AMC 8

2010 AMC 8

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

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Problem 1 · 2010 AMC 8 Easy
Arithmetic & Operations addition

At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are 11 students in Mrs. Germain's class, 8 students in Mr. Newton's class, and 9 students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest?

Show hint
Just add the three class counts.
Show solution
  1. 11 + 8 + 9 = 28.
C 28.
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Problem 2 · 2010 AMC 8 Easy
Algebra & Patterns operator-definition

If a@b = a × ba + b for a, b positive integers, then what is 5@10?

Show hint
Just substitute.
Show solution
  1. 5@10 = (5 · 10)/(5 + 10) = 50/15 = 10/3.
D 10/3.
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Problem 3 · 2010 AMC 8 Easy
Fractions, Decimals & Percents percent-increase

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?

Show hint
Percent more = (high − low) / low × 100.
Show solution
  1. Highest = 17, lowest = 10.
  2. (17 − 10) / 10 = 0.7 = 70%.
C 70%.
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Problem 4 · 2010 AMC 8 Easy
Arithmetic & Operations mean-median-mode

What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

Show hint
Sort first, then read off mode and median; compute mean from the sum.
Show solution
  1. Sorted: 0, 0, 1, 2, 3, 3, 3, 4. Sum = 16 ⇒ mean = 16/8 = 2.
  2. Median = avg of 4th and 5th = (2 + 3)/2 = 2.5.
  3. Mode = 3.
  4. Total: 2 + 2.5 + 3 = 7.5.
C 7.5.
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Problem 5 · 2010 AMC 8 Easy
Arithmetic & Operations unit-conversion

Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

Show hint
Convert everything to centimeters. Stool height = light-bulb height − (Alice + reach).
Show solution
  1. Bulb height above floor: 240 − 10 = 230 cm.
  2. Alice's reach (height + arm): 150 + 46 = 196 cm.
  3. Stool: 230 − 196 = 34 cm.
B 34 cm.
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Problem 6 · 2010 AMC 8 Easy
Geometry & Measurement lines-of-symmetry

Which of the following figures has the greatest number of lines of symmetry?

Show hint
Triangle: 3. Rhombus (non-square): 2. Rectangle (non-square): 2. Isosceles trapezoid: 1. Square: 4.
Show solution
  1. Equilateral triangle: 3 lines. Non-square rhombus: 2. Non-square rectangle: 2. Isosceles trapezoid: 1. Square: 4 lines (two diagonals + two midpoint lines).
  2. Most lines: square.
E Square (4 lines).
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Problem 7 · 2010 AMC 8 Medium
Logic & Word Problems greedy-coin

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?

Show hint (soft nudge)
Greedy: start with pennies (need 4 to cover 1–4 cents), then nickels, dimes, quarters — one of each value up to the next.
Show hint (sharpest)
After 4 pennies + 1 nickel, you can pay anything up to 9 cents. Each new coin should extend the reachable range as much as possible.
Show solution
  1. 4 pennies (cover 0–4¢). +1 nickel ⇒ reach 9¢. +1 dime ⇒ reach 19¢. +1 nickel ⇒ reach 24¢. +1 quarter ⇒ reach 49¢. +1 quarter ⇒ 74¢. +1 quarter ⇒ 99¢.
  2. Coins used: 4 + 1 + 1 + 1 + 1 + 1 + 1 = 10. (Specifically 4 pennies, 2 nickels, 1 dime, 3 quarters.)
  3. Smallest count: 10.
B 10 coins.
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Problem 8 · 2010 AMC 8 Medium
Ratios, Rates & Proportions relative-speed

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/2 mile in front of her. After she passes him, she can see him in her rear mirror until he is 1/2 mile behind her. Emily rides at a constant rate of 12 miles per hour, and Emerson skates at a constant rate of 8 miles per hour. For how many minutes can Emily see Emerson?

Show hint
Use the gap-closing speed: Emily − Emerson = 4 mph. She sees him while the gap shrinks from 1/2 ahead to 1/2 behind — a total relative shift of 1 mile.
Show solution
  1. Relative speed: 12 − 8 = 4 mph.
  2. Emily must cover 1 mile of relative displacement (from 1/2 ahead to 1/2 behind).
  3. Time: 1 / 4 hour = 15 minutes.
D 15 minutes.
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Problem 9 · 2010 AMC 8 Easy
Fractions, Decimals & Percents weighted-average

Ryan got 80% of the problems correct on a 25-problem test, 90% on a 40-problem test, and 70% on a 10-problem test. What percent of all the problems did Ryan answer correctly?

Show hint
Count correct on each test, then divide by the total number of problems.
Show solution
  1. Correct: 0.8 · 25 + 0.9 · 40 + 0.7 · 10 = 20 + 36 + 7 = 63.
  2. Total: 25 + 40 + 10 = 75.
  3. 63 / 75 = 84%.
D 84%.
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Problem 10 · 2010 AMC 8 Medium
Geometry & Measurement area-ratio-of-circles

Six pepperoni circles will exactly fit across the diameter of a 12-inch pizza when placed. If a total of 24 circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?

Show hint
Each pepperoni has diameter 12/6 = 2, so its area is π, compared to the pizza's 36π. Each pepperoni is 1/36 of the pizza.
Show solution
  1. Pepperoni radius: 1. Pizza radius: 6. Area ratio: (1/6)2 = 1/36 per pepperoni.
  2. 24 pepperonis: 24/36 = 2/3.
B 2/3.
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Problem 11 · 2010 AMC 8 Easy
Algebra & Patterns ratio-and-difference

The top of one tree is 16 feet higher than the top of another tree. The heights of the two trees are in the ratio 3 : 4. In feet, how tall is the taller tree?

Show hint
The ratio 3:4 means the difference (1 part) corresponds to 16 feet. So 1 part = 16 ft.
Show solution
  1. 4 − 3 = 1 part = 16 ft.
  2. Taller (4 parts) = 4 · 16 = 64 ft.
B 64 feet.
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Problem 12 · 2010 AMC 8 Medium
Fractions, Decimals & Percents fix-the-invariant

Of the 500 balls in a large bag, 80% are red and the rest are blue. How many of the red balls must be removed from the bag so that 75% of the remaining balls are red?

Show hint
Blue balls don't change. 75% red ⇒ 25% blue, so the 100 blue balls represent 25% of the new total.
Show solution
  1. Initial: 400 red, 100 blue.
  2. After removal, 25% blue means total = 100 / 0.25 = 400 balls.
  3. Removed: 500 − 400 = 100 red balls.
D 100 red balls.
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Problem 13 · 2010 AMC 8 Medium
Algebra & Patterns consecutive-integers percent-equation

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30% of the perimeter. What is the length of the longest side?

Show hint
Let the smallest side be s. Then perimeter = 3s + 3. Set s = 0.3 · (3s + 3).
Show solution
  1. Sides: s, s+1, s+2. Perimeter: 3s + 3.
  2. s = 0.3(3s + 3) ⇒ s = 0.9s + 0.9 ⇒ 0.1s = 0.9 ⇒ s = 9.
  3. Longest = s + 2 = 11.
E 11 inches.
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Problem 14 · 2010 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the prime factors of 2010?

Show hint
Factor out small primes first: 2010 / 2 / 3 / 5 = ?
Show solution
  1. 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67 (67 is prime).
  2. Sum: 2 + 3 + 5 + 67 = 77.
C 77.
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Problem 15 · 2010 AMC 8 Medium
Fractions, Decimals & Percents find-total-from-percent

A jar contains five different colors of gumdrops: 30% are blue, 20% are brown, 15% red, 10% yellow, and the other 30 gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

Show hint (soft nudge)
Green = 100% − 30 − 20 − 15 − 10 = 25%, so 30 gumdrops = 25% ⇒ total = 120.
Show hint (sharpest)
Brown starts at 20% · 120 = 24. Add half the blue gumdrops (which switch color).
Show solution
  1. Green % = 25 ⇒ total = 30 / 0.25 = 120.
  2. Blue count: 30% · 120 = 36. Brown count: 20% · 120 = 24.
  3. Half of blue (18) become brown: 24 + 18 = 42.
C 42.
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Problem 16 · 2010 AMC 8 Easy
Geometry & Measurement area-equation

A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Show hint
Set s2 = πr2 and solve for s/r.
Show solution
  1. s2 = πr2 ⇒ (s/r)2 = π.
  2. s/r = √π.
B √π.
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Problem 17 · 2010 AMC 8 Hard
Geometry & Measurement area-bisector similar-triangles

The diagram shows an octagon consisting of 10 unit squares. The portion below PQ is a unit square and a triangle with base 5. If PQ bisects the area of the octagon, what is the ratio XQQY?

Show hint (soft nudge)
Total area = 10, so each half = 5. Below PQ: a unit square + a triangle of base 5. So the triangle has area 4 and base 5.
Show hint (sharpest)
Triangle area = (1/2) · 5 · ZQ = 4 ⇒ ZQ = 1.6. With XY = 2, QY = ZQ − 1 = 0.6 and XQ = 2 − ZQ = 0.4.
Show solution
  1. Total area = 10, half = 5. Below PQ = unit square + triangle = 5, so the triangle has area 4 with base 5.
  2. Triangle height = 2 · 4 / 5 = 1.6. That's the height ZQ.
  3. XY spans 2 units. With Q on it at 1.6 above the base: QY = 1.6 − 1 = 0.6 and XQ = 2 − 1.6 = 0.4.
  4. Ratio XQ/QY = 0.4/0.6 = 2/3.
D 2/3.
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Problem 18 · 2010 AMC 8 Medium
Geometry & Measurement area-ratio

A decorative window is made up of a rectangle with semicircles on either end. The ratio of AD to AB is 3 : 2, and AB is 30 inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles?

Show hint (soft nudge)
Two semicircles of diameter 30 combine to a single circle of diameter 30 ⇒ area 225π.
Show hint (sharpest)
Rectangle: 30 × 45 = 1350. Form the ratio.
Show solution
  1. AD = (3/2) · 30 = 45. Rectangle area: 30 · 45 = 1350.
  2. Two semicircles of diameter 30 ⇒ one circle of radius 15: area = 225π.
  3. Ratio: 1350 : 225π = 6 : π.
C 6 : π.
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Problem 19 · 2010 AMC 8 Medium
Geometry & Measurement tangent-radius annulus-area

The two circles pictured have the same center C. Chord AD is tangent to the inner circle at B, AC is 10, and chord AD has length 16. What is the area between the two circles?

Show hint (soft nudge)
CBAD at the tangent point and B bisects AD, so AB = 8.
Show hint (sharpest)
Annulus area = π(AC2CB2) = π · AB2 by Pythagoras.
Show solution
  1. Tangent ⇒ CBAD at B, so B is the midpoint of AD: AB = 8.
  2. Annulus area: πAC2 − πCB2 = π(AC2CB2) = π · AB2 = 64π.
C 64π.
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Problem 20 · 2010 AMC 8 Hard
Counting & Probability inclusion-exclusion divisibility

In a room, 2/5 of the people are wearing gloves, and 3/4 of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?

Show hint (soft nudge)
Total people must be divisible by 5 and 4 ⇒ multiple of 20. Smallest is 20.
Show hint (sharpest)
Inclusion-exclusion gives the minimum overlap: gloves + hats − 1 (cap).
Show solution
  1. Total people: multiple of 20. Take 20 (the smallest).
  2. Gloves: 2/5 · 20 = 8. Hats: 3/4 · 20 = 15.
  3. Min(both) = max(0, gloves + hats − total) = max(0, 8 + 15 − 20) = 3.
A 3.
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Problem 21 · 2010 AMC 8 Hard
Algebra & Patterns work-backwards

Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read 1/5 of the pages plus 12 more, and on the second day she read 1/4 of the remaining pages plus 15 pages. On the third day she read 1/3 of the remaining pages plus 18 pages. She then realized that there were only 62 pages left to read, which she read the next day. How many pages are in this book?

Show hint (soft nudge)
Work backwards from the 62 pages left. Day 3 left her with 62 pages after reading (1/3)R + 18 from R; so 2R/3 − 18 = 62.
Show hint (sharpest)
Repeat the inversion for day 2 and day 1.
Show solution
  1. After day 3 there are 62 pages left. If R3 = pages at start of day 3: (2/3)R3 − 18 = 62 ⇒ R3 = 120.
  2. Start of day 2: (3/4)R2 − 15 = 120 ⇒ R2 = 180.
  3. Start of day 1: (4/5)R1 − 12 = 180 ⇒ R1 = 240.
C 240 pages.
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Problem 22 · 2010 AMC 8 Medium
Algebra & Patterns place-value-difference

The hundreds digit of a three-digit number is 2 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

Show hint (soft nudge)
Let units = u, tens = t, hundreds = u + 2. Original − reversed simplifies to a constant.
Show hint (sharpest)
Reversing flips the hundreds and units digits. Difference = 99 · (hundreds − units).
Show solution
  1. Original − reversed = 100(u+2) + 10t + u − (100u + 10t + u+2) = 99(u+2) − 99u = 198.
  2. Units digit of 198 = 8.
E 8.
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Problem 23 · 2010 AMC 8 Hard
Geometry & Measurement circle-radii pythagorean

Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

Show hint (soft nudge)
Each small semicircle has diameter PQ (or RS) = 2; radius 1.
Show hint (sharpest)
The big circle has radius OQ = √2 (Pythagoras on (1,1)). Compute both areas and divide.
Show solution
  1. Small semicircles: radius 1, each area π/2; combined π.
  2. Big circle: radius √(12+12) = √2, area 2π.
  3. Ratio: π / (2π) = 1/2.
B 1/2.
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Problem 24 · 2010 AMC 8 Medium
Algebra & Patterns match-bases-or-exponents

What is the correct ordering of the three numbers, 108, 512, and 224?

Show hint (soft nudge)
Rewrite each as a product of equal-exponent powers. E.g., 108 = 28 · 58, 224 = 28 · 48, 512 = 58 · 54.
Show hint (sharpest)
Compare pairwise using shared factors.
Show solution
  1. Compare 224 vs 108: 224 = 28 · 48, 108 = 28 · 58. Since 4 < 5: 224 < 108.
  2. Compare 108 vs 512: 108 = 44 · 58, 512 = 54 · 58. Since 44 = 256 < 625 = 54: 108 < 512.
  3. Therefore 224 < 108 < 512.
A 2^24 &lt; 10^8 &lt; 5^12.
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Problem 25 · 2010 AMC 8 Hard
Counting & Probability recurrence compositions

Everyday at school, Jo climbs a flight of 6 stairs. Jo can take the stairs 1, 2, or 3 at a time. For example, Jo could climb 3, then 1, then 2. In how many ways can Jo climb the stairs?

Show hint (soft nudge)
Let f(n) be the number of ways to climb n stairs. Each climb ends in a 1, 2, or 3 step: f(n) = f(n−1) + f(n−2) + f(n−3).
Show hint (sharpest)
Start with f(1)=1, f(2)=2, f(3)=4.
Show solution
  1. f(1) = 1, f(2) = 2, f(3) = 4.
  2. f(4) = 1 + 2 + 4 = 7.
  3. f(5) = 2 + 4 + 7 = 13.
  4. f(6) = 4 + 7 + 13 = 24.
E 24 ways.
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