HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES (A) 3 (B) 5 (C) 9 (D) 15 (E) 25

Size: px

Start display at page:

Download "HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES (A) 3 (B) 5 (C) 9 (D) 15 (E) 25"

Russell Francis Bell
5 years ago
Views:

1 HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE For ech of the following 5 questions, crefull lcken the pproprite o on the nswer sheet with # pencil. o not fold, end, or write str mrks on either side of the nswer sheet. Ech correct nswer is worth 6 points. Two points re given if no o, or more thn one o, is mrked. Zero points re given for n incorrect nswer. Note tht wild guessing is likel to lower our score. When the em is over, give our nswer sheet to our proctor. You m keep our cop of the questions. NO CALCULATORS 90 MINUTES. Suppose tht for ech non-negtive rel numer, represents the lrgest positive integer k such tht k. Compute the vlue of (A) (B) 5 (C) 9 () 5 (E) 5. Tom treted his est friend to ride in his new cr. However, the cr rn out of gs nd the hd to wlk ck home. If the cr verged 5 miles per hour, nd the wlked t the rte of miles per hour, how mn miles did the ride if the rrived home 8 hours fter the strted? (Assume the took the sme route in oth directions nd mde no stops.) (A) 9.5 (B) (C) 5 () 8.5 (E) Let,, c 0. If,, nd c re rrnged from smllest to lrgest, then which of the following is correct? (A) < < c (B) < c < (C) c < < () < c < (E) < < c. Two mrks re mde on n ordinr inch ruler, one on ech side of the ruler s midpoint. If one mrk divides the ruler into two prts in the rtio :5, nd the other mrk divides the ruler into prts in the rtio 5:, how mn inches seprte these two mrks? (A) (B) (C) () 5 5 (E) 6 5. You cut squre from the pge of clendr. All the squres contin dte nd the sum of the 9 dtes is multiple of. Wht numer is in the lower left corner of the squre? Sun Mon Tue Wed Thu Fri St (A) 5 (B) 7 (C) 9 () (E)

2 6. In the circle, chord is prllel to dimeter. B how mn degrees does the mesure of ngle C eceed the mesure of ngle? (A) 50 (B) 60 (C) 75 () 90 (E) 00 A C B 7. on nd eie were one of four mrried couples t Hlloween prt which no other people ttended. At the prt, individuls greeted ech other with hndshkes, ut one guest did not shke none s hnd. At most one hndshke took plce etween n pir of persons, nd no one shook hnds with his or her spouse nd of course no one shook their own hnd. After ll the introductions hd een mde, on sked the other seven people how mn hnds the ech shook. Surprisingl, the ll gve different nswer. How mn hnds did on shke? (A) (B) (C) () 5 (E) 6 8. The ordered pir of rel numers (, ) stisfies the following sstem of equtions: = log log =. Compute (A) 0 99 (B) 0 (C) 0 99 () (E) 0 9. Jenn is more thn 0 ers old. The (nonzero) product of the digits of Jenn s current ge is equl to the product of the digits of her ge si ers go. Wht is the product of the digits of Jenn s current ge? (A) 8 (B) (C) 6 () 8 (E) 0. All of the permuttions (rrngements) of the letters of the nme FERMAT re formed nd then listed in lpheticl order. The position tht the nme FERMAT occupies in the list is (A) 55 th (B) 67 th (C) 79 th () 0 rd (E) None of these. On r. Grner s lst mth test, on which grdes rnged from 0 to 00 nd were ll integers, the grdes were: 6,, 8, 89,, 8, 9, 56, 67, 68, 57, 5, 96, A, B, C,, E Grdes A, B, C,, nd E ech hve n interesting propert. The re ll different nd when the digits of n one of them re reversed, the clss verge is ectl two points higher thn it ctull ws. Compute the sum of A, B, C,, nd E. (A) 6 (B) 8 (C) 59 () 70 (E) 85. Compute the sum of ll positive integers n such tht n + divides n + 0. (A) 08 (B) 8 (C) 759 () 786 (E) 968

3 . The positive odd integers re grouped into sets of s follows: Set : {,, 5}; Set : {7, 9, }; Set : {, 5, 7}, etc. Wht is the sum of the numers in Set 0? (A) 6,95 (B) 6,07 (C) 6,5 () 6,9 (E) 6,7. Let,, nd c e the roots of the polnomil eqution Compute the vlue of c. (A) 0 (B) (C) 7 () 5 (E) Ech of the integers to 9 is written on different slip of pper, nd ll nine slips of pper re plced in jr. You pick one of the slips t rndom, record the numer nd return the slip to the jr. You pick second slip from the jr. The digit which is most likel to e the units digit of the sum of the two numers tht ou picked is: (A) 0 (B) (C) 5 () 9 (E) All digits re equll likel 6. In tringle ABC, the isector of ngle A meets side BC t point such tht A nd C re the sme length. If the lengths of AB nd BC re nd 6 inches, respectivel, compute the cosine of ngle ACB. (A) (B) (C) () 5 (E) A B C 7. A group of 5 of Hrr Potter s friends t Hogwrts School for Witchcrft nd Wizrdr were sked which of three sujects the liked: Potions, Herolog, nd efense Aginst the rk Arts. Of the 5 students, 80% liked t lest one of the three sujects. Twent of the students liked t lest Potions, 5 liked t lest Herolog, nd liked t lest efense Aginst the rk Arts. Twelve of the students liked t lest Potions nd Herolog, fourteen liked t lest Herolog nd efense Aginst the rk Arts, nd eleven liked t lest Potions nd efense Aginst the rk Arts. How mn of the students liked ll three sujects? (A) (B) 7 (C) 9 () (E) 6 8. The first three terms of geometric sequence re the vlues,, nd z, in tht order. The first three terms of n rithmetic sequence re the vlues,, nd z, in tht order. If,, nd z re distinct rel numers, compute the rtio of the fifth term of the geometric sequence to the fifth term of the rithmetic sequence. (A) (B) 7 (C) 6 7 () (E)

4 9. If > 0 nd > 0 nd (, ) is solution of the sstem of equtions nd, compute (A) 5 (B) 6 9 (C) 75 8 () (E) 0. In right tringle, the squre of the hpotenuse is four times the product of the legs. If is the smllest ngle of the tringle, compute tn. (A) (B) (C) () (E). When the numer is written in se, the vlue is. When the numer in se, the vlue is. Compute +. (A) 7 (B) 9 (C) () (E) 5 is written. A digonl of the regulr pentgon shown divides the pentgon into qudrilterl nd tringle. Which of the following represents the rtio of the re of the qudrilterl to the re of the tringle? (A) (B) (C) () (E). If nd +, compute the vlue of. (A).75 (B).65 (C).50 ().875 (E).5 z. A point (,, z) is rndoml chosen inside the unit cue shown. Wht is the proilit tht the coordintes of this point stisf the reltionship z? (0,0,) (A) (B) (C) 6 () 8 (E) 9 (,0,0) (0,,0) 5. A line drwn from the origin to the center of circle hs slope of, nd tngent line to the circle drwn from the origin hs slope of. Compute the slope of the other tngent line drwn from the origin. (A) 9 (B) 6 (C) 5 9 () 5 (E)

5 THE 0-0 KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION SOLUTIONS. A = = = 7 =. + 8 = + =.. E Let = distnce one w. Then 5 = 8. Solving gives =.5 miles B Rewrite the three numers s follows: , nd c order is < c <.. B + 5 = =, 5 + = = indicted. The mrks re.75 = 00, inches prt.. Therefore, correct. The ruler ppers elow with the mrks C The squre must hve dtes s shown. The sum of ll the dtes is 9n. In order for 9n to e multiple of, n must equl. Therefore, the dte in the lower left corner is n + 6 = 9. n - 8 n - 7 n - 6 n - n n + n + 6 n + 7 n Since prllel chords intercept congruent rcs of circle, represent the mesures of rcs AC nd B, s shown. Since nd re inscried ngles, their mesures re hlf the mesures of their intercepted rcs. Therefore, m C m = (80 + ) = 90. A C B 7. B Since ech person shook hnds with t most 6 people, the nswers on received were 0,,,,, 5, 6. The spouse of the individul who hd 6 hndshkes is the onl person who could hve zero hndshkes. Of the remining people, the spouse of the person who hd 5 hndshkes is the onl person who could hve hndshke. Likewise, the nd the hndshke individuls re mrried, nd the memers of the lst couple oth hd hndshkes. on is the onl person who could hve hd the sme numer of hndshkes s someone else, so he shook hnds.

6 8. = = log log = log = log00 = 00 = 00( ) = nd = 99. Therefore, (, ) = ( 00 0, ), nd + = A enote Jenn s current ge. Suppose tht 6, then = (-6), which implies = 0, which violtes the nonzero ssumption. Thus, 5, nd = (-)(+), or - = nd the onl nonzero solution is =, =. Jenn s current ge is, nd the product of the digits is C There re 5! =0 lpheticl rrngements eginning with A nd nother 0 eginning with E. There re! = rrngements eginning with FA, nd! = 6 rrngements eginning with FEA nd FEM. There re rrngements eginning with FERA. Since FERMAT is the net one, it is in the ( ) = 79 th plce.. E Since there re 8 scores, the difference etween n one of the five scores in question nd the numer with its digits reversed must e (8) = 6. Representing n of the missing numers s 0t + u, tht sme difference is 0u + t (0t + u) = 9u 9t = 9(u t) = 6, nd u t =. From this, it is es to see tht the five scores re 5, 6, 7, 8, nd 59 (notice the re ech prt), nd their sum is 85.. E Perform polnomil long division to see tht n + 0 = (n + ) (n + ) + 0. Thus, if n + divides n + 0, it lso divides 0 = ()()(6). Therefore, n + =,,,, 6,8, 67, or 0. Thus n = 0,, 0,, 60, 8, 670, or 0, nd the desired sum is B First of ll, the n th odd positive integer is given the epression n. One pttern for this prolem is to oserve tht Set # contins the rd ( ) odd positive integer, nmel 5 = (). Set # contins the 6 th ( ) odd positive integer, nmel = (6). Generlizing, the 0 th set contins the 606 th odd positive integer or (606) =,07. Therefore we wnt the sum,067 +,069 +,07 = 6,07.. E Since = 7 + 9, = 7 + 9, nd c = 7c + 9, we hve + + c = 7( + + c) Since + + c is the negtive of the coefficient of the term of 7 9, + + c = 0 so tht + + c = 57.

5. A If we mke tle of outcomes, listing the possile numers on the first slip on the left, the possile numers on the second slip on the top, nd the units digit of the sum in the tle we otin 5 6 7 8 9

We find tht 0 occurs nine times while ech of the other digits occurs onl eight times. Thus the most prole digit is 0 with proilit 8 P(0) =, while P() = P() = = P(9) =. 9 8 B 6.

7 5. A If we mke tle of outcomes, listing the possile numers on the first slip on the left, the possile numers on the second slip on the top, nd the units digit of the sum in the tle we otin where ech of the 8 entries is equll likel. We find tht 0 occurs nine times while ech of the other digits occurs onl eight times. Thus the most prole digit is 0 with proilit 8 P(0) =, while P() = P() = = P(9) =. 9 8 B 6. B Using the Lw of Sines on tringle ABC nd noting tht ngles BA, AC, nd ACB re ll congruent, sin sin 6 sin cos sin cos A 6 7. B Consider the digrm t the right, where the three circles represent the students who like Potions (P), Herolog (H), nd efense Aginst the rk Arts (). We hve used the letters through g to represent the H P vrious susets of the students who like different comintions of sujects. We re interested in the vlue e of e. We know tht 80% of the 5 friends, or 6 students, like t lest one of the suject. From the remining informtion in the prolem we cn conclude tht + + c + d + e + f + g = 6 + c + e + f = d + e = 5 d + e + f + g = + e = d + e = e + f = Adding the second, third, nd fourth equtions ove nd sutrcting the first, we get + d + e + f = 0, while dding the lst three equtions together ields + d + e + f = 7. Comining these, we see tht e = 7. (The remining vlues re = 6, = 5, c =, d = 7, f =, nd g = ). C

8 8. From the geometric sequence, z = = nd from the rithmetic sequence, z = + ( ) =. Therefore, = = = 0. Fctoring, we otin ( + )( ) = 0 nd (since nd re distinct) = -. Thus, the rithmetic sequence is,,, 7, 0, nd the geometric sequence is, -,, -8, 6. Therefore, the rtio of the fifth term of the geometric sequence to the fifth term of the 6 8 rithmetic sequence is E Method : Adding the two equtions gives so tht 5 5. The first given eqution ecomes, from which Similrl, the second eqution ecomes from which Therefore,. Method : Adding the frctions in ech eqution we otin nd. 9 6 ividing the first of these the second gives (i) 6 9 gives (ii). Sustituting (i) into (ii) gives 9 nd = From this, so tht.. Multipling the first eqution C We re given c =. If the smllest ngle is opposite side then < nd tn =. B the Pthgoren Theorem, c = +. Thus + = + = 0 6 = ( ). Since <, = ( ) nd tn. c. We re given = + + = nd = + =. Then, ( + ) + + =. ( + ) + 8 =. ivide oth sides ( + ) = = 6 = ( + )( ). Noting tht is positive integer greter thn, the onl possiilit is + = 6 nd = 5. Thus = 8, nd + =

9 . rw the other digonl from point P. This digonl lso hs length. It is es to show tht the ech of the three ngles formed where the digonls meet mesures 6. Thus Are of tringle I = ( )( )sin6 sin6 nd Are of tringle II = ( )( )sin6 Since the re of the qudrilterl is the sum of the res of tringles I nd II, The required rtio is sin6 sin6 sin6 P II I. A Lel the equtions () nd () +. If 0, then eqution () gives =. However, if =, then eqution () gives =, contrdiction. Therefore, > 0 nd eqution () ecomes () + =. If 0, then eqution () gives =. However, if =, then eqution () gives = 6, contrdiction. Therefore, < 0 nd eqution () ecomes () = Solving equtions () nd (), =, = nd the required rtio C The points in ck (nd to the right) of the shded plne section (shown elow, left) stisf >. This region represents of the cue. The points elow (nd in front of) the shded plne section (shown elow, right) stisf > z. z z (0,0,) (0,0,) (,0,0) (0,,0) (,0,0) (0,,0) The overlp in these two regions is shown t the right. It is tringulr prmid whose se is the fce of the cue nd z (0,0,) whose height is equl to the height of the cue. Therefore, its volume represents of the cue nd this is the desired proilit. 6 (,0,0) (0,,0)

10 5. A Method : The slope of line is the tngent of the ngle it mkes with the positive -is. Therefore, tn(po) =, tn(ao) =, nd let tn(bo) = k. Noting tht AOP BOP, tn(aop) = tn(bop) = tn(ao PO) = tn( AO) tn( PO) tn( AO) tn( PO ()() 7 m = A P m = B Then = tn(po) = tn(bo+bop) = k tn( BO)+ tn( BOP) 7 - tn( BO)tn( BOP) ( k) 7 7k Therefore, = nd k =. 7 k 9 7k. 7 k O Method : As in the digrm, let A e the point of tngenc for the line with slope, nd let B e the other point of tngenc. Let M e the midpoint of segment AB. B rescling the digrm, we cn ssume the coordintes of A re (, ). Let e the horizontl distnce etween A nd M, nd let e the verticl distnce etween A nd M. Then, ( +, ) re the coordintes of M, nd we hve Now, the slope of the line through A nd B is, ecuse it is perpendiculr to line OP whose slope is. This implies nd therefore nd 5. Hence the coordintes point B re, nd the slope of the other tngent line is..8 9

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES THE 08 09 KENNESW STTE UNIVERSITY HIGH SHOOL MTHEMTIS OMPETITION PRT I MULTIPLE HOIE For ech of the following questions, crefully blcken the pproprite box on the nswer sheet with # pencil. o not fold,