AMC 8

2012 AMC 8

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

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Problem 1 · 2012 AMC 8 Easy
Ratios, Rates & Proportions proportion

Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighborhood picnic?

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Answer: E — 9 pounds.
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Hint 1
24 hamburgers is 3 times as many as 8, so the meat triples.
Show solution
Approach: scale by 3
  1. 24 / 8 = 3, so triple the meat: 3 × 3 = 9 pounds.
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Problem 2 · 2012 AMC 8 Easy
Ratios, Rates & Proportions unit-rate

In the country of East Westmore, statisticians estimate there is a baby born every 8 hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?

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Answer: B — About 700.
Show hint
Hint 1
Births per day = 24/8 = 3. Net growth per day = 3 − 1 = 2. Multiply by 365.
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Approach: net per day × 365
  1. Births per day: 24/8 = 3. Net daily growth: 3 − 1 = 2.
  2. Yearly: 2 × 365 = 730 ≈ 700.
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Problem 3 · 2012 AMC 8 Easy
Arithmetic & Operations time-arithmetic

On February 13 The Oshkosh Northwester listed the length of daylight as 10 hours and 24 minutes, the sunrise was 6:57 AM, and the sunset as 8:15 PM. The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?

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Answer: B — 5:21 PM.
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Hint 1
Sunset = sunrise + length of daylight.
Show solution
Approach: add the daylight length to sunrise
  1. 6:57 AM + 10 hours = 4:57 PM.
  2. Add the remaining 24 minutes: 4:57 + 0:24 = 5:21 PM.
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Problem 4 · 2012 AMC 8 Easy
Fractions, Decimals & Percents fraction-of-whole

Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?

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Answer: C — 1/8.
Show hint
Hint 1
Peter ate 1 full slice + half a slice = 1.5 slices out of 12.
Show solution
Approach: slices eaten / total slices
  1. Peter's slices: 1 + 1/2 = 1.5 = 3/2.
  2. Fraction: (3/2) / 12 = 3/24 = 1/8.
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Problem 5 · 2012 AMC 8 Easy
Geometry & Measurement match-total-heights
amc8-2012-05
Show answer
Answer: E — X = 5.
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Hint 1
Sum of vertical segments along the left side must equal the sum along the right (same overall height).
Show solution
Approach: total height left = total height right
  1. Right side total: 1 + 2 + 1 + 6 = 10.
  2. Left side total: 1 + 1 + 1 + 2 + X = 5 + X.
  3. 5 + X = 10 ⇒ X = 5.
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Problem 6 · 2012 AMC 8 Easy
Geometry & Measurement outer-minus-inner

A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches?

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Answer: E — 88 square inches.
Show hint
Hint 1
Outer dimensions add 2 inches on each of two sides ⇒ 4 inches in each direction. Subtract the photo area.
Show solution
Approach: outer area minus photo area
  1. Outer: (8 + 4) × (10 + 4) = 12 × 14 = 168.
  2. Photo: 8 × 10 = 80.
  3. Border: 168 − 80 = 88.
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Problem 7 · 2012 AMC 8 Easy
Arithmetic & Operations average-budgetmin-with-max-on-other

Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?

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Answer: B — 92.
Show hints
Hint 1 of 2
Total points needed = 4 × 95 = 380. Subtract the first two scores to find what's left for tests 3 + 4.
Still stuck? Show hint 2 →
Hint 2 of 2
To minimize the 3rd test, maximize the 4th (cap = 100).
Show solution
Approach: budget the remaining two test scores
  1. Needed total: 4 · 95 = 380. Used: 97 + 91 = 188. Remaining: 380 − 188 = 192.
  2. Max possible on the 4th test = 100, so min on the 3rd = 192 − 100 = 92.
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Problem 8 · 2012 AMC 8 Easy
Fractions, Decimals & Percents successive-percentages

A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?

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Answer: D — 60%.
Show hint
Hint 1
Multiply the surviving fractions: 50% × 80% = 40% of original. Customer saves 100% − 40%.
Show solution
Approach: multiply remaining fractions
  1. After 50% off: 0.5 of original. After 20% off that: 0.5 × 0.8 = 0.4 of original.
  2. Saved: 1 − 0.4 = 60%.
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Problem 9 · 2012 AMC 8 Easy
Algebra & Patterns system-of-equationshead-leg-trick

The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?

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Answer: C — 139 birds.
Show hint
Hint 1
If every animal had 4 legs, you'd see 800 legs. The shortage (800 − 522) comes from the 2-leg birds, each missing 2 legs.
Show solution
Approach: leg-shortage shortcut
  1. If all 200 had 4 legs: 200 · 4 = 800 legs.
  2. Actual: 522 ⇒ shortage = 800 − 522 = 278 legs.
  3. Each bird is short 2 legs ⇒ birds = 278 / 2 = 139.
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Problem 10 · 2012 AMC 8 Medium
Counting & Probability permutations-with-repeats

How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?

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Answer: D — 9.
Show hint
Hint 1
Multiset of digits: {0, 1, 2, 2}. Total arrangements with repetition: 4!/2! = 12. Subtract those starting with 0.
Show solution
Approach: multiset permutations minus leading-zero cases
  1. Digits {0, 1, 2, 2}: arrangements = 4! / 2! = 12.
  2. Leading zero (then arrange {1, 2, 2}): 3! / 2! = 3 such arrangements.
  3. Valid 4-digit numbers: 12 − 3 = 9.
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Problem 11 · 2012 AMC 8 Easy
Arithmetic & Operations mean-median-mode

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and x are all equal. What is the value of x?

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Answer: D — 11.
Show hints
Hint 1 of 2
"Unique mode" is the only repeated value — that's 6 in this list. So mode = mean = median = 6.
Still stuck? Show hint 2 →
Hint 2 of 2
Force the mean to be 6: solve for x from the 7-term sum.
Show solution
Approach: lock mode = 6, then force mean = 6
  1. Mode must remain 6 (every other value appears once), so x ≠ 3, 4, 5, 7.
  2. Mean = 6 ⇒ sum = 7 · 6 = 42.
  3. 3 + 4 + 5 + 6 + 6 + 7 + x = 31 + x = 42 ⇒ x = 11. (Sorted list: 3, 4, 5, 6, 6, 7, 11 — median 6 ✓.)
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Problem 12 · 2012 AMC 8 Medium
Number Theory units-digit-cycle

What is the units digit of 132012?

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Answer: A — 1.
Show hint
Hint 1
Only the units digit of 13 matters: same as 32012. Powers of 3 cycle: 3, 9, 7, 1, 3, 9, 7, 1, … (period 4).
Show solution
Approach: powers of 3 units-digit cycle (3, 9, 7, 1)
  1. Units digit of 13n equals units digit of 3n.
  2. Cycle length 4: 31→3, 32→9, 33→7, 34→1, then repeats.
  3. 2012 ≡ 0 (mod 4) ⇒ units digit = 1.
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Problem 13 · 2012 AMC 8 Medium
Number Theory gcd

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?

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Answer: C — 4 more pencils.
Show hint
Hint 1
Working in cents: the price (in cents) divides both 143 and 187. Take their gcd; since price > 1 cent, only one option survives.
Show solution
Approach: gcd of the two amounts
  1. gcd(143, 187) = 11 (since 143 = 11 · 13, 187 = 11 · 17). Price > 1 cent ⇒ price = 11 cents.
  2. Difference Sharona − Jamar = (187 − 143) / 11 = 44 / 11 = 4.
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Problem 14 · 2012 AMC 8 Medium
Counting & Probability handshake-counting

In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?

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Answer: B — 7 teams.
Show hint
Hint 1
Round-robin: number of games = C(N, 2) = N(N − 1)/2.
Show solution
Approach: solve C(N, 2) = 21
  1. N(N − 1)/2 = 21 ⇒ N(N − 1) = 42.
  2. 7 · 6 = 42 ⇒ N = 7.
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Problem 15 · 2012 AMC 8 Medium
Number Theory lcm

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

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Answer: D — Between 61 and 65.
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Hint 1
If x ≡ 2 (mod each of 3, 4, 5, 6), then x − 2 is a multiple of all of them, so a multiple of lcm(3, 4, 5, 6) = 60.
Show solution
Approach: shift to a multiple-of-LCM problem
  1. x − 2 divisible by 3, 4, 5, 6 ⇒ divisible by lcm = 60.
  2. Smallest such x > 2 is x = 60 + 2 = 62, which lies between 61 and 65.
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Problem 16 · 2012 AMC 8 Medium
Algebra & Patterns place-value-greedy

Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?

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Answer: C — 87431.
Show hint
Hint 1
Higher place values dominate. To maximize the sum, put the two biggest digits in the ten-thousands place (one each), the next two in the thousands place, etc.
Show solution
Approach: split digits into matched pairs by place value
  1. Pair digits by descending size for each place: {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
  2. Each number gets one from each pair. So the digits of each number (left to right) come from {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
  3. Only 87431 matches: 8∈{9,8}, 7∈{7,6}, 4∈{5,4}, 3∈{3,2}, 1∈{1,0}. ✓
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Problem 17 · 2012 AMC 8 Medium
Geometry & Measurement area-boundconstruction

A square with an integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?

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Answer: B — Side 4.
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Hint 1 of 2
Lower bound: total area ≥ 10 (each piece has integer side ≥ 1). So side ≥ √10 ⇒ side ≥ 4.
Still stuck? Show hint 2 →
Hint 2 of 2
Upper bound: find an explicit dissection of a 4×4 square into 10 integer-side squares, with 8 of them 1×1.
Show solution
Approach: lower bound + explicit construction
  1. Side ≥ 4: total area is at least 10 (each of 10 pieces ≥ 1), so side2 ≥ 10 ⇒ side ≥ 4.
  2. Construction for side 4: cover the top half (a 4×2 strip) with two 2×2 squares; tile the bottom half (4×2 strip) with 8 unit squares. Total: 2 + 8 = 10 squares, eight of area 1. ✓
  3. So smallest side is 4.
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Problem 18 · 2012 AMC 8 Medium
Number Theory product-of-distinct-primes

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?

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Answer: A — 3127.
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Hint 1 of 2
Not prime & no factor < 50 ⇒ product of at least two primes, all ≥ 50. Not a square ⇒ the primes are distinct.
Still stuck? Show hint 2 →
Hint 2 of 2
Smallest two primes ≥ 50 are 53 and 59.
Show solution
Approach: smallest two primes ≥ 50
  1. First primes ≥ 50: 53, 59, 61, …
  2. Smallest product of two distinct such primes: 53 · 59 = 3127.
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Problem 19 · 2012 AMC 8 Easy
Algebra & Patterns all-but-trick

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?

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Answer: C — 9 marbles.
Show hint
Hint 1
"All but 6 are red" means green + blue = 6. Get all three two-color sums and add them up.
Show solution
Approach: add the three 'all but' equations
  1. Green + Blue = 6, Red + Blue = 8, Red + Green = 4.
  2. Add: 2(Red + Green + Blue) = 18 ⇒ Total = 9.
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Problem 20 · 2012 AMC 8 Medium
Fractions, Decimals & Percents fraction-comparisonrewrite-as-1-minus

What is the correct ordering of the three numbers 519, 721, and 923, in increasing order?

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Answer: B — 5/19 < 7/21 < 9/23.
Show hint
Hint 1
Each fraction is of the form (n)/(n + 14) for n = 5, 7, 9. Write as 1 − 14/(n + 14). Bigger denominator on the subtracted piece ⇒ bigger overall fraction.
Show solution
Approach: rewrite as 1 − (constant)/(denominator)
  1. 5/19 = 1 − 14/19, 7/21 = 1 − 14/21, 9/23 = 1 − 14/23.
  2. Since 14/19 > 14/21 > 14/23, subtracting the largest gives the smallest fraction.
  3. Order: 5/19 < 7/21 < 9/23.
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Problem 21 · 2012 AMC 8 Medium
Geometry & Measurement surface-area-split

Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?

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Answer: D — 50 square feet.
Show hint
Hint 1
Total surface area is 6 × 102. Split it into the green and the white parts.
Show solution
Approach: total surface area − green
  1. Total: 6 · 100 = 600 sq ft. Green covers 300, so white covers 300.
  2. Six congruent white squares share that 300: each is 300 / 6 = 50 sq ft.
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Problem 22 · 2012 AMC 8 Hard
Counting & Probability median-window

Let R be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of R?

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Answer: D — 7 possible values.
Show hints
Hint 1 of 2
Median of 9 distinct integers = 5th smallest. To make some value m the median, we need exactly 4 elements < m and 4 elements > m.
Still stuck? Show hint 2 →
Hint 2 of 2
Already given: three elements below 6 (namely 2, 3, 4) and two above (9, 14), plus 6 itself. Add 3 more integers strategically.
Show solution
Approach: decide which values can sit as the 5th element
  1. Median is the 5th smallest. Sort the six known: 2, 3, 4, 6, 9, 14.
  2. Any candidate median m needs 4 elements < m and 4 > m in the final 9. So m must be reachable by adding 3 integers strategically below/above.
  3. If m < 3: already 5 elements (3, 4, 6, 9, 14) are > m; can't balance. So m ≥ 3.
  4. If m > 9: already 5 elements (2, 3, 4, 6, 9) are < m; can't balance. So m ≤ 9.
  5. Each integer m with 3 ≤ m ≤ 9 works (build the set so 4 are below, 4 above). That's 9 − 3 + 1 = 7 values.
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Problem 23 · 2012 AMC 8 Hard
Geometry & Measurement hexagon-decompositionscaling-area

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?

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Answer: C — Area 6.
Show hints
Hint 1 of 2
Equal perimeters ⇒ triangle side = 2 × hexagon side. The hexagon is made of 6 equilateral triangles of its own side.
Still stuck? Show hint 2 →
Hint 2 of 2
Each of those mini-triangles is a 1/2-scale copy of the big triangle, so its area is 4 · (1/2)2 = 1.
Show solution
Approach: hexagon = 6 equilateral triangles of half-side
  1. Let triangle side = s. Perimeter 3s = 6 · hexagon-side ⇒ hexagon-side = s/2.
  2. Hexagon splits into 6 equilateral triangles of side s/2.
  3. Each is a (1/2)-scale of the original (area scales by 1/4), so each has area 4/4 = 1.
  4. Hexagon area = 6 · 1 = 6.
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Problem 24 · 2012 AMC 8 Hard
Geometry & Measurement rearrangementbounding-square
amc8-2012-24
Show answer
Answer: A — (4 − π)/π.
Show hints
Hint 1 of 2
The star fits snugly inside a square. What about the leftover regions in the corners?
Still stuck? Show hint 2 →
Hint 2 of 2
Inside a 4×4 square, the four corner 'bite' regions are exactly the four arcs that originally made up the circle — so they total to the circle's area.
Show solution
Approach: fit the star and circle pieces in a 4x4 square
  1. Inscribe the star in a 4×4 square (its four points touch the four sides). The square's area is 16.
  2. The four regions outside the star but inside the square are precisely the four arc-pieces from the circle — together their area equals the circle's area π(2)2 = 4π.
  3. Star area = 16 − 4π.
  4. Ratio = (16 − 4π) / 4π = (4 − π)/π.
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Problem 25 · 2012 AMC 8 Hard
Geometry & Measurement inscribed-squarearea-of-corner-triangles
amc8-2012-25
Show answer
Answer: C — 1/2.
Show hints
Hint 1 of 2
The big square is sliced by the small square into 4 congruent right triangles plus the small square. Total leftover area = 5 − 4 = 1, split equally among the 4 triangles.
Still stuck? Show hint 2 →
Hint 2 of 2
Each triangle has legs a and b, area (1/2)ab.
Show solution
Approach: leftover area gives ab
  1. Big square area 5 = small square area 4 + total triangle area ⇒ triangles total 1.
  2. By symmetry, the 4 triangles are congruent, each with area 1/4. Each has legs a and b, area (1/2)ab.
  3. (1/2)ab = 1/4 ⇒ ab = 1/2.
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