswer{25.}{c} &\setanswer{35.}{d} \\ \setanswer{6.}{d} &\setanswer{16.}{c} &\setanswer{26.}{b} &\setanswer{36.}{d} \\ \setanswer{7.}{c} &\setanswer{17.}{c} &\setanswer{27.}{a} &\setanswer{37.}{e} \\ \setanswer{8.}{b} &\setanswer{18.}{a} &\setanswer{28.}{d} &\setanswer{38.}{c} \\ \setanswer{9.}{b} &\setanswer{19.}{b} &\setanswer{29.}{c} &\setanswer{39.}{d} \\ \setanswer{10.}{b} &\setanswer{20.}{b} &\setanswer{30.}{d} &\setanswer{40.}{c} \\ \hline \end{tabular} \medskip \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 9} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 9 \setanswer{1.}{c} &\setanswer{11.}{e} &\setanswer{21.}{c} &\setanswer{31.}{a} \\ \setanswer{2.}{b} &\setanswer{12.}{c} &\setanswer{22.}{a} &\setanswer{32.}{a} \\ \setanswer{3.}{c} &\setanswer{13.}{d} &\setanswer{23.}{a} &\setanswer{33.}{d} \\ \setanswer{4.}{d} &\setanswer{14.}{c} &\setanswer{24.}{a} &\setanswer{34.}{b} \\ \setanswer{5.}{c} &\setanswer{15.}{b} &\setanswer{25.}{c} &\setanswer{35.}{d} \\ \setanswer{6.}{a} &\setanswer{16.}{c} &\setanswer{26.}{b} &\setanswer{36.}{d} \\ \setanswer{7.}{d} &\setanswer{17.}{c} &\setanswer{27.}{c} &\setanswer{37.}{e} \\ \setanswer{8.}{a} &\setanswer{18.}{d} &\setanswer{28.}{a} &\setanswer{38.}{c} \\ \setanswer{9.}{e} &\setanswer{19.}{b} &\setanswer{29.}{c} &\setanswer{39.}{d} \\ \setanswer{10.}{a} &\setanswer{20.}{b} &\setanswer{30.}{e} &\setanswer{40.}{c} \\ \hline \end{tabular} \medskip \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 10} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 10 \setanswer{1.}{d} &\setanswer{11.}{a} &\setanswer{21.}{b} &\setanswer{31.}{e} \\ \setanswer{2.}{a} &\setanswer{12.}{a} &\setanswer{22.}{b} &\setanswer{32.}{d} \\ \setanswer{3.}{c} &\setanswer{13.}{c} &\setanswer{23.}{d} &\setanswer{33.}{a} \\ \setanswer{4.}{a} &\setanswer{14.}{c} &\setanswer{24.}{b} &\setanswer{34.}{b} \\ \setanswer{5.}{d} &\setanswer{15.}{c} &\setanswer{25.}{a} &\setanswer{35.}{d} \\ \setanswer{6.}{b} &\setanswer{16.}{b} &\setanswer{26.}{c} &\setanswer{36.}{d} \\ \setanswer{7.}{c} &\setanswer{17.}{c} &\setanswer{27.}{a} &\setanswer{37.}{d} \\ \setanswer{8.}{b} &\setanswer{18.}{c} &\setanswer{28.}{e} &\setanswer{38.}{b} \\ \setanswer{9.}{b} &\setanswer{19.}{b} &\setanswer{29.}{a} &\setanswer{39.}{c} \\ \setanswer{10.}{b} &\setanswer{20.}{b} &\setanswer{30.}{c} &\setanswer{40.}{e} \\ \hline \end{tabular} \medskip \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 11} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 11 \setanswer{1.}{e} &\setanswer{11.}{b} &\setanswer{21.}{e} &\setanswer{31.}{e} \\ \setanswer{2.}{a} &\setanswer{12.}{a} &\setanswer{22.}{e} &\setanswer{32.}{d} \\ \setanswer{3.}{a} &\setanswer{13.}{c} &\setanswer{23.}{d} &\setanswer{33.}{a} \\ \setanswer{4.}{b} &\setanswer{14.}{d} &\setanswer{24.}{e} &\setanswer{34.}{b} \\ \setanswer{5.}{a} &\setanswer{15.}{e} &\setanswer{25.}{b} &\setanswer{35.}{d} \\ \setanswer{6.}{d} &\setanswer{16.}{d} &\setanswer{26.}{c} &\setanswer{36.}{d} \\ \setanswer{7.}{b} &\setanswer{17.}{a} &\setanswer{27.}{d} &\setanswer{37.}{d} \\ \setanswer{8.}{e} &\setanswer{18.}{d} &\setanswer{28.}{d} &\setanswer{38.}{b} \\ \setanswer{9.}{a} &\setanswer{19.}{e} &\setanswer{29.}{e} &\setanswer{39.}{c} \\ \setanswer{10.}{a} &\setanswer{20.}{c} &\setanswer{30.}{a} &\setanswer{40.}{e} \\ \hline \end{tabular} \medskip \begin{tabular}{|l|l|l|l|} \hline \multicolumn{4}{|c|}{\bf Answers to}\\ \multicolumn{4}{|c|}{\bf 1995 Exam Grade 12} \\ \hline %\setanswer{1.}{c} &\setanswer{11.}{c} & \setanswer{21.}{a} & \setanswer{31.}{d} \\ %grade 12 \setanswer{1.}{c} &\setanswer{11.}{a} &\setanswer{21.}{e} &\setanswer{31.}{e} \\ \setanswer{2.}{a} &\setanswer{12.}{d} &\setanswer{22.}{b} &\setanswer{32.}{d} \\ \setanswer{3.}{c} &\setanswer{13.}{c} &\setanswer{23.}{d} &\setanswer{33.}{a} \\ \setanswer{4.}{e} &\setanswer{14.}{e} &\setanswer{24.}{b} &\setanswer{34.}{b} \\ \setanswer{5.}{a} &\setanswer{15.}{a} &\setanswer{25.}{b} &\setanswer{35.}{d} \\ \setanswer{6.}{d} &\setanswer{16.}{c} &\setanswer{26.}{a} &\setanswer{36.}{d} \\ \setanswer{7.}{e} &\setanswer{17.}{a} &\setanswer{27.}{a} &\setanswer{37.}{d} \\ \setanswer{8.}{e} &\setanswer{18.}{b} &\setanswer{28.}{b} &\setanswer{38.}{b} \\ \setanswer{9.}{b} &\setanswer{19.}{b} &\setanswer{29.}{c} &\setanswer{39.}{c} \\ \setanswer{10.}{d} &\setanswer{20.}{b} &\setanswer{30.}{a} &\setanswer{40.}{e} \\ \hline \end{tabular} \end{center} \end{multicols} \end{document}