3. One pencil costs 25 cents, and we have 5 pencils, so the cost is 25 5 = 125 cents. 60 =

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1 JHMMC 0 Grade Solutios October, 0. By coutig, there are 7 words i this questio Oe pecil costs cets, ad we have pecils, so the cost is cets. 4. A cube has edges % of a umber is. Thus the umber itself is 4, or % percet of 48 is the 48.7 or There are 7 uit squares i the first rectagle, ad uit squares i the secod, so there are + 4 squares i total. 8. A pecil costs 0 cets, ad a eraser costs cets. So, pecils ad erasers cost cets. 9. For the first hours of his hours, he catches fish per hour, so he catches 6 fish. For the latter hours, he catches fish every hour, so he catches fish. I total, he will catch fish. 0. Sice he ca swallow or mice at a time, ad must swallow the 7 mice i the least umber of gulps, it is optimal to swallow mice at time as much as possible. thus, if he swallows mice a time times, ad mice at time oe time, he will have swallowed 7 mice. This takes + 6 gulps.. To fid out how may miutes are i a week, ote that there are 60 miutes i a hour, 4 hours i a day, ad 7 days i a week. Therefore, a week has miutes. Sice Arthur ca make a fried every miutes, he ca make a maximum of frieds.. There are primes betwee 40 ad 0. They are 4, 4, ad 47.. Jay Pee has choices of texture, 4 choices of dye, ad choices of legth. I total, therefore, he has 4 6 choices of hairstyle, sice for each differet outfit he ca choose exactly oe of each texture, dye, ad legth. 4. 4slaps 4sleps slaps slips 8sleps slops 6 slops. 9slips. It is impossible for two differet circles to itersect ay more tha times. 6. 8x + 7 meas that either 8x + 7 or 8x + 7. Fidig the solutios to both of these, x or 9. These are both perfectly valid, so there are solutios The umber of sweets Amy has must be a perfect square, sice we ca arrage the sweets i a square. It must also be divisible by 4 ad, sice we ca arrage the sweets i that may rows. Sice the umber is a perfect square, must also divide it, ad so 4 00 is the least such umber. 8. Sice the letters that the crowd chat repeat every three letters, every third positio will be A. 0 is divisible by, so the 0 th positio will be A. 9. We kow that if is a eve umber, the a ad a +. So for ay eve, a + a + + ( ) 0. We ca thus simplify a 0 + a + + a 0 (a 0 + a ) + (a + a ) + + (a 0 + a 0 )

2 JHMMC 0 Grade Solutios October, 0 0. Sice x +, x + 4 x x 0. By the same reasoig, x + 4 x , x 7, ad. The area of the square is d 4. Next, we see that the diameter of the iscribed circle is equal to the side legth of the square. So, the radius is equal to r d, ad so the area of the iscribed circle is r π π π. Hece, the area i the square but ot i the circle is 4 π.. The possible sums from the two dice rolls are,,...,. Out of these umbers, oly 4, 6, 8, 9, 0, are composite. There are three ways of gettig a 4 as the sum amely, if we roll ad, ad, or ad. Similarly we ca also see that there are ways of gettig a 6, ways of gettig a 8, 4 ways of gettig a 9, ways of gettig a 0, ad way of gettig a. Lastly, sice each die has 6 sides, there are total possible outcomes. Therefore, the aswer is Note that Kelvi claims Alex ate the cake. If he is tellig the truth, the both Alex ad AJ must be lyig. Sice there is exactly oe liar, Kelvi must be tellig a lie. Therefore AJ is tellig the truth, ad Kelvi ate the cake. 4. The sum of the umbers from 0 to 0 is 0, sice for every iteger k we are also addig the opposite umber k. Next, the sum of the umbers 04, 0, ad 06 is 604. Thus, the sum of the umbers from 0 to 06 is 604, meaig that 06.. Note that if is i the set, so is, ad so the sum of the umbers i the set is equal to 0. Therefore, the average is also Josephie eats a quarter of the 48 jelly beas. Thus, she eats of them, leavig 6. She ow spills of 6, or 4 of them. This leaves 6 4 for Jared. 7. Sice ABC ad DEF are similar, AB BC DE EF. Therefore, 4 EF EF The dimesios of the sheet of paper ca be coverted to 6 iches by 60 iches. So, if both the sheet of paper ad the cards are positioed vertically, we ca completely cover the paper by placig 6 cards across ad cards dow. Therefore, our desired aswer is Sice 74 is eve, it is divisible by. Ad sice , we see that 74 ca be writte as 7. (Note that ) The differece betwee the smallest ad the largest is We cout off umbers from is oly divisible by two primes, ad. 98 is oly divisible by primes, ad is prime. 96 is oly divisible by two primes, ad. 9 is oly divisible by two primes, ad is oly divisible by two primes, ad is oly divisible by two primes, ad. 9 is oly divisible by primes, ad. 9 is oly divisible by primes, 7 ad. 90 is divisible by three primes,,, ad. Thus 90 is our aswer.. There is a equal umber of quarters, dimes ad ickels totalig $4.80. Let the umber of quarters be. The Sice there are cois, there are 6 cois i total i the jar.. Let x The, we also get that 0000x x. Solvig this for x yields x

3 JHMMC 0 Grade Solutios October, 0. We eed each camera to record a equal umber of miutes ad we ca record a maximum of 60 miutes per camera. 6 hours is equal to miutes. 90, so the greatest umber less tha 60 dividig 90 evely is 9, ad so with each camera we ca record 9 miutes. Therefore, we eed at least 0 cameras. S 4. Let S be the sum of the remaiig umbers. Sice their average is 0, we get that 0 0 S 0 0. So, sice the sum of the umbers i the origial collectio is just 0 + S, we get that the origial average is 0 + S ( + 0) Dividig 000 by gives 8 ad a remaider < 000, but 990 is divisible by. Therefore, the aswer we seek is 990 9, as 9 is ot divisible by. 6. After drawig the diagram, we see that CE CD DE 0 8. Next, by the Power of a Poit Theorem, AE EB CE DE 8 6. Sice AE EB, AE 6 AE EB 6 4. Therefore, AB AE + EB I the prime factorizatio of a perfect square, each prime factor must have a eve expoet. Sice 8! The miimum umber that eeds to be multiplied to this to result i a factorizatio with all eve expoets is Assume without loss of geerality that circle A has a smaller or equal radius tha circle B. Call the ceter of A poit D, ad the ceter of B poit E. The DE a + b, sice the two circles are exterally taget. Draw the perpedicular to lie segmet N E from D. Call the foot of the perpedicular F. The, sice MNF D is a rectagle, we get that EF EN NF EN DM b a. Lastly, by the Pythagorea Theorem o right triagle DEF, DF DE EF (a + b) (b a) ab 7. Thus 4ab 7, ad so ab We ote that the umbers of katz ad frogs icrease expoetially per year. So, Kelvi will have 80 katz after years, ad Alex will have 40 frogs after years. Therefore, to fid the umber of years that eeds to pass for Kelvi ad Alex to have the same umber of katz as frogs is the solutio to the equatio We ca simplify this to the equatio ( ), which yields 4 as the solutio. 40. O each edge of the cube, there are 0 cubes. The first ad last cubes are corer cubes, ad have faces paited. The middle 8 have two faces paited. There are 8 such cubes per edge ad edges, thus 96 cubes i total. The oly other cubes are the cubes part of the 8 8 ceter square o each face, which oly have oe face paited. Thus, the aswer is There are 9 digits to be writte from to 9. There are digits from 0 to 99. There are digits writte from 00 to 999, which is too large. Therefore the umber of pages i the book is some three-digit umber. Sice there are 0 digits i the book, we have where deotes the umber of -digit umbers i the book. Solvig yields 608 -digit pages. Therefore, there are pages i the book. 4. If goslig is bor out of every 0 duck eggs, the gosligs are bor out of 0404 duck eggs. Oe-third of these is a ugly ducklig, so there are 0 4 ugly duckligs.

4 JHMMC 0 Grade Solutios October, 0 4. Whe Johy bought the y mugs, he spet $0y ad was $ short, meaig that 0y x +. Whe he put back half the mug, thus spedig a total of $y, he had $ left over, so y x. Subtractig the secod equatio from the first gives y 40 y 8. Pluggig this value for y ito the first equatio results i 80 x + x 6. The sum of x ad y is therefore We remember that i a regular polygo with, side, each iterior agle has a measure of 80( ) Sice the exterior agle is the agle supplemetary to this iterior agle, we get that the exterior agles all have a measure of 80 (80 60 ) 60. Specifically, i the problem we are give that they are each degrees, ad so 60. Thus, Let the sum of the umbers o the 8 corers be S. Whe we take the sum of the umbers o the 6 faces, we see that we couted the umber o each vertex times, sice a vertex lies o exactly faces. Therefore, S Arthur ca sow a field i hours, so his rate of sowig is per hour. James ca sow a field i hours, so his rate of sowig is per hour. Let the umber of hours i which Kevi ca sow a field be hours. The his rate is, ad their combied rate is + + Sice we are give that together they ca fiish i hours, we get that the rate combied is also equal to Solvig yields The formula for T ca be derived as follows: T +T (+)+(+( ))+ +(+) (+ ( + ) ) T. (This is the famous formula for the sum of the first positive itegers.) Therefore, T 0 + T We cosider two cases. Case is that Arthur fiished secod. Sice James fiished before Deis, the possibilities are James-Arthur-Deis-Wag, James-Arthur-Wag-Deis, ad Wag-Arthur- James-Deis, givig a total of. Case is that Arthur fiished third. The possibilities are the James-Deis-Arthur-Wag, Wag-James-Arthur-Deis, ad James-Wag-Arthur-Deis, givig a total of. Therefore, there are + 6 ways the race could have eded. 49. If we wat to fid the probability that evet A happes give that evet B happes, we have to divide the probability of both evets happeig by the probability of evet B happeig. (This is how we ca calculate such coditioal probabilities.) So, i this case, evet A is pickig box A, ad evet B is removig a gree ball. Sice we choose either box with equal probability, we see that the chace of both choosig box A ad choosig a gree ball from there is equal to 7. However, there are actually two ways to choose a gree ball by either pickig box A or by pickig box B. The first case has the same probability 7, but the secod has probability. Therefore, our aswer is equal to Chace of A ad B Chace of B

5 JHMMC 0 Grade Solutios October, 0 0. Suppose there are a red marbles, b blue marbles, c gree marbles, ad d orage marbles. The we have the system of equatios b + c + d a + c + d 0 a + b + d a + b + c 7. Addig these gives (a + b + c + d) 87, ad so a + b + c + d 9. Subtractig this equatio by the first equatio yields a 4.

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =