An amusement park has a collection of scale models, with a ratio of 1 : 20, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?
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Scale 1:20 means the replica is 1/20 of the original. Divide.
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
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List the early 'bad' numbers: 7, 14, 17, 21, 27, 28, 35, … Track who lands on each as the circle shrinks.
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After each elimination, continue counting from the next person in the shrunken circle.
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Counts that eliminate: 7, 14, 17, 21, 27, 28, … (7 itself, 14, multiples of 7, or any number with a 7 digit).
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
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Coastal speed = 10 mi / 30 min = 1/3 mile/minute. Highway is 3× faster = 1 mile/minute.
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Highway time = 50 miles ÷ 1 mile/min = 50 min. Add the 30 min coastal.
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Coastal: 10 miles in 30 min ⇒ 1/3 mile per minute. Highway is 3× that: 1 mile/min.
Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?
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Border (1×1 tiles): walk around the rectangle — total = 2×(12+16) − 4 for shared corners.
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Inner area is 10 × 14 (one foot off each side), filled with 2×2 tiles.
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Border: 2(12) + 2(16) − 4 = 52 unit tiles (subtract 4 because each corner is shared by two sides).
Interior: 10 ft × 14 ft = 140 sq ft, filled by 2×2 tiles → 140 / 4 = 35 tiles.
The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?
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Three steps: take reciprocals, average, take reciprocal again.
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Reciprocals: 1, 1/2, 1/4. Sum = 7/4. Average = 7/12. Final reciprocal: 12/7.
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture. If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
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Count unordered Abby-Bridget seat pairs that are adjacent. Total pairs = C(6,2) = 15.
The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?
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30 real minutes correspond to 35 car-clock minutes. Ratio: actual / car = 30/35 = 6/7.
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Car clock shows 7 hours = 420 car-minutes past noon. Actual = 420 × 6/7.
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30 real minutes = 35 car-clock minutes ⇒ actual = (6/7) × car-clock time.
Car shows 7:00, which is 420 car-minutes past noon. Actual = 420 × 6/7 = 360 minutes = 6 hours.
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
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Set 4f + l = 410, with f < l ≤ 100. Get bounds on l first.
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Since 4f is divisible by 4 and 410 ≡ 2 (mod 4), we need l ≡ 2 (mod 4).
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Average 82 over 5 tests ⇒ total = 410. Let f = first-four score, l = last. Then 4f + l = 410 with f < l ≤ 100.
l > 82 (else 4f + l would need f ≥ l).
Mod 4: 4f ≡ 0, 410 ≡ 2 ⇒ l ≡ 2 (mod 4). In (82, 100]: l ∈ {86, 90, 94, 98}.
Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
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Glue the Arabic books into one block and the Spanish books into another. Then arrange 5 objects (block-A, 3 German, block-S).
Treat the 2 Arabic books as one block and the 4 Spanish books as one block. Together with the 3 German books, we have 5 objects to order: 5! = 120 ways.
Inside the Arabic block: 2! = 2 orderings. Inside the Spanish block: 4! = 24 orderings.
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
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Bella : Ella speed ratio 1 : 5, so Bella covers 1/6 of the total 10,560 feet by the time they meet.
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Bella's distance = 1760 feet. Divide by 2.5 ft/step.
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Combined speed ratio Bella : Ella = 1 : 5, so Bella covers 1/6 of the 10,560-foot distance.
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
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Each remainder is exactly 4 less than the divisor (2 = 6−4, 5 = 9−4, 7 = 11−4). So x ≡ −4 mod each of them.
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x + 4 is divisible by lcm(6, 9, 11) = 198. Count multiples of 198 with x + 4 in [104, 1003].
Use similar triangles. F lies on diagonal AC, where BE crosses it. The ratio AF:FC follows from the parallel sides.
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AB ∥ DE? No, but consider triangles ▵BFC and ▵EFA on the diagonal AC — they're similar with ratio AB:EC = 2:1. So AF/AC = 2/3.
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Let the square have side s. EC = s/2 (E midpoint of CD), and AB = s.
Triangles ▵AFB and ▵CFE share vertex F on diagonal AC, with AB ∥ EC. Similar with ratio AB : EC = 2 : 1, so AF : FC = 2 : 1, meaning F is 2/3 of the way from A to C.
Area of ▵CEF: take ▵BCE (area (s/2)·s/2 = s2/4) and subtract ▵BCF (similar argument). Net ▵CEF area = s2/12.
Complement: count triangles with NO octagon-side. The three chosen vertices then have gaps ≥ 1 around the octagon.
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Stars and bars: with three gaps summing to 5 and each ≥ 1, there are C(4, 2) = 6 ways (with one vertex fixed). Total triangles through that fixed vertex: C(7, 2) = 21.
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Fix a vertex A. Choose the other two vertices around the octagon, leaving gaps x, y, z between consecutive chosen vertices (each gap = number of skipped octagon vertices), with x + y + z = 5.
Total ways: C(7, 2) = 21 (choose 2 of the remaining 7 vertices).
No-side cases: each gap ≥ 1, i.e. x, y, z ≥ 1 and sum 5. Stars-and-bars: C(4, 2) = 6.
P(no side on octagon) = 6/21 = 2/7. P(at least one side) = 1 − 2/7 = 5/7.
