Barney Schwinn notices that the odometer on his bicycle reads 1441, a palindrome, because it reads the same forward and backward. After riding 4 more hours that day and 6 the next, he notices that the odometer shows another palindrome, 1661. What was his average speed in miles per hour?
In 2005 Tycoon Tammy invested 100 dollars for two years. During the first year her investment suffered a 15% loss, but during the second year the remaining investment showed a 20% gain. Over the two-year period, what was the change in Tammy's investment?
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
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Answer: D — 34.
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Hint 1
Combined average = total ages / total people. Each room's total = avg × count.
Each of the 39 students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and 26 students have a cat. How many students have both a dog and a cat?
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Answer: A — 7.
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Hint 1
|A ∪ B| = |A| + |B| − |A ∩ B|. Everyone has at least one, so |A ∪ B| = 39.
A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 23 of the height it reached in the previous bounce. On which bounce will it rise to a height less than 0.5 meters?
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Answer: C — 5th bounce.
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Hint 1
After the nth bounce, height = 3 · (2/3)n. Test small n until it drops below 1/2.
Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than 100 pounds or more than 150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122, 125 and 127 pounds. What is the combined weight in pounds of the three boxes?
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Answer: C — 187 pounds.
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Hint 1
Each box is in 2 of the 3 pair-sums. Adding all three pair-sums double-counts each weight.
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Approach: sum the pair-sums and halve
Sum of pair-sums: 122 + 125 + 127 = 374 = 2(a + b + c).
Place B in the second row (2 choices for column) and then in the third row (constrained). C is then forced.
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Approach: case-split on B's placement
Row 1 is fixed up to permutation of B, C (2 ways). Row 2 starts with B or C (2 choices), then is determined. Each row 2 case constrains row 3 to one arrangement.
In Theresa's first 8 basketball games, she scored 7, 4, 3, 6, 8, 3, 1 and 5 points. In her ninth game, she scored fewer than 10 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 10 points and her points-per-game average for the 10 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
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Answer: B — 40.
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Hint 1 of 2
Sum of first 8: 37. After game 9 (score < 10), total is between 37 and 47; must be a multiple of 9 (mean integer).
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Hint 2 of 2
Then after game 10 the total is < 56 and a multiple of 10.
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Approach: fit each total to the divisibility condition
Sum after 8: 37.
After 9: total in [38, 47], divisible by 9 ⇒ 45. Game 9 = 8.
After 10: total in [46, 55], divisible by 10 ⇒ 50. Game 10 = 5.
Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 50 units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
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Answer: D — 132.
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Hint 1
l + w = 25, both positive integers. Area = l(25 − l); max near l = 12 or 13, min at l = 1.
The path is built from circular arcs plus straight pieces — separate the two.
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Hint 2 of 2
Arc total = (fraction of each circle traversed) × (its circumference); straight total = lengths of the radial / diameter segments.
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Approach: split the path into arcs and straight segments
Arcs: the path traces a half-arc of the big circle (radius 20), giving ½ · 2π · 20 = 20π.
Straights: two radial segments of length 10 (each crossing the ring between circles) plus a diameter of the small circle of length 20. That's 10 + 10 + 20 = 40.
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 3/4 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
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Answer: B — 17.
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Hint 1 of 2
Let p be the common count passing. Boys = (3/2)p, girls = (4/3)p; total = (17/6)p.
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Hint 2 of 2
Total must be a positive integer; smallest p making it integer is p = 6.
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Approach: introduce a common-count variable
Boys = (3/2)p, girls = (4/3)p. Total = (3/2 + 4/3)p = (17/6)p.
Smallest positive integer total requires p = 6 ⇒ total = 17.
△BFD doesn't sit nicely in the square — but the three corner triangles around it do.
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Hint 2 of 2
Triangle area = square area − (the three right triangles cut off in the corners).
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Approach: subtract three corner right triangles from the square
Take side 1, so AF = 2/3, FE = 1/3, CD = 2/3, DE = 1/3. The three corner triangles cut off around △BFD are △ABF (legs 1, 2/3, area 1/3), △BCD (legs 1, 2/3, area 1/3), and △FED (legs 1/3, 1/3, area 1/18).
Ten tiles numbered 1 through 10 are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
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Answer: C — 11/60.
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Hint 1
Case on the die roll d; for each, count tiles t in 1–10 with dt a perfect square.
