AMC 8

2008 AMC 8

25 problems — read each, give it a real try, then peek at the hints.

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Problem 1 · 2008 AMC 8 Easy
Arithmetic & Operations word-problem

Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?

Show hint
Rides: 2 · 12 = 24. Subtract food + rides from 50.
Show solution
  1. Total spent: 12 + 2 · 12 = 36.
  2. Left: 50 − 36 = 14.
B $14.
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Problem 2 · 2008 AMC 8 Easy
Logic & Word Problems letter-position-decode

The ten-letter code "BEST OF LUCK" represents the ten digits 0–9, in order. What 4-digit number is represented by the code word "CLUE"?

Show hint
Position 1 = digit 0, position 2 = digit 1, …, position 10 = digit 9. Look up each letter.
Show solution
  1. B=0, E=1, S=2, T=3, O=4, F=5, L=6, U=7, C=8, K=9.
  2. CLUE = 8671.
A 8671.
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Problem 3 · 2008 AMC 8 Easy
Arithmetic & Operations modular-days

If February is a month that contains Friday the 13th, what day of the week is February 1?

Show hint
Days of the week repeat every 7. Step back from Friday Feb 13 by 12 days.
Show solution
  1. Feb 13 is Friday. 12 days earlier is Feb 1. 12 mod 7 = 5, so 5 days before Friday.
  2. Going back: Thu, Wed, Tue, Mon, Sun.
A Sunday.
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Problem 4 · 2008 AMC 8 Easy
Geometry & Measurement area-decomposition

In the figure, the outer equilateral triangle has area 16, the inner equilateral triangle has area 1, and the three trapezoids are congruent. What is the area of one of the trapezoids?

Show hint
Trapezoids share 16 − 1 = 15 of area; they're congruent.
Show solution
  1. Trapezoid combined area: 16 − 1 = 15.
  2. Each: 15 / 3 = 5.
C 5.
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Problem 5 · 2008 AMC 8 Easy
Ratios, Rates & Proportions average-speed

Barney Schwinn notices that the odometer on his bicycle reads 1441, a palindrome, because it reads the same forward and backward. After riding 4 more hours that day and 6 the next, he notices that the odometer shows another palindrome, 1661. What was his average speed in miles per hour?

Show hint
Distance = 1661 − 1441 = 220 over 10 hours.
Show solution
  1. Distance: 220 miles. Time: 4 + 6 = 10 hours.
  2. Average: 220 / 10 = 22 mph.
E 22 mph.
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Problem 6 · 2008 AMC 8 Easy
Geometry & Measurement unit-area-count

In the figure, what is the ratio of the area of the gray squares to the area of the white squares?

Show hint
Count gray and white unit squares. 16 total.
Show solution
  1. Gray: 6, white: 10.
  2. Ratio: 6 : 10 = 3 : 5.
D 3 : 5.
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Problem 7 · 2008 AMC 8 Easy
Fractions, Decimals & Percents equivalent-fractions

If 35 = M45 = 60N, what is M + N?

Show hint
Cross-multiply each equation.
Show solution
  1. 3 · 45 = 5MM = 27.
  2. 3N = 5 · 60 ⇒ N = 100.
  3. Sum: 27 + 100 = 127.
E 127.
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Problem 8 · 2008 AMC 8 Easy
Arithmetic & Operations average

Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?

Show hint
Sum the four months, divide by 4.
Show solution
  1. Sum: 100 + 60 + 40 + 120 = 320.
  2. Average: 320 / 4 = 80.
D $80.
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Problem 9 · 2008 AMC 8 Easy
Fractions, Decimals & Percents successive-percentages

In 2005 Tycoon Tammy invested 100 dollars for two years. During the first year her investment suffered a 15% loss, but during the second year the remaining investment showed a 20% gain. Over the two-year period, what was the change in Tammy's investment?

Show hint
Multiplier: 0.85 × 1.20. Compare to 1.
Show solution
  1. Net multiplier: 0.85 · 1.20 = 1.02 ⇒ 2% gain.
D 2% gain.
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Problem 10 · 2008 AMC 8 Easy
Arithmetic & Operations weighted-average

The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?

Show hint
Combined average = total ages / total people. Each room's total = avg × count.
Show solution
  1. Room A total: 6 · 40 = 240. Room B: 4 · 25 = 100.
  2. Combined: 340 / 10 = 34.
D 34.
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Problem 11 · 2008 AMC 8 Easy
Counting & Probability inclusion-exclusion

Each of the 39 students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and 26 students have a cat. How many students have both a dog and a cat?

Show hint
|A ∪ B| = |A| + |B| − |A ∩ B|. Everyone has at least one, so |A ∪ B| = 39.
Show solution
  1. 20 + 26 = 46 = 39 + (both).
  2. Both = 46 − 39 = 7.
A 7.
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Problem 12 · 2008 AMC 8 Medium
Algebra & Patterns geometric-sequence

A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 23 of the height it reached in the previous bounce. On which bounce will it rise to a height less than 0.5 meters?

Show hint
After the nth bounce, height = 3 · (2/3)n. Test small n until it drops below 1/2.
Show solution
  1. After bounce 4: 3 · (2/3)4 = 16/27 ≈ 0.593 (above 0.5).
  2. After bounce 5: 3 · (2/3)5 = 32/81 ≈ 0.395 (below 0.5).
  3. Answer: 5.
C 5th bounce.
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Problem 13 · 2008 AMC 8 Medium
Algebra & Patterns system-of-equations

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than 100 pounds or more than 150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122, 125 and 127 pounds. What is the combined weight in pounds of the three boxes?

Show hint
Each box is in 2 of the 3 pair-sums. Adding all three pair-sums double-counts each weight.
Show solution
  1. Sum of pair-sums: 122 + 125 + 127 = 374 = 2(a + b + c).
  2. Total: 374 / 2 = 187.
C 187 pounds.
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Problem 14 · 2008 AMC 8 Medium
Counting & Probability latin-square

Three A's, three B's, and three C's are placed in the nine spaces so that each row and column contain one of each letter. If A is placed in the upper left corner, how many arrangements are possible?

Show hint
Place B in the second row (2 choices for column) and then in the third row (constrained). C is then forced.
Show solution
  1. Row 1 is fixed up to permutation of B, C (2 ways). Row 2 starts with B or C (2 choices), then is determined. Each row 2 case constrains row 3 to one arrangement.
  2. Total: 2 · 2 = 4.
C 4 arrangements.
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Problem 15 · 2008 AMC 8 Medium
Number Theory divisibility average-as-integer

In Theresa's first 8 basketball games, she scored 7, 4, 3, 6, 8, 3, 1 and 5 points. In her ninth game, she scored fewer than 10 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 10 points and her points-per-game average for the 10 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

Show hint (soft nudge)
Sum of first 8: 37. After game 9 (score < 10), total is between 37 and 47; must be a multiple of 9 (mean integer).
Show hint (sharpest)
Then after game 10 the total is < 56 and a multiple of 10.
Show solution
  1. Sum after 8: 37.
  2. After 9: total in [38, 47], divisible by 9 ⇒ 45. Game 9 = 8.
  3. After 10: total in [46, 55], divisible by 10 ⇒ 50. Game 10 = 5.
  4. Product: 8 · 5 = 40.
B 40.
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Problem 16 · 2008 AMC 8 Medium
Geometry & Measurement volume-surface-3d

A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?

Show hint
Volume: 7 unit cubes ⇒ 7. Surface: the 6 outer cubes each have 5 faces exposed; the center cube is fully hidden.
Show solution
  1. Volume: 7 cubic units.
  2. Each outer cube: 5 exposed faces ⇒ 6 · 5 = 30. Center cube: all faces touched, 0 exposed.
  3. Ratio: 7 : 30.
D 7 : 30.
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Problem 17 · 2008 AMC 8 Medium
Algebra & Patterns fixed-perimeter-area max-min

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 50 units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

Show hint
l + w = 25, both positive integers. Area = l(25 − l); max near l = 12 or 13, min at l = 1.
Show solution
  1. Largest: l = 12, w = 13 ⇒ area 156.
  2. Smallest: l = 1, w = 24 ⇒ area 24.
  3. Difference: 156 − 24 = 132.
D 132.
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Problem 18 · 2008 AMC 8 Medium
Geometry & Measurement arc-length path-decomposition

Two circles that share the same center have radii 10 meters and 20 meters. An aardvark runs along the path shown, starting at A and ending at K. How many meters does the aardvark run?

Show hint
Decompose the path: quarter-arcs of the big circle, quarter-arcs of the small circle, the diameter of the small circle, and radial segments between the two circles.
Show solution
  1. Big-circle quarter-arc (one piece): (1/4)(2π)(20) = 10π. Two such arcs ⇒ 20π.
  2. Small-circle quarter-arc (one piece): (1/4)(2π)(10) = 5π. Two such arcs ⇒ 10π... but the path uses just enough small arcs and a small-circle diameter to traverse the inner ring.
  3. Per the figure: 2 big-quarter arcs (total 20π) + 2 radial segments of length 10 (total 20) + a diameter of the small circle of length 20.
  4. Total: 20π + 20 + 20 = 20π + 40.
E 20&pi; + 40.
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Problem 19 · 2008 AMC 8 Medium
Counting & Probability symmetric-counting

Eight points are spaced around at intervals of one unit around a 2 × 2 square, as shown. Two of the 8 points are chosen at random. What is the probability that the two points are one unit apart?

Show hint
Each point has exactly 2 neighbors at distance 1 (its left and right neighbors on the perimeter).
Show solution
  1. Fix one point. Of the remaining 7, exactly 2 are 1 unit away.
  2. Probability: 2/7 = 2/7.
B 2/7.
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Problem 20 · 2008 AMC 8 Medium
Algebra & Patterns common-numerator smallest-integer-total

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 3/4 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?

Show hint (soft nudge)
Let p be the common count passing. Boys = (3/2)p, girls = (4/3)p; total = (17/6)p.
Show hint (sharpest)
Total must be a positive integer; smallest p making it integer is p = 6.
Show solution
  1. Boys = (3/2)p, girls = (4/3)p. Total = (3/2 + 4/3)p = (17/6)p.
  2. Smallest positive integer total requires p = 6 ⇒ total = 17.
B 17.
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Problem 21 · 2008 AMC 8 Medium
Geometry & Measurement cylinder-volume pi-approximation

Jerry cuts a wedge from a 6-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?

Show hint
The wedge is exactly half the cylinder. Cylinder volume = πr2h.
Show solution
  1. Cylinder: radius 4, height 6. Volume = π · 16 · 6 = 96π.
  2. Wedge: half ⇒ 48π ≈ 48 · 3.14 = 150.8 ≈ 151.
C About 151.
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Problem 22 · 2008 AMC 8 Medium
Number Theory range-of-integers

For how many positive integer values of n are both n3 and 3n three-digit whole numbers?

Show hint
Let x = n/3 (so n = 3x). Then 3n = 9x. Both x and 9x are 3-digit.
Show solution
  1. 100 ≤ x ≤ 999 and 100 ≤ 9x ≤ 999.
  2. The second is the binding constraint: x ≤ 111.
  3. Combined with x ≥ 100: x ∈ {100, 101, …, 111} — 12 values.
A 12.
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Problem 23 · 2008 AMC 8 Hard
Geometry & Measurement coordinate-bash shoelace

In square ABCE, AF = 2FE and CD = 2DE. What is the ratio of the area of ▵BFD to the area of square ABCE?

Show hint (soft nudge)
Set up coordinates with side s; A=(0,s), B=(s,s), C=(s,0), E=(0,0).
Show hint (sharpest)
F is on AE with AF/FE = 2 ⇒ F = (0, s/3). D is on CE with CD/DE = 2 ⇒ D = (s/3, 0).
Show solution
  1. B=(s,s), F=(0,s/3), D=(s/3,0).
  2. Area ▵BFD = (1/2) |s(s/3 − 0) + 0(0 − s) + (s/3)(ss/3)| = (1/2)(s2/3 + 2s2/9) = 5s2/18.
  3. Ratio: 5/18 = 5/18.
C 5/18.
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Problem 24 · 2008 AMC 8 Hard
Counting & Probability case-on-die-roll

Ten tiles numbered 1 through 10 are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?

Show hint
Case on the die roll d; for each, count tiles t in 1–10 with dt a perfect square.
Show solution
  1. d = 1: t ∈ {1, 4, 9} ⇒ 3.
  2. d = 2: t = 2, 8 ⇒ 2.
  3. d = 3: t = 3 ⇒ 1.
  4. d = 4: t = 1, 4, 9 ⇒ 3.
  5. d = 5: t = 5 ⇒ 1.
  6. d = 6: t = 6 ⇒ 1.
  7. Total: 11 successes / 60 outcomes = 11/60.
C 11/60.
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Problem 25 · 2008 AMC 8 Hard
Geometry & Measurement annulus-area

Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?

Show hint
Black regions: disk of radius 2, annulus 4–6, annulus 8–10. Use area = π(R2r2) for each annulus.
Show solution
  1. Black areas: π(22) + π(62 − 42) + π(102 − 82) = 4π + 20π + 36π = 60π.
  2. Total: π(12)2 = 144π.
  3. Fraction: 60/144 = 5/12 ≈ 41.7% ⇒ closest to 42%.
A About 42%.
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