AMC 8

2012 AMC 8

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

Practice: Take as test →
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?

Show hint
24 hamburgers is 3 times as many as 8, so the meat triples.
Show solution
  1. 24 / 8 = 3, so triple the meat: 3 × 3 = 9 pounds.
E 9 pounds.
Mark: · log in to save
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?

Show hint
Births per day = 24/8 = 3. Net growth per day = 3 − 1 = 2. Multiply by 365.
Show solution
  1. Births per day: 24/8 = 3. Net daily growth: 3 − 1 = 2.
  2. Yearly: 2 × 365 = 730 ≈ 700.
B About 700.
Mark: · log in to save
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?

Show hint
Sunset = sunrise + length of daylight.
Show solution
  1. 6:57 AM + 10 hours = 4:57 PM.
  2. Add the remaining 24 minutes: 4:57 + 0:24 = 5:21 PM.
B 5:21 PM.
Mark: · log in to save
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?

Show hint
Peter ate 1 full slice + half a slice = 1.5 slices out of 12.
Show solution
  1. Peter's slices: 1 + 1/2 = 1.5 = 3/2.
  2. Fraction: (3/2) / 12 = 3/24 = 1/8.
C 1/8.
Mark: · log in to save
Problem 5 · 2012 AMC 8 Easy
Geometry & Measurement match-total-heights

In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is X, in centimeters?

Show hint
Sum of vertical segments along the left side must equal the sum along the right (same overall height).
Show solution
  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.
E X = 5.
Mark: · log in to save
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?

Show hint
Outer dimensions add 2 inches on each of two sides ⇒ 4 inches in each direction. Subtract the photo area.
Show solution
  1. Outer: (8 + 4) × (10 + 4) = 12 × 14 = 168.
  2. Photo: 8 × 10 = 80.
  3. Border: 168 − 80 = 88.
E 88 square inches.
Mark: · log in to save
Problem 7 · 2012 AMC 8 Easy
Arithmetic & Operations average-budget min-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?

Show hint (soft nudge)
Total points needed = 4 × 95 = 380. Subtract the first two scores to find what's left for tests 3 + 4.
Show hint (sharpest)
To minimize the 3rd test, maximize the 4th (cap = 100).
Show solution
  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.
B 92.
Mark: · log in to save
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?

Show hint
Multiply the surviving fractions: 50% × 80% = 40% of original. Customer saves 100% − 40%.
Show solution
  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%.
D 60%.
Mark: · log in to save
Problem 9 · 2012 AMC 8 Easy
Algebra & Patterns system-of-equations head-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?

Show hint
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
  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.
C 139 birds.
Mark: · log in to save
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?

Show hint
Multiset of digits: {0, 1, 2, 2}. Total arrangements with repetition: 4!/2! = 12. Subtract those starting with 0.
Show solution
  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.
D 9.
Mark: · log in to save
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?

Show hint (soft nudge)
"Unique mode" is the only repeated value — that's 6 in this list. So mode = mean = median = 6.
Show hint (sharpest)
Force the mean to be 6: solve for x from the 7-term sum.
Show solution
  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 ✓.)
D 11.
Mark: · log in to save
Problem 12 · 2012 AMC 8 Medium
Number Theory units-digit-cycle

What is the units digit of 132012?

Show hint
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
  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.
A 1.
Mark: · log in to save
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?

Show hint
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
  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.
C 4 more pencils.
Mark: · log in to save
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?

Show hint
Round-robin: number of games = C(N, 2) = N(N − 1)/2.
Show solution
  1. N(N − 1)/2 = 21 ⇒ N(N − 1) = 42.
  2. 7 · 6 = 42 ⇒ N = 7.
B 7 teams.
Mark: · log in to save
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?

Show hint
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
  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.
D Between 61 and 65.
Mark: · log in to save
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?

Show hint
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
  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}. ✓
C 87431.
Mark: · log in to save
Problem 17 · 2012 AMC 8 Medium
Geometry & Measurement area-bound construction

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?

Show hint (soft nudge)
Lower bound: total area ≥ 10 (each piece has integer side ≥ 1). So side ≥ √10 ⇒ side ≥ 4.
Show hint (sharpest)
Upper bound: find an explicit dissection of a 4×4 square into 10 integer-side squares, with 8 of them 1×1.
Show solution
  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.
B Side 4.
Mark: · log in to save
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?

Show hint (soft nudge)
Not prime & no factor < 50 ⇒ product of at least two primes, all ≥ 50. Not a square ⇒ the primes are distinct.
Show hint (sharpest)
Smallest two primes ≥ 50 are 53 and 59.
Show solution
  1. First primes ≥ 50: 53, 59, 61, …
  2. Smallest product of two distinct such primes: 53 · 59 = 3127.
A 3127.
Mark: · log in to save
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?

Show hint
"All but 6 are red" means green + blue = 6. Get all three two-color sums and add them up.
Show solution
  1. Green + Blue = 6, Red + Blue = 8, Red + Green = 4.
  2. Add: 2(Red + Green + Blue) = 18 ⇒ Total = 9.
C 9 marbles.
Mark: · log in to save
Problem 20 · 2012 AMC 8 Medium
Fractions, Decimals & Percents fraction-comparison rewrite-as-1-minus

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

Show hint
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
  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.
B 5/19 &lt; 7/21 &lt; 9/23.
Mark: · log in to save
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?

Show hint
Total surface area is 6 × 102. Split it into the green and the white parts.
Show solution
  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.
D 50 square feet.
Mark: · log in to save
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?

Show hint (soft nudge)
Median of 9 distinct integers = 5th smallest. To make some value m the median, we need exactly 4 elements < m and 4 elements > m.
Show hint (sharpest)
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
  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.
D 7 possible values.
Mark: · log in to save
Problem 23 · 2012 AMC 8 Hard
Geometry & Measurement hexagon-decomposition scaling-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?

Show hint (soft nudge)
Equal perimeters ⇒ triangle side = 2 × hexagon side. The hexagon is made of 6 equilateral triangles of its own side.
Show hint (sharpest)
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
  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.
C Area 6.
Mark: · log in to save
Problem 24 · 2012 AMC 8 Hard
Geometry & Measurement rearrangement bounding-square

A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

Show hint
Imagine the star inside a 4×4 square. The four 'bite' regions outside the star are exactly the four pieces that, when rearranged, form the original circle.
Show solution
  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.
A (4 &minus; &pi;)/&pi;.
Mark: · log in to save
Problem 25 · 2012 AMC 8 Hard
Geometry & Measurement inscribed-square area-of-corner-triangles

A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?

Show hint (soft nudge)
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.
Show hint (sharpest)
Each triangle has legs a and b, area (1/2)ab.
Show solution
  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.
C 1/2.
Mark: · log in to save