AMC 8 · Test Mode

2006 AMC 8

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Problem 1 · 2006 AMC 8 Easy
Arithmetic & Operations estimation

Mindy made three purchases for $1.98, $5.04, and $9.89. What was her total, to the nearest dollar?

Show hint
Round each to the nearest dollar before adding (each is already close to a whole).
Show solution
  1. $1.98 + $5.04 + $9.89 ≈ 2 + 5 + 10 = 17.
D $17.
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Problem 2 · 2006 AMC 8 Easy
Arithmetic & Operations scoring-rule

On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score? (right = +1, wrong or N/A = +0)

Show hint
Only correct answers count.
Show solution
  1. 13 correct · 1 = 13.
C 13.
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Problem 3 · 2006 AMC 8 Easy
Ratios, Rates & Proportions minutes-per-lap

Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?

Show hint
Minutes per lap before and after; subtract.
Show solution
  1. Before: 25/10 = 2.5 min/lap. Now: 24/12 = 2 min/lap.
  2. Improvement: 0.5 = 1/2 minute per lap.
A 1/2 minute.
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Problem 4 · 2006 AMC 8 Easy
Fractions, Decimals & Percents net-rotation

Initially, a spinner points west. Chenille moves it clockwise 214 revolutions and then counterclockwise 334 revolutions. In what direction does the spinner point after the two moves?

Show hint
Net rotation = clockwise − counterclockwise. Reduce mod 1 revolution.
Show solution
  1. Net counterclockwise: 3.75 − 2.25 = 1.5 revolutions counterclockwise.
  2. 1.5 mod 1 = 0.5 ⇒ spinner turns half-revolution from west ⇒ east.
B East.
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Problem 5 · 2006 AMC 8 Easy
Geometry & Measurement midpoint-square

Points A, B, C and D are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?

Show hint
Folding the four outer triangles inward exactly covers the inner square ⇒ inner = half outer.
Show solution
  1. Inner square = (1/2) outer square = 60 / 2 = 30.
D 30.
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Problem 6 · 2006 AMC 8 Easy
Geometry & Measurement perimeter-subtract-overlap

The letter T is formed by placing two 2 × 4 inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?

Show hint
Sum each rectangle's perimeter, then subtract the buried edges. The two rectangles share a 2-inch segment internally; that hides 2 from each rectangle's perimeter.
Show solution
  1. Each 2 × 4 perimeter: 12. Two rectangles: 24.
  2. Junction hides 2 inches from each ⇒ subtract 4.
  3. 24 − 4 = 20.
C 20.
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Problem 7 · 2006 AMC 8 Easy
Geometry & Measurement circle-formulas

Circle X has a radius of π. Circle Y has a circumference of 8π. Circle Z has an area of 9π. List the circles in order from smallest to largest radius.

Show hint
Y: C = 2πr = 8π ⇒ r = 4. Z: πr2 = 9π ⇒ r = 3. X: r = π ≈ 3.14.
Show solution
  1. X: r = π ≈ 3.14. Y: r = 4. Z: r = 3.
  2. Order: 3 < 3.14 < 4 ⇒ Z, X, Y.
B Z, X, Y.
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Problem 8 · 2006 AMC 8 Easy
Fractions, Decimals & Percents two-way-table

The table shows some of the results of a survey by radio station KACL. What percentage of the males surveyed listen to the station? (Total surveyed: 200. Females: 96. Females who listen: 58. Males who don't listen: 26. Total listeners: 136. Total non-listeners: 64.)

Show hint
Total males = 200 − 96 = 104. Of those, 26 don't listen.
Show solution
  1. Males total: 200 − 96 = 104. Males who listen: 104 − 26 = 78.
  2. Percentage: 78 / 104 = 75%.
E 75%.
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Problem 9 · 2006 AMC 8 Easy
Algebra & Patterns telescoping-product

What is the product of 32 × 43 × 54 × … × 20062005?

Show hint
Each numerator cancels with the next denominator. Only the first denominator and last numerator survive.
Show solution
  1. Product collapses to 2006 / 2 = 1003.
C 1003.
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Problem 10 · 2006 AMC 8 Easy
Algebra & Patterns inverse-proportion

Jorge's teacher asks him to plot all the ordered pairs (w, l) of positive integers for which w is the width and l is the length of a rectangle with area 12. What should his graph look like?

Show hint
wl = 12 means l = 12/w: as w increases, l decreases hyperbolically.
Show solution
  1. l = 12/w: w and l are inversely proportional — plot decreases.
  2. Only graph A shows this shape.
A Graph A.
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Problem 11 · 2006 AMC 8 Medium
Number Theory digit-sum casework

How many two-digit numbers have digits whose sum is a perfect square?

Show hint (soft nudge)
Digit sums of two-digit numbers range 1–18. Perfect squares in range: 1, 4, 9, 16.
Show hint (sharpest)
Count two-digit numbers with each of these digit sums.
Show solution
  1. Sum = 1: {10} ⇒ 1.
  2. Sum = 4: {13, 22, 31, 40} ⇒ 4.
  3. Sum = 9: {18, 27, 36, 45, 54, 63, 72, 81, 90} ⇒ 9.
  4. Sum = 16: {79, 88, 97} ⇒ 3.
  5. Total: 1 + 4 + 9 + 3 = 17.
C 17.
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Problem 12 · 2006 AMC 8 Easy
Fractions, Decimals & Percents weighted-percentage

Antonette gets 70% on a 10-problem test, 80% on a 20-problem test and 90% on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?

Show hint
Total correct / 60.
Show solution
  1. Correct: 7 + 16 + 27 = 50.
  2. 50 / 60 ≈ 0.833 ⇒ closest to 83%.
D 83%.
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Problem 13 · 2006 AMC 8 Medium
Ratios, Rates & Proportions closing-speed

Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?

Show hint
By 9:00, Cassie has covered 6 miles. Then 56 miles remain at combined speed 28 mph ⇒ 2 hours.
Show solution
  1. By 9:00 AM Cassie has biked (1/2)(12) = 6 miles. Gap: 62 − 6 = 56 miles.
  2. Combined: 12 + 16 = 28 mph. Time: 56 / 28 = 2 hours.
  3. Meeting: 9:00 + 2:00 = 11:00 AM.
D 11:00 AM.
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Problem 14 · 2006 AMC 8 Easy
Ratios, Rates & Proportions time-difference

Problems 14, 15 and 16 involve Mrs. Reed's English assignment. A Novel Assignment. The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?

Show hint
Difference per page: 45 − 30 = 15 sec. Multiply by 760.
Show solution
  1. Per page: 45 − 30 = 15 sec.
  2. 760 · 15 = 11,400.
B 11,400 seconds.
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Problem 15 · 2006 AMC 8 Medium
Ratios, Rates & Proportions equal-time-split

Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?

Show hint
30x = 45(760 − x). Solve for x.
Show solution
  1. 30x = 45(760 − x) ⇒ 75x = 45 · 760 ⇒ x = 45 · 760 / 75 = 456.
  2. Chandra reads through page 456.
C Page 456.
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Problem 16 · 2006 AMC 8 Medium
Ratios, Rates & Proportions page-time-ratio

Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?

Show hint (soft nudge)
Equal time means pages-per-reader are in inverse proportion to seconds-per-page (rates).
Show hint (sharpest)
Reading rates: Alice 1/20, Bob 1/45, Chandra 1/30. Ratio of pages = 1/20 : 1/45 : 1/30 = 9 : 4 : 6 (multiply by 180).
Show solution
  1. Reading rates: 1/20, 1/45, 1/30 pages/sec for Alice, Bob, Chandra.
  2. Pages each reads in equal time: ratio 9 : 4 : 6 (sum 19). Total 760 pages ⇒ Alice 360, Bob 160, Chandra 240.
  3. Time each: e.g., Bob 160 · 45 = 7200 sec.
E 7200 seconds.
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Problem 17 · 2006 AMC 8 Medium
Counting & Probability parity-of-sum

Jeff rotates spinners P, Q and R and adds the resulting numbers. What is the probability that his sum is an odd number?

Show hint
Q is always even, R is always odd. The parity of the sum depends only on P: sum is odd ⇔ P is even.
Show solution
  1. Q contributes an even number; R contributes an odd. Their sum is odd. Adding P: total odd iff P is even.
  2. P is even (lands on 2) with probability 1/3.
B 1/3.
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Problem 18 · 2006 AMC 8 Medium
Geometry & Measurement corner-cubes surface-fraction

A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?

Show hint
Symmetry: each face has the same pattern. Count white unit squares on one face.
Show solution
  1. Each face is 3 × 3 = 9 unit squares. Corners (4 of them) show the black corner cubes ⇒ 4 black, 5 white.
  2. Fraction: 5/9.
D 5/9.
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Problem 19 · 2006 AMC 8 Medium
Geometry & Measurement congruent-triangles midpoint

Triangle ABC is an isosceles triangle with AB = BC. Point D is the midpoint of both BC and AE, and CE is 11 units long. Triangle ABD is congruent to triangle ECD. What is the length of BD?

Show hint
From the congruence, AB = EC = 11. Isosceles gives BC = AB = 11. D is the midpoint of BC.
Show solution
  1. ABD ≅ ▵ECDAB = EC = 11.
  2. Isosceles: BC = AB = 11.
  3. BD = (1/2)BC = 5.5.
D 5.5.
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Problem 20 · 2006 AMC 8 Easy
Counting & Probability round-robin

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica (the sixth player) win?

Show hint
Total games = C(6, 2) = 15 ⇒ total wins = 15. Subtract the five known.
Show solution
  1. Total games: 15. Sum of known wins: 4 + 3 + 2 + 2 + 2 = 13.
  2. Monica: 15 − 13 = 2.
C 2 games.
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Problem 21 · 2006 AMC 8 Easy
Geometry & Measurement displacement-volume

An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm3 is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?

Show hint
Rise = displaced volume / base area = 1000 / (100 · 40).
Show solution
  1. Rise = 1000 / 4000 = 0.25 cm.
A 0.25 cm.
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Problem 22 · 2006 AMC 8 Medium
Algebra & Patterns pyramid-formula

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

Show hint (soft nudge)
Top = a + 2b + c where a, b, c are bottom-left, middle, bottom-right.
Show hint (sharpest)
Use 1, 2, 3 (min) and 7, 8, 9 (max), with the largest digit in the middle.
Show solution
  1. Top = a + 2b + c. Middle has double weight.
  2. Min: a=2, b=1, c=3 ⇒ 2 + 2 + 3 = 7.
  3. Max: a=8, b=9, c=7 ⇒ 8 + 18 + 7 = 33.
  4. Diff: 33 − 7 = 26.
D 26.
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Problem 23 · 2006 AMC 8 Medium
Number Theory chinese-remainder-by-listing

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

Show hint
List numbers ≡ 4 (mod 6) and find the first one that's also ≡ 3 (mod 5).
Show solution
  1. ≡ 4 (mod 6): 4, 10, 16, 22, 28, …
  2. ≡ 3 (mod 5): 3, 8, 13, 18, 23, 28, …
  3. First common: 28. 28 / 7 = 4 remainder 0.
A 0.
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Problem 24 · 2006 AMC 8 Hard
Number Theory factor-1010 cryptarithm

In the multiplication problem below, A, B, C, D are different digits. ABA × CD = CDCD. What is A + B?

Show hint
CDCD = CD × 101. So ABA = 101.
Show solution
  1. CDCD = CD · 101 ⇒ ABA = 101.
  2. So A = 1, B = 0; A + B = 1.
A 1.
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Problem 25 · 2006 AMC 8 Hard
Number Theory parity-of-prime

Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? (Visible sides: 44, 59, 38.)

Show hint (soft nudge)
Even-visible cards (44, 38) need an odd prime to make an odd sum; the 59 card needs an even prime to make an odd sum. Equal sums ⇒ the 59 card pairs with the only even prime, 2.
Show hint (sharpest)
Common sum = 59 + 2 = 61.
Show solution
  1. Sum behind 59 + 59 = 61 (odd) only if hidden is even ⇒ hidden = 2 (only even prime).
  2. Common sum 61. Behind 44: 61 − 44 = 17 (prime ✓). Behind 38: 61 − 38 = 23 (prime ✓).
  3. Average: (2 + 17 + 23)/3 = 42/3 = 14.
B 14.
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