Focused Practice

Across all years

Tick one or more bands and topics — problems are pulled from every authored year.

Showing 20 of 205 matching
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Problem 11 · 2026 AMC 8 Medium
Geometry & Measurement arc-length
amc8-2026-11
Show hint (soft nudge)
A quarter circle in a square of side s has arc length one-fourth of 2πs.
Show hint (sharpest)
Add up the arcs for all five squares.
Show solution
  1. Each square of side s contributes a quarter-circle of length ¼ · 2πs = πs/2.
  2. Total = (π/2)(1 + 1 + 2 + 3 + 5) = (π/2)(12) = .
B 6π.
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Problem 12 · 2026 AMC 8 Hard
Logic & Word Problems constraint-propagation
amc8-2026-12
Show hint (soft nudge)
Start at the most restrictive sum and see which digits can make it.
Show hint (sharpest)
A sum of 10 between two circles can only be 4 and 6.
Show solution
  1. The edge summing to 10 forces those two circles to be 4 and 6; taking the upper one as 4 makes the top circle 9 − 4 = 5 and the bottom-left 6.
  2. Then the remaining sums give 8 − 6 = 2, 5 − 2 = 3, 4 − 3 = 1, and 5 + 1 = 6 checks out — using each digit 1–6 once.
  3. So the top circle must be 5.
D 5.
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Problem 13 · 2026 AMC 8 Hard
Geometry & Measurement tilted-square pythagorean
amc8-2026-13
Show hint (soft nudge)
The square is tilted, so its area equals the square of its side length.
Show hint (sharpest)
Read one side as a step across the lattice and use the Pythagorean theorem.
Show solution
  1. Each side of the shaded square is the hypotenuse of a right triangle with legs 3 and 1, measured across the tiling.
  2. So the area is the side squared: 3² + 1² = 10.
A 10.
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Problem 14 · 2026 AMC 8 Medium
Algebra & Patterns arithmetic-sequence

Jami picked three equally spaced integers on the number line. The sum of the first and second is 40, and the sum of the second and third is 60. What is the sum of all three numbers?

Show hint (soft nudge)
For equally spaced numbers, the middle one is the average of the other two.
Show hint (sharpest)
Add the two given sums and see how many times the middle number appears.
Show solution
  1. Since the numbers are equally spaced, first + third = 2 × second. Adding the two given sums: 40 + 60 = first + 2·second + third = 4·second, so the middle is 25.
  2. The total is 3 × 25 = 75.
B 75.
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Problem 15 · 2026 AMC 8 Hard
Geometry & Measurement spatial-reasoning
amc8-2026-15
Show hint (soft nudge)
A cube's two shaded faces share an edge, so both must be glued to neighbors to hide them.
Show hint (sharpest)
Look for an arrangement where every cube has two glued faces that meet at an edge.
Show solution
  1. To hide a cube's two shaded faces — which meet at an edge — it must be glued to neighbors on two faces sharing an edge.
  2. Four cubes arranged in a 2 × 2 square give each cube exactly two such adjacent glued faces; with three or fewer, some cube has only one glued face (or two opposite ones), so a shaded face shows.
  3. The fewest is 4.
A 4 cubes.
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Problem 11 · 2025 AMC 8 Medium
Geometry & Measurement spatial-reasoning area-decomposition
amc8-2025-11
Show hint (soft nudge)
Place the S tile first — it's the most awkward. What's left over?
Show hint (sharpest)
If S sits in a corner, the remaining 8 squares form an L-shape that splits into two L tetrominoes.
Show solution
  1. The 3 × 4 rectangle has 12 squares; three tetrominoes (4 squares each) must cover them all.
  2. Place the S piece against an edge so it doesn't block too much. The remaining 8 squares form an L-shape.
  3. That L-shape splits cleanly into two L tetrominoes — answer L and L.
C L and L.
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Problem 12 · 2025 AMC 8 Medium
Geometry & Measurement area spatial-reasoning
amc8-2025-12
Show hint (soft nudge)
The biggest circle that fits is limited by whichever inward corners of the region poke closest to the center.
Show hint (sharpest)
Each inward corner sits 1 unit horizontally and 2 units vertically from the center. Use the Pythagorean theorem to get the radius.
Show solution
  1. By symmetry, the largest inscribed circle is centered at the region's center. Its radius is the distance from there to the nearest inward-poking corner.
  2. Each such corner is 1 unit across and 2 units up (or down) from the center: distance = √(12 + 22) = √5.
  3. Area = π × (√5)2 = .
C 5π square centimeters.
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Problem 13 · 2025 AMC 8 Medium
Number Theory divisibility spiral-pattern
amc8-2025-13
Show hint (soft nudge)
Don't compute 25 separate remainders — even numbers' remainders mod 7 repeat in a pattern.
Show hint (sharpest)
The cycle is 2, 4, 6, 1, 3, 5, 0 (period 7). 25 numbers = 3 full cycles + 4 extras.
Show solution
  1. The remainders mod 7 of 2, 4, 6, 8, 10, 12, 14, … cycle through 2, 4, 6, 1, 3, 5, 0 with period 7.
  2. 25 even numbers = 3 full cycles (21 numbers, hitting each remainder 3 times) plus 4 extras: 2, 4, 6, 1.
  3. So remainders 1, 2, 4, 6 occur 4 times each; remainders 0, 3, 5 occur 3 times each — the pattern in choice A.
A Histogram (A).
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Problem 14 · 2025 AMC 8 Medium
Arithmetic & Operations work-backward substitution

A number N is inserted into the list 2, 6, 7, 7, 28. The mean is now twice as great as the median. What is N?

Show hint (soft nudge)
With 6 numbers, the median is the average of the middle two. Where in the sorted list does N land?
Show hint (sharpest)
Every answer choice is at least 7, so the middle two are always 7 and 7 — the median locks in at 7 no matter what.
Show solution
  1. All answer choices are ≥ 7, so when N is inserted into the sorted list, the middle two stay 7 and 7. The median is always 7.
  2. Mean = 2 × 7 = 14, so the six numbers must sum to 6 × 14 = 84.
  3. The original five sum to 2 + 6 + 7 + 7 + 28 = 50, so N = 84 − 50 = 34.
E 34.
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Problem 15 · 2025 AMC 8 Stretch
Counting & Probability careful-counting casework
amc8-2025-15
Show hint (soft nudge)
First count the gold squares. Then think about the 18 pairs the fold creates.
Show hint (sharpest)
For the minimum, spread golds across pairs first (one per pair). For the maximum, pair golds together first.
Show solution
  1. Gold squares: 36 − 13 = 23. Folding pairs up the 36 squares into 18 overlap pairs.
  2. Minimum m: spread golds so each pair gets one gold first — that uses 18 of them, leaving 23 − 18 = 5 to double up. So m = 5.
  3. Maximum M: pair golds 2-at-a-time. 23 = 2 × 11 + 1, so M = 11 gold-on-gold pairs (1 lone gold left).
  4. m + M = 5 + 11 = 16.
C 16.
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Problem 11 · 2024 AMC 8 Medium
Geometry & Measurement area
amc8-2024-11
Show hint (soft nudge)
Pick the side that's easiest — AB is horizontal. Use that as the base.
Show hint (sharpest)
Base AB has length 6 and lies on y = 7. The height is just (y − 7) since C is above that line.
Show solution
  1. AB lies on the line y = 7 with length 11 − 5 = 6.
  2. Height from C to line y = 7 is y − 7. Area = 12 · 6 · (y − 7) = 12.
  3. So y − 7 = 4 ⇒ y = 11.
D y = 11.
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Problem 12 · 2024 AMC 8 Medium
Algebra & Patterns substitution sum-constraint

Rohan keeps a total of 90 guppies in 4 fish tanks.

  • There is 1 more guppy in the 2nd tank than in the 1st tank.
  • There are 2 more guppies in the 3rd tank than in the 2nd tank.
  • There are 3 more guppies in the 4th tank than in the 3rd tank.

How many guppies are in the 4th tank?

Show hint (soft nudge)
Stack the increments from tank 1: tank 2 = tank 1 + 1, tank 3 = tank 1 + 3, tank 4 = tank 1 + 6.
Show hint (sharpest)
Add all four expressions: 4·(tank 1) + 10 = 90, so tank 1 = 20.
Show solution
  1. Let tank 1 = x. Then tank 2 = x + 1, tank 3 = x + 3, tank 4 = x + 6.
  2. Sum: 4x + 10 = 90, so x = 20.
  3. Tank 4 = 20 + 6 = 26.
E 26 guppies.
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Problem 13 · 2024 AMC 8 Medium
Counting & Probability careful-counting casework

Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of 6 hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)

Show hint (soft nudge)
Each sequence needs 3 ups and 3 downs. But Buzz can never go below the ground — the running count of downs can never exceed ups.
Show hint (sharpest)
Start with U, end with D. Enumerate carefully without breaking the rule.
Show solution
  1. The sequence has 3 U's and 3 D's, with D's never exceeding U's at any point (otherwise Buzz goes below the ground).
  2. All valid sequences: UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD.
  3. 5 sequences in total.
B 5 sequences.
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Problem 14 · 2024 AMC 8 Hard
Logic & Word Problems work-backward casework
amc8-2024-14
Show hint (soft nudge)
Build up the shortest distance to each town in turn, always coming from the cheapest predecessor seen so far.
Show hint (sharpest)
Go A → X → M → Y → C → Z, each time choosing the cheapest way in.
Show solution
  1. Shortest A→X = 5 (direct).
  2. Shortest A→M = min(8 direct, 5 + 2 via X) = 7.
  3. Shortest A→Y = min(5 + 10 via X, 7 + 6 via M) = 13.
  4. Shortest A→C = min(7 + 14 via M, 13 + 5 via Y) = 18.
  5. Shortest A→Z = min(7 + 25 via M, 13 + 17 via Y, 18 + 10 via C) = 28.
A 28 km.
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Problem 15 · 2024 AMC 8 Hard
Number Theory factorization casework work-backward

Let the letters F, L, Y, B, U, G represent distinct digits. Suppose FLYFLY is the greatest number that satisfies the equation

8 · FLYFLY = BUGBUG.

What is the value of FLY + BUG?

Show hint (soft nudge)
A six-digit number that repeats a three-digit block (like ABCABC) has a hidden common factor.
Show hint (sharpest)
FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation just says 8 · FLY = BUG.
Show solution
  1. FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation reduces to 8 · FLY = BUG.
  2. BUG is a 3-digit number, so 8 · FLY < 1000 → FLY ≤ 124. So F = 1, and L ≤ 2 (else 8 · FLY ≥ 1040).
  3. Maximize: L = 2 (can't be 1, F took it). Then 8 · 12Y needs all 6 digits distinct. Y = 4: 8 · 124 = 992 (two 9s, fails). Y = 3: 8 · 123 = 984 — digits {1, 2, 3, 9, 8, 4} all distinct ✓.
  4. FLY + BUG = 123 + 984 = 1107.
C 1107.
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Problem 11 · 2023 AMC 8 Medium
Ratios, Rates & Proportions distance-speed-time estimate-and-pick

NASA's Perseverance Rover was launched on July 30, 2020. After traveling 292,526,838 miles, it landed on Mars in Jezero Crater about 6.5 months later. Which of the following is closest to the Rover's average interplanetary speed in miles per hour?

Show hint (soft nudge)
The choices span orders of magnitude, so round generously. Distance ≈ 3 × 108 miles.
Show hint (sharpest)
6.5 months ≈ 200 days ≈ 5000 hours. Then speed ≈ distance ÷ time.
Show solution
  1. Distance ≈ 3 × 108 miles.
  2. 6.5 months × 30 days/month ≈ 195 days ≈ 200 days ≈ 200 × 24 = 4800 hours ≈ 5000 hours.
  3. Speed ≈ (3 × 108) ÷ 5000 = 60,000 mph.
C About 60,000 mph.
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Problem 12 · 2023 AMC 8 Hard
Geometry & Measurement area area-decomposition
amc8-2023-12
Show hint (soft nudge)
Sum the shaded areas, subtract the white circles that sit inside the big shaded disk, divide by the area of the outer circle.
Show hint (sharpest)
Outer circle area = 9π. Three small shaded circles (radius 1/2): total 3π/4. Big shaded disk (radius 2) minus 2 inner whites (radius 1 each): 4π − 2π = 2π.
Show solution
  1. Outer white circle area: π · 32 = 9π.
  2. Three small shaded circles (radius 12): each has area π4; together 4.
  3. Big shaded disk (radius 2): area 4π, minus two white inner circles (radius 1 each, total 2π): net 2π.
  4. Total shaded: 2π + 4 = 11π4.
  5. Fraction: 11π/4 = 1136.
B 11/36.
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Problem 13 · 2023 AMC 8 Medium
Fractions, Decimals & Percents proportion fraction-to-decimal
amc8-2023-13
Show hint (soft nudge)
Where, as a fraction of the route, are the 1st repair station and the 3rd water station?
Show hint (sharpest)
Repair stations split into thirds: 1st is at 1/3. Water stations split into eighths: 3rd is at 3/8. Difference is 2 miles.
Show solution
  1. Let L be the race length. 2 repair stations evenly spaced between start and finish divide the route into thirds → the 1st repair is at L/3.
  2. 7 water stations evenly spaced divide the route into eighths → the 3rd water is at 3L/8.
  3. Their gap: 3L/8 − L/3 = (9L − 8L)/24 = L/24 = 2 miles.
  4. So L = 48.
D 48 miles.
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Problem 14 · 2023 AMC 8 Hard
Number Theory complementary-counting careful-counting

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 5-cent, 10-cent, and 25-cent stamps, with exactly 20 of each type. What is the greatest number of stamps Nicolas can use to make exactly $7.10 in postage?

Show hint (soft nudge)
Maximizing stamps used = minimizing stamps removed from his whole collection. What does the whole collection total?
Show hint (sharpest)
Total of all 60 stamps = $8. He needs to remove $0.90. Minimize the number of stamps that sum to $0.90.
Show solution
  1. Total value of all 60 stamps: 20·($0.05 + $0.10 + $0.25) = 20 · $0.40 = $8.00.
  2. He needs to make $7.10, so he removes $8.00 − $7.10 = $0.90 worth. Maximizing stamps used ≡ minimizing stamps removed.
  3. Minimum stamps summing to $0.90: three 25¢ (75¢) + one 10¢ + one 5¢ = $0.90 in 5 stamps.
  4. Stamps used = 60 − 5 = 55.
E 55 stamps.
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Problem 15 · 2023 AMC 8 Hard
Ratios, Rates & Proportions distance-speed-time proportion
amc8-2023-15
Show hint (soft nudge)
Find Viswam's usual speed in mph, then figure out how many blocks vs minutes he has left after the detour starts.
Show hint (sharpest)
After 5 blocks he has 5 min left. Detour makes the remaining distance 5 + (3 − 1) = 7 blocks instead of 5. Distance per block: 0.05 mile.
Show solution
  1. Usual: 10 blocks = 0.5 mile in 10 min = 1/6 hr → speed = 3 mph. Each block is 0.05 mile.
  2. After 5 blocks, 5 minutes are left. Detour replaces 1 block with 3, so remaining distance becomes 5 + 2 = 7 blocks = 0.35 mile.
  3. 5 minutes = 1/12 hr. Speed = 0.35 ÷ (1/12) = 0.35 × 12 = 4.2 mph.
B 4.2 mph.
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