AMC10 2014 B

Leah has 7 pennies and 6 nickels, which are worth 37 cents.

Note that $\frac{2^{3} + 2^{3}}{2^{-3} + 2^{-3}} = 2 \cdot \frac{2^{3}}{2 \cdot 2^{-3}} = 2^{6} = 64$.

The fraction of Randy’s trip driven on pavement was $1 - \frac{1}{3} - \frac{1}{5} = \frac{7}{15}$. Therefore the entire trip was $20 \div \frac{7}{15} = \frac{300}{7}$ miles.

Let a muffin cost $m$ dollars and a banana cost $b$ dollars. Then $2(4m + 3b) = 2m + 16b$, and simplifying gives $m = \frac{5}{3}b$.

Denote the height of a pane by $5x$ and the width by $2x$. Then the square window has height $2 \cdot 5x + 6$ inches and width $4 \cdot 2x + 10$ inches. Solving $2\cdot5x+6 = 4\cdot2x+10$ gives $x = 2$. The side length of the square window is 26 inches.

The special allows Orvin to purchase balloons at $\frac{1 + 2/3}{2} = \frac{5}{6}$ times the regular price. Because Orvin had just enough money to purchase 30 balloons at the regular price, he may now purchase $30 \cdot \frac{6}{5} = 36$ balloons.

The fraction by which $A$ is greater than $B$ is simply the positive difference $(A - B)$ divided by $B$. The percent difference is 100 times this, or $100\frac{A-B}{B}$.

The truck travels for $3 \cdot 60 = 180$ seconds, at a rate of $\frac{b}{6t} \cdot \frac{1}{3}$ yards per second. Hence the truck travels $180 \cdot \frac{b}{6t} \cdot \frac{1}{3} = 10\frac{b}{t}$ yards.

English (OCR): Answer (A): Note that $2014=\dfrac{\dfrac{1}{w}+\dfrac{1}{z}}{\dfrac{1}{w}-\dfrac{1}{z}}=\dfrac{\dfrac{w+z}{wz}}{\dfrac{z-w}{wz}}=\dfrac{w+z}{z-w}.$ Because $\dfrac{w+z}{z-w}=-\dfrac{w+z}{w-z}$, the requested value is $-2014$.

Answer (C): As indicated by the leftmost column $A+B\le 9$. Then both the second and fourth columns show that $C=0$. Because $A$, $B$, and $C$ are distinct digits, $D$ must be at least 3. The following values for $(A,B,C,D)$ show that $D$ may be any of the 7 digits that are at least 3: $(1,2,0,3)$, $(1,3,0,4)$, $(2,3,0,5)$, $(2,4,0,6)$, $(2,5,0,7)$, $(2,6,0,8)$, $(2,7,0,9)$.

11. Answer (C): If $P$ is the price paid for an item, then the discounted prices with the three given discounts are given by the following calculations: (1) $(0.85)^2P = 0.7225P$ for a discount of 27.75% (2) $(0.9)^3P = 0.729P$ for a discount of 27.1% (3) $(0.75)\cdot(0.95)P = 0.7125P$ for a discount of 28.75% The smallest integer greater than 27.75, 27.1, and 28.75 is 29.

By inspection, the five smallest positive divisors of 2,014,000,000 are 1, 2, 4, 5, and 8. Therefore the fifth largest divisor is $\frac{2,014,000,000}{8} = 251,750,000$.

Label points E and F as shown in the figure, and let D be the midpoint of BE. Because $\triangle BFD$ is a 30–60–90 triangle with hypotenuse 1, the length of BD is $\sqrt{3}/2$, and therefore BC = $\sqrt{3}$. It follows that the area of $\triangle ABC$ is $\frac{\sqrt{3}}{4} \cdot (\sqrt{3})^2 = 3\sqrt{3}$.

Let m be the total mileage of the trip. Then m must be a multiple of 55. Also, because m = cba − abc = 99(c − a), it is a multiple of 9. Therefore m is a multiple of 495. Because m is at most a 3-digit number and a is not equal to 0, m = 495. Therefore c − a = 5. Because a + b + c ≤7, the only possible abc is 106, so a² + b² + c² = 1 + 0 + 36 = 37.

15. Answer (A): Let $AD=\sqrt{3}$. Because $\angle ADE=30^\circ$, it follows that $AE=1$ and $DE=2$. Now $\angle EDF=30^\circ$ and $\angle DEF=120^\circ$, so $\triangle DEF$ is isosceles and $EF=2$. Thus the area of $\triangle DEF$ (with $EF$ viewed as the base) is $\frac12\cdot2\cdot\sqrt{3}=\sqrt{3}$, and the desired ratio is $\frac{\sqrt{3}}{\sqrt{3}\cdot2\cdot\sqrt{3}}=\frac{\sqrt{3}}{6}$.

If exactly three of the four dice show the same number, then there are 6 possible choices for the repeated value and 5 possible choices for the non-repeated value. The non-repeated value may appear on any one of the 4 dice, so there are 6·5·4=120 possible ways for such a result to occur. There are 6 ways for all four dice to show the same value. There are 6^4 total possible outcomes for the four dice. The probability of the desired result is $\frac{120+6}{6^4} = \frac{7}{72}$.

Answer (D): Note that $10^{1002}-4^{501}=2^{1002}\cdot 5^{1002}-2^{1002}$ $=2^{1002}(5^{1002}-1)$ $=2^{1002}(5^{501}-1)(5^{501}+1)$ $=2^{1002}(5-1)(5^{500}+5^{499}+\cdots+5+1)(5+1)(5^{500}-5^{499}+\cdots-5+1)$ $=2^{1005}(3)(5^{500}+5^{499}+\cdots+5+1)(5^{500}-5^{499}+\cdots-5+1).$ Because each of the last two factors is a sum of an odd number of odd terms, they are both odd. The greatest power of 2 is $2^{1005}$.

Answer (E): The numbers in the list have a sum of $11\cdot 10=110$. The value of the 11th number is maximized when the sum of the first ten numbers is minimized subject to the following conditions. - If the numbers are arranged in nondecreasing order, the sixth number is 9. - The number 8 occurs either 2, 3, 4, or 5 times, and all other numbers occur fewer times. If 8 occurs 5 times, the smallest possible sum of the first 10 numbers is $8+8+8+8+8+9+9+9+9+10=86$. If 8 occurs 4 times, the smallest possible sum of the first 10 numbers is $1+8+8+8+8+9+9+9+10+10=80$. If 8 occurs 3 times, the smallest possible sum of the first 10 numbers is $1+1+8+8+8+9+9+10+10+11=75$. If 8 occurs 2 times, the smallest possible sum of the first 10 numbers is $1+2+3+8+8+9+10+11+12+13=77$. Thus the largest possible value of the 11th number is $110-75=35$.

Answer (D): Let $A$ be the first point chosen on the outer circle, let chords $\overline{AB}$ and $\overline{AC}$ on the outer circle be tangent to the inner circle at $D$ and $E$, respectively, and let $O$ be the common center of the two circles. Triangle $ADO$ has a right angle at $D$, $OA=2$, and $OD=1$, so $\angle OAD=30^\circ$. Similarly, $\angle OAE=30^\circ$, so $\angle BAC=\angle DAE=60^\circ$, and minor arc $BC=120^\circ$. If $X$ is the second point chosen on the outer circle, then chord $\overline{AX}$ intersects the inner circle if and only if $X$ is on minor arc $BC$. Therefore the requested probability is $\frac{120^\circ}{360^\circ}=\frac{1}{3}$.

Answer (C): Note that $x^4 - 51x^2 + 50 = (x^2 - 50)(x^2 - 1)$, so the roots of the polynomial are $\pm 1$ and $\pm \sqrt{50}$. Arranged from least to greatest, these roots are approximately $-7.1$, $-1$, $1$, $7.1$. The polynomial takes negative values on the intervals $(-7.1, -1)$ and $(1, 7.1)$, which include 12 integers: $-7$, $-6$, $-5$, $-4$, $-3$, $-2$, $2$, $3$, $4$, $5$, $6$, $7$.

Answer (B): Assume without loss of generality that $DA=10$ and $BC=14$. Let $M$ and $N$ be the feet of the perpendicular segments to $\overline{AB}$ from $D$ and $C$, respectively. The four points $A,M,N,B$ appear on $\overline{AB}$ in that order. Let $x=DM=CN$, $y=AM$, and $z=NB$. Then $x^2+y^2=10^2=100$, $x^2+z^2=14^2=196$, and $y+21+z=33$. Therefore $z=12-y$, and it follows that $\sqrt{196-x^2}=12-\sqrt{100-x^2}$. Squaring and simplifying gives $24\sqrt{100-x^2}=48$, so $x^2=96$ and $y=\sqrt{100-96}=2$. The square of the length of the shorter diagonal, $\overline{AC}$, is $(y+21)^2+x^2=23^2+96=625$, so $AC=25$.

Answer (B): Let $O$ be the center of the circle and choose one of the semicircles to have center point $B$. Label the point of tangency $C$ and point $A$ as in the figure. In $\triangle OAB$, $AB=\frac{1}{2}$ and $OA=1$, so $OB=\frac{\sqrt{5}}{2}$. Because $BC=\frac{1}{2}$, $OC=\frac{\sqrt{5}}{2}-\frac{1}{2}=\frac{\sqrt{5}-1}{2}$.

Answer (E): Assume without loss of generality that the radius of the top base of the truncated cone (frustum) is $1$. Denote the radius of the bottom base by $r$ and the radius of the sphere by $a$. The figure on the left is a side view of the frustum. Applying the Pythagorean Theorem to the triangle on the right yields $r=a^2$. The volume of the frustum is $$\frac{1}{3}\pi(r^2+r\cdot1+1^2)\cdot2a=\frac{1}{3}\pi(a^4+a^2+1)\cdot2a.$$ Setting this equal to twice the volume of the sphere, $\frac{4}{3}\pi a^3$, and simplifying gives $$a^4-3a^2+1=0,\ \text{or}\ r^2-3r+1=0.$$ Therefore $r=\frac{3+\sqrt5}{2}$.

Answer (B): The circular arrangement 14352 is bad because the sum $6$ cannot be achieved with consecutive numbers, and the circular arrangement 23154 is bad because the sum $7$ cannot be so achieved. It remains to show that these are the only bad arrangements. Given a circular arrangement, sums $1$ through $5$ can be achieved with a single number, and if the sum $n$ can be achieved, then the sum $15-n$ can be achieved using the complementary subset. Therefore an arrangement is not bad as long as sums $6$ and $7$ can be achieved. Suppose $6$ cannot be achieved. Then $1$ and $5$ cannot be adjacent, so by a suitable rotation and/or reflection, the arrangement is $1bc5e$. Furthermore, $\{b,c\}$ cannot equal $\{2,3\}$ because $1+2+3=6$; similarly $\{b,c\}$ cannot equal $\{2,4\}$. It follows that $e=2$, which then forces the arrangement to be $14352$ in order to avoid consecutive $213$. This arrangement is bad. Next suppose that $7$ cannot be achieved. Then $2$ and $5$ cannot be adjacent, so again without loss of generality the arrangement is $2bc5e$. Reasoning as before, $\{b,c\}$ cannot equal $\{3,4\}$ or $\{1,4\}$, so $e=4$, and then $b=3$ and $c=1$, to avoid consecutive $421$; therefore the arrangement is $23154$, which is also bad. Thus there are only two bad arrangements up to rotation and reflection.

Answer (C): First note that once the frog is on pad 5, it has probability $\frac{1}{2}$ of eventually being eaten by the snake, and a probability $\frac{1}{2}$ of eventually exiting the pond without being eaten. It is therefore necessary only to determine the probability that the frog on pad 1 will reach pad 5 before being eaten. Consider the frog’s jumps in pairs. The frog on pad 1 will advance to pad 3 with probability $\frac{9}{10}\cdot\frac{8}{10}=\frac{72}{100}$, will be back at pad 1 with probability $\frac{9}{10}\cdot\frac{2}{10}=\frac{18}{100}$, and will retreat to pad 0 and be eaten with probability $\frac{1}{10}$. Because the frog will eventually make it to pad 3 or make it to pad 0, the probability that it ultimately makes it to pad 3 is $$ \frac{\frac{72}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{36}{41}, $$ and the probability that it ultimately makes it to pad 0 is $$ \frac{\frac{10}{100}}{\left(\frac{72}{100}+\frac{10}{100}\right)}=\frac{5}{41}. $$ Similarly, in a pair of jumps the frog will advance from pad 3 to pad 5 with probability $\frac{7}{10}\cdot\frac{6}{10}=\frac{42}{100}$, will be back at pad 3 with probability $\frac{7}{10}\cdot\frac{4}{10}+\frac{3}{10}\cdot\frac{8}{10}=\frac{52}{100}$, and will retreat back to pad 1 with probability $\frac{3}{10}\cdot\frac{2}{10}=\frac{6}{100}$. Because the frog will ultimately make it to pad 5 or pad 1 from pad 3, the probability that it ultimately makes it to pad 5 is $$ \frac{\frac{42}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{7}{8}, $$ and the probability that it ultimately makes it to pad 1 is $$ \frac{\frac{6}{100}}{\left(\frac{42}{100}+\frac{6}{100}\right)}=\frac{1}{8}. $$ The sequences of pairs of moves by which the frog will advance to pad 5 without being eaten are $$ 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 5,\; 1\to 3\to 1\to 3\to 1\to 3\to 5, $$ and so on. The sum of the respective probabilities of reaching pad 5 is then $$ \frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots $$ $$ =\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right) $$ $$ =\frac{63}{82}\cdot\left(\frac{1}{1-\frac{9}{82}}\right) $$ $$ =\frac{63}{73}. $$ Therefore the requested probability is $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}$.