AMC 8 · Test Mode

2007 AMC 8

Test mode — hints and solutions are locked. Pick an answer for each, then submit at the bottom to see your score.

← Exit test mode No hints or solutions until you submit.
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?

Show hint
Target total = 6 · 10 = 60. Subtract what she's already done.
Show solution
  1. Done so far: 8 + 11 + 7 + 12 + 10 = 48.
  2. Needed: 60 − 48 = 12.
D 12 hours.
Mark: · log in to save
Problem 2 · 2007 AMC 8 Easy
Arithmetic & Operations ratio-from-graph

650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

Show hint
Read off counts; simplify the ratio.
Show solution
  1. Spaghetti / Manicotti = 250 / 100 = 5/2.
E 5/2.
Mark: · log in to save
Problem 3 · 2007 AMC 8 Easy
Number Theory prime-factorization

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

Show hint
250 = 2 · 53.
Show solution
  1. 250 = 2 · 53. Two smallest (and only) primes: 2 and 5.
  2. Sum: 2 + 5 = 7.
C 7.
Mark: · log in to save
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?

Show hint
6 choices to enter, 5 to leave (different).
Show solution
  1. 6 · 5 = 30.
D 30 ways.
Mark: · log in to save
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?

Show hint
Subtract birthday total from 500, then divide by 16.
Show solution
  1. Birthday: 50 + 35 + 15 = 100. Needed from route: 400.
  2. Weeks: 400 / 16 = 25.
B 25 weeks.
Mark: · log in to save
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.

Show hint
Drop / original. 34 / 41 ≈ 0.83.
Show solution
  1. Drop: 41 − 7 = 34. Original: 41.
  2. 34 / 41 ≈ 0.83 ⇒ closest to 80%.
E About 80%.
Mark: · log in to save
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?

Show hint
Total before: 5 · 30 = 150. After: 150 − 18 = 132. Divide by 4.
Show solution
  1. Old total: 150. New total: 132.
  2. Average: 132 / 4 = 33.
D 33.
Mark: · log in to save
Problem 8 · 2007 AMC 8 Easy
Geometry & Measurement right-triangle-area

In trapezoid ABCD, AD is perpendicular to DC, AD = AB = 3, and DC = 6. In addition, E is on DC, and BE is parallel to AD. Find the area of ▵BEC.

Show hint
BE = AD = 3 (perpendicular distance). EC = DCDE = 6 − 3 = 3.
Show solution
  1. BEC is right-angled at E with legs 3 and 3.
  2. Area = (1/2)(3)(3) = 4.5.
B 4.5.
Mark: · log in to save
Problem 9 · 2007 AMC 8 Easy
Logic & Word Problems latin-square-deduction

To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?

Show hint
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
  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.
B 2.
Mark: · log in to save
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]].

Show hint
[11] first; 11 is prime so [11] = 1 + 11 = 12. Then compute [12].
Show solution
  1. [11] = 1 + 11 = 12 (since 11 is prime).
  2. [12] = 1 + 2 + 3 + 4 + 6 + 12 = 28.
D 28.
Mark: · log in to save
Problem 11 · 2007 AMC 8 Medium
Logic & Word Problems matching-puzzle

Tiles I, II, III and IV are translated so one tile coincides with each of the rectangles A, B, C and D. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle C?

Show hint (soft nudge)
Look for unique numbers that anchor placement. Tile III is the only tile with 0 and 5 ⇒ it must be on an edge of the arrangement and matched accordingly.
Show hint (sharpest)
Then match adjacent tiles by their shared edge numbers.
Show solution
  1. Tile III is the only tile with 0 and 5. These numbers must be on the outer perimeter (no other tile has them). The arrangement forces III into rectangle D.
  2. Tile IV's left edge (9) matches III's right edge (5)? No — check edges. By matching: IV's edges (top 2, right 1, bottom 6, left 9) match III's (top 7, right 5, bottom 0, left 1) where IV's left 9 ... actually the matching is best traced from LIVE: IV ends up in rectangle C.
  3. Tile IV goes to Rectangle C.
D Tile IV.
Mark: · log in to save
Problem 12 · 2007 AMC 8 Medium
Geometry & Measurement hexagon-decomposition

A unit hexagram is composed of a regular hexagon of side length 1 and its 6 equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

Show hint
A regular hexagon of side 1 decomposes into 6 equilateral triangles of side 1 — the same triangles as the extensions.
Show solution
  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.
A 1 : 1.
Mark: · log in to save
Problem 13 · 2007 AMC 8 Medium
Counting & Probability inclusion-exclusion

Sets A and B, shown in the Venn diagram, have the same number of elements. Their union has 2007 elements and their intersection has 1001 elements. Find the number of elements in A.

Show hint
|A ∪ B| = |A| + |B| − |A ∩ B|, with |A| = |B|.
Show solution
  1. 2007 = 2|A| − 1001 ⇒ 2|A| = 3008 ⇒ |A| = 1504.
C 1504.
Mark: · log in to save
Problem 14 · 2007 AMC 8 Easy
Geometry & Measurement isosceles-altitude pythagorean-triple

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

Show hint
Altitude to the base = 2 · area / base = 5. Half-base = 12. Then 5-12-13 right triangle.
Show solution
  1. Height = 2 · 60 / 24 = 5.
  2. Each congruent side = √(52 + 122) = √169 = 13.
C 13.
Mark: · log in to save
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?

Show hint
a > 0 and c > b imply a + c > b.
Show solution
  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).
A a + c &lt; b is impossible.
Mark: · log in to save
Problem 16 · 2007 AMC 8 Medium
Algebra & Patterns quadratic-vs-linear

Amanda draws five circles with radii 1, 2, 3, 4 and 5. Then for each circle she plots the point (C, A), where C is its circumference and A is its area. Which of the following could be her graph?

Show hint
C = 2πr, A = πr2. Eliminating r: A = C2/(4π). Look for a graph that's an increasing concave-up curve.
Show solution
  1. A grows like C2: both increasing, concave-up.
  2. Only graph A shows that pattern.
A Graph A.
Mark: · log in to save
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?

Show hint
Yellow: 0.30 · 30 = 9 L. After adding 5: 14 L out of 35 L.
Show solution
  1. Original yellow: 9 L. New yellow: 9 + 5 = 14 L.
  2. Total volume: 30 + 5 = 35 L.
  3. Fraction: 14/35 = 40%.
C 40%.
Mark: · log in to save
Problem 18 · 2007 AMC 8 Medium
Number Theory last-digits-only

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?

Show hint
The last 4 digits of each factor are 0303 and 0505. Multiply those mod 10000.
Show solution
  1. 303 · 505 = 153015.
  2. Last 4 digits: 3015 ⇒ thousands digit A = 3, units digit B = 5.
  3. A + B = 8.
D 8.
Mark: · log in to save
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?

Show hint
(x+1)2x2 = 2x + 1 = sum of the two integers. The sum < 100 and is odd.
Show solution
  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.
C 79.
Mark: · log in to save
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?

Show hint
Let pre-district games be x. Pre-district wins: 0.45x. Final wins: 0.45x + 6 = (x + 8)/2.
Show solution
  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.
A 48 games.
Mark: · log in to save
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?

Show hint
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
  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.
D 4/7.
Mark: · log in to save
Problem 22 · 2007 AMC 8 Hard
Geometry & Measurement invariant interior-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?

Show hint
For any point inside a square of side 10, distance to opposite sides always sums to 10.
Show solution
  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.
C 5.
Mark: · log in to save
Problem 23 · 2007 AMC 8 Hard
Geometry & Measurement subtract-from-whole

What is the area of the shaded pinwheel shown in the 5 × 5 grid?

Show hint
Compute the unshaded area (4 unit squares + 4 congruent triangles of base 3 and height 5/2) and subtract from 25.
Show solution
  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.
B 6.
Mark: · log in to save
Problem 24 · 2007 AMC 8 Medium
Counting & Probability divisibility-by-3 subset-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?

Show hint
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
  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.
C 1/2.
Mark: · log in to save
Problem 25 · 2007 AMC 8 Hard
Counting & Probability area-weighted-probability parity-sum

On the dart board shown in the figure, the outer circle has radius 6 and the inner circle has a radius of 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?

Show hint (soft nudge)
Score is odd iff exactly one dart hits a 1 and the other hits a 2.
Show hint (sharpest)
Find probability of hitting a 1 and probability of hitting a 2, then 2 · P(1) · P(2).
Show solution
  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.
B 35/72.
Mark: · log in to save