The 2004 AMC 10 Papers A and B balanced algebraic evaluation with geometric counting, spanning average-speed, coin-outcome, units-digit and combinatorial-counting problem types, with the geometry block stressing quick perimeter and area computation. Scan for the full papers and answer keys.
2004 AMC 10 A/B paper overview and exam highlights.
The 2004 AMC 10 was held in November 2004 in two papers (A and B), fairly balanced between algebraic evaluation and geometric counting.
Format: 25 multiple-choice questions, 75 minutes, 150 points max; a blank answer earns 1.5 points and a wrong answer earns 0. The AIME cut-off has recently been around 100 (Paper A).
The 2004 paper set more problems on average speed, coin outcomes, units digits of powers, and combinatorial counting, combining formulas with careful reading.
2004 AMC 10 main topic areas, weights and emphasis.
Evaluation, quadratics, word and speed problems; about 29%.
Regular polygons, rectangle perimeter, circles and cylinder surface area; about 23%.
Divisibility, remainders, GCD and divisor counts; about 18%.
Coin outcomes, combinations and ordered counting; about 20%.
Classical probability and dice sums; about 10%.
2004 AMC 10-style mock questions (Basic · Medium · Hard) — submit to grade and view step-by-step solutions.
If x = 3, what is the value of x2 + 4x - 2?
Correct answer: A
Substitute x = 3: 32 + 4(3) - 2 = 9 + 12 - 2 = 19. Answer A. Key: direct substitution.The sum of three consecutive integers is 129. What is the largest?
Correct answer: B
Let the middle integer be n: 3n = 129, so n = 43 and the largest is 44. Answer B. Key: model consecutive integers by their middle.How many positive divisors does 288 have?
Correct answer: C
288 = 25×32, so the number of divisors is (5+1)(2+1) = 18. Answer C. Key: divisor-count formula.A regular pentagon has side length 8. What is its perimeter?
Correct answer: D
All five sides of a regular pentagon are equal, so the perimeter = 5×8 = 40. Answer D. Key: regular-polygon perimeter = sides × side length.An item is marked up 50% above cost to a price of
If x2 - 8x + 15 = 0, what is the sum of all possible values of x?
Correct answer: A
Factor: (x-3)(x-5) = 0, so x = 3 or 5, and the sum is 8. Answer A. Key: Vieta — sum of roots = -b/a = 8.What is the greatest common divisor of 45 and 75?
Correct answer: B
45 = 32×5 and 75 = 3×52, so GCD = 3×5 = 15. Answer B. Key: take the smallest power of each shared prime.A committee of 2 is chosen from 24 people. How many different committees are possible?
Correct answer: C
C(24,2) = (24×23)/(2×1) = 276. Answer C. Key: combinations are order-independent, C(n,k) = n!/(k!(n-k)!).In a triangle, two angles are 48° and 72°. What is the third angle?
Correct answer: D
Triangle angles sum to 180°, so the third angle = 180 - 48 - 72 = 60°. Answer D. Key: triangle angle-sum theorem.The average of 8 numbers is 12. After removing one number, the average of the remaining 7 is 10. What was the removed number?
Correct answer: E
Sum of 8 = 8×12 = 96; sum of 7 = 7×10 = 70; the removed number = 96 - 70 = 26. Answer E. Key: average = sum / count.What is the units digit of 92004?
Correct answer: A
The units digit of powers of 9 cycles as 9, 1 (period 2). Since 2004 = 2×1002, the remainder is 0, corresponding to the 2nd value, so the units digit is 1. Answer A. Key: cyclic units digits of powers.One worker can finish a job in 6 hours and another also in 6 hours. How long does it take them together?
Correct answer: B
Rates: 1/6 + 1/6 = 2/6 = 1/3 per hour, so together they need 1 / (1/3) = 3 hours. Answer B. Key: combined rate = sum of individual rates.A square and an equilateral triangle have the same perimeter. If the triangle side is 68, what is the square side?
Correct answer: C
Triangle perimeter = 3×68 = 204, so the square side = 204 / 4 = 51. Answer C. Key: equal perimeters split by side count.Two fair six-sided dice are rolled. What is the probability the sum is 12?
Correct answer: D
The only way to sum to 12 is (6,6) — 1 outcome out of 36, so 1/36. Answer D. Key: enumerate ordered dice pairs.A car travels 90 km at 30 km/h and returns at 45 km/h. What is the average speed for the whole trip (km/h)?
Correct answer: E
Outbound time = 90/30 = 3 h; return = 90/45 = 2 h. Total distance 180 km over 5 h gives 180/5 = 36 km/h. Answer E. Key: average speed = total distance / total time, not the average of speeds.How many positive integers less than 100 are divisible by both 9 and 15?
Correct answer: A
Divisible by both 9 and 15 means divisible by LCM(9,15) = 45. The multiples of 45 below 100 are 45, 90 — 2 of them. Answer A. Key: divisible by both ⟺ divisible by their LCM.A committee of 2 is chosen from 14 people. How many different committees are possible?
Correct answer: B
C(14,2) = (14×13)/(2×1) = 91. Answer B. Key: combinations are order-independent, C(n,k) = n!/(k!(n-k)!).A right cylinder has radius 7 and height 2. What is its total surface area? (in terms of π)
Correct answer: C
Surface area = 2πr2 + 2πrh = 2π(72) + 2π(7)(2) = 98π + 28π = 126π. Answer C. Key: cylinder surface = two bases + lateral area.Given x + y = 27 and xy = 180, what is x2 + y2?
Correct answer: D
x2 + y2 = (x+y)2 - 2xy = 272 - 2×180 = 729 - 360 = 369. Answer D. Key: identity linking sum of squares to sum and product.In how many ways can 31 be written as a sum of three positive integers, where order does not matter?
Correct answer: E
Let a ≤ b ≤ c with a+b+c = 31. For a=1: 15 ways — (1,1,29) to (1,15,15); a=2: 13 ways — (2,2,27) to (2,14,15); a=3: 12 ways — (3,3,25) to (3,14,14); a=4: 10 ways — (4,4,23) to (4,13,14); a=5: 9 ways — (5,5,21) to (5,13,13); a=6: 7 ways — (6,6,19) to (6,12,13); a=7: 6 ways — (7,7,17) to (7,12,12); a=8: 4 ways — (8,8,15) to (8,11,12); a=9: 3 ways — (9,9,13) to (9,11,11); a=10: 1 way — (10,10,11). Total 80. Answer E. Key: ordered enumeration to avoid duplicates.Unified AMC 10 specs, consistent across years.
Papers A and B are comparable in difficulty and scope. Students may take either, or both — the higher score is used for awards and AIME qualification.
Four steps · Overview / Topics / Practice / Prep
Read the 2004 AMC 10 A/B overview and format specs to grasp the year’s style.
Review the key topics and weights, and locate your weak areas.
Complete the 25 mock questions (75-minute limit); submit for auto-grading and right/wrong marks.
Open the step-by-step solutions, learn from mistakes, and scan for full papers to push for AIME.
Prep tips and score-boosting advice for the 2004 AMC 10 A/B paper.
Finish all 25 within 75 minutes; skip when stuck, bank the solvable ones, and return to the hard ones.
Average speed = total distance / total time — never the arithmetic mean of the two speeds.
Ordered uses P(n,k); unordered uses C(n,k); check order and repeatability.
Units digits cycle by the base; locate the result via the exponent mod the period.
Aiming above 100? Lock the basics and medium questions, then take the last 5 step by step; scan for full papers and a plan.
Frequent questions about the 2004 AMC 10 papers and answers.
Scan to add the advisor for 2004 AMC 10 A/B past papers, answer keys and solutions, plus a custom prep plan.
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