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2016 State Competition Solutions Are you wondering how we could have possibly thought that a Mathlete® would be able to answer a particular Sprint Round problem without a calculator? Are you wondering how we could have possibly thought that a Mathlete would be able to answer a particular Target Round problem in less 3 minutes? Are you wondering how we could have possibly thought that a particular Team Round problem would be solved by a team of only four Mathletes? The following pages provide solutions to the Sprint, Target and Team Rounds of the 2016 MATHCOUNTS® State Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches to solving these fun and challenging MATHCOUNTS problems. Special thanks to solutions author Mady Bauer for graciously and voluntarily sharing her solutions with the MATHCOUNTS community for many, many years! 2016 State Competition Sprint Round 1. Given: ππ@ππ = Find: [email protected] 5 ππ 2ππ+ππ [email protected] = (2×5)+3 = 5 13 Ans. There are 2 rectangles that are two rows high by two columns wide. 1 here 2. Find: the number of rectangles displayed in the diagram. and 1 here There are 2 rectangles that are three rows high by one column wide. There are 6 rectangles that are one row high by one column wide. Finally, there is 1 rectangle three rows high by 2 columns wide. There are 3 rectangles that are one row high by two columns wide. There are 4 rectangles that are two rows high by one column wide. 2 here and 2 here The total number of rectangles is 6 + 3 + 4 + 2 + 2 + 1 = 18 Ans. 3. Given: 7π₯π₯ + 13 = 328 Find: the value of 14π₯π₯ + 13 Multiply the equation by 2. 14π₯π₯ + 26 = 656 14π₯π₯ + 26 β 13 = 656 β 13 = 643 Ans. 4. Find: the median of the positive perfect squares less than 250. The largest square less than 250 is 225 or 152 . That would be 15 integers so the median is the eighth one. 82 = 64 Ans. 5. Given: π₯π₯+5 π₯π₯β2 = 2 3 Find: π₯π₯ π₯π₯ + 5 2 = π₯π₯ β 2 3 3π₯π₯ + 15 = 2π₯π₯ β 4 π₯π₯ = β19 Ans. 6. Given: WX=4, XY=2, YV=1 and UV=6. Find: the absolute difference between the areas of triangles TXZ and UYZ. The area of triangle TXZ is equal to the area of triangle TXU minus TZU. The area of triangle UYZ is equal to the area of triangle TYU minus TZU. This means the difference in the area of triangles TXZ and UYZ is equal to the difference between TXU and TYU. The base of both triangles is TU. The height of both triangles is the same β a perpendicular from WV to TU. Therefore, the area of each triangle is the same. The difference between the two areas is 0. Ans. 7. Given: A bag has 4 blue, 5 green and 3 red marbles. Find: How many green marbles must be added to the bag so that 75% of the marbles are green? There are a total of 4 + 5 + 3 = 12 marbles. Let π₯π₯ = the number of green marbles added. 3 5 + π₯π₯ = 12 + π₯π₯ 4 20 + 4π₯π₯ = 36 + 3π₯π₯ π₯π₯ = 16 Ans. 8. Given: MD swaps out the tires on his trike so that all four will have been used for the same distance as he drives 25,000 miles. Find: How many miles does each tire drive. 3 4 Each tire will be on for of the trip. 3 4 × 25000 = 18750 Ans. 9. Given: Lucy and her dad have the same birthday. When Lucy was 15 her father was 3 times her age. On this birthday, Lucyβs father turned twice as old and Lucy did. Find: How old is Lucy? When Lucy was 15 her father was 45. Currently, her Dad is twice her age. Let π₯π₯ represent the number of years since Lucy was 15. Then 2 × (15 + π₯π₯) = 45 + π₯π₯ 30 + 2π₯π₯ = 45 + π₯π₯ π₯π₯ = 15 Her Dad is now 45 + 15 = 60, and Lucy is 15 + 15 = 30 Ans. 10. Given: The sum of three distinct 2-digit primes is 53. Two of the primes have a units digit of 3 and the other prime has a units digit of 7. Find: the greatest of the three primes. Letβs list the 2-digit primes less than 53 that end in 3: 13, 23, 43. Now list the 2-digit primes less than 53 that end in 7: 17, 37, 47. Remove 43 and 47 because they are too large to be one of the three primes. 13 + 23 = 36 53 β 36 = 17 So the primes are 13, 17 and 23. The greatest of the primes is 23. Ans. 11. Given: Ross and Max weigh 184 pounds. Ross and Seth weigh 197 pounds. Max and Seth weigh 189 pounds. Find: Rossβ weight. Let ππ = Rossβ weight. Let ππ = Maxβs weight. Let π π = Sethβs weight. ππ + ππ = 184 ππ + π π = 197 ππ + π π = 189 Adding these three equations yields 2ππ + 2ππ + 2π π = 184 + 197 + 189 = 570 Doubling the third equations, we get 2ππ + 2π π = 189 × 2 = 378 Substituting this into the equation, 2r + 2m + 2s = 570, we get 378 + 2ππ = 570 2ππ = 570 β 378 = 192 ππ = 96 Ans. 12. Find: the least possible denominator of a positive rational number 0. οΏ½οΏ½οΏ½οΏ½ π΄π΄π΄π΄ where π΄π΄ and π΅π΅ are distinct digits. οΏ½οΏ½οΏ½οΏ½ Let π₯π₯ = 0. π΄π΄π΄π΄ Then 100π₯π₯ = π΄π΄π΄π΄. οΏ½οΏ½οΏ½οΏ½ π΄π΄π΄π΄ 99π₯π₯ = π΄π΄π΄π΄ π΄π΄π΄π΄ π₯π₯ = 99 99 = 11 × 9 Suppose the denominator is 9. An example 11 1 οΏ½οΏ½οΏ½ is = β 0. οΏ½11 99 9 No, this doesnβt work since π΄π΄ = π΅π΅ and that violates the requirements. What about 11? An example is: 9 1 οΏ½οΏ½οΏ½ = β 0. οΏ½09 99 11 Yes, in this case π΄π΄ β π΅π΅. 11 Ans. 13. Given: It costs $3.25 for the first mile and $0.45 for each 1 additional 4 mile. Find: How many miles can be travelled for $13.60 13.60 β 3.25 = 10.35 So, after the first mile there is $10.35 left. 10.35 = 23 0.45 So thatβs an additional 23 quarter miles. 1+ 23 4 3 4 =1+5 =6 3 4 Ans. 14. Given: 2 m3 of soil containing 35% sand is mixed with 6 m3 of soil containing 15% sand. Find: the percent of the new mixture that is sand Of the 2 m3 mixture, 0.35 × 2 = 0.7 m3 is sand, and of the 6 m3 mixture, 0.15 × 6 = 0.9 m3 is sand. Putting the two mixtures together gives us 2 + 6 = 8 m3 of mixture, of which 0.7 + 0.9 = 1.6 m3 is sand. 1.6 8 1 5 = = 20% Ans. 15. Given: Alex runs a lap in 1 minute, 28 seconds. Becky runs a complete lap in 1 minute, 16 seconds. Find: How many complete laps will Alex have run when Becky passes him for the first time? Alex runs at a rate of 88 seconds per lap and Becky at a rate of 76 seconds per lap. Let ππ = the length of the lap. After one second Alex has run 1 88 of a lap. 1 Becky has run of a lap. The difference 76 1 1 88β76 12 3 3 β = = = = 76 88 76×88 76×88 19×88 1672 3 laps per second it At a separation of 1672 1672 will take seconds to reach one lap 3 difference. 1672 3 557 88 1 3 1 3 = 557 seconds 1 3 = 6 laps. Therefore, Alex will have run 6 complete laps when Becky passes him. 6 Ans. is 16. Given: Four teams, the Beavers, Ducks, Platypuses and Narwhals, are in a singleelimination tournament. Find: The probability that the Ducks and the Beavers will play each other in one of the two rounds. Letβs start out with the Ducks and the Beavers playing each other in the first round. There are 2 possibilities for the outcome of the Ducks versus Beavers game and 2 possibilities for the outcome of the Platypuses versus Narwhals game. There are 2 × 2 = 4 match-up possibilities for the final round. In 4 out of 4 of these possibilities the Ducks and the Beavers will play since they meet in the first round. If the Ducks and the Beavers do not play in the first round, there are two different first round scenarios. First, letβs look at the Beavers playing the Platypuses and the Ducks playing the Narwhals in the first round. Again, there are 2 × 2 = 4 possible outcomes of these matchups. Of these possibilities, only 1 will the Beavers and the Ducks play each other, shown here. In the other first round scenario the Beavers play the Narwhals and the Ducks play the Platypuses. Again, there are 4 possible outcomes and only 1, shown here, where the Beavers play the Ducks. Of the three possible first round lineups, there are 4 + 4 + 4 = 12 possible second round matchups. Out of these matchups the Ducks and the Beavers will end up playing each other in 4 + 1 + 1 = 6 of the scenarios. So the probability is 6 12 = 1 2 Ans. 17. Given: ππ(π₯π₯) is defined for positive integers and ππ(ππ) + ππ(ππ) = ππ(ππππ) and ππ(3) = 5. Find: ππ(27) ππ(3) + ππ(3) = ππ(3 × 3) ππ(9) = 5 + 5 = 10 ππ(3) + ππ(9) = ππ(3 × 9) ππ(27) = 5 + 10 = 15 Ans. 18. Given: In rectangle ABCD, AB = 6 and AD = 5. AC is extended to point E so that οΏ½οΏ½οΏ½οΏ½ οΏ½οΏ½οΏ½οΏ½ . π΄π΄π΄π΄ is congruent to πΆπΆπΆπΆ Find: the length of οΏ½οΏ½οΏ½οΏ½ π·π·π·π· Since we extend C to E and π΄π΄π΄π΄ = πΆπΆπΆπΆ, then CE is just another diagonal of a 5 × 6 rectangle. it. This tells us that 2 must be added, which will be carried over from the thousands column. So B + 7B + N = 21, where N represents a single digit integer number carried over from the hundreds column. Rearranging we get 8B + N = 21 If B is 1, N would be 13, if B is 2, N would be 5, and if B is 3, N would be β3. The only way to get N to be a single digit positive integer is if B = 2. This means our multiplication would be 127 × 721 which equals 91567. 721 Ans. 20. Given: Diagonal XZ of rectangle WXYZ is divided into 3 segments each of length 2. MW and NY are parallel and perpendicular to XZ. οΏ½οΏ½οΏ½οΏ½ is 12 and the length of οΏ½οΏ½οΏ½οΏ½ The length of π·π·π·π· πΉπΉπΉπΉ is 5. Triangle DEF is a 5-12-13 right triangle. οΏ½οΏ½οΏ½οΏ½ is Therefore, the length of π·π·π·π· 13. Ans. 19. Given: The digits of a 3-digit number are reversed. The product of the new and original integers is 91567. Find: the new integer. Since the last digit of our product is a 7, letβs look at multiply a 7 and a 1. The product will look like this: Looking at the far left column, the tenthousands place, the product of our two numbers has a 9 here but we see a 7 above Find: the area of WXYZ. First letβs draw in a line parallel to XZ from point R to point P, as shown, to create rectangle MRPN and triangle RPY. We know mβ MXR β mβ PRY, RP = MN = 2 and MR = NP. Triangles XMR and RPY are congruent (by ASA), and, therefore, XR = RY and MR = PY. We also can determine, by AAA similarity, that triangle RPY is similar to triangle YNZ. Using this information, we can set up two ratios to solve for unknown distances. 2/NY = PY/2 and NY = 2PY Solving we get: 4 = NY × PY 4 = 2PY × PY 2 = PY2 PY = β2 And therefore NY = 2β2 Now we can use the Pythagorean theorem to find XY and YZ. PY2 + 22 = RY2 (β2)2 + 4 = RY2 6 = RY2 RY = β6 Therefore, XY = 2RY = 2β6. NY2 + 22 = YZ2 (2β2)2 + 4 = YZ2 12 = YZ2 YZ = 2β3. The area of rectangle WXYZ is XY × YZ = 2β6 × 2β3 = 4β18 = 12β2 Ans. 21. Given: A spinner is divided into 5 sectors. The angles of sections 1 through 3 measure 60°. The angles of sections 4 and 5 measure 90°. Find: the probability that at least one spin lands on an even number if the spinner is spun twice. There are two even numbers, 2 and 4. The section labeled 2 is 60°, and the section labeled 4 is 90°. The probability of one spin landing on an even number is, therefore, 60+90 360 = 150 360 5 . 12 = The probability of one spin landing on an odd number is 1 β 5 12 = 7 . 12 If you are spinning twice, there are three ways to get at least one even. You can spin two evens, an even and then and odd or an odd and then an even. The probability of spinning two evens is 5 12 × 5 12 = 25 . 144 The probability of spinning an even then an odd is 5 12 × 7 12 = 35 . 144 The probability of spinning an odd then an even is 7 12 × 5 12 = 35 . 144 So the probability of spinning at least one even is 25 35 35 + + 144 144 144 = 95 144 Ans. 22. Given: The student council has 8 members consisting of 4 officers and 4 other members. The 4 officers must sit together at the circular table. Find: How many distinguishable circular seating orders are possible? Let o1 through o4 be the officers and m1 through m4 be the members. There are 4! = 24 ways that o1 through o4 can be ordered. Similarly, there are 24 ways that m1 through m4 can be ordered. 24 × 24 = 576 Ans. 23. Given: a chip is placed in the upper-left corner square of a 15 × 10 grid. The chip can move in an L-shaped pattern. Find: the minimum number of L-shaped moves to move the chip to the square marked βXβ. You need to move 9 down and 14 to the right. Moving 14 to the right will take at least 7 L-shaped patterns that move 2 to the right and 1 down. But, if we do 7 of those we end up at (15, 8) given that we started at (1, 1) and must end up at (15, 10). So, can we do it with 8 L-shaped moves? We can do 6 moves of (2 right, 1 down) to get us to (13, 7). One move of (1 right, 2 down) will get us to (14, 9) leaving us with having to go (1, 1) more but thereβs no L-shaped move to get us there. Again, after 6 moves, we can move (1 left, 2 down) to (12, 9). From there we are still going to need 2 more moves because 15 β 12 = 3. It looks like it will take us 9 moves to get this to work. We start at (1, 1). Now letβs do 6 (2 right, 1 down) to get to (13, 7) This has positioned us to move 2 more Lshaped moves of 2 right and 1 down. 9 Ans. 24. Given: On line segment π΄π΄π΄π΄, π΅π΅ is the midpoint of segment π΄π΄π΄π΄ and π·π· is the midpoint of segment πΆπΆπΆπΆ. π΄π΄π΄π΄ = 17 and π΅π΅π΅π΅ = 21. Find: the length of segment π΄π΄π΄π΄. π΄π΄π΄π΄ = π΅π΅π΅π΅ πΆπΆπΆπΆ = π·π·π·π· Let π₯π₯ = π΄π΄π΄π΄ Let π¦π¦ = πΆπΆπΆπΆ Then the length of segment π΄π΄π΄π΄ is 2π₯π₯ + π¦π¦ and the length of π΅π΅π΅π΅ is π₯π₯ + 2π¦π¦. Since AD = 17 and BE = 21, we have 2π₯π₯ + π¦π¦ = 17 π₯π₯ + 2π¦π¦ = 21 Adding these two equations yields 3π₯π₯ + 3π¦π¦ = 38 3(π₯π₯ + π¦π¦) = 38 38 π₯π₯ + π¦π¦ = 3 The length of π΄π΄π΄π΄ is 2π₯π₯ + 2π¦π¦. 2π₯π₯ + 2π¦π¦ = 2(π₯π₯ + π¦π¦) = 2 × Now the 7th L-shaped move is 2 left and 1 down getting us to (11, 8). 38 3 = 76 Ans. 3 25. Given: There are 12 different mixed numbers that can be created by substituting three of the numbers 1, 2, 3 and 5 for a, b, ππ ππ and c in the expression ππ where ππ < ππ. Find: the mean of the 12 numbers. These are the 12 numbers: 2 2 3 1 ,1 ,1 3 5 5 1 1 3 2 ,2 ,2 3 5 5 1 1 2 3 ,3 ,3 2 5 5 1 1 2 5 ,5 ,5 2 3 3 Add up all the mixed numbers whose denominator is 2. 1 1 3 +5 =9 2 2 Add up all the mixed numbers whose denominator is 3. 2 1 1 2 6 1 + 2 + 5 + 5 = 13 = 15 3 3 3 3 3 Add up all the mixed numbers whose denominator is 5. 2 3 1 3 1 2 1 +1 +2 +2 +3 +3 = 5 5 5 5 5 5 2 12 = 14 12 5 5 Now, add them all up. 2 2 192 9 + 15 + 14 = 38 = 5 5 5 Now determine the mean. 192 5 12 = 192 5×12 = 192 60 =3 12 60 =3 1 5 Ans. 26. Given: 738 consecutive integers are added. The 178th number in the sequence is 4,256,815. Find: the remainder when the sum is divided by 6. First, notice 738 is evenly divisible by 6 because it is even and its digits sum to 18, a multiple of 3. The formula for adding the first n consecutive integers is (n)(n + 1)/2. Let x represent the first integer in this sequence. The sum of these 738 integers will be (738)(738 + 1)/2 + 738x Because we know 738 is divisible by 6, we know 738x will also be divisible by 6 and therefore have remainder 0. We donβt need to even find the value of x. Now lets look at what still remains in the sum (738)(738 + 1)/2 = (369)(738 + 1) = 369 × 738 + 369 We know 6 evenly divides 369 × 738 So the remainder for this integer sum is equal to the remainder when 369 is divided by 6. 369 ÷ 6 = 61 remainder 3. Ans. 27. Given: A(-5,0) and B(5,0) Find: How many points X are there such that XA and XB are both positive integer distances, each less than or equal to 10? Letβs assume that the point X is not on the x-axis. Then itβs all about creating a triangle where one side is 10 and the other two are 10 or smaller. Letβs start with one of the other two sides being 10. The max length for the third side would be 19 but we are limited to lengths β€ 10. So the third side can be 1 through 10 for a total of 10 triangles. Now let one of the other two sides be 9. The max length for the third side is 18. Again, with the limitations, and taking into account that the first two lines differ by one, the third side can be 2 through 10 or 9 possibilities. Similarly for the second side having a length of 8, the third side can be 3 through 10 or 8 possibilities. Thereβs the pattern: Second side is 7; third side is 4 through 10. Second side is 6; third side is 5 through 10. Second side is 5; third side is 6 through 10. Second side is 4; third side is 7 through 10. Second side is 3; third side is 8 through 10. Second side is 2; third side is 9 through 10. Second side is 1, third side is 10. 1 + 2 + β― + 9 + 10 = 55 All of these triangles have been above the x-axis. But we can have the same triangles reflected over the x-axis. That gives us, so far, 55 + 55 = 110 possibilities. All of these triangles assumed that X was not on the x-axis. But if it is, you could have X at (β4,0), (β3,0), β¦ , (3,0), (4,0) Thatβs another 9 points. 110 + 9 = 119 Ans. 28. Given: ππ(ππ) = ππ β ππ! + ππ, where ππ and ππ are positive integers. The range of ππ contains two numbers that differ by 20. Find: the least possible value of ππ(1). Two values, f(y) and f(x) must differ by 20. f(y) β f(x) = a β y! + b β (a β x! + b) f(y) β f(x) = a β (y! β x!) 20 = a β (y! β x!) Since all the variable are positive integers, letβs look at pairs integers that multiply to be 20: 1 × 20, 2 × 10, 4 × 5. We want to minimize the value of a in order to minimize the value of f(1). Letβs look at the value of the first 5 factorials: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120 If a is 1, then we need two factorials that differ by 20. There arenβt two. If a is 2, then we need two factorials that differ by 10. There arenβt two. If a is 4, then the factorials 6! β 1! = 5 will work. So we know a is 4, b can be any positive integer but we want to minimize f(1) so lets pick b to be 1. f(1) = 4 β 1! + 1 = 5 Ans. 29. Given: A list of numbers 1, 2, β¦ 9999 where the digits 0 through 9 are replaced with the letters A through J. The resulting list of 9999 strings is sorted alphabetically. Find: How many strings appear before βCHAIβ? Alphabetic sorting is just what you find in the dictionary. - Single digit numbers 0 (A) canβt be considered so we start with 1. 1 and 2 (i.e., B and C) will be before CHAI. Thatβs two numbers. - Double digit numbers. Again, we canβt start with A but we can start with B. 10-19 (BA-BJ) will appear before CHAI. For double digit numbers starting with C you have 20-27 (CA-CH) that will appear before CHAI. Thatβs a total of 18 numbers. - Three digit numbers. Again, we canβt start with A, but we can start with B. 100-199 (BAA-BJJ) will appear before CHAI. 200-269 (CAA-CGJ) will appear before CHAI. Moving to 270, thatβs the only one we can have (CHA). Thatβs a total of 100 + 70 + 1 = 171 - Four digit numbers. As usual we start with B β we have 10001999 (BAAA-BJJJ) for 1000 numbers. Then we have 2000-2699 (CAAA-CGJJ) or 700 numbers. Finally we have 2700-2707) (CHAA-CHAH) or 8 numbers. Thatβs a total of 1000 + 700 + 8 = 1708 Now letβs add them all up: 2 + 18 + 171 + 1708 = 1899 Ans. 30. Given: A 12-sided die has the shape of a hexagonal bipyramid. Each has a hexagonal base with side 1 and height 1. The die is rolled and lands on one of its triangular faces. Find: How high off the ground is the opposite face? This is what a hexagonal bipyramid looks like. β=1 π₯π₯ is just half of the height of a hexagon, or Now the problem says tht the bipyramid has landed on one of its sides. So letβs rotate it onto its side. And what do you notice about this image? Doesnβt it look, in 2-dimensional space, like a rhombus? So what we have is a rhombus and what we are looking for is the height of the rhombus. If you think of one of the triangles of one of β3 2 So now we have to find π¦π¦, which is the height of the triangles of the pyramid and the hypotenuse of the triangle formed with 1 and π₯π₯. 12 + π₯π₯ 2 = π¦π¦ 2 2 β3 1 + οΏ½ οΏ½ = π¦π¦ 2 2 3 7 π¦π¦ 2 = 1 + = 4 4 7 β7 π¦π¦ = οΏ½ = 2 4 The sides of the rhombus are, therefore, the pyramids lying flat on the surface the height of the triangle is the length of the top and bottom of the rhombus. The triangles that connect the bottom triangle (lying flat on the surface) to its opposite side also use the height of the triangle as the length of the other two sides. Thatβs why this is a rhombus. So we are given that the height of the hexagonal pyramid is 1 and that the hexagonal βconnectorβ between the two pyramids has sides of length 1. β7 2 Great! Now we just need the height of the rhombus. But thatβs not so easy. Or is it. We do know of two equations to get the area of the rhombus. π΄π΄ = ππβ where ππ is the side length of the rhombus and β is the height. 1 π΄π΄ = ππ1 ππ2 2 where ππ1 and ππ2 are the diagonals of the rhombus. Hmm, maybe we do know those. One of those diagonals, if you look back at the picture of the hexagonal bipyramid, is just the heights of each pyramid or 2×1=2 The other one is just the height of the hexagon or β3. Now we can finally find the height! 1 ππβ = ππ1 ππ2 2 1 β7 β = × 2 × β3 = β3 2 2 2 × β3 2 × β3 × β7 = β= β7 β7 × β7 β= 2β21 7 Ans. Target Round 1. Given: The bug crawls 1 unit up, 2 units right, 3 units down and 4 units left. The bug repeats this set of moves 2015 more times. Find: If the coordinates of the bugβs final location are (ππ, ππ), find ππ + ππ. Starting at (0, 0) moving up 1 unit puts the bug at (0, 1). Moving to the right 2 units puts the bug at (2,1). Moving down three units puts the bug at (2, β2). Moving 4 units to the left puts the bug at (β2, β2). So the bug started at (0, 0) and ended at (β2, β2). Since the bug will continue to move in the same pattern, we can see that for each move β2 is added to both the x and y coordinates. At the end of 2016 moves the bug can be found at the coordinates (β4032, β4032). ππ + ππ = β4032 + β4032 = β8064 Ans. 2. Given: A rectangular piece of cardboard measures 6 by 8. It is trimmed identically on all four corners. Each corner is a quarter-circle. Find: the perimeter of the resulting figure. The four quarter-circles form a complete circle with radius 3. Its perimeter is 2ππππ = 6ππ Since the length of the longer side is 8 and we have 2 quarter-circles with radius 3, what remains of the original length is 8 β (2 × 3) = 2 Therefore, the perimeter of the resulting figure is 2 + 2 + 6ππ = 6ππ + 4 Ans. 3. Given: A coin is flipped four times. Find: the probability of getting 2 heads and two tails. This one is simple enough to just list the variations. HHTT, HTHT, HTTH, TTHH, THTH, THHT 1 4 2 6×οΏ½ οΏ½ = 6 24 = 6 16 3 8 = = 37.5% Ans. 4. Given: The areas of three faces of a rectangular prism are 54, 24 and 36. Find: the length of the space diagonal. Let ππ = the length of the prism. Let π€π€ = the width of the prism. Let β = the height of the prism. Then ππππ = 54 π€π€β = 24 ππβ = 36 ππ 2 π€π€ 2 β2 = 54 × 24 × 36 = 9×6×4×6×9×4= 92 × 62 × 42 So, we have a 9 × 6 × 4 rectangular prism. There are less ways to get sums of 10 or more than less than 10, so letβs look at the probability of drawing those. We can draw 4 + 6, 5 + 5, 5 + 6 or 6 + 6 The probability of drawing two of the same specific card is First letβs find the hypotenuse of the rectangle with sides 4 and 9. Let π₯π₯ = the length of the red line. 42 + 92 = π₯π₯ 2 π₯π₯ 2 = 16 + 81 = 97 Now we can find the length of the space diagonal. Let π¦π¦ = the length of the space diagonal. π¦π¦ 2 = π₯π₯ 2 + 62 π¦π¦ 2 = 97 + 36 = 133 π¦π¦ = β133 Ans. 5. Given: Two 8 by 10 inch sheets are placed on top of a 2 by 3 foot table. The area of the table not covered by the paper is 708 in2. Find: the area of the overlap of the two sheets of paper. A 2 foot by 3 foot rectangular table is a 24 inch by 36 inch rectangular table. The area of the table is 24 × 36 = 864 in2. The area not covered by the paper is 708 in2 leaving 864 β 708 = 156 in2 that is covered by the paper. If each of the sheets of paper did not overlap each other, the amount of area taken by the paper would be 8 × 10 × 2 = 160 in2. That is 4 more than 156. 4 Ans. 6. Given: Each card in a deck of cards contains a number between 2 and 6. The deck has 4 cards of each value. Two of the cards are chosen at random without replacement. Find: the probability that the sum of their values is less than 10. 4 20 × 3 19 = 3 95 The probability of drawing a specific pair of 8 20 × 4 19 8 95 = 22 95 two different cards is = 8 95 So for the four possible sums of 10 or more, we have two that are the same number and two that are different numbers. The probability of picking one of these four sums is 2 × 3 95 +2× So the probability of drawing two cards whose sum is less than 10 is 1β 22 95 = 73 Ans. 95 7. Given: (π₯π₯ 2 + 3π₯π₯ + 6)(π₯π₯ 2 + ππππ + ππ) = π₯π₯ 4 + πππ₯π₯ 2 + ππ for integers ππ, ππ, ππ, ππ. Find: the product of ππ and ππ. (π₯π₯ 2 + 3π₯π₯ + 6)(π₯π₯ 2 + ππππ + ππ) = π₯π₯ 4 + πππ₯π₯ 3 + πππ₯π₯ 2 + 3π₯π₯ 3 + 3πππ₯π₯ 2 + 3π₯π₯π₯π₯ + 6π₯π₯ 2 + 6ππππ + 6ππ = π₯π₯ 4 + (ππ + 3)π₯π₯ 3 + (ππ + 3ππ + 6)π₯π₯ 2 + (3ππ + 6ππ)π₯π₯ + 6ππ = π₯π₯ 4 + πππ₯π₯ 2 + ππ The coefficient of π₯π₯ 3 must be 0. Therefore, ππ + 3 = 0 ππ = β3 The coefficient of π₯π₯ 2 is ππ. Therefore, ππ = ππ + 3ππ + 6 = ππ + (3 × β3) + 6 = ππ β 9 + 6 = ππ β 3 The coefficient of π₯π₯ must be 0. Therefore, 3ππ + 6ππ = 0 3ππ + (6 × β3) = 0 3ππ β 18 = 0 3ππ = 18 ππ = 6 ππ = ππ β 3 = 6 β 3 = 3 Finally, ππ = 6ππ = 36 ππ × ππ = 3 × 36 = 108 Ans. 8. Given: The polygon is constructed from two squares and six equilateral triangles, each of side length 6. The polygon can be folded into a polyhedron by creasing along the dotted lines and joining adjacent edges as indicated by the arrows. Find: the volume of the resulting polyhedron. 2 β2 + οΏ½3β2οΏ½ = 62 β2 + 18 = 36 β2 = 18 β = 3β2 The created polyhedron is twice the volume of the right square pyramid. β 3β2 = 2(6)2 = 3 3 72β2 Ans. 2ππ = 2ππ 2 Team Round When you fold the polygon what you get is two square pyramids connected at an angle. The volume of the polyhedron is the sum of the volume of two identical square pyramids. The volume of a square pyramid is ππ = ππ 2 β 3 where ππ is a base edge and β is the height. The base edge, ππ, is just 6. To find β, we need to find half the diagonal of the square. Let π₯π₯ = half the diagonal of the square. Then 62 + 62 = (2π₯π₯)2 36 + 36 = 4π₯π₯ 2 4π₯π₯ 2 = 72 π₯π₯ 2 = 18 π₯π₯ = 3β2 Now, we can determine the height. β2 + π₯π₯ 2 = 62 1. Given: The end of each blade of a fan is two feet from the center of the fan. Each blade makes three full rotations per second. Find: the speed, in miles per hour, of the end of one of the blades. One rotation is 2ππππ = 4ππ feet. The blade travels 12ππ feet in three rotations (one second) or 720ππ feet in one minute. It travels 43,200ππ feet in one hour. In miles per hour this is 43,200ππ β 25.7 5280 25.7 Ans. 2. Given: π΄π΄, π΅π΅, πΆπΆ, π·π·, πΈπΈ in 0. π΄π΄π΄π΄π΄π΄ and 0. π·π·π·π· represent the digits 1, 2, 3, 4 and 5, in some order. Find: the least possible absolute difference between the numbers 0. π΄π΄π΄π΄π΄π΄ and 0. π·π·π·π·. To minimize the difference, we need to make 0.ABC as close to 0.DE as close in value as possible. The smallest possible difference between A and D is 1. If D < A, then we want to maximize the value of E and minimize the value of BC. The largest possible value for E is 5 and the smallest value for BC is 12. This makes the absolute difference 0.412 β 0.35 = 0.062. Similarly, if A < D, then we want to maximize BC and minimize E. The largest possible value for BC is 54 and the smallest possible value for E is 1. This makes the absolute difference 0.31 β 0.254 = 0.056. The least possible absolute difference is 0.056 Ans. 3. Given: 70.8% of Earthβs surface is covered with water. Mars is entirely covered by land and the amount is approximately the same as the amount of land on Earth. Find: What percent of Earthβs radius is Marsβ radius? The surface area of a sphere is 4ππππ 2 Let πππΈπΈ = the radius of Earth. Let ππππ = the radius of Mars. Then 0.292 × 4ππ(πππΈπΈ )2 = 4ππ(ππππ )2 0.292(πππΈπΈ )2 = (ππππ )2 ππππ β 0.54037πππΈπΈ As a percentage, the radius of Mars is 54% of the radius of earth. 54% Ans. 4. Given: Billy wrote a sequence of 5 numbers, each an integer between 0 and 4. Penny wrote a sequence of five numbers that included the number of 0βs, 1βs, 2βs, 3βs and 4βs in Billyβs sequence. Pennyβs sequence is the same as Billyβs. Find: the sequence. The sum of all 5 numbers must be 5 because the sequence is a count of how many times each number appears and we know there are only 5 numbers total present in the sequence. You canβt have 5 of any number because 5 is not one of the numbers. You also canβt have 4 of any number since only one other number could be used. And weβre going to have to have some values that we will have none of. For example, 11112 would translate to 04100. Similarly, you canβt have 3 of one number and 2 of another number or 3 of one number, one of another number and one of a third number. For example, 01112 would translate to 13100. Letβs look at 2 of one number, 2 of another number and 1 of a third number. 2 1 2 0 0 would translate to 2 1 2 0 0. 2 1 2 0 0 Ans. 5. Given: 3π₯π₯ + 3π₯π₯+2 = 9π₯π₯ + 9π₯π₯+2 Find: 3π₯π₯ 3x + (3x32) = (3 × 3)x + (3 × 3)x(3 × 3)2 3x(1 + 32) = 3x3x + 3x3x3232 3x(10) = 3x3x(1 + 34) 10 = 3x(82) 3x = 10/82 = 5/41 Ans. 6. Given: A dime has a mass of 2,268 grams. A nickel has a mass of 5,000 grams. Diane has a bag of dimes and Nick has a bag of nickels. Both bags of coins have the same mass. Find: how many times the total value of Nickβs coins is Dianeβs coins. Let π₯π₯ = the number of dimes. Let π¦π¦ = the number of nickels. 2268π₯π₯ = 5000π¦π¦ π₯π₯ 5000 = β 2.2045 π¦π¦ 2268 There are 2.2045 times as many dimes as there are nickels. The value of dimes to nickels is 10π₯π₯ 5π¦π¦ π₯π₯ π¦π¦ = 2 β 4.409 β 4.4 Ans. 7. Given: For each positive integer ππ, ππππ = 9ππ + 2 and ππππ = 7ππ + 3 The values common to both sequences are written as a sequence. The nth term can be expressed as pn+q. Find: the value of ππ β ππ Since an is multiples of 9 shifted by 2 and bn is multiples of 7 shifted by 3. We just need to find the first common term, then we know all common terms following that will just be 9 × 7 = 63 more than the previous. Letβs compare the first few values of ππππ : 11, 20, 29, 38, 47 to the first few values of ππππ : 10, 17, 24, 31, 38, 45. The first common value is 38. So the sequence of common values will be represented by the formula 38 + 63(n β 1) = 38 + 63n β 63 = 63n β 25. Therefore, the value of p is 63 and the value of q is β25, and p β q = 63 β (β25) = 88 Ans. 8. Given: Quadrilateral APBQ has vertices π΄π΄(0,0) and π΅π΅(8,0). Vertices P and Q lie on the line 4π₯π₯ + 3π¦π¦ = 19. ππππ = 3 Find: the area of quadrilateral π΄π΄π΄π΄π΄π΄π΄π΄. The slope of the line 4x + 3y = 19 is β4/3. If we draw in a right triangle with hypotenuse PQ, then we know the ratio of its height to its base is equal to the slope, rise over run, of the line (note: we can ignore the negative because we are talking about distance without direction) or h/b = 4/3 β h = 4/3 × b. We also know, since it is a right triangle, that b2 + h2 = 32. We have two equations and two variables, so we can solve for the values of h and b. b2 + (4/3 × b)2 = 9 b2 + 16/9 × b2 = 9 25/9 × b2 = 9 b2 = 81/25 b = 9/5 Therefore, h = 4/3 × 9/5 = 12/5. The area of the quadrilateral is the area of the triangle APB minus the area of triangle AQB. The difference in the height of these two triangles heights is h. So the area of the quadrilateral is 1/2 × 8 × h = 4 × h = 4 × 12/5 = 48/5 Ans. 9. Given: 8 blue and 5 orange tiles are arranged in an ordered line such that the tile on the left must be blue and every tile must be adjacent to at least one tile of the same color. Find: How many different arrangements are possible if all 13 tiles must be used? The requirements are that B (for blue) must me the first tile (starting from the left) and that Bβs and Oβs (for orange) must always come in a minimum of two tiles of the same color. We have 5 orange tiles so weβll be able to have them in one group of 2 and one group of 3 (since we canβt have a group of 1) and, of course, a group of 5. So letβs start with a group of 5 orange tiles. - BBBBBBBBOOOOO (#1) - BBBBBBOOOOOBB (#2) - BBBBBOOOOOBBB (#3) - BBBBOOOOOBBBB (#4) - BBBOOOOOBBBBB (#5) - BBOOOOOBBBBBB (#6) So much for the group of 5 tiles. Now letβs work with 2 orange tiles always come before the group of 3 orange tiles. - BBOOBBOOOBBBB (#7) - BBOOBBBOOOBBB (#8) - BBOOBBBBOOOBB (#9) - BBOOBBBBBBOOO (#10) Move the 2 orange tiles to the right by 1. - BBBOOBBOOOBBB (#11) - BBBOOBBBOOOBB (#12) - BBBOOBBBBBOOO (#13) And one more move to the right. - BBBBOOBBOOOBB (#14) - BBBBOOBBBBOOO (#15) And one more to the right - BBBBBOOBBBOOO (#16) And one more - BBBBBBOOBBOOO (#17) Now letβs work with 3 orange tiles always before the 2 orange tiles. Inste - BBOOOBBOOBBBB (#18) - BBOOOBBBOOBBB (#19) - BBOOOBBBBOOBB (#20) - BBOOOBBBBBBOO (#21) Now move the group of 3 Oβs to the right by one tile. - BBBOOOBBOOBBB (#22) - BBBOOOBBBOOBB (#23) - BBBOOOBBBBBOO (#24) Now move the three tiles one more position to the right. - BBBBOOOBBOOBB (#25) - BBBBOOOBBBBOO (#26) And one more to the right. - BBBBBOOOBBBOO (#27) And one more - BBBBBBOOOBBOO (#28) 28 Ans. 10. Given: A fountainβs diameter is 10 feet. The grass extends out from the edge of the fountain 20 feet. Grass seed comes in bags that cover 300 sq. ft. of grass. Find: How many whole bags of grass seed are needed to seed around the fountain? The radius of the fountain itself is 10 2 = 5. The radius of the entire area including the grass around the fountain is 20+10+20 2 = 50 2 = 25 The area of the fountain is ππππ 2 = ππ52 = 25ππ The area of the fountain and the grass is ππ252 = 625ππ This makes the area of the grass 625ππ β 25ππ = 600ππ β 1963.49375 square feet. Each bag of grass covers 300 square feet. 1963.49375 300 β 6.554979 bags. This means we must have 7 bags to cover the area completely. 7 Ans.