AMC 8

2011 AMC 8

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

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Problem 1 · 2011 AMC 8 Easy
Arithmetic & Operations subtraction

Margie bought 3 apples at a cost of 50 cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?

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Answer: E — $3.50.
Show hint
Hint 1
Total cost = 3 × $0.50; change = $5 − that.
Show solution
Approach: compute then subtract
  1. Cost: 3 · $0.50 = $1.50.
  2. Change: $5.00 − $1.50 = $3.50.
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Problem 2 · 2011 AMC 8 Easy
Geometry & Measurement area-comparison

Karl's rectangular vegetable garden is 20 feet by 45 feet, and Makenna's is 25 feet by 40 feet. Which of the following statements are true?

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Answer: E — Makenna's garden is larger by 100 square feet.
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Hint 1
Compute each area; subtract.
Show solution
Approach: multiply and subtract
  1. Karl: 20 × 45 = 900 sq ft.
  2. Makenna: 25 × 40 = 1000 sq ft.
  3. Difference: 1000 − 900 = 100 ⇒ Makenna's is larger by 100 sq ft.
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Problem 3 · 2011 AMC 8 Easy
Geometry & Measurement border-count
amc8-2011-03
Show answer
Answer: D — 32 : 17.
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Hint 1
Border around an n × n square has 4n + 4 tiles. White count is unchanged.
Show solution
Approach: count the new border tiles
  1. Original is 5 × 5 = 25 tiles (8 black + 17 white). New 7 × 7 has 49 tiles.
  2. Added border tiles: 49 − 25 = 24, all black. New black total: 8 + 24 = 32. White still 17.
  3. Ratio: 32 : 17.
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Problem 4 · 2011 AMC 8 Easy
Arithmetic & Operations mean-median-mode

Here is a list of the numbers of fish that Tyler caught in nine outings last summer: 2, 0, 1, 3, 0, 3, 3, 1, 2. Which statement about the mean, median, and mode is true?

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Answer: C — mean < median < mode.
Show hint
Hint 1
Sort the list. Mode = most common value, median = middle (5th of 9), mean = sum/9.
Show solution
Approach: compute each
  1. Sorted: 0, 0, 1, 1, 2, 2, 3, 3, 3.
  2. Mode = 3, median = 2, mean = 15/9 = 5/3 ≈ 1.67.
  3. 5/3 < 2 < 3 ⇒ mean < median < mode.
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Problem 5 · 2011 AMC 8 Easy
Arithmetic & Operations time-conversion

What time was it 2011 minutes after midnight on January 1, 2011?

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Answer: D — January 2 at 9:31 AM.
Show hint
Hint 1
Divide 2011 by 60 to convert to hours and minutes. Then subtract 24 hours to advance to the next day.
Show solution
Approach: convert minutes to hours, modular over 24
  1. 2011 / 60 = 33 hr 31 min (since 33 · 60 = 1980 and 2011 − 1980 = 31).
  2. 33 hr = 24 hr + 9 hr ⇒ one day later, 9 hr 31 min after midnight = 9:31 AM on January 2.
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Problem 6 · 2011 AMC 8 Easy
Counting & Probability complementary-counting

In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own motorcycles, how many of the car owners do not own a motorcycle?

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Answer: D — 306.
Show hint
Hint 1
Everyone owns at least one. So car-only count = total − motorcycle-owners.
Show solution
Approach: everyone has at least one ⇒ car-only = total − motorcycle-owners
  1. Each non-motorcycle-owner must own a car (since every adult has at least one).
  2. Car-only = 351 − 45 = 306.
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Problem 7 · 2011 AMC 8 Easy
Geometry & Measurement fraction-of-area
amc8-2011-07
Show answer
Answer: C — 25%.
Show hint
Hint 1
Compute each square's shaded fraction; add; divide by 4 (since there are 4 squares of equal area).
Show solution
Approach: sum fractions, divide by number of squares
  1. Shaded fractions of the four squares: 1/4, 1/8, 3/8, 1/4.
  2. Sum: 1/4 + 1/8 + 3/8 + 1/4 = 2/8 + 1/8 + 3/8 + 2/8 = 8/8 = 1.
  3. Of 4 total squares: 1 / 4 = 25%.
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Problem 8 · 2011 AMC 8 Easy
Counting & Probability enumerate-sums

Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?

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Answer: B — 5 different values.
Show hint
Hint 1
All sums are odd + even = odd. List the 9 sums and count distinct values.
Show solution
Approach: list sums, dedupe
  1. Possible sums: 1+2, 1+4, 1+6, 3+2, 3+4, 3+6, 5+2, 5+4, 5+6 = 3, 5, 7, 5, 7, 9, 7, 9, 11.
  2. Distinct: {3, 5, 7, 9, 11} ⇒ 5 values.
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Problem 9 · 2011 AMC 8 Easy
Ratios, Rates & Proportions average-speed
amc8-2011-09
Show answer
Answer: E — 5 mph.
Show hint
Hint 1
Average speed = total distance / total time. The graph's endpoints give you both.
Show solution
Approach: total miles / total hours
  1. Total: 35 miles in 7 hours.
  2. Average speed: 35 / 7 = 5 mph.
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Problem 10 · 2011 AMC 8 Medium
Fractions, Decimals & Percents piecewise-rate

The taxi fare in Gotham City is $2.40 for the first 12 mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?

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Answer: C — 3.3 miles.
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Hint 1
Subtract the $2 tip from $10. Then subtract the $2.40 flag-drop for the first half-mile. Convert the rest at $0.20 per 0.1 mile = $2 per mile.
Show solution
Approach: peel off fixed costs, then divide the remainder
  1. Available for fare: $10 − $2 = $8.
  2. After the first 1/2 mile costing $2.40: $8 − $2.40 = $5.60 left.
  3. Additional rate: $0.20 / 0.1 mile = $2 per mile. So $5.60 buys 2.80 miles.
  4. Total: 0.5 + 2.80 = 3.3 miles.
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Problem 11 · 2011 AMC 8 Easy
Arithmetic & Operations average-of-differences
amc8-2011-11
Show answer
Answer: A — 6 minutes.
Show hint
Hint 1
Add the daily differences (Sasha − Asha) and divide by 5.
Show solution
Approach: sum of differences / 5
  1. Daily Sasha − Asha: +10, −10, +20, +30, −20. Sum: 30.
  2. Average over 5 days: 30 / 5 = 6.
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Problem 12 · 2011 AMC 8 Easy
Counting & Probability fix-one-position

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

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Answer: B — 1/3.
Show hint
Hint 1
Fix Angie's seat. Carlos lands in any of the 3 remaining seats with equal probability; only 1 is opposite.
Show solution
Approach: fix one seat, count Carlos's options
  1. Fix Angie in any seat. Carlos has 3 equally likely seats among the remaining 3.
  2. Exactly 1 is opposite Angie ⇒ probability 1/3.
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Problem 13 · 2011 AMC 8 Medium
Geometry & Measurement overlap-regionarea-ratio
amc8-2011-13
Show answer
Answer: C — 20%.
Show hints
Hint 1 of 2
Together the two squares cover area 2 · 152, but the rectangle is only 25 · 15 — the overlap is counted twice.
Still stuck? Show hint 2 →
Hint 2 of 2
Overlap area = (sum of both squares) − (rectangle).
Show solution
Approach: overlap by double-counting
  1. Each square has area 225, total 450. Rectangle area = 25 × 15 = 375.
  2. Overlap = 450 − 375 = 75.
  3. Fraction of rectangle: 75 / 375 = 1/5 = 20%.
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Problem 14 · 2011 AMC 8 Medium
Fractions, Decimals & Percents ratio-totals

There are 270 students at Colfax Middle School, where the ratio of boys to girls is 5 : 4. There are 180 students at Winthrop Middle School, where the ratio of boys to girls is 4 : 5. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?

Show answer
Answer: C — 22/45.
Show hint
Hint 1
Each ratio uses 9 parts. Compute girls at each school, then total girls / total students.
Show solution
Approach: ratio-of-9 parts per school
  1. Colfax: girls = (4/9)(270) = 120.
  2. Winthrop: girls = (5/9)(180) = 100.
  3. Total girls: 220 of 450 students ⇒ 22/45.
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Problem 15 · 2011 AMC 8 Easy
Number Theory exponent-rewrite

How many digits are in the product 45 · 510?

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Answer: D — 11 digits.
Show hint
Hint 1
Rewrite 45 as 210. Then 210 · 510 = 1010.
Show solution
Approach: rewrite into 10n
  1. 45 · 510 = (22)5 · 510 = 210 · 510 = 1010.
  2. 1010 = 1 followed by ten zeros ⇒ 11 digits.
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Problem 16 · 2011 AMC 8 Medium
Geometry & Measurement isosceles-altitudepythagorean-triple

Let A be the area of the triangle with sides of length 25, 25, and 30. Let B be the area of the triangle with sides of length 25, 25, and 40. What is the relationship between A and B?

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Answer: C — A = B.
Show hints
Hint 1 of 2
Each triangle is isosceles. Drop the altitude to the unequal side and use the Pythagorean theorem.
Still stuck? Show hint 2 →
Hint 2 of 2
Look for 15-20-25 (3-4-5 scaled) in both pictures.
Show solution
Approach: drop altitudes, both reveal a 15-20-25 triangle
  1. Triangle with base 30: half-base 15, hypotenuse 25 ⇒ height = √(252 − 152) = 20. Area A = (1/2)(30)(20) = 300.
  2. Triangle with base 40: half-base 20, hypotenuse 25 ⇒ height = √(252 − 202) = 15. Area B = (1/2)(40)(15) = 300.
  3. A = B.
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Problem 17 · 2011 AMC 8 Medium
Number Theory prime-factorization

Let w, x, y, and z be whole numbers. If 2w · 3x · 5y · 7z = 588, then what does 2w + 3x + 5y + 7z equal?

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Answer: A — 21.
Show hint
Hint 1
Factor 588 into primes: 588 = 4 · 147 = 4 · 3 · 49.
Show solution
Approach: prime factorization
  1. 588 = 22 · 31 · 50 · 72w = 2, x = 1, y = 0, z = 2.
  2. 2w + 3x + 5y + 7z = 4 + 3 + 0 + 14 = 21.
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Problem 18 · 2011 AMC 8 Medium
Counting & Probability symmetryprobability

A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

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Answer: D — 7/12.
Show hint
Hint 1
P(first > second) = P(second > first) by symmetry. P(equal) = 6/36 = 1/6.
Show solution
Approach: split into 'equal' and 'greater'
  1. P(first = second) = 6/36 = 1/6.
  2. By symmetry, P(first > second) = P(second > first) = (1 − 1/6)/2 = 5/12.
  3. P(first ≥ second) = 1/6 + 5/12 = 2/12 + 5/12 = 7/12.
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Problem 19 · 2011 AMC 8 Medium
Counting & Probability systematic-counting
amc8-2011-19
Show answer
Answer: D — 11 rectangles.
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Hint 1 of 2
The three big rectangles drawn count, plus every rectangle assembled from the small pieces created where they overlap.
Still stuck? Show hint 2 →
Hint 2 of 2
Three originals; then look for every small piece (alone or glued to a neighbour) that is itself a rectangle.
Show solution
Approach: count originals, atoms, and rectangular unions
  1. The three drawn rectangles themselves contribute 3.
  2. The overlapping lines cut the figure into smaller atomic rectangles. The center is where all three overlap. Each individual atom is a rectangle (there are several), and pairs of adjacent atoms that share a full side form another rectangle.
  3. Adding up all distinct rectangles — the three originals plus every smaller rectangle formed by the cuts — gives 11.
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Problem 20 · 2011 AMC 8 Medium
Geometry & Measurement trapezoid-areadrop-altitudes
amc8-2011-20
Show answer
Answer: D — 750.
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Hint 1 of 2
Drop altitudes from A and B to DC. Each leg becomes the hypotenuse of a right triangle of height 12.
Still stuck? Show hint 2 →
Hint 2 of 2
Use 9-12-15 and 12-16-20 (3-4-5 scaled) to find the base extensions.
Show solution
Approach: drop perpendiculars to find the longer base
  1. Drop altitudes from A and B. On the left: 15-12-? right triangle ⇒ horizontal piece 9. On the right: 20-12-? ⇒ horizontal piece 16.
  2. Longer base DC = 9 + 50 + 16 = 75.
  3. Area = (1/2)(50 + 75)(12) = (1/2)(125)(12) = 750.
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Problem 21 · 2011 AMC 8 Medium
Logic & Word Problems constraint-satisfaction

Students guess that Norb's age is 24, 28, 30, 32, 36, 38, 41, 44, 47, and 49. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

Show answer
Answer: C — 37.
Show hints
Hint 1 of 2
"At least half too low" with 10 guesses means age > 5th-smallest guess = 36, so age ≥ 37.
Still stuck? Show hint 2 →
Hint 2 of 2
"Two off by one" means age is squeezed between two guesses that differ by 2. The only such pair above 36 is 36 and 38, or 47 and 49.
Show solution
Approach: apply each clue in turn
  1. Sorted guesses: 24, 28, 30, 32, 36, 38, 41, 44, 47, 49. "At least half too low" ⇒ age > 36.
  2. "Two are off by one" ⇒ age sits between two guesses 2 apart. Candidates: 37 (between 36 and 38) or 48 (between 47 and 49).
  3. Age is prime ⇒ 37 (prime) wins; 48 isn't prime.
  4. Norb is 37.
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Problem 22 · 2011 AMC 8 Hard
Number Theory tens-digit-cyclemod-100

What is the tens digit of 72011?

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Answer: D — Tens digit 4.
Show hints
Hint 1 of 2
Tens digit depends only on the value mod 100. Compute 7k mod 100 for small k and look for a cycle.
Still stuck? Show hint 2 →
Hint 2 of 2
71 = 07, 72 = 49, 73 = 343 (43), 74 = 2401 (01). Cycle length 4.
Show solution
Approach: powers of 7 mod 100 cycle in 4
  1. Last two digits cycle: 07, 49, 43, 01, repeating with period 4.
  2. 2011 = 4 · 502 + 3 ⇒ 72011 ≡ 73 ≡ 43 (mod 100).
  3. Tens digit = 4.
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Problem 23 · 2011 AMC 8 Hard
Counting & Probability caseworkpermutations-with-restrictions

How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

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Answer: D — 84.
Show hints
Hint 1 of 2
All digits come from {0, 1, 2, 3, 4, 5}, with 5 present (it's the largest). Units digit is 0 or 5 (divisible by 5).
Still stuck? Show hint 2 →
Hint 2 of 2
Split into two cases by units digit.
Show solution
Approach: casework on the units digit
  1. Case A: units = 0. The remaining three slots contain 5 and two distinct digits chosen from {1, 2, 3, 4}: C(4, 2) = 6 ways to pick the other two; 3! = 6 ways to arrange them. Subtotal: 6 × 6 = 36.
  2. Case B: units = 5. The remaining three slots use three distinct digits from {0, 1, 2, 3, 4}. Choose and arrange: 5 · 4 · 3 = 60. Subtract leading-zero arrangements: 4 · 3 = 12 with 0 first. Subtotal: 60 − 12 = 48.
  3. Total: 36 + 48 = 84.
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Problem 24 · 2011 AMC 8 Medium
Number Theory parityprimes

In how many ways can 10001 be written as the sum of two primes?

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Answer: A — 0 ways.
Show hint
Hint 1
Odd sum ⇒ one prime must be even ⇒ that prime is 2. Then check 10001 − 2 = 9999.
Show solution
Approach: force one prime to be 2 and check primality of 9999
  1. 10001 is odd; primes summing to odd require one to be even, so it must be 2.
  2. Then the other is 10001 − 2 = 9999 = 3 · 3333 (digit sum 36 divisible by 3). Not prime.
  3. So 0 representations.
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Problem 25 · 2011 AMC 8 Hard
Geometry & Measurement inscribed-circumscribedapproximation
amc8-2011-25
Show answer
Answer: A — Closest to 1/2.
Show hints
Hint 1 of 2
Inner square has diagonal = diameter = 2, so its side is √2 and area is 2.
Still stuck? Show hint 2 →
Hint 2 of 2
Shaded area inside the circle = π · 12 − 2 = π − 2. Area between the two squares = outer area − inner area = 4 − 2 = 2.
Show solution
Approach: compute the two areas and form the ratio
  1. Outer square: side 2, area 4. Inner square: diagonal = 2 (diameter), so side √2 and area 2.
  2. Shaded (inside circle, outside inner square): π(1)2 − 2 = π − 2.
  3. Between the squares: 4 − 2 = 2.
  4. Ratio: (π − 2)/2 ≈ (3.14 − 2)/2 ≈ 0.57. Closest to 1/2.
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