AMC8 1995

The total value of the coins is $0.01 + 0.05 + 0.10 + 0.25 = 0.41$, which is $0.41/1.00$ or $41/100 = 41\%$ of a dollar.

Zack must be $15 + 3 = 18$ years old, so Jose must be $18 - 4 = 14$ years old.

Since $\frac{3}{4} \div \frac{3}{5} = \frac{3}{4} \times \frac{5}{3} = \frac{5}{4}$, multiplying by $\frac{5}{4}$ has the same effect.

Ben adds 1 to 6 to get 7, and then doubles 7 to get 14. Sue subtracts 1 from 14 to get 13, and then doubles 13 to get 26.

The sum of the fractions adds between 1 and 2 to the sum of the whole numbers, which is $2 + 3 + 4 + 5 = 14$. Thus the overall sum is between 15 and 16, so the smallest whole number larger than the sum is 16.

Since the perimeter of I is 12, its side is 3. Similarly, the side of II is 6. Hence the side of III is $3 + 6 = 9$. Thus the perimeter of III is $4 \times 9 = 36$.

One half ($\frac{1}{2}$) go by bus, one fourth ($\frac{1}{4}$) by automobile, and one tenth ($\frac{1}{10}$) by bicycle. So $100\% - (50\% + 25\% + 10\%) = 15\%$ or $\frac{3}{20}$ walk home.

$1.00 = \frac{3000\ \text{lire}}{\$1.60} = 1875$ lire.

The length of BC is the same as the diameter of the circle with center Q, so BC = 4. The diameters of the circles with centers P and R are also 4. The sum of the diameters of the circles with centers P and R gives the length of AB, so AB = 4 + 4 = 8. Hence the area of the rectangle is 8 × 4 = 32.

Answer (A): For the jacket, 40% of \$80 is a savings of \$32. For the shirt, 55% of \$40 is a savings of \$22. The total savings is \$32 + \$22 = \$54. The total of the original prices is \$120. Thus, \$54/\$120 = 0.45 = 45%.

Jane covers twice as much distance as Hector in the same time. When they meet, they will have covered 18 blocks together. Since 12+6 = 18 and 12 is twice 6, they must meet 6 blocks counterclockwise from the start, which is point D.

The last two digits of 1994 can only be factored as 94 = 2 × 47. All the other choices have at least one date that makes them lucky: (A) 9/10/90 (B) 7/13/91 (C) 4/23/92 (D) 3/31/93.

Answer (E): In $\triangle BDE$, $\angle BED + \angle BDE + \angle B = 180^\circ$. Since $\angle BED = \angle BDE$ and $\angle B = 90^\circ$, it follows that $\angle BED = \angle BDE = 45^\circ$. In $\triangle AEF$, $\angle A + \angle AEF + \angle AFE = 180^\circ$. Since $\angle A = 90^\circ$ and $\angle AEF = 40^\circ$, it follows that $\angle AFE = 50^\circ$. Consequently $\angle BFG = 50^\circ$ in $\triangle BFG$ and, since $\angle B = 90^\circ$, it follows that $\angle BGF = 40^\circ$. Consequently $\angle CGD = 40^\circ$ in $\triangle CDG$, and since $\angle C = 90^\circ$, it follows that $\angle CDG = 50^\circ$. Thus $\angle CDE = 50^\circ + 45^\circ = 95^\circ$.

The total number of games for the season is 50+40 = 90 games. Since 70% of 90 games is 63, it follows that 63 − 40 = 23 more wins are needed.

Since 4/37 = 0.108108..., it follows that the 3rd, 6th, 9th,..., 99th digits are 8. Thus the 100th digit must be 1.

Answer (C): Tabulate the data: \[ \begin{array}{lcl} \text{Allen School: } 7 \text{ students for } 3 \text{ days} & = & 21 \text{ worker days} \\ \text{Balboa School: } 4 \text{ students for } 5 \text{ days} & = & 20 \text{ worker days} \\ \text{Carver School: } 5 \text{ students for } 9 \text{ days} & = & 45 \text{ worker days} \\ \hline \textbf{Total:} & & 86 \text{ worker days} \end{array} \] Hence, $774 \div 86 = 9$ per worker day. Thus the students from Balboa School earned \$9 per worker day, for a total of \$180.

Answer (D): In Annville 11% of 100, or 11 students are in the 6th grade. In Cleona 17% of 200, or 34 students are in the 6th grade. Thus in the two schools combined, 45 out of 300 students are in the 6th grade, so 45/300 = 0.15 = 15% of all the students are in the 6th grade.

Answer (C): The four L-shaped regions account for $4\times\frac{3}{16}=\frac{12}{16}=\frac{3}{4}$ of the total area. That leaves $\frac{1}{4}$ of the total area for the center square, which yields $\left(\frac{1}{4}\right)(100\times100)=2500$ square inches. Thus the length of the side of the center square is $\sqrt{2500}=50$ inches.

Answer (D): Putting the number of children in each family in order from least to greatest yields $1, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5.$ The median is the middle, or $7^{\text{th}}$ value, which is $4$.

Answer (B): There are $6\times 6=36$ possible outcomes of rolling the dice. Since Diana and Apollo roll the same number in 6 of these, there are 30 in which the numbers on the two dice are different. By symmetry, Diana’s number is larger than Apollo’s number in exactly half of these. Thus the requested probability is $\frac{15}{36}=\frac{5}{12}$.

Answer (B): Using one, two or three cubes always leaves one protruding snap showing. The smallest number of cubes is four, arranged as shown(view from above).

Answer (A): The prime factorization of 6545 is $5 \times 7 \times 11 \times 17$. Since the product of any three of these primes is a three-digit number and since $7 \times 17$ and $11 \times 17$ are both three-digit numbers, it follows that the only pair of two-digit numbers with product 6545 is $5 \times 17 = 85$ and $7 \times 11 = 77$. Thus the answer is $85 + 77 = 162$.

Answer (B): There are 5 odd and 5 even digits that can be used for the two leftmost digits in the number. Once an odd and even digit have been selected and since all four digits are different, there are 8 choices remaining for the third digit, and then 7 choices for the fourth digit. Thus there are $5\times5\times8\times7=1400$ such whole numbers.

Answer (C): Since opposite sides of a parallelogram are equal, $AB = 12$. Then $AE = 12 - 4 = 8$. Using the Pythagorean Theorem gives $AD = \sqrt{8^2 + 6^2} = 10$, and then $BC = 10$ also. The area of a parallelogram is base $\times$ altitude. Using base $AB = 12$ and altitude $DE = 6$ gives an area of $12 \times 6 = 72$. Using base $BC = 10$ and altitude $DF$ must also give an area of $72$. Thus $DF = 72/10 = 7.2$.

A Houston-bound bus leaving Dallas at 6:00pm will arrive at 11:00pm, passing buses that left Houston at 1:30pm, 2:30pm, ..., 10:30pm: 10 buses.