AMC8 2015

Answer (A): The floor is 12 ft = 4 yards long and 9 ft = 3 yards wide, so it will take $4\times 3=12$ square yards of carpet to cover it.

Answer (D): The octagon can be divided into 8 congruent triangles, 3 of which are BOC, ACOD, and ADOE. The area of triangle XOB is half the area of triangle BOC, so the fraction of the area of the octagon that is shaded is $3\cdot \frac{1}{8} + \frac{1}{2}\cdot \frac{1}{8} = \frac{7}{16}$.

Jill takes $\frac{1}{10}$ of an hour, or 6 minutes, to get to the pool, and Jack takes $\frac{1}{4}$ of an hour, or 15 minutes, so Jill arrives $15 - 6=9$ minutes before Jack.

Answer (E): There are 2 ways to seat the boys, one on each end, and $(3\cdot2\cdot1=6)$ ways to seat the three girls in the middle. So there are $(2\cdot6=12)$ possible arrangements.

Answer (A): The range is the high score minus the low score, so the range changes from 31 to 33. The range is the only listed statistic that will increase. Because 40 is the lowest score for the season, it will cause the mean to decrease. The median value of the first 11 games is the 6th highest score, or 58. The median value of the first 12 games will be the average of the 6th highest and 7th highest scores, or $(58+58)/2=58$, so no change will occur in the median. Similarly, the score that occurs most frequently in either situation is 58, so the mode will not change. The mid-range is the average of the highest score and the lowest score. The mid-range of the first 11 games is $(73+42)/2=57.5$. The mid-range of the first 12 games is 56.5, a decrease from 57.5.

Answer (B): Let $D$ be the midpoint of side $AC$. Then $BD$ is the altitude to $AC$ and $\triangle BDC$ is a right triangle with $BC=29$ and $DC=21$. So $BD=\sqrt{29^2-21^2}=\sqrt{400}=20$. The area of $\triangle ABC$ is $\frac{1}{2}\cdot 20\cdot 42=420$.

Answer (E): The nine possible equally likely outcomes are: $(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)$ In five of the nine outcomes the product is even. Therefore the probability is $\frac{5}{9}$.

Answer (D): Let $t$ be the length of the third side of the triangle. By the Triangle Inequality, $t<5+19=24$. So the perimeter $5+19+t<5+19+(5+19)=48$.

Answer (D): Note that Janabel has sold a total of $1$ widget after $1$ day, $1+3=4=2^2$ after $2$ days and $1+3+5=3^2$ widgets after $3$ days. It can be shown that this pattern continues so that after $20$ days, Janabel has sold a total of $20^2=400$.

Answer (B): The thousands position can be filled by the digits $1$ through $9$ ($0$ is excluded). Without repetition, the hundreds position can be filled with any of the remaining $9$ digits (including $0$). Similarly without repetition, the tens position and ones position can be filled with the remaining $8$ and $7$ digits, respectively. Thus there are $9\cdot9\cdot8\cdot7=4536$ integers between $1000$ and $9999$ that have distinct digits.

Answer (B): The first symbol can be any of the 5 vowels, the second can be any of the 21 consonants, the third can be any of the 20 other consonants, and the fourth can be any of the 10 digits. The total number of possible license plates is $5\cdot21\cdot20\cdot10=21{,}000$. Only one plate will read "AMC8", so the probability is $\frac{1}{21{,}000}$.

Each of the 12 edges is parallel to 3 other edges giving 36 possible pairs of parallel edges. But each pair of parallel edges is counted twice in this process, so there are 18 pairs of parallel edges.

Answer (D): If the average of the remaining $9$ numbers is $6$, then their sum is $54$. Because the sum of the numbers in the original set is $66$, the sum of the two numbers removed must be $12$. There are five such subsets: $\{1,11\}$, $\{2,10\}$, $\{3,9\}$, $\{4,8\}$, and $\{5,7\}$.

Answer (D): The sum of $4$ consecutive odd integers is always a multiple of $8$, \[ (2n-3)+(2n-1)+(2n+1)+(2n+3)=8n. \] Among the given choices, only $100$ is not a multiple of $8$. The other four numbers can each be written as the sum of four consecutive odd numbers: $16=1+3+5+7$ $40=7+9+11+13$ $72=15+17+19+21$ $20=47+49+51+53$

Answer (D): The sum 149 + 119 is related to the Venn diagram shown. The top number in the left set is the number who voted for the first issue but not the second; and the number at the bottom in the left set is the number who voted for both issues. Similarly, the top number in the right set is the number who voted for the second issue but not the first; and 29 students voted against both issues. The sum of the numbers in the diagram must be 198, so \[ (149 - x) + x + x + (119 - x) + 29 = 198, \] \[ 297 - 2x = 198, \] \[ x = 99. \]

Answer (B): 2 out of every 5 sixth graders are paired with a ninth grade buddy, and 2 out of every 6 ninth graders are paired with a sixth grade buddy. (The ratios are now expressed so that the number of sixth graders matches the number of ninth graders.) So 4 out of every 11 students are in the mentoring program. The fraction is $\frac{4}{11}$.

Answer (D): Because the new time is $\frac{12}{20}=\frac{3}{5}$ of the original time, the new speed must be $\frac{5}{3}=1\frac{2}{3}$ of the original speed. Then the additional $18$ miles per hour must be $\frac{2}{3}$ of the original speed, which is then $27$ mph. In $20$ minutes, Jeremy’s father travels $\frac{1}{3}\cdot27=9$ miles.

Answer (B): The middle number in an arithmetic sequence with $5$ terms is the average of the first and last numbers. The average of $1$ and $25$ is $13$. The average of $17$ and $81$ is $49$. Thus, $X$ is the average of $13$ and $49$, or $31$. Alternatively, $X=\frac{9+53}{2}=31$. In fact $X$ is the average of the four corner entries.

Answer (A): The triangle is inscribed in a $4\times3$ rectangle with vertices at $(1,1)$, $(1,4)$, $(5,4)$, and $(5,1)$. Three triangular regions are inside the $4\times3$ rectangle but outside $\triangle ABC$. The area of the lower-left triangle is $\frac{1}{2}\cdot4\cdot2=4$ square units. The area of the upper-left triangle is $\frac{1}{2}\cdot1\cdot3=\frac{3}{2}$ square units. The area of the third triangle is also $\frac{1}{2}\cdot1\cdot3=\frac{3}{2}$ square units. So the area of $\triangle ABC$ is $12-4-\frac{3}{2}-\frac{3}{2}=5$ square units. The area of the $6\times5$ grid is $30$ square units. Thus, the fraction covered by the triangle is $\frac{5}{30}=\frac{1}{6}$.

Answer (D): If Ralph buys 6 pairs of \$1 socks, then the other 6 pairs of socks would cost at least \$19 making the total cost more than \$24. Buying fewer than 6 pairs of \$1 socks would make Ralph’s cost even higher. If he bought 8 pairs of \$1 socks, then the other 4 pairs would cost less than \$16 making the total cost less than \$24. Buying more than 8 pairs of \ $1 socks would make his total cost even lower. So Ralph bought 7 pairs of \$1 socks, 3 pairs of \$3 socks, and 2 pairs of \$4 socks.

Answer (C): The area of the square $ABJI$ is $18$ and $\triangle KJB$ is equilateral, so $KB = JB = \sqrt{18} = 3\sqrt{2}$. The area of the square $FEHG$ is $32$, so $BC = FE = \sqrt{32} = 4\sqrt{2}$. Each interior angle of the hexagon is $120^\circ$, so $\angle KBC = 360^\circ - 60^\circ - 90^\circ - 120^\circ = 90^\circ$ and $\triangle KBC$ is a right triangle. Its area is $\frac{1}{2}\cdot 3\sqrt{2}\cdot 4\sqrt{2} = 12$.

Answer (C): The number of students must be a multiple of 6 and also a multiple of 15. So the number of students must be divisible by the least common multiple of 6 and 15 which is 30. The divisors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30, so there are only 8 divisors. The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. So 60 has 12 divisors and 60 is the smallest possible number of students.

The sum of the numbers is $35$. So the $5$ consecutive numbers in the cups must be $5$, $6$, $7$, $8$, and $9$. It is impossible to get a sum of $5$ or $7$ using the slip with $3.5$. Cup $B$ needs a sum of $6$, but it already has a slip with $3$ on it so the slip with a $3.5$ can’t go there. Cup $E$ needs a sum of $9$, but with a slip with $2$ in it the slip with $3.5$ can’t go there. The only place the slip with $3.5$ on it can go is Cup $D$. One possibility is:

Answer (B): The number of games played by a team is $3N + 4M = 76$. Because $M > 4$ and $N > 2M$ it follows that $N > 8$. Because $4$ divides both $76$ and $4M$, $4$ must divide $3N$ and hence $N$. If $N = 12$ then $M = 10$ and the condition $N > 2M$ is not satisfied. If $N \ge 20$ then $M \le 4$ and the condition $M > 4$ is not satisfied. So the only possibility is $N = 16$ and $M = 7$. So each team plays $3 \cdot 16 = 48$ games within its division and $4 \cdot 7 = 28$ games against the other division.

Answer (C): Let $EQ = c$ and $TQ = s$ as indicated in the figure. Triangles $QUV$ and $FEQ$ are similar since $\angle FQE$ and $\angle QVU$ are congruent because both are complementary to $\angle VQU$. So $\frac{QU}{UV}=\frac{FE}{EQ}$ and thus $QU=\frac{1}{c}$. Then $AB=1+c+\frac{1}{c}+1=5$ and so $c+\frac{1}{c}=3$. Since the area of square $ABCD$ equals the sum of areas of square $QRST$, four unit squares, four $1\times c$ triangles, and four $\frac{1}{c}\times 1$ triangles, it follows that $25=s^2+4\left(1+\frac{c}{2}+\frac{1}{2c}\right)$ $=s^2+4+2\left(c+\frac{1}{c}\right)$ $=s^2+4+2\cdot 3$ Therefore, the area of square $QRST=s^2=15$.