Math Counts Test Review 2008 State

1. Math Counts Test Review2008 State

2. 3. What is the greatest number of interior right angles a convex octagon can have? Total angle = 6*180 = 1080 Let X be number of right angles, all other angles must be less than 180: 90 X + (8 – X) * 180 > 1080 90 X < 8*180 – 1080 90 X < 360 X < 4 Answer: 3

3. 7. The tri-sectors of angles B and C of scalene triangle ABC meet at points P and Q, as shown. Angle A measures 39 degrees and angle QBP measures 14 degrees. What is the measure of angle BPC? QBP = 14  ABC = 14*3 = 42 A = 39  ACB = 180 - 42- 39 = 99 In triangle PBC, CBP = QBP = 14, BCP = 99/3 = 33 BPC = 180 - 14 - 33 = 133

4. 9. A clock loses 5 seconds every 12 minutes. At this rate, how many minutes will the clock lose in a 24-hour period? Every 12 min loses 5 seconds Every hour loses 5*5 = 25 seconds 24 hours will lose 25 * 24 = 2400/4 = 600 seconds = 10 min

5. 10. Elliott Farms has a silo for storage. The silo is a right circular cylinder topped by a right circular cone, both having the same radius. The height of the cone is half the height of the cylinder. The diameter of the base of the silo is 10 meters and the height of the entire silo is 27 meters. What is the volume, in cubic meters, of the silo? Express your answer in terms of . Cone height = 1/3 * 27 = 9 Cylinder height = 2/3 * 27 = 18 Cone Radius = Cylinder Radius = 5 Cone Vol = 5*5*  *9 /3 = 25*3  Cylinder Vol = 5*5*  *18 = 25*18  Silo Vol = 25*  *(3+18) = 25*21  = 525 

6. 15. The integers from 1 through 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first? Consider the #s are printed on a round table. If we start with 1, we get: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 2, 3, 5, 6, 8, 9, 11, 12, 14, 15 3, 5, 8, 9, 12, 14 3, 5, 9, 12 3, 5, 12 3, 12 ---> Leaving the 3rd and the 12th element in the end. The pattern suggests that, if we rotate the table and startwith 9, as shown below: 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8 … …, … …, … …, …, 11, 5 (ends with 3rd and 12th elements) Answer: 9

7. 21. A circle with a radius of 2 units has its center at (0, 0). A circle with a radius of 7 units has its center at (15, 0). A line tangent to both circles intersects the x-axis at (x, 0) to the right of the origin. What is the value of x? Express your answer as a common fraction. Note that OAX CBX X / (15 – X) = 2/7 B 7 X = 2 (15 – X) = 30 – 2X O x C(15,0) A 9 X = 30 X = 10/3

8. 23. A circular cylindrical post with a circumference of 4 feet has a string wrapped around it, spiraling from the bottom of the post to the top of the post. The string evenly loops around the post exactly four full times, starting at the bottom edge and finishing at the top edge. The height of the post is 12 feet. What is the length, in feet, of the string? For one round of loop, we can find: length of string = 5 3 For the 4 loops, 4 The string length = 5*4 = 20

9. 26. Kevin will start with the integers 1, 2, 3 and 4 each used exactly once and written in a row in any order. Then he will find the sum of the adjacent pairs of integers in each row to make a new row, until one integer is left. For example, if he starts with 3, 2, 1, 4, and then takes sums to get 5, 3, 5, followed by 8, 8, he ends with the final sum 16. Including all of Kevin’s possible starting arrangements of the integers 1, 2, 3 and 4, how many possible final sums are there? From any a1, a2, a3, a4, we can generate the sequence: a1+a2, a2+a3, a3+a4; a1+a2+a2+a3, a2+a3+a3+a4; a1+3a2+3a3+a4; From 1, 2, 3, 4, there are 6 ways to pick different a1 & a4 Hence the maximum # of possible sums are 6. Note that 1, 2, 3, 4 and 2, 1, 4, 3 generate the same sum We have the final answer: 6 – 1 = 5

10. 27. A rectangular band formation is a formation with m band members in each of r rows, where m and r are integers. A particular band has less than 100 band members. The director arranges them in a rectangular formation and finds that he has two members left over. If he increases the number of members in each row by 1 and reduces the number of rows by 2, there are exactly enough places in the new formation for each band member. What is the largest number of members the band could have? Let X be total number of members X = r m + 2 ---- (1) X = (m + 1) *(r -2) = r m + r – 2m – 2 ---- (2) (2) – (1): r – 2m – 4 = 0  r = 2m + 4 ---- (3) (3) to (2): X = (m+1)(2m+4-2)=2(m+1)2 ---- (4) Since X < 100, (m+1)2 = X/2 < 50 Largest possible integer m is 7. Answer: X = 98

11. 28. A right cylinder with a base radius of 3 units is inscribed in a sphere of radius 5 units. What is the total volume, in cubic units, of the space inside the sphere and outside the cylinder? Express your answer as a common fraction in terms of . Note that the height of the cylinder is 4*2 = 8 3 Vc = (3)2  * 8 = 72  o 4 5 Vs = 4/3 * 53 = 500/3  Answer = 500/3  – 72  = 284/3 

12. 29. Given that x2 – 5x + 8 = 1, what is the positive value of x4 – 10x3 + 25x2 – 9? From x2 –5x + 8 = 1  x2 – 5x = -7 (x2 – 5x ) 2 = 49 x4 – 10x3 +25x2= 49 Hence x4 – 10x3 +25x2- 9 = 49 – 9 = 40

13. 30. An o-Pod MP3 player stores and plays entire songs. Celeste has 10 songs stored on her o-Pod. The time length of each song is different. When the songs are ordered by length, the shortest song is only 30 seconds long and each subsequent song is 30 seconds longer than the previous song. Her favorite song is 3 minutes, 30 seconds long. The o-Pod will play all the songs in random order before repeating any song. What is the probability that she hears the first 4 minutes, 30 seconds of music - there are no pauses between songs - without hearing every second of her favorite song? Express your answer as a common fraction. It’s easier to find the reverse possibility here. That is, to hear the entire favorite song. That leaves 4m30sec – 3min30sec = 1 min to hear other songs. She can only hear (1) her favorite song first or (2) the first song first, and then her favorite song or (3) the second song first, and then her favorite song Prob1 = 1/10; Prob2 = 1/10 * 1/9; Prob3 = 1/10 * 1/9 Answer = 1 – 1/10 – 1/90 – 1/90 = 9/10 – 2/90 = 79/90

14. G3. What is the shortest distance from the point (6, 0) to the line y = 2x – 2? Express your answer in simplest radical form. C(6,10) D Note that the distance from A to CD is AD O A(6,0) B(1,0) Area(OAC) = ½ OA * AC = ½ * 5 * 10 = ½ 50 Area(OAC) = ½ OC * AD = ½ 5 √5 * AD AD = 50 / (5 √5) = 10/√5 = 2 √5

15. G4. Kendra has an unlimited supply of unbreakable sticks of length 2, 4 and 6 inches. Using these sticks, how many non-congruent triangles can she make? Two sticks can be joined only at a vertex of the triangle. (A triangle with sides of length 4, 6, 6 is an example of one such triangle to be included.) Listing all 10 possible non-congruent combinations: 2, 2, 2; 2, 2, 4; 2, 2, 6; 2, 4, 4; 2, 4, 6; 2, 6, 6; 4, 4, 4; 4, 4, 6; 4, 6, 6;6, 6, 6 Cross out the ones that don’t fit triangle inequality:2, 2, 2; 2, 2, 4; 2, 2, 6; 2, 4, 4; 2, 4, 6; 2, 6, 6; 4, 4, 4; 4, 4, 6; 4, 6, 6;6, 6, 6 Answer: 7

16. G8. To transmit a positive integer less than 1000, the Networked Number Node offers two options. Option 1: Pay $d to send each digit d. Therefore, 987 would cost $9 + $8 + $7 = $24 to transmit. Option 2: Encode integer into binary (base 2) first, and then pay $d to send each digit d. Therefore, 987 becomes 1111011011 and would cost $1 + $1 + $1 + $1 + $0 + $1 + $1 + $0 + $1 + $1 = $8. What is the largest integer less than 1000 that costs the same whether using Option 1 or Option 2? Note: Calculators are allowed in the target round! 1 111 111 1112 = 1024 – 1 = 1023 > 1000 We know the largest $ in option 2 will be < 10. Listing all the $9 pay in Option 2, we get: 1 011 111 111; 1 101 111 111; 1 110 111 111; 1 111 011 111; 1 111 101 111; 1 111 110 111; 1 111 111 011; 1 111 111 101; 1 111 111 110, 111 111 111 767, 895, 959, 991, 1007, … -> none equal to $9 pay in option 1 & < 1K. Now listing all $8 pay with option 2, from big to small, we get: 111 111 110; 111 111 101; 111 111 011; 111 110 111; 111 101 111; 111 011 111; 110 111 111; 101 111 111; 11 111 111  111 110 111 = 503

17. M3. At a particular speed, a jumbo jet can travel its own length in 20 seconds. At this same speed, the jumbo jet taxied completely past a 710-foot-long hangar in 70 seconds, as shown. What is the length of the jumbo jet? Let X be the length of the jet (X + 710) / 70 = X/20 = speed of the jet 70 X = 20X + 710 * 20 50 X = 710 * 20 X = 71 * 20 / 5 = 284

18. M7. One night two cylindrical wax candles of different heights and different diameters were lit. One of the candles was 20 cm taller than the other. They were both lit at the same time and each burned at a steady rate. Five hours after they were lit they were both the same height. The taller one burned all of its wax six hours after it was lit, and the shorter one burned all of its wax 10 hours after it was lit. What was the ratio of the original height of the shorter candle to the original height of the taller candle? Express your answer as a common fraction. Let H and h be the heights of two candles Let R and r be their speed of burning (in/hr) H = 6 * R h = 10 r H – 5 R = h – 5 r H – 5 * H/6 = h – 5 * h/10 1/6 H = ½ h  h/H = 1/3

19. M8. Each of three, standard, six-sided dice is rolled once. What is the probability that the three numbers rolled are the lengths of the sides of an obtuse triangle? Express your answer as a common fraction. Let a, b, c be the lengths of any obtuse triangle, with a < c & b < c. For the triangle to be obtuse, we have: a2 + b2 < c2 and c < a + b List of triplets meeting the criteria include: 1,1,(3,4,5,6); 1,2,(3,4,5,6); 1,3,(4,5,6); 1,4,(5,6); 1,5,6; 1,6,(none) 2,2,(3,4,5,6);2,3,(4,5,6); 2,4,(5,6); 2,5,6; 2,6,(none) 3,1,(4,5,6);3,2,(4,5,6); 3,3,(5,6); 3,4,6; 3,5,6; 3,6,(none) 4,1,(5,6);4,2,(5,6); 4,3,6; 4,4,6; 4,(5,6),(none) 5,1,6; 5,2,6; 5,3,6; 5,(4,5,6),(none); 6,(1,2,3,4,5,6), (none) # of triplets fitting the requirement: 13 * 3 = 39 (since c can be any of the thee dices) Total # of triplets: 6*6*6 Answer: 39/(6*6*6) = 13/72

20. M10. A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector is to be painted with one of four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.) 1) # of ways with 1 color: 4 (Reason: 4 colors to choose, each with 1 painting) 2) # of ways with 2 different colors: 6 * 4 = 24 Reason: # of ways to pick 2colors: 4*3/2 = 6 For any pick of 2 colors, there are 4 ways: ABAB, AABB,AAAB, ABBB 3) # of ways with 3 different colors: 6 * 6 = 36 Reason: # of ways to pick 3 colors (= # of ways to drop 1 color): 4 For each pick of 3 colors, there are 9 ways to paint: AABC, AACB, BBAC, BBCA, CCAB, CCBA, ABAC, BABC, CACB 4) # of ways with 4 different colors: 6 Reason: # of ways to color without rotation: 4! = 24 Each way can be repeated 4 times with the rotation, hence 24/4 = 6 Total ways of painting = 1) + 2) + 3) + 4) = 4 + 24 + 36 + 6 = 70

21. Advanced Math Practice

22. A. In rectangle ABCD, AB = 12 and BC = 10. Points E and F lie inside rectangle so that BE = 9, DF = 8, BE // DF, EF // AB, and line BE intersects segment AD. The length EF can be expressed in the form m n - p, where m, n, and p are positive integers and n is not divisible by the square of any prime. Find m + n + p. B. Suppose that a parabola has vertex (1/4, -9/8) and equation Y = aX2 + bX + c, where a > 0 and a + b + c is an integer. The minimum possible value of a can be written in the form p/q, where p and q are relatively prime positive integers. Find p + q.

23. A. In rectangle ABCD, AB = 12 and BC = 10. Points E and F lie inside rectangle so that BE = 9, DF = 8, BE // DF, EF // AB, and line BE intersects segment AD. The length EF can be expressed in the form m n - p, where m, n, and p are positive integers and n is not divisible by the square of any prime. Find m+n+p. A H 12 B Extend BE to get parallelogram DGEF. 9 We get: EF = DG & CF = EG = 8 E F 10 Assume EF = DG = X 8 From right triangle BGH, we have: G D C 102 + (12+X)2 = (8 + 9)2 (12+X)2 = 172 - 102 = 27*7 = 32*21 X + 12 = 3 21  X = 3 21 - 12 Answer: m + n + p = 3 + 21 + 12 = 36

24. Modula Arithmetic A + B (mod X) = A (mod X) + B (mod X) A - B (mod X) = A (mod X) - B (mod X) A * B (mod X) = A (mod X) * B (mod X) A / B (mod X) = A (mod X) / B (mod X)

25. B. Suppose that a parabola has vertex (1/4, -9/8) and equation Y = aX2 + bX + c, where a > 0 and a + b + c is an integer. The minimum possible value of a can be written in the form p/q, where p and q are relatively prime positive integers. Find p + q. Recall that the parabola equation with vertex (m, n) : Y – n = a (X – m)2 We have: Y + 9/8 = a (X – ¼) 2 Hence: Y = a X 2 – ½ a X + 1/16 a – 9/8  b = - ½ a; c = 1/16 a – 9/8 = (a – 18)/16  a + b + c = a – ½ a + (a-18)/16 = (9a – 18)/16 Since a + b + c is integer, need to find the smallest positive a: (9a – 18)/16 = integer In other words: (9a – 18) = 0 (mod 16) That is: 9a (mod 16) = 18 (mod 16) = 2 (mod 16) 9a = 16K+2  a = 16/9 K + 2/9  smallest positive a = 2/9 Answer: 2 + 9 = 11