Sample AMC 12. Detailed Solutions MOCK EXAMINATION. Test Sample. American Mathematics Contest 12

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1 MOCK EXAMINATION AMC 12 American Mathematics Contest 12 Test Detailed Solutions Make time to take the practice test. It s one of the best ways to get ready for the AMC.

2 AMC Mock Test Detailed Solutions Problem 1 (E) Solution 1 Note that there is more than 1 four-legged table. So there are at least 2 four-legged tables. Since there are 23 legs in total, there must be fewer than 6 four-legged tables, which have = legs. Thus, there are between 2 and 5 four-legged tables. If there are 2 four-legged tables, then these tables account for = legs, leaving = legs for the three-legged tables, which implies that there are 5 three-legged tables. (We can check that if there are 3 or 4 four-legged tables, then the number of remaining legs is not divisible by 3, and if there are 5 four-legged tables, then there is only 1 three-legged table, which is not allowed.) Solution 2 Since there is more than 1 table of each type, then there are at least 2 three-legged tables and 2 four-legged tables. These tables account for + = legs. There are = more legs that need to be accounted for. These must come from a combination of three-legged and four-legged tables. The only way to make 9 from 3s and 4s is to use three 3s. Therefore, there are + = three-legged tables and 2 four-legged tables. chiefmathtutor@gmail.com Page 1

3 Problem 2 (B) Note that = + = =. So Problem 3 =. (E) The darkened curve intersects line l in three points. Let be the point of intersection that lies on, as shown in the figure below. Then is a radius of the semicircle on the left, is a radius of the semicircle on the right, and + = =. Since the darkened curve consists of the two semicircles, its length is the sum of the lengths of the two semicircles: π + π = π + = π. chiefmathtutor@gmail.com Page 2

4 Problem 4 (D) By the given conditions, + + =, + =. Thus, = = Problem 5 (A) If,, and are the roots of + =, then by Vieta's theorem, Thus, + + =, + + =, and = = = =. Problem 6 (D) Brad could have traveled at most = miles. Each 5-digit palindrome is uniquely determined by its first three digits. The next palindromes after are 39093, 39193, 39293, and chiefmathtutor@gmail.com Page 3

5 Note that So this number and all larger ones are too large. Since = >. = >, the difference is too far to drive in 5 hours without exceeding the speed limit of 70 miles per hour. Brad could have driven miles during the 5 hours for an average speed of miles per hour. Problem 7 (D) = = Since = + is the inverse function of = Thus, = =. =. 7, we have: Problem 8 (C) Since Alex needs 12 hours to shovel all of his snow, he shovels of his snow per hour. Since Bob needs 8 hours to shovel all of Alex's snow, he shovels of Alex's snow per hour. chiefmathtutor@gmail.com Page 4

6 Similarly, Carl shovels of Alex's snow every hour, and Dick shovels of Alex's snow per hour. Together, Alex, Bob, Carl, and Dick can shovel of Alex's snow per hour. Therefore, together they can shovel of Alex's snow per minute = = Thus, by shoveling of Alex's snow per minute, together they will shovel all of Alex's snow in 96 minutes. Problem 9 (C) Squaring the both sides of the equation sin + cos = gives: Using we have: sin + sin cos + cos =. sin + cos =, cos + sin cos + sin =, The equation can be simplified to which is equivalent to It follows that which implies that sin sin cos + cos =, sin cos =. sin = cos chiefmathtutor@gmail.com Page 5

7 tan =. Problem 10 (A) Together, Hose X and Hose Y fill the pool in 10 hours. Thus, it must take Hose X more than 6 hours to fill the pool when used by itself. Therefore,, since is a positive integer. Similarly, it must take Hose Y more than 10 hours to fill the pool when used by itself. Therefore,, since is a positive integer. When used by itself, the fraction of the pool that Hose X fills in 10 hours is. When used by itself, the fraction of the pool that Hose Y fills in 10 hours is. When used together, Hose X and Hose Y fill the pool once in 10 hours. Thus, which is equivalent to Completing the rectangle gives: Note that the prime factorization of 100 is Thus, 100 has distinct factors. + =, =. =. =. + + = Correspondingly,, has 9 positive solutions. Hence, there are only 9 different possible values for. In fact, 9 positive solutions are: chiefmathtutor@gmail.com Page 6

8 , =,,,,,,,,,,,,,,,,,. That is,, =,,,,,,,,,,,,,,,,,. Problem 11 (D) From the equation we have 3 cases: or or + + =, + =, + =, + = and + is even. Case 1: We get: + =. = or =. Case 2: We get: + =. = or =. chiefmathtutor@gmail.com Page 7

9 Case 3: + = and + is even. The following 2 solutions satisfy both conditions: = or =. Therefore, the equation has 5 integer solutions. Problem 12 (B) Plugging in = into + = gives: for every real number. Note that So for any integer n >, we have: Hence, + = =. n = n = n = = = =. =. Problem 13 Note that Thus, (B) log + log + log + log = log + log + log + log = log. Raising both sides to the tenth power gives: log =. =. chiefmathtutor@gmail.com Page 8

10 So Hence, = =, = = =. Problem 14 (D) Solution 1 Label the vertices of the new figure as shown below. When the paper is folded in this way, the portion of the original bottom face of the paper that is visible has the same area as the original portion of the top side of the paper to the right of the fold. This is quadrilateral. Of the portion of the original sheet to the left of the fold, the part, which is hidden and thus not included in the area of the new figure, is the triangular portion under the folded part. This is the section under, which is an isosceles right triangle with side lengths of 8. The hidden triangle is congruent to. Thus, the area of the portion of the original top face of the paper that is visible is the area to the left of the fold, minus the area of the hidden triangle. Therefore, the area of the new figure equals the area of the original rectangle minus the area of : chiefmathtutor@gmail.com Page 9

11 =. Solution 2 Let =. Since we have + + =, = = =. Note that is a right triangle with base and height both equal to 8. So the area of the new figure is equal to: AreaRectangle + Area + AreaRectangle = + + =. Problem 15 Solution 1 Note that and (A) + i ( ) = + i + i + i + i = chiefmathtutor@gmail.com Page 10

12 Since we have Solution 2 Note that Thus, + = =. i ( ) = i + i i + i =. = +, and = +, + i + i + + i + i = sin π = cos π. k = cos π + i sin π k + cos π i sin π k Using De Moivre's formula, we have: = cos π + i sin π k + cos π + i sin π k. k = (cos kπ + i sin kπ ) + (cos ( kπ ) + i sin ( kπ )) = cos kπ. Hence, + = cos π + cos π = cos ( π + π ) + cos π + π = cos ( π ) + cos π =. chiefmathtutor@gmail.com Page 11

13 Problem 16 (B) Using the sum formula for tangent, we have: Note that Thus, which implies that Then Problem 17 (D) tan + = tan + tan tan tan. tan + = tan =. tan + tan tan tan =, tan tan = tan + tan. + tan + tan = + tan + tan + tan tan = + tan tan + tan tan =. Denote by the first point that is picked. Let and be the points on the circle which are exactly 1 unit away from. Then where denotes the center of the circle. = = = = =, chiefmathtutor@gmail.com Page 12

14 The triangles and are equilateral and the arc has angle. Hence, two-thirds of the points on the circle are at least 1 unit away from. Therefore, the probability that the chord joining the two points has length at least 1 is Problem 18 Solution 1 (D). Let the millipede try to put on all n things in a random order. Each of the (n! permutations is equally probable. For any fixed leg, the probability that the millipede first puts on the sock and only then the shoe is clearly. Then the probability that the millipede correctly puts things on all legs is n. Therefore, the number of correct permutations must be n! n. chiefmathtutor@gmail.com Page 13

15 Solution 2 Number the millipede's legs from 1 through n, and let k and k denote the sock and shoe that will go on leg k. A possible arrangement of the socks and shoes is a permutation of the n symbols,,, n, n in which k precedes k for k n. There are (2 n! permutations of the n symbols, and precedes in exactly half of these, or permutations. n! Similarly, precedes in exactly half of those, or n! permutations. Continuing, we can conclude that k precedes k for k n in exactly permutations. Problem 19 (B) Since is rectangular, then Also, = = and = =. n! = =. Note that and are two congruent right triangles with legs of 15 and 20. So the hypotenuse length is =. Draw perpendiculars from and to and, respectively, on. Also, join to. chiefmathtutor@gmail.com Page 14

16 By the AA similarity postulate, right is similar to right because they share. So is also a right triangle with hypotenuse of 15, and thus, =, which implies that =, Similarly, we have: =, Thus, =. =. = = =. Using the 3-D Pythagorean Theorem, we obtain: = + + = + + =. Problem 20 (E) Solution 1 Alan gives 24 bars that account for 45% of the total weight to Bob. Thus, each of these 24 bars accounts for an average of chiefmathtutor@gmail.com Page 15

17 of the total weight. Unauthorized copying or reuse of any part of this page is illegal! % = % Alan gives 13 bars that account for 26% of the total weight to Carl. Thus, each of these 13 bars accounts for an average of of the total weight. Note that the bars given to Dale account for of the total weight. % = % % % % = % Let be the number of bars that Dale received. Then each of these bars accounts for an average of of the total weight. % = % Since each of the bars that she gives to Dale is heavier than each of the bars given to Bob (which were the 24 lightest bars) and is lighter than each of the bars given to Carl (which were the 13 heaviest bars), then the average weight of the bars given to Dale must be larger than % and smaller than 2%. Thus, or % % %, = = Hence, there is only one solution: =. chiefmathtutor@gmail.com Page 16

18 Solution 2 Without loss of generality, assume that the total weight of all of the bars is 100. Then the bars given to Bob weigh 45 and the bars given to Carl weigh 26. Suppose that Alan gives n bars to Dale. These bars weigh Let be the weights of the 24 bars given to Bob. Let be the weights of the 13 bars given to Carl. Let be the weights of the n bars given to Dale. Note that =. < < < < < < < < < n < < < < < < < < < < < n since the lightest bars are given to Bob and the heaviest to Carl. Also, =, =, n =. Now is the heaviest of the bars given to Bob, so and so = < > =. Also, is the lightest of the bars given to Carl, so and so = n >. chiefmathtutor@gmail.com Page 17

19 <. But each of the n bars given to Dale is heavier than and each is lighter than. Thus, or It follows that and so and n < < n, n < < n. n < n < < n < n, n < = n > =. Note that n is an integer. So we have only one solution: So Dale receives 15 bars. Problem 21 n =. =, (D) The maximum is obtained when no three planes pass through the same diameter of the sphere. Let denote the surface of the sphere. One plane divides into two regions. The second plane intersects the first plane in 2 points on, thus adding two more regions. The third plane intersects each of the previous planes on S at two points which means we will get additional regions. The nth plane intersects the previous n planes at n points, which means it would add n regions on. Hence, the total number of regions is = + =. chiefmathtutor@gmail.com Page 18

20 Problem 22 (A) Let be the probability that the sum will reach ± if it is presently. If the sum is zero to start, then there is a probability of that it will be 1 after one trial, and that it will be. By symmetry, Expanding probabilities gives =. = + =, = +, = +. Solving the about 3 equations simultaneously gives =. Thus, the probability that the sum never reaches 3 is: =. Problem 23 (D) Note that the grid is a 5 by 5 grid of squares and each square has side length 10 units, then the whole grid is 50 by chiefmathtutor@gmail.com Page 19

21 Since the diameter of the coin is 8 units, the radius of the coin is 4 units. We consider where the center of the coin lands when the coin is tossed, since the location of the center determines the position of the coin. Since the coin lands so that no part of it is off of the grid, then the center of the coin must land at least 4 units away from each of the outer edges of the grid. This means that the center of the coin lands anywhere in the region extending from 4 units from the left edge to 4 units from the right edge (a length of = units) and from 4 units from the top edge to 4 units to the bottom edge (a width of = units). Thus, the center of the coin must land in a square that is 42 by 42 in order to land so that no part of the coin is off the grid. Therefore, the total admissible area in which the center can land is. Consider one of the 25 squares. For the coin to lie completely inside the square, its center must land at least 4 units from each edge of the square. As illustrated above, it must land in a region of length and of width = =. There are 25 possible such regions (one for each square) so the area in which the center of the coin can land to create a winning position is. Therefore, the probability that the coin lands in a winning position is equal to the area of the region in which the center lands giving a winning position, divided by the area of the region in which the coin may land, or = =. Problem 24 (A) chiefmathtutor@gmail.com Page 20

22 Squaring the both sides of the equation = + + gives: which implies that Thus, = + + = = + + +, + + =. + + = + +. Rearranging and multiplying both sides by 2, we have: Completing the squares yields Hence, =. + + =. = =, which implies that the expression has only one possible value: + + =. The number of possible values of the expression is 1. Problem 25 (E) Let un denote the units digit of the positive integer n. We make three important notes: It is the final position on the circle for which we are looking, not the total number of times travelled around the circle. Therefore, moving 63 steps, for example, around the circle is equivalent to moving 3 steps around the circle because in both cases the counter ends up in the same position. Since 10 steps gives one complete trip around the circle, then we only care about the units digit of the number of steps. That is, un n. chiefmathtutor@gmail.com Page 21

23 To determine the final position, we want to find the sum of the number of steps for each of the 1234 moves; that is, we want to determine = In fact, we only need to calculate the units digit of the sum of the numbers of steps u, which is equal to u(u + u + + u + u ). To calculate un n, we only need to detect the units digit of the base n, un. Note that un n = u((un) n ). To actually perform this calculation, we can always truncate to the units digit at each step because only the units digits affect the units digits. After that, we consider the different possible values of un and determine a pattern of the units digits of powers of n: If un is 0, 1, 5, or 6, then un k is 0, 1, 5, or 6, respectively, for any positive integer k. If un =, then the units digits of powers of n alternate 4, 6, 4, 6, and so on. If un =, then the units digits of powers of n alternate 9, 1, 9, 1, and so on. If un =, then the units digits of powers of n cycle as 2, 4, 8, 6, 2, 4, 8, 6, and so on. If un =, then the units digits of powers of n cycle as 8, 4, 2, 6, 8, 4, 2, 6, and so on. If un =, then the units digits of powers of n cycle as 3, 9, 7, 1, 3, 9, 7, 1, and so on. If un =, then the units digits of powers of n cycle as 7, 9, 3, 1, 7, 9, 3, 1, and so on. Next, we determine un n, based on u(n): If u(n) is 0, 1, 5, or 6, then un n is 0, 1, 5, or 6, respectively. If un =, then un n =, since the exponent is even so the units digit will be that occurring in even positions in the pattern. If un =, then un n =, since the exponent is odd so the units digit will be that occurring in odd positions in the pattern. If un =, then un n will be either 4 or 6, depending on the exponent n, because the exponent is certainly even, but the pattern of units digits cycles with length 4 chiefmathtutor@gmail.com Page 22

24 If un =, then un n will be either 4 or 6, depending on the exponent n, because the exponent is certainly even, but the pattern of units digits cycles with length 4. Since the units digits of the base n repeat in a cycle of length 10 and the units digits of the powers of n for a fixed n repeat every 1, 2, or 4 powers, then un n repeats in a cycle of length 20 (the least common multiple of 10, 1, 2, and 4). So u(u + u + + u + u ) = u = u =. To calculate the total for 1234 steps, we note that 61 cycles of 20 bring us to 1220 steps. After 1220 steps, the units digit of the sum is u = u =. We then add the units digit of the sum of 14 more steps, starting at the beginning of the cycle, to obtain a final position equal to the units digit of u = u =. Therefore, the final position is 7. These problems are copyright Ivy League Education Center () chiefmathtutor@gmail.com Page 23