Answers1. 256 (c)2. 1.35 x 105 (C)3. t7 (F, C)4. 80 (C, F, M)

(r, c,E, G)(c)

(M, F, c, P)

8.79. x,2*10. 36

(r ,c,F,G)(E,6, F, T)(r, P, E, s)

The situotion in this problem con be represented with the eguationTotaf Points = 50 + 4C - 2W,where C is fhe number of correct onswersond W is the number of wrong ones. Since we dre looking of thesituotion where Dolene eorns 100 points, the eguotion weneed to grophis 100 : 50 + 4C - 2W or W : 2C - 25. Since Dolene hod 16 correctonswers, look of the W-volua on the groph when C = 16. On o grophingcolculotor, using the Toble function or Troce function con help you

Warm-Up 25. L46. 8647. t2

Solution - Problem #7Seporoting the shope into 4 triongles,we see thot eoch of the

triongles is holf of a rectangle. Theref ore the oreo of the originolregion will be holf of the lorgest rectangulor region circumscribedobout the shoded oreo. Just by counting, we con see thot there are24 square centimeters within the four smoll rectongulor regions.Toking holf of this omount yields the answer of tZ square centimetersfor theoreo of the shoded region.

Representation - Problem #8

t6,7)

locote the exoct volue for W when C = 16. We see thot W = 7. Finally, we need tobe sure thot W + C < 25, since there ore only 25 guestions on the exom. This

condition is met, ond we con olso determine now how mony problems were leftunonswered.

Gonnection to ... Angle measures in polygons (Problem #41Meosuring the cenfralangle in a circle can be used to find the ongle meosures of o regulor

pofygon. A regular nsided polygon con be inscribed in o circle. A regular hexogon is shown here.Notice that the centrol ongle (stor) i. t*o " for ony regulor n-gon. Since thetriongles in the polygon areisosceles, the sum of the meosures of the boseongles (dots) is (180 - # )". An interior angle of the polygon is composed oftwo of these bose angles,so its meosure willolso eguol (180 - #)".Therefore, the meosure of on interior angle of this regulor hexogon is eguolto (180 - * l = !2Oo.

48 MATHCOUNTS 200243 MATHCOT

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Warm-Up 1

Problem 1. 120 is the least common multiple (LCM) of 6, 8 and 10. All the prime factors of the three numbers must be present in the prime

factor izat ionoftheLCM. Considerthepr imefactor izat ionof6(2"3),8(23)andl0(2x5). TheLCMof6,8andl0musthavethreefactors

of2,afactorof3andafactorof5. Theproductof thesenecessaryfactorsis23x3xS=120.

Problem2. I f thebookwerelaidf latandopenedtothecenter, thebottomsheetofpaperhasthenumbersl '2, l5and16;thesecondsheethas

thenumbers3,4, l3and 14;thethirdsheethasthenumbers5,6, l l and 12;andthefourthsheethasthenumbersT,8,9and l0. Thustheothet

three page numbers on the same sheet ofpaper as page 5 are 6, ll and 12.

problem 3. The number 4 has the three factors I , 2 and 4. Only the perfect squares have an odd number of factors.

problem4.Therat ionalnumber- l32.48isbetweentheintegers- l32and-133. I t is0.48uni tsawayfrom-l32and0'52uni tsawayfrom-133.

Thus the integer on the number line closest to -l 32.48 is -132.

problem5.Theleastpossibleper imeterofhexagonABCDEFwouldoccul i fasmanysidesaspossiblearetheminimumlengthof5cm. The

length of segmentAB mustbe 7 cm, butthe otherfive sides can be 5 cm. Theperimeteris thus 5 x 5 +7 =32cm.

problem 6. To purchase 30 calculators at $90 each, they will need 30 x $90 = $2700. Since they make $10 profit on each $90 calculator that

they sell for $100, they will need to sell $2700 + $10 = 270 calculators'

Probfem 7. Problem solved in the Answer Key to llarm'Up l.

probfemg.The2}%odiscountontheslacksmeansyoupayS}Voofthe$25pr ice,or0.8x25=$20each. The25olodiscountonshir tsmeansyou

pay75%oofthe$l8pr ice,or0.75x18=$13.50each. Forthreepairsofs lacksandthreeshir ts, thetotalpr icewi l lbe(3x$20)+(3x$13.50)=

$100.50.

problem 9. Each of the eight vertices of a cube can be connected to just one other vertex to form a space diagonal. To avoid double counting,

we divide 8 by 2 and get 4 space diagonals.

problem 10. To find the mean of { and { , we add the two fractions and divide by 2. The sum of the fractions is}+{ -f

+f =$. Dividing

by 2 is the same as multiplying by f , so $ x f =.{}. Representation provided in the Answer Key to ll'arm'Up l.

Warm-Up 2

probtem l. Working backward, we can find the number that is double the value of 8, which is 16. Squaring 16, we get 256, which is the number

whose square root is I 6.

problem 2. A flight of 1.5 hours is 90 minutes. If the hummingbird flaps its wings 1500 times per minute for 90 minutes, it will flap its wings

1500 x 90 = l35.000t imes. tn scient i f icnotat ion, 135,000= 1.35 x 105.

Problem3.Rewri t ingtheexpressionv(1,2,3)+v(3,2,1)+\ , (3,1,2)accordingtotherule,weget(1 x23)+(3x2t)+(3xl '?) . This

becomesS+6+3=17.

probtem 4. The larger angle and the smaller angle together form a straight angle and are therefore supplementary, i.e. they have a sum of

lg0 degrees. We have two angles, one 20 degrees greater than the other, and their sum is 180 degrees. lfwe subtract 20 degrees from the larger,

we have two equal angles with a sum of I 60 degtees, so the smaller angle must measure 80 degrees.

problem5.Thenumber2003isbetweentheperfectsquares44,-1936and 451 =2025. Thismeans.6OOl i tsomewherebetween44and45.

Todeterminewhether2003ispr ime,wemusttest fordiv is ib i l i tybyeverypr imelessthan45. These14pr imesarc2,3,5,7 ' l l '13 '17,19,23'

29,31,37,41 and43.

problem6. I f thefarmerplantsl2poundsofseedperacreonal lT5acres,hewi l lneedl2xT5=900poundsofseed. I f thereare25poundsof

seed per bag, then he will need 900 + 25 = 36 bags of seed. 36 bags of seed at $24 per bag is 36 x 24 = $864 to seed the fields'

Problem 7. Problem solved in the Answer Key to llrarn'Up 2.

Problem twould be 5inconect oto Wann-L

Problem 9possible nrand2'= tfsolutions anegative?greatil tha

Problem Ihow rnanyplace. Likrgreater tha

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Problem ?442,500=

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Problem 9multiplyinl

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Warm-

Problem 1

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Problem 3

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Problem 8. For the 16 questions Dalene answered conectly she eams 4 x l6 = 64 points. This combined with the 50 points she starts withwould be 50 + 64 = I 1 4 points. Since she only got I 00 points, we know she must have lost I 4 points for inconect answers. At 2 points perincorrect answer, she must have answered 7 questions inconectly. (She did not answer 2 questions.) Repres entation provi(led in the Answer Keyto ll/arm-Up 2.

Problem 9. There are six ways to choose two items from a group of four items. For the five pairs that follow, onepossible natural numbersolut ion is given: x=x: whenx= l ,x=.r f whenx= 1, . r3= 2'whenx =2,x2 = f whenx= 2,and 2' = I when x - 2. The only pair that has no natural-number solution is x = 2'. In fact, there are no real numbersolutions at all. The graph shows that the line y =

-r and the points along the curve y = 2, never cross. When -r isnegative 2' is a positive number between 0 and 1 . When -r

= 0,2" = 2(' = I . Finally, for every positive value of x, 2* isgreater than r.

Problem 10. Since our mountain numbers must be greater than 70,000, we know that the first three digits are 78,9_ _. The question becomeshow many ways can we descend from 9 in the last two digits. If the tens digit is an 8, the eight digits 0 through 7 are available for the onesplace. Likewise, i f thetensdigi t is T,wecanmakesevenmountain numbers. Thereare 8 + 7 +6+ 5 +4+ 3+2+ 1=36mountain numbersgreater than 70,000.

Workout I

Probleml.Thecommonfract ion=fc isequal to-18f ,whichisbetweentheintegers- l8and- lg. l t isonlyf uni tsfrom-lg,whereasi t is

f units from-18. Therefore it is closest to -19.

Problem 2. The 400 pounds ofprime rib steak that sold at $9.98 per pound brought in 400 x 9.98 = $3992. The 120 pounds ofrib-eye steak thatsoldat$6.49perpoundbroughtinl20x6.49=$778.80. That 'satotalof3992+778.80=$4770.80for400+120=520poundsofbeef,whichgives an averuge of 4770.80 - 520 = $9.17 per pound.

Problem 3. Evaluating Ray Mercedes' eamed run average according to the rule, we get 48 = 164 x 9 = 2.63.

Problem4.Fromthecluesgiven,wecanwri tethefol lowingsystemoftwoequat ionswithtwounknowns: d+2b= l5anda+5b=3. Not icehow the second equation just has three more D's than the first and the result is | 2 less. If three 6's account for a drop of 12, each b is equal to -4Subst i tut ingb=-4backintothef i rstequat ion,wegetd+2(4)= 15ora-8=15,sod=23. Thevalueofa+bisthus23+1-11=19.

Problem 5. There are 128 ounces in a gallon, so 60 gallons ofsoft drinks is 60 x 128 = 7680 ounces. There were 365 days in 1994, so7680 ounces for the year comes to about 7680

- 365 = 2l ounces of soft drinks per day, to the nearest whole number.

Problem 6. Choosing among the highest primes that are less than 100, we are looking for a sum ofthree ofthem that ends in zero or five. Theprimes 73, 79 and 83 have a sum of 235, which is the greatest possible sum.

Problcm 7. Ifthe sale price of the desk was $442,500 and this is 8850% of the pre-auction estimated value, we can write the equation442,500 - 88.5x. Solving for r, we divide 442,500 by 88.5 and get the pre-auction estimated value of $5000.

Problem 8. Problem solved dnd Representotion providetl in the Answer Key to Ll/orkout L

Problem 9. Paul's total eamings are his wage plus his benefits $28.80 + $8.1 I = $36.91. Dividing his benefits by his total eamings andmultiplying by 100, we find that his benefits are $8.I I + $36.91 - .2197 or 22Vo.

Probfeml0'AddinglTtobothsidesoftheinequal i tyrx-17<20,wegettheinequal i tynx<37. Sincer isal i t t lemorethan3and3x12=36,ourf i rstguessmightbethat12isthegreatest integersolut ion. However,nx12* 37.699,so12istoomuchand11 isthegreatest integer solution.

Warm-Up 3

Problem 1. If last month was July, this month is August and twenty-four months from now is also August. Two months before August is June.

Problem 2. To find the positive difference in the ramp slopes, we subtract the fractions as follows: S-11=*-* =*

Problem 3. Problem solved in the Answer Key to llarm-Up 3.

$olutions MATHCOUNTS 2002-2003 107