AMC 8 · Test Mode

2007 AMC 8

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Problem 1 · 2007 AMC 8 Easy
Arithmetic & Operations average-target

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?

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Answer: D — 12 hours.
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Hint 1
Target total = 6 · 10 = 60. Subtract what she's already done.
Show solution
Approach: target total minus actual
  1. Done so far: 8 + 11 + 7 + 12 + 10 = 48.
  2. Needed: 60 − 48 = 12.
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Problem 2 · 2007 AMC 8 Easy
Arithmetic & Operations ratio-from-graph
amc8-2007-02
Show answer
Answer: E — 5/2.
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Hint 1
Read off counts; simplify the ratio.
Show solution
Approach: read and reduce
  1. Spaghetti / Manicotti = 250 / 100 = 5/2.
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Problem 3 · 2007 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the two smallest prime factors of 250?

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Answer: C — 7.
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Hint 1
250 = 2 · 53.
Show solution
Approach: factor 250
  1. 250 = 2 · 53. Two smallest (and only) primes: 2 and 5.
  2. Sum: 2 + 5 = 7.
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Problem 4 · 2007 AMC 8 Easy
Counting & Probability multiplication-principle

A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

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Answer: D — 30 ways.
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Hint 1
6 choices to enter, 5 to leave (different).
Show solution
Approach: multiplication
  1. 6 · 5 = 30.
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Problem 5 · 2007 AMC 8 Easy
Arithmetic & Operations target-amount

Chandler wants to buy a 500 dollar mountain bike. For his birthday, his grandparents send him 50 dollars, his aunt sends him 35 dollars and his cousin gives him 15 dollars. He earns 16 dollars per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?

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Answer: B — 25 weeks.
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Hint 1
Subtract birthday total from 500, then divide by 16.
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Approach: remaining / weekly earning
  1. Birthday: 50 + 35 + 15 = 100. Needed from route: 400.
  2. Weeks: 400 / 16 = 25.
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Problem 6 · 2007 AMC 8 Easy
Fractions, Decimals & Percents percent-decrease

The average cost of a long-distance call in the USA in 1985 was 41 cents per minute, and the average cost of a long-distance call in the USA in 2005 was 7 cents per minute. Find the approximate percent decrease in the cost per minute of a long-distance call.

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Answer: E — About 80%.
Show hint
Hint 1
Drop / original. 34 / 41 ≈ 0.83.
Show solution
Approach: compute drop, divide by original
  1. Drop: 41 − 7 = 34. Original: 41.
  2. 34 / 41 ≈ 0.83 ⇒ closest to 80%.
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Problem 7 · 2007 AMC 8 Easy
Arithmetic & Operations average-update

The average age of 5 people in a room is 30 years. An 18-year-old person leaves the room. What is the average age of the four remaining people?

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Answer: D — 33.
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Hint 1
Total before: 5 · 30 = 150. After: 150 − 18 = 132. Divide by 4.
Show solution
Approach: total × count update
  1. Old total: 150. New total: 132.
  2. Average: 132 / 4 = 33.
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Problem 8 · 2007 AMC 8 Easy
Geometry & Measurement right-triangle-area
amc8-2007-08
Show answer
Answer: B — 4.5.
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Hint 1
BE = AD = 3 (perpendicular distance). EC = DCDE = 6 − 3 = 3.
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Approach: right triangle at E
  1. BEC is right-angled at E with legs 3 and 3.
  2. Area = (1/2)(3)(3) = 4.5.
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Problem 9 · 2007 AMC 8 Easy
Logic & Word Problems latin-square-deduction
amc8-2007-09
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Answer: B — 2.
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Hint 1
Column 4 already has a 4. Row 1 col 4 must be 3 (not 4, not 1 or 2 elsewhere). Then row 2 col 4 must be 1. So col 4 needs a 2 in row 4.
Show solution
Approach: rule out values along column 4
  1. Column 4 already contains 4 in row 3. Row 1's missing digits are 3 and 4 (col 2 and col 4), and col 4 can't take 4 ⇒ row 1 col 4 = 3.
  2. Row 2 needs 1 and 4 in cols 3 and 4. Col 4 can't take 4 ⇒ row 2 col 4 = 1.
  3. Column 4 has 3, 1, 4 so far ⇒ row 4 col 4 = 2.
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Problem 10 · 2007 AMC 8 Easy
Number Theory sigma-function

For any positive integer n, define [n] to be the sum of the positive factors of n. For example, [6] = 1 + 2 + 3 + 6 = 12. Find [[11]].

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Answer: D — 28.
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Hint 1
[11] first; 11 is prime so [11] = 1 + 11 = 12. Then compute [12].
Show solution
Approach: apply twice
  1. [11] = 1 + 11 = 12 (since 11 is prime).
  2. [12] = 1 + 2 + 3 + 4 + 6 + 12 = 28.
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Problem 11 · 2007 AMC 8 Medium
Logic & Word Problems matching-puzzle
amc8-2007-11
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Answer: D — Tile IV.
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Hint 1 of 2
Look for numbers that appear on only one tile — those edges must sit on the outside boundary.
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Hint 2 of 2
Once you anchor a tile, propagate by matching shared-edge numbers to its neighbors.
Show solution
Approach: anchor on outside-only numbers, then propagate matches
  1. Some edge numbers appear on only one tile (no other tile carries that number), so those edges must lie on the outer boundary of the 2 × 2 arrangement. Pinning those tiles into their forced corners removes most of the freedom.
  2. From the anchored tile, walk to its neighbors by matching the shared edge number. The chain forces tile IV into rectangle C.
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Problem 12 · 2007 AMC 8 Medium
Geometry & Measurement hexagon-decomposition
amc8-2007-12
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Answer: A — 1 : 1.
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Hint 1
A regular hexagon of side 1 decomposes into 6 equilateral triangles of side 1 — the same triangles as the extensions.
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Approach: decompose hexagon into 6 unit triangles
  1. Hexagon = 6 equilateral triangles of side 1.
  2. Extensions = 6 equilateral triangles of side 1 (one on each edge).
  3. Same total area ⇒ ratio 1 : 1.
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Problem 13 · 2007 AMC 8 Medium
Counting & Probability inclusion-exclusion
amc8-2007-13
Show answer
Answer: C — 1504.
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Hint 1
|A ∪ B| = |A| + |B| − |A ∩ B|, with |A| = |B|.
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Approach: inclusion-exclusion
  1. 2007 = 2|A| − 1001 ⇒ 2|A| = 3008 ⇒ |A| = 1504.
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Problem 14 · 2007 AMC 8 Easy
Geometry & Measurement isosceles-altitudepythagorean-triple

The base of isosceles ▵ABC is 24 and its area is 60. What is the length of one of the congruent sides?

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Answer: C — 13.
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Hint 1
Altitude to the base = 2 · area / base = 5. Half-base = 12. Then 5-12-13 right triangle.
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Approach: altitude, then Pythagoras
  1. Height = 2 · 60 / 24 = 5.
  2. Each congruent side = √(52 + 122) = √169 = 13.
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Problem 15 · 2007 AMC 8 Medium
Algebra & Patterns inequality-reasoning

Let a, b and c be numbers with 0 < a < b < c. Which of the following is impossible?

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Answer: A — a + c < b is impossible.
Show hint
Hint 1
a > 0 and c > b imply a + c > b.
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Approach: compare to b
  1. Since c > b and a > 0, a + c > b — never less.
  2. All other choices have explicit examples (e.g., a = 1/3, b = 1/2, c = 1 gives ac = 1/3 < 1/2 = b).
  3. Impossible: (A).
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Problem 16 · 2007 AMC 8 Medium
Algebra & Patterns quadratic-vs-linear
amc8-2007-16
Show answer
Answer: A — Graph A.
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Hint 1
C = 2πr, A = πr2. Eliminating r: A = C2/(4π). Look for a graph that's an increasing concave-up curve.
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Approach: identify the shape of A vs C
  1. A grows like C2: both increasing, concave-up.
  2. Only graph A shows that pattern.
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Problem 17 · 2007 AMC 8 Easy
Fractions, Decimals & Percents mixture-update

A mixture of 30 liters of paint is 25% red tint, 30% yellow tint and 45% water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?

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Answer: C — 40%.
Show hint
Hint 1
Yellow: 0.30 · 30 = 9 L. After adding 5: 14 L out of 35 L.
Show solution
Approach: track yellow / total
  1. Original yellow: 9 L. New yellow: 9 + 5 = 14 L.
  2. Total volume: 30 + 5 = 35 L.
  3. Fraction: 14/35 = 40%.
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Problem 18 · 2007 AMC 8 Medium
Number Theory last-digit

The product of the two 99-digit numbers 303,030,303,…,030,303 and 505,050,505,…,050,505 has thousands digit A and units digit B. What is the sum of A and B?

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Answer: D — 8.
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Hint 1
The last 4 digits of each factor are 0303 and 0505. Multiply those mod 10000.
Show solution
Approach: compute the last four digits
  1. 303 · 505 = 153015.
  2. Last 4 digits: 3015 ⇒ thousands digit A = 3, units digit B = 5.
  3. A + B = 8.
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Problem 19 · 2007 AMC 8 Easy
Algebra & Patterns difference-of-squares

Pick two consecutive positive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

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Answer: C — 79.
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Hint 1
(x+1)2x2 = 2x + 1 = sum of the two integers. The sum < 100 and is odd.
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Approach: factor the difference
  1. (x+1)2x2 = 2x + 1 = (x) + (x+1).
  2. Difference equals the sum of the two integers — less than 100, and odd.
  3. Only 79 is odd and less than 100.
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Problem 20 · 2007 AMC 8 Medium
Algebra & Patterns percent-equation

Before district play, the Unicorns had won 45% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

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Answer: A — 48 games.
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Hint 1
Let pre-district games be x. Pre-district wins: 0.45x. Final wins: 0.45x + 6 = (x + 8)/2.
Show solution
Approach: set up an equation
  1. 0.45x + 6 = (x + 8)/2.
  2. Multiply by 10: 4.5x + 60 = 5x + 40 ⇒ 0.5x = 20 ⇒ x = 40.
  3. Total games: 40 + 8 = 48.
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Problem 21 · 2007 AMC 8 Medium
Counting & Probability fix-first-card

Two cards are dealt from a deck of four red cards labeled A, B, C, D and four green cards labeled A, B, C, D. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

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Answer: D — 4/7.
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Hint 1
Fix the first card. Of the 7 remaining cards, count those that win against it: 3 of the same color and 1 of the same letter (different color).
Show solution
Approach: fix one card and count winners
  1. Same color: 3 of the remaining 7.
  2. Same letter (different color): 1 of the remaining 7.
  3. Probability: (3 + 1)/7 = 4/7.
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Problem 22 · 2007 AMC 8 Hard
Geometry & Measurement invariantinterior-of-rectangle

A lemming sits at a corner of a square with side length 10 meters. The lemming runs 6.2 meters along a diagonal toward the opposite corner. It stops, makes a 90° right turn and runs 2 more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?

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Answer: C — 5.
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Hint 1
For any point inside a square of side 10, distance to opposite sides always sums to 10.
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Approach: use the inside-the-square invariant
  1. Lemming stays inside the square.
  2. Distance to left + right walls = 10. Distance to top + bottom walls = 10. Total: 20.
  3. Average: 20/4 = 5.
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Problem 23 · 2007 AMC 8 Hard
Geometry & Measurement subtract-from-whole
amc8-2007-23
Show answer
Answer: B — 6.
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Hint 1
Compute the unshaded area (4 unit squares + 4 congruent triangles of base 3 and height 5/2) and subtract from 25.
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Approach: total − unshaded
  1. Unit squares in corners: 4 · 1 = 4.
  2. Four triangles, each base 3 and height 5/2: total area 4 · (1/2)(3)(5/2) = 15.
  3. Unshaded: 4 + 15 = 19. Total: 25. Shaded: 25 − 19 = 6.
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Problem 24 · 2007 AMC 8 Medium
Counting & Probability divisibility-by-3subset-counting

A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3, or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

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Answer: C — 1/2.
Show hint
Hint 1
Divisible by 3 iff digit-sum divisible by 3. The chosen 3 digits' sum is what matters; the order is irrelevant for divisibility.
Show solution
Approach: count 3-element subsets with sum divisible by 3
  1. Subsets of size 3 from {1, 2, 3, 4}: 4 total ({1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}).
  2. Digit sums: 6 ✓, 7, 8, 9 ✓ ⇒ 2 subsets give a multiple of 3.
  3. Probability: 2/4 = 1/2.
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Problem 25 · 2007 AMC 8 Hard
Counting & Probability area-weighted-probabilityparity-sum
amc8-2007-25
Show answer
Answer: B — 35/72.
Show hints
Hint 1 of 2
Score is odd iff exactly one dart hits a 1 and the other hits a 2.
Still stuck? Show hint 2 →
Hint 2 of 2
Find probability of hitting a 1 and probability of hitting a 2, then 2 · P(1) · P(2).
Show solution
Approach: compute P(1), P(2), then 2 P(1) P(2)
  1. Outer ring area: 36π − 9π = 27π. Each outer sector: 9π ⇒ prob 9π/36π = 1/4.
  2. Inner sectors: 3π each ⇒ prob 1/12 each.
  3. Inner has one 1 and two 2s. Outer has two 1s and one 2.
  4. P(1) = (1)(1/12) + (2)(1/4) = 1/12 + 6/12 = 7/12.
  5. P(2) = (2)(1/12) + (1)(1/4) = 2/12 + 3/12 = 5/12.
  6. P(odd) = 2 · P(1) · P(2) = 2 · (7/12)(5/12) = 35/72.
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