Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of 10 miles per hour. Jack walks to the pool at a constant speed of 4 miles per hour. How many minutes before Jack does Jill arrive?
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Compute each one's minutes-per-mile (60 / speed) and subtract.
The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
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Two independent decisions: order the 2 boys at the ends (2!), and order the 3 girls in the middle (3!). Multiply.
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Boys at the two ends: 2! = 2 arrangements.
Girls in the middle three spots: 3! = 6 arrangements.
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?
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40 is below the previous minimum (42), so it becomes the new low. Which statistic depends on the spread between min and max?
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Original range: 73 − 42 = 31. New range: 73 − 40 = 33 → increases.
Median (lowering by adding a small value) decreases or stays. Mean drops (40 is below current mean). Mode stays 58. Mid-range = (max+min)/2 decreases (min drops, max unchanged).
Each of two boxes contains three chips numbered 1, 2, 3. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?
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Complement: product is odd iff both chips are odd. Two odd values out of {1, 2, 3} each → (2/3)(2/3).
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?
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The first 20 positive odd numbers: 1, 3, 5, …, 39. The sum of the first n odd numbers is n2.
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Day k sales: 2k − 1. Days 1 to 20 sum: 1 + 3 + 5 + … + 39.
In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?
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Count the total number of allowed plates by multiplying choices for each slot. The probability of AMC8 is 1 over that total.
How many subsets of two elements can be removed from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} so that the mean (average) of the remaining numbers is 6?
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Original sum is 1 + 2 + ... + 11 = 66. After removing 2 elements, 9 remain with mean 6, so the remaining sum is 54.
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The removed pair must sum to 66 − 54 = 12. Count two-element subsets of {1, ..., 11} with sum 12.
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Sum of 1 through 11 is 66. After removing 2 numbers, 9 remain; mean 6 means remaining sum is 9 × 6 = 54.
So the removed pair sums to 66 − 54 = 12.
Pairs from {1, …, 11} summing to 12: {1,11}, {2,10}, {3,9}, {4,8}, {5,7}. That is 5 pairs.
At Euler Middle School, 198 students voted on two issues in a school referendum with the following results: 149 voted in favor of the first issue and 119 voted in favor of the second issue. If there were exactly 29 students who voted against both issues, how many students voted in favor of both issues?
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Voters in favor of at least one issue = total − voted against both.
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 13 of all the ninth graders are paired with 25 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
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Pairs are one-to-one, so the number of paired ninth-graders equals the number of paired sixth-graders. Set that up as one equation and pick a convenient size.
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Let there be 15 ninth-graders. Then 5 of them are paired (one-third). So 5 sixth-graders are paired too, which is two-fifths of all sixth-graders ⇒ 12 sixth-graders total.
Show solution
Pairs are one-to-one, so paired ninth-graders = paired sixth-graders. Pick small numbers that make both fractions whole: ninth = 6, sixth = 5.
Then paired ninth = (1/3)(6) = 2 and paired sixth = (2/5)(5) = 2. ✓
Total students = 6 + 5 = 11. Buddied students = 2 + 2 = 4. Fraction = 4/11.
B
4/11.
Another way: algebra
Let n = ninth-graders, s = sixth-graders. (1/3)n = (2/5)s ⇒ 5n = 6s.
Buddied / total = ((1/3)n + (2/5)s) / (n + s) = (2 · (1/3)n) / (n + s) (since the two numerators are equal).
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
Show hint (soft nudge)
Set the two distance expressions equal: rush-hour distance = no-traffic distance. Convert 20 min and 12 min to hours (1/3 and 1/5).
Show hint (sharpest)
Solve s · (1/3) = (s + 18) · (1/5) for the rush-hour speed s, then plug back.
Show solution
Let rush-hour speed be s mph. Distance: s · (1/3) = (s + 18) · (1/5).
Multiply by 15: 5s = 3(s + 18) ⇒ 2s = 54 ⇒ s = 27.
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, 2, 5, 8, 11, 14 is an arithmetic sequence with five terms, in which the first term is 2 and the constant added is 3. Each row and each column in this 5 × 5 array is an arithmetic sequence with five terms. The square in the center is labelled X. What is the value of X?
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In a 5-term arithmetic sequence the middle (3rd) term equals the average of the 1st and 5th terms.
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Find the middle of the top row (avg of 1 and 25), middle of the bottom row (avg of 17 and 81), then average those two.
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Middle of top row = (1 + 25)/2 = 13.
Middle of bottom row = (17 + 81)/2 = 49.
X is the middle of the middle column, which is the average of those two: (13 + 49)/2 = 31.
Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?
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Let a, b, c be the counts of $1, $3, $4 pairs. Write the two equations and subtract.
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After subtracting, you get 2b + 3c = 12 with b, c ≥ 1. Use parity (2b is even) to pin down c.
Show solution
a + b + c = 12 and a + 3b + 4c = 24.
Subtract: 2b + 3c = 12. So 3c is even, meaning c is even; and 0 < c < 4 ⇒ c = 2.
In the given figure hexagon ABCDEF is equiangular, ABJI and FEHG are squares with areas 18 and 32 respectively, ▵JBK is equilateral and FE = BC. What is the area of ▵KBC?
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Find BK from the square ABJI and the equilateral ▵JBK. Find BC from the square FEHG and the given FE = BC.
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Use angles at B: 90° (square) + 60° (equilateral) + ∠KBC + 120° (equiangular hexagon) = 360°. What does that make ∠KBC?
Show solution
BK = JB = √18 = 3√2 (square side, then equilateral).
On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
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Each day's row count is a divisor of n, all different. Twelve days ⇒ n has at least 12 divisors.
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n is divisible by both 15 and 6, so by lcm(6, 15) = 30. Smallest multiple of 30 with exactly 12 divisors?
Show solution
Each day's number of students per row is a divisor of n, and all 12 are different ⇒ n has exactly 12 divisors (no 13th option exists).
n is a multiple of 15 and 6, hence of lcm(6, 15) = 30 = 2 · 3 · 5. That has only (1+1)(1+1)(1+1) = 8 divisors.
Double one prime exponent to get 12: 22 · 3 · 5 = 60 has (2+1)(1+1)(1+1) = 12 divisors. The other options (32·2·5 = 90, 2·3·52=150) are larger.
Tom has twelve slips of paper which he wants to put into five cups labeled A, B, C, D, E. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from A to E. The numbers on the papers are 2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, and 4.5. If a slip with 2 goes into cup E and a slip with 3 goes into cup B, then the slip with 3.5 must go into what cup?
Show hint (soft nudge)
Find the five cup totals. The slips sum to 35, so the five consecutive integers sum to 35 ⇒ their average (and middle) is 7.
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Cups must be 5, 6, 7, 8, 9 for A through E. Now place the 3.5 by elimination using the remaining slips.
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Cup B already has a 3 and needs a total of 6, so it also has another 3. Cup E has a 2 and needs 7 more.
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Sum of slips: 2+2+2+2.5+2.5+3+3+3+3+3.5+4+4.5 = 35. Five consecutive integers summing to 35 must be 5, 6, 7, 8, 9, so A=5, B=6, C=7, D=8, E=9.
B has a 3 and needs 6 total ⇒ the other slip in B is another 3.
Try placing the 3.5 in each cup; the rest of that cup must come from the leftover slips {2, 2, 2.5, 2.5, 3, 3.5, 4, 4.5} minus the 3.5 itself.
A (need 5): 5 − 3.5 = 1.5 — no slip equals 1.5 and no combo of leftovers sums to 1.5. ✗
C (need 7): 7 − 3.5 = 3.5 from leftovers — no such combo without using the only 3.5. ✗
E (already has 2, need 7 more): 7 − 3.5 = 3.5 — same issue. ✗
D (need 8): 8 − 3.5 = 4.5 — pair with the 4.5 slip. ✓ The remaining slips fill the other cups, e.g. A = {2.5, 2.5}, C = {3, 4}, E = {2, 2, 3}.
A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N > 2M and M > 4. Each team plays a 76 game schedule. How many games does a team play within its own division?
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A team faces 3 division rivals (N games each) and 4 outside teams (M games each): 3N + 4M = 76.
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Use the bounds M > 4 and N > 2M; also 4M ≡ 76 (mod 3) forces a residue on M.
Show solution
3N + 4M = 76.
Mod 3: 4M ≡ 76 ⇒ M ≡ 1 (mod 3). So M ∈ {7, 10, 13, …}.
N > 2M ⇒ 76 = 3N + 4M > 10M ⇒ M < 7.6.
Combining M > 4 and M < 7.6 and M ≡ 1 (mod 3): M = 7. Then N = (76 − 28)/3 = 16.
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fit into the remaining space?
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An axis-aligned square fits at best with side 3 (the unblocked center column/row), giving area 9. Try a tilted square instead.
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Tilt a square so each vertex lies on one straight edge of the cut-out shape. Decompose it into a central 3×3 square plus 4 right triangles that stick out into the unblocked middles of the original sides.
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After cutting the corner unit squares, the middle 3 units of each side of the 5×5 are clear. Inscribe a tilted square whose four vertices each sit on the middle of one of those sides.
Decompose this square as a central axis-aligned 3×3 square (area 9) plus four congruent right triangles, one extending toward the middle of each original side.
Each triangle has legs 3 and 1, area (1/2)(3)(1) = 3/2.
