Mary s top book shelf holds five books with the 1. following widths, in centimeters: 6,

Transcription

1 Mary s top book shelf holds five books with the 1 following widths, in centimeters: 6, 2, 1, 2.5, and 5. What is the average book width, in centimeters? (A) 1 (B) 2 (C) 3 (D) 4 (E) AMC 10 A, Problem #1 The average of five values = the sum of the values 5. Answer (C): The average of the five values is = 15 5 = 3. Difficulty: Easy NCTM Standard: Number and Operations Standard for Grades 9 12: compute fluently and make reasonable estimates.

2 Three identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? (A) 5 4 (B) 4 3 (C) 3 2 (D) 2 (E) AMC 10 A, Problem #2 The length of the rectangle is equal to three times the side length of the smaller square. Answer (C): Let s be the side length of the smaller square. Then the length of the rectangle is 3s, and the width is 3s s = 2s. Hence the rectangle length is 3s = 3 times as large as its width. 2s 2 Difficulty: Easy NCTM Standard: Geometry Standard for Grades 9 12: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them.

3 A ferry boat shuttles tourists to an island every hour starting at 9 am until its last trip starting at 3 pm. One day the boat captain notes that on the 9 am trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day? (A) 585 (B) 594 (C) 672 (D) 679 (E) AMC 10 A, Problem # AMC 12 A, Problem #2 The ferry boat makes 7 trips to the island everyday. Answer (D): The ferry boat makes 7 trips to the island. The number of tourists shuttled was (100 1) + (100 2) + (100 3) + (100 4) + (100 5) + (100 6) = ( ) = = 679. Difficulty: Easy NCTM Standard: Number and Operations Standard for Grades 9 12: compute fluently and make reasonable estimates.

4 The area of a circle whose circumference is 12π is kπ. What is the value of k? (A) 6 (B) 12 (C) 24 (D) 36 (E) AMC 10 A, Problem # AMC 12 A, Problem #1 The circumference of a circle with radius r is 2πr. Answer (D): Because the circumference is 2πr = 12π, the radius r is 6. Therefore the area is πr 2 = 36π, and k = 36. Difficulty: Easy NCTM Standard: Algebra Standard for Grades 9 12: use symbolic algebra to represent and explain mathematical relationships.

5 Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. Her next n shots are bulleyes, which guarantees her victory. What is the minimum possible value for n? (A) 40 (B) 42 (C) 44 (D) 46 (E) AMC 10 A, Problem #5 Find out how many points Chelsea needs to score to guarantee victory. Answer (B): The second place archer could score a maximum of = 500 points with the remaining shots. Therefore Chelsea needs to score more than = 450 points to guarantee victory. Chelsea s next n shots will score 10n points, and her remaining 50 n shots will score at least 4(50 n) points. To guarantee victory, n 10 + (50 n) 4 > 450 6n > 450 n > Therefore Chelsea needs at least 42 bullseyes to guarantee victory. Difficulty: Medium-easy NCTM Standard: Number and Operations Standard for Grades 9 12: judge the effects of such operations as multiplication, division, and computing powers and roots on the magnitudes of quantities.

6 Point A = (4, 3) is rotated counterclockwise 90 around the origin to point A. The point A is reflected across the x-axis to point A. What is the distance between A and A? (A) 2 (B) 8 (C) 7 2 (D) 10 (E) AMC 10 A, Problem #6 Visualize the transformations by drawing. Answer (C): The figure shows that A = ( 3, 4) and A = ( 3, 4). The distance between A and A is (4 ( 3))2 + (3 ( 4)) 2 = = 7 2. A' ( 3,4) A (4,3) A'' ( 3, 4) Difficulty: Medium-easy NCTM Standard: Geometry Standard for Grades 9 12: apply transformations and use symmetry to analyze mathematical situations.

7 A box 2 centimeters high, 3 centimeters wide, and 5 centimeters long can hold 40 grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold n grams of clay. What is n? (A) 120 (B) 160 (C) 200 (D) 240 (E) AMC 10 A, Problem # AMC 12 A, Problem #3 Compare the volume of the two boxes. Answer (D): The volume of the second box is 2 3 = 6 times the volume of the first box. Hence it can hold 6 40 = 240 grams of clay. Difficulty: Medium NCTM Standard: Geometry Standard for Grades 9 12: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them.

8 If x < 0, then which of the following must be positive? (A) x 2 (B) 2 x (C) x 1 (D) 3 x (E) x x 2010 AMC 10 A, Problem # AMC 12 A, Problem #4 Let x be any negative number, and plug it into all five choices. Answer (C): Choice (C) may be written as 1. If x is negative, choice x (C) is positive. To see that the other choices need not be positive, let x = 1 and then (A) ( 1) 2 = 1, (B) 2 1 = 1 2, (D) 3 1 = 1, (E) 1 1 = 1. Difficulty: Medium NCTM Standard: Number and Operations Standard for Grades 9 12: judge the effects of such operations as multiplication, division, and computing powers and roots on the magnitudes of quantities.

9 Let x represent the greatest integer less than or equal to x. What is the value of ? (A) 30 (B) 32 (C) 34 (D) 36 (E) AMC 10 A, Problem # AMC 12 A, Problem #7 3 2 < 10 < 4 2 and 4 2 < 20 < 5 2. Answer (E): The integers 10, 11, 12, 13, 14, and 15 are each greater than 3 2 = 9 and less than 4 2 = 16. Hence = 6 3 = 18. Similarly, the integers 16, 17, 18, 19, and 20 are each greater than or equal to 4 2 = 16 and less than 5 2 = 25. Hence = 5 4 = 20. The sum is = 38. Difficulty: Medium NCTM Standard: Number and Operations Standard for Grades 9 12: judge the effects of such operations as multiplication, division, and computing powers and roots on the magnitudes of quantities.

10 Let x > 0. If A is x percent of B, then B is what percent of A? (A) 10,000 x (B) 100 x (C) 100x (D) 100 x 1 (E) 1 x AMC 10 A, Problem # AMC 12 A, Problem #5 Write a equation to show the relationship. Answer (A): percent of A. If A = x 100 B, then B = 100 x ,000 A. Hence B is 100 = x x Difficulty: Medium-hard NCTM Standard: Algebra Standard for Grades 9 12: write equivalent forms of equations, inequalities, and systems of equations and solve them with fluencymentally or with paper and pencil in simple cases and using technology in all cases.

11 Two squares have areas whose positive difference is numerically equal to the sum of their perimeters. What is the difference of their perimeters? (A) 2 (B) 6 (C) 8 (D) 12 (E) AMC 10 A, Problem # AMC 12 A, Problem #6 Write a equation to show the relationship. Answer (E): If the larger square has side length a and the smaller square has side length b, then a 2 b 2 = 4a + 4b. Factoring gives (a + b)(a b) = 4(a + b). Because a + b does not equal 0, it follows that a b = 4. The difference of the perimeters is 4a 4b = 4(a b) = 16. Difficulty: Medium-hard NCTM Standard: Algebra Standard for Grades 9 12: use symbolic algebra to represent and explain mathematical relationships.

12 A right circular cylinder is circumscribed by a cube as shown. The surface area of the cube is 12 square units. What is the volume of the cylinder? (A) 2 (B) 2 π 2 (C) 6 (D) 2π (E) 3π 2010 AMC 10 A, Problem #12 Find the radius of the cylinder by first find out the edge length of the cube. Answer (B): Let s be the edge length of the cube. The surface area of the cube is 6s 2 = 12, so s = 2. Thus the radius of the cylinder is r = 2 2 and the height of the cylinder is h = 2. The volume of the cylinder is then ( ) 2 2 V = πr 2 h = π 2 2 = 2 2 π. Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them.

13 In a certain town bicycle license plates are to have 3 different letters, each letter chosen from A through F, followed by 3 different digits, each digit chosen from 1 through 6. The letters will appear in alphabetical order and the digits in numerical order. For example, ACE236 is allowed and ADB124 is not allowed. How many such license plates are possible? (A) 36 (B) 72 (C) 120 (D) 240 (E) AMC 10 A, Problem #13 There are ( 6 3) = 20 letter combinations to choose from. Answer (E): There are ( ) 6 = = 20 letter combinations to choose from, and each particular choice results in exactly one possible configuration for the letters in a license plate. Similarly, there are ( 6 3) = 20 possible configurations for the digits. Hence there are = 400 possible license plates. Difficulty: Medium NCTM Standard: Algebra Standard for Grades 9 12: understand relations and functions and select, convert flexibly among, and use various representations for them.

14 A square of side length K is composed of K 2 unit squares. Exactly one third of the unit squares are to be colored red. Then exactly one fourth of the remaining uncolored squares are to be colored blue. What is the sum of the smallest three values of K for which this is possible? (A) 12 (B) 14 (C) 18 (D) 21 (E) AMC 10 A, Problem #14 Explore the relationships with the given information. Answer (E): If 1 3 k2 unit squares are colored red, then 2 3 k2 unit squares are left uncolored. Subsequently k2 = 1 6 k2 unit squares are colored blue. Hence k must be a multiple of 6. The three smallest multiples of 6 are 6, 12, and 18; their sum is 36. Difficulty: Medium-hard NCTM Standard: Algebra Standard for Grades 9 12: use symbolic algebra to represent and explain mathematical relationships.

15 The game of reverse tic-tac-toe is similar to the standard game of tic-tac-toe. Two players take turns marking the spaces in a 3 3 grid; the first player to complete 3 X s or 3 O s in a line (horizontally, vertically, or diagonally) loses the game. Xavier and Oliver are playing reverse tic-tac-toe. Xavier is using X and Oliver is using O. Oliver always makes the best choice in the placement of his O s. The current state of their game is shown below, and it is Xavier s turn. X O O X In how many of the remaining five spaces can Xavier put an X on this turn and not eventually lose the game? (A) 0 (B) 1 (C) 2 (D) 3 (E) AMC 10 A, Problem #15 Try-and-error. Answer (A): If Xavier plays in the first column he loses. If Xavier plays in the second or third column Oliver responds by playing in the remaining spot in the same column. By doing so, Oliver will never lose, and eventually Xavier will have to play in the first column and lose the game. Thus the number of places that Xavier can put an X and not eventually lose the game is 0. Difficulty: Medium NCTM Standard: Algebra Standard for Grades 9 12: understand patterns, relations, and functions.

16 Three children have a foot race in which the leader after 1 minute is declared the winner. Each child runs at a constant speed. At the end of the minute, Alicia has run 720 feet. Ben, who started 40 feet in front of Alicia, has run 600 feet from his starting point. Cheryl, who started 40 feet in front of Ben, has run 400 feet from her starting point. For how many seconds during the race was Ben in the lead? (A) 8 (B) 10 (C) 12 (D) 15 (E) AMC 10 A, Problem # AMC 12 A, Problem #9 Use inequalities to represent the relationships. Answer (A): The distance of Alicia, Ben, and Cheryl from Alicia s starting point after t minutes are, respectively, 720t, t, and t feet. For Ben to be in the lead we must have t > 720t and t > t. These inequalities are equivalent to or 40 > 120t and 200t > > t and t > 1 5. Therefore Ben is in the lead when 1 > t > 1, a time interval of 1 1 = minutes, or 2 60 = 8 seconds Difficulty: Medium-hard NCTM Standard: Algebra Standard for Grades 9 12: write equivalent forms of equations, inequalities, and systems of equations and solve them with fluency-mentally or with paper and pencil in simple cases and using technology in all cases.

17 Equilateral triangle ABC has side length 1. From center point O perpendiculars are drawn to sides AB and BC. These perpendiculars are then extended to D and E so that DE is parallel to AC and passes through B. What is the area of DOE? (A) 3 4 (B) 3 3 (C) 3 2 (D) (E) AMC 10 A, Problem #17 Let M be the intersection of OD and AB triangle with BM = 1. 2 Then BOM is a Answer (B): Let M be the intersection of OD and AB. Then BOM is a triangle with BM = 1, so OB = Also, DOB is a triangle with OB = 1 3, so BD = 1. The area of DOE is ( ) 1 twice the area of DOB, which is = OR F A D O M C B E Extend EC and AD so that they meet at point F. Triangle DEF has four times the area of ABC, which is = 3. Also DOE has one third the area of DEF, so the area of DOE is 3 3. Difficulty: Medium-hard NCTM Standard: Geometry Standard for Grades 9 12: analyze properties and determine attributes of two- and three-dimensional objects.

18 Suppose that ABCD is a square, M is the midpoint of AB, and E is the intersection of CM and BD. What is EB EM? (A) 1 (B) 5 4 (C) (D) 15 3 (E) AMC 10 A, Problem #18 Let F be the foot of the altitude to AB from E. Answer (C): Let F be the foot of the altitude to AB from E and let EF = x. Then EB = 2x because BEF is a triangle. Triangle EMF is similar to triangle CMB, so F M = x. By the 2 Pythagorean Theorem applied to EM F, ( x ) 2 5 EM = x 2 + = 2 2 x. Thus EB EM = 2x 5 x = 2 2 = C D E B F M A Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: analyze properties and determine attributes of two- and three-dimensional objects.

19 Isabella has a set of interlocking building blocks. Disregarding the small knobs used to join blocks together, each block is a cube with a side length of 1 inch. Isabella uses her blocks to build a hollow rectangular house with a flat roof and no floor. The four walls and the roof are each one block thick. The house has outside dimensions of 8 inches, 10 inches, and 12 inches in no particular order. The inside has a volume of V cubic inches. What is the difference between the largest and smallest possible values of V? (A) 12 (B) 20 (C) 24 (D) 32 (E) AMC 10 A, Problem # AMC 12 A, Problem #8 The inside length and width of the house are 2 inches less than the outside length and width. Answer (D): The inside length and width of the house are 2 inches less than the outside length and width. The inside height is 1 inch less than the outside height. Therefore the inside volume V is = , or = , or = The difference between the largest and smallest possible values of V is ( ) ( ) = = 32. Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: analyze properties and determine attributes of two- and three-dimensional objects.

20 Rhombus ABCD has side length 1. Let M be the midpoint of AB, and suppose CM = 1. What is the area of ABCD? (A) (B) 15 4 (C) (D) 35 6 (E) AMC 10 A, Problem #20 Triangle BCM is isosceles. Answer (B): Because BCM is isosceles, the altitude from C meets the base MB at its midpoint P. Then by the Pythagorean Theorem CP = ( ) MC 2 MP 2 = 1 = 4 4. The area of ABCD is AB CP = A M P B 1 1 D C Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: analyze properties and determine attributes of two- and three-dimensional objects.

21 Let m, n, and p be positive integers such that m 2010, n 3, and p is prime. How many triples (m, n, p) satisfy the equation 2010 n + m n = p 2? (A) 4 (B) 5 (C) 6 (D) 7 (E) AMC 10 A, Problem # AMC 12 A, Problem #21 We divide into cases according to the three possible values of n, namely n = 1, 2, or 3. Answer (A): We divide into cases according to the three possible values of n, namely n = 1, 2, or 3. For n = 1 the equation becomes m = p 2. Because 1 m 2010, it follows that 2011 p and 45 p 63. The only primes in this range are 47, 53, 59, and 61. These yield the following solution triples: (199, 1, 47), (799, 1, 53), (1471, 1, 59), and (1711, 1, 61). For n = 2 the equation becomes p 2 m 2 = Because p is a prime and cannot equal 2, it follows that p and m are both odd. So p 2 m 2 1 (mod 8) and p 2 m 2 0 (mod 8). Because (mod 8) there are no solutions in this case. For n = 3 the given equation becomes m 3 = p 2. Now, m 3 = ( m)( m + m 2 ). For this product to be the square of a prime, either one of the factors is 1 and the other factor is p 2, or both factors are equal to p. Because 1 m 2010, we have Therefore m(m 1) > 2010m m + m 2 > m(m 1) + m 2 = m > 1. Thus ( m)( m + m 2 ) cannot be the square of a prime, so there are no solutions in this case. The total number of solutions for all cases is 4. Difficulty: Hard NCTM Standard: Number and Operations Standard for Grades 9 12: understand meanings of operations and how they relate to one another.

22 Three distinct vertices are selected at random from a regular 14-sided polygon (tetradecagon) and used as vertices to form a triangle. What is the probability that the triangle is a right triangle? (A) 1 13 (B) 2 13 (C) 3 13 (D) 4 13 (E) AMC 10 A, Problem #22 Circumscribe the regular tetradecagon with a circle. The triangle will be a right triangle if and only if two of the vertices are antipodal points of the circle. Answer (C): Circumscribe the regular tetradecagon with a circle. The triangle will be a right triangle if and only if two of the vertices are antipodal points of the circle (points on the opposite ends of a diameter). There are 7 pairs of antipodal points, and for each pair, any one of the other 12 vertices would result in a right triangle. Hence there are 7 12 = 84 possible combinations of vertices that produce a right triangle. There are a total of ( ) 14 3 possible vertex combinations, so the probability is 84 ( 14 ) = OR Again note that the triangle will be a right triangle if and only if two of the vertices are antipodal. After the first point is selected, there is a 1 chance 13 that the second point chosen is antipodal to the first point. If the second point is not antipodal to the first, there is a 2 chance that the third point 12 is antipodal to one of the first two points. Therefore the probability is = Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them.

23 Triangle ABC has vertices at A(0, 0), B(8, 14) and C(20, 0). The line y = mx divides ABC into two triangles of equal area. What is m? (A) 1 3 (B) 1 2 (C) 3 5 (D) 2 3 (E) AMC 10 A, Problem #23 The line y = mx passes through vertex A. Answer (B): The line y = mx passes through vertex A. Note that a median divides a triangle into two triangles of equal area. The midpoint of BC is (14, 7) which is on the line y = 1x, so m = OR A line passing through a triangle s centroid divides the triangle into two equal areas. The centroid of a triangle is equal to the average of the vertex coordinates. The centroid for ABC is ( , which lies on the line y = 1 2 x, so m = 1 2. ) = ( 28 3, 14 ) 3 Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: use Cartesian coordinates and other coordinate systems, such as navigational, polar, or spherical systems, to analyze geometric situations.

24 A train, traveling at constant speed, passes through two tunnels. In the graph below, t is elapsed time in seconds, and L is the length in meters of the portion of the train that is inside a tunnel. What is the sum, in meters, of the lengths of the two tunnels? 240 L t (A) 360 (B) 420 (C) 480 (D) 560 (E) AMC 10 A, Problem # AMC 12 A, Problem #17 The train is at least 240 meters long and the first tunnel is 80 meters long. Answer (D): Because 240 meters of the train were inside the second tunnel for 36 t 48, the train is at least 240 meters long and the first tunnel is 80 meters long. The front of the train enters the first tunnel at t = 0 and exits at t = 4, so the speed of the train is 80 = 20 meters 4 per second. The rear of the train enters the first tunnel at t = 12, so the train takes 12 seconds to pass the tunnel entrance. Hence the train is = 240 meters long, and it was entirely within the second tunnel for 36 t 48. The front of the train entered the second tunnel at t = 24 and left it at t = 48, so it took = 24 seconds to pass through the second tunnel. Therefore the second tunnel is = 480 meters long, and the requested sum is = 560 meters. Difficulty: Hard NCTM Standard: Algebra Standard for Grades 9 12: identify essential quantitative relationships in a situation and determine the class or classes of functions that might model the relationships.

25 Tom drives in a city where there are 11 two-way streets running north-south and 11 two-way streets running east-west. Unfortunately, the steering wheel in his car is damaged, so at intersections he can only make a left turn or keep going straight. Tom begins a drive by traveling one block east from the center of the city. Assume he does not pass by the same spot twice. How many different ways can he reach the northeast corner of the city? (A) 4826 (B) 5226 (C) 5626 (D) 8626 (E) AMC 10 A, Problem # AMC 12 A, Problem #22 Model the intersections of the streets by the points (x, y) with integer coordinates, 5 x 5 and 5 y 5. Answer (E): Model the intersections of the streets by the points (x, y) with integer coordinates, 5 x 5 and 5 y 5. Tom needs to travel along the horizontal and vertical lines, from (0, 0) to (5, 5). Let (x 0, y 0 ) = (0, 0), and let (x 1, y 0 ), (x 1, y 1 ), (x 2, y 1 ), (x 2, y 2 ),..., (x 2k+1, y 2k+1 ) = (5, 5) be the intersections at which left turns are made, in order. Because Tom starts heading east, only turns left at intersections, and never passes by the same spot twice, it follows that 5 x 2k < x 2k 2 < < x 2 < x 0 = 0 < x 1 < x 3 < < x 2k+1 = 5, and 5 y 2k < y 2k 2 < < y 2 < y 0 = 0 < y 1 < y 3 < < y 2k 1 y 2k+1 = 5. Moreover, if the sequence of x i s and y i s satisfies these inequalities, then the sequence of intersections previously defined indicates a valid path. There are ( 5 k) ways of choosing the k-tuple (x2, x 4,..., x 2k ) and ( 4 k) ways of choosing the k-tuple (x1, x 3,..., x 2k 1 ) subject to the required inequalities. Similarly, there are ( 5 k) ways of choosing the k-tuple (y2, y 4,..., y 2k ) and because y 2k 1 could be equal to y 2k+1 = 5, there are ( 5 k) ways of choosing the k-tuple (y1, y 3,..., y 2k 1, y 2k+1 ). Therefore the total number of possible paths is 4 ( 5 )3( 4 = 1 k k) k=0 = = Difficulty: Hard NCTM Standard: Geometry Standard for Grades 9 12: use Cartesian coordinates and other coordinate systems, such as navigational, polar, or spherical systems, to analyze geometric situations.