Mock Test

Mock Test (random)

25 random problems — one per position, pulled from random authored years. Hints and solutions are locked until you submit. Retake as often as you want — every attempt is saved to your test history (if you're logged in).

25 problems Reshuffle
Paper:
← Back Tip: real AMC 8 is 40 minutes — set a timer if you want the full pressure.
Problem 1 · 1992 AJHSME Medium
Arithmetic & Operations pair-terms
ajhsme-1992-01
Show hint (soft nudge)
Pair consecutive terms in the top, and in the bottom, to add them up quickly.
Show hint (sharpest)
Both the numerator and the denominator come out to 5.
Show solution
  1. Top: (10−9)+(8−7)+(6−5)+(4−3)+(2−1) = 5. Bottom: (1−2)+(3−4)+(5−6)+(7−8)+9 = −4+9 = 5.
  2. So the value is 5/5 = 1.
B 1.
Mark: · log in to save
Problem 2 · 2007 AMC 8 Easy
Arithmetic & Operations ratio-from-graph

650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

Show hint
Read off counts; simplify the ratio.
Show solution
  1. Spaghetti / Manicotti = 250 / 100 = 5/2.
E 5/2.
Mark: · log in to save
Problem 3 · 2001 AMC 8 Easy
Arithmetic & Operations division

Granny Smith has $63. Elberta has $2 more than Anjou, and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?

Show hint
Work down the chain: find Anjou's amount first, then Elberta's.
Show solution
  1. Anjou has one-third of $63, which is $21.
  2. Elberta has $2 more: $21 + $2 = $23.
E $23.
Mark: · log in to save
Problem 4 · 1994 AJHSME Medium
Geometry & Measurement rotation
ajhsme-1994-04
Show hint (soft nudge)
A 120° clockwise turn moves each shape to the next position clockwise.
Show hint (sharpest)
Track where the triangle, circle, and diamond each land.
Show solution
  1. Clockwise 120° sends top → lower-right → lower-left → top.
  2. So the circle moves to the top, the diamond to the lower-left, and the triangle to the lower-right — matching choice B.
B Choice B.
Mark: · log in to save
Problem 5 · 2005 AMC 8 Easy
Arithmetic & Operations greedy-packing

Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?

Show hint
Use as many 24-packs as possible, then fill in with 12s and 6s.
Show solution
  1. 3 · 24 = 72 leaves 18. Then one 12 + one 6 = 18.
  2. Packs used: 3 + 1 + 1 = 5.
B 5 packs.
Mark: · log in to save
Problem 6 · 1992 AJHSME Medium
Algebra & Patterns custom-operation
ajhsme-1992-06
Show hint (soft nudge)
Each triangle means top + bottom-left − bottom-right.
Show hint (sharpest)
Evaluate both triangles, then add.
Show solution
  1. First triangle: 1 + 3 − 4 = 0. Second triangle: 2 + 5 − 6 = 1.
  2. Their sum is 0 + 1 = 1.
D 1.
Mark: · log in to save
Problem 7 · 2020 AMC 8 Medium
Counting & Probability careful-counting casework

How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2347 is one integer.)

Show hint (soft nudge)
Pin down the leading digits first. The thousands digit is forced; the hundreds digit has only one valid value.
Show hint (sharpest)
First digit = 2. For the digits to be increasing and the number ≤ 2400, the second digit must be 3 (since it must exceed 2 and be ≤ 3). Then the last two digits are any two distinct digits from {4,5,6,7,8,9}.
Show solution
  1. Digits must be increasing, so they're strictly ascending. First digit = 2 (number is between 2020 and 2400). Second digit must be > 2 and ≤ 3 (else the number exceeds 2400): second digit = 3.
  2. Last two digits: any 2 distinct values from {4, 5, 6, 7, 8, 9}, arranged in increasing order. That's C(6, 2) = 15.
C 15 integers.
Mark: · log in to save
Problem 8 · 2007 AMC 8 Easy
Geometry & Measurement right-triangle-area

In trapezoid ABCD, AD is perpendicular to DC, AD = AB = 3, and DC = 6. In addition, E is on DC, and BE is parallel to AD. Find the area of ▵BEC.

Show hint
BE = AD = 3 (perpendicular distance). EC = DCDE = 6 − 3 = 3.
Show solution
  1. BEC is right-angled at E with legs 3 and 3.
  2. Area = (1/2)(3)(3) = 4.5.
B 4.5.
Mark: · log in to save
Problem 9 · 2014 AMC 8 Easy
Geometry & Measurement isosceles-triangle linear-pair

In ▵ABC, D is a point on side AC such that BD = DC and ∠BCD measures 70°. What is the degree measure of ∠ADB?

Show hint (soft nudge)
BDC is isosceles (BD = DC), so its base angles are equal.
Show hint (sharpest)
ADB and ∠BDC are supplementary (linear pair along AC).
Show solution
  1. Since BD = DC, ∠DBC = ∠DCB = 70°.
  2. BDC = 180° − 70° − 70° = 40°.
  3. ADB = 180° − 40° = 140°.
D 140 degrees.
Another way: exterior angle
  1. ADB is an exterior angle of ▵BDC at D, so it equals the sum of the two remote interior angles: 70° + 70° = 140°.
Mark: · log in to save
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.
Mark: · log in to save
Problem 11 · 2015 AMC 8 Medium
Counting & Probability multiplication-principle probability

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

Show hint (soft nudge)
Count the total number of allowed plates by multiplying choices for each slot. The probability of AMC8 is 1 over that total.
Show hint (sharpest)
Slot 1: 5 vowels. Slot 2: 21 non-vowels. Slot 3: 20 (different from slot 2). Slot 4: 10 digits.
Show solution
  1. Total plates = 5 × 21 × 20 × 10 = 21,000.
  2. AMC8 is one specific plate, so probability = 1/21,000.
B 1/21,000.
Mark: · log in to save
Problem 12 · 1999 AMC 8 Medium
Fractions, Decimals & Percents ratio-to-percent

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is 114. To the nearest whole percent, what percent of its games did the team lose?

Show hint (soft nudge)
Turn the ratio into parts: 11 won and 4 lost make 15 games in all.
Show hint (sharpest)
The lost fraction is 4 out of 15.
Show solution
  1. Treat the ratio as 11 wins and 4 losses, so 15 games total.
  2. Losses are 4/15 ≈ 26.7%, which rounds to 27%.
B 27%.
Mark: · log in to save
Problem 13 · 2015 AMC 8 Medium
Counting & Probability sum-mean-relationship pair-counting

How many subsets of two elements can be removed from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} so that the mean (average) of the remaining numbers is 6?

Show hint (soft nudge)
Original sum is 1 + 2 + ... + 11 = 66. After removing 2 elements, 9 remain with mean 6, so the remaining sum is 54.
Show hint (sharpest)
The removed pair must sum to 66 − 54 = 12. Count two-element subsets of {1, ..., 11} with sum 12.
Show solution
  1. Sum of 1 through 11 is 66. After removing 2 numbers, 9 remain; mean 6 means remaining sum is 9 × 6 = 54.
  2. So the removed pair sums to 66 − 54 = 12.
  3. Pairs from {1, …, 11} summing to 12: {1,11}, {2,10}, {3,9}, {4,8}, {5,7}. That is 5 pairs.
D 5 pairs.
Mark: · log in to save
Problem 14 · 1992 AJHSME Medium
Fractions, Decimals & Percents fraction-of

When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is

Show hint (soft nudge)
The 4 gallons raised the level from 1/3 to 1/2 — what fraction is that?
Show hint (sharpest)
Then scale up to the full tank.
Show solution
  1. 1/2 − 1/3 = 1/6 of the tank equals 4 gallons.
  2. So the full tank is 6 × 4 = 24 gallons.
D 24.
Mark: · log in to save
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.
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 · 1993 AJHSME Hard
Geometry & Measurement surface-area net
ajhsme-1993-17
Show hint (soft nudge)
After cutting 5×5 corners, the base is (20−10) by (30−10) and the height is 5.
Show hint (sharpest)
The open box has a bottom and four inner walls — no top.
Show solution
  1. The base is 10 × 20 = 200 and the height is 5, so the four walls add 2(10·5) + 2(20·5) = 300.
  2. The interior surface (no top) is 200 + 300 = 500.
B 500.
Mark: · log in to save
Problem 18 · 1999 AMC 8 Medium
Fractions, Decimals & Percents percent-decrease round-up

Cookies for a Crowd. The recipe makes a pan of 15 cookies, and only full recipes are made. Normally 108 students each eat 2 cookies, but a concert cuts attendance by 25%. How many recipes should Walter and Gretel make for the smaller party?

Show hint (soft nudge)
A 25% drop leaves three-fourths of the 108 students.
Show hint (sharpest)
Find their cookies, then round up to whole pans of 15.
Show solution
  1. Three-fourths of 108 is 81 students, eating 81 × 2 = 162 cookies.
  2. That needs 162 ÷ 15 = 10.8 → 11 full recipes.
E 11 recipes.
Mark: · log in to save
Problem 19 · 2004 AMC 8 Medium
Number Theory lcm

A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5, and 6. The smallest such number lies between which two numbers?

Show hint
x − 2 is divisible by lcm(3, 4, 5, 6) = 60.
Show solution
  1. Smallest x > 2 with x − 2 = 60 ⇒ x = 62.
  2. 62 lies between 60 and 79.
B Between 60 and 79.
Mark: · log in to save
Problem 20 · 1988 AJHSME Hard
Fractions, Decimals & Percents percent-of-whole
ajhsme-1988-20
Show hint (soft nudge)
45 cups is 36% of the full capacity.
Show hint (sharpest)
Divide 45 by 0.36.
Show solution
  1. If 45 = 0.36 · Full, then Full = 45 ⁄ 0.36 = 4500 ⁄ 36.
  2. = 125 cups.
C 125.
Mark: · log in to save
Problem 21 · 1990 AJHSME Stretch
Algebra & Patterns work-backward

A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.

Show hint (soft nudge)
Each term is the product of the two before it, so divide to step backward.
Show hint (sharpest)
From 16, 64, 1024 work back: the term before 16 is 64 ÷ 16, and so on.
Show solution
  1. Since 1024 = 16 · 64, the term before 16 is 64 ÷ 16 = 4, then 16 ÷ 4 = 4, then 4 ÷ 4 = 1, then 4 ÷ 1 = 4, and finally 1 ÷ 4 = 1/4.
  2. So the first number is 1/4.
B 1/4.
Mark: · log in to save
Problem 22 · 2001 AMC 8 Stretch
Counting & Probability careful-counting

On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point, and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?

Show hint (soft nudge)
Start from the top: what's the maximum score, and the next one just below it?
Show hint (sharpest)
There's a gap right under the maximum that no score can land in.
Show solution
  1. All 20 correct scores 20 × 5 = 100. The next-best is 19 correct plus 1 unanswered: 95 + 1 = 96.
  2. So 97, 98, 99 are unreachable — 97 is the impossible score (90, 91, 92, 95 are all attainable).
E 97.
Mark: · log in to save
Problem 23 · 2003 AMC 8 Stretch
Algebra & Patterns
amc8-2003-23
Show hint (soft nudge)
The cat and mouse move independently — handle each one's cycle on its own.
Show hint (sharpest)
The cat repeats every 4 moves and the mouse every 8; use the remainders of 247.
Show solution
  1. The cat's position repeats every 4 moves and the mouse's every 8, so only the remainder of 247 matters for each.
  2. 247 = 4·61 + 3, so the cat is where it is after 3 moves: the bottom-right square.
  3. 247 = 8·30 + 7, so the mouse is where it is after 7 moves: the bottom-left segment.
  4. The picture matching both is A.
A the cat in the bottom-right square, the mouse on the bottom-left segment.
Mark: · log in to save
Problem 24 · 2007 AMC 8 Medium
Counting & Probability divisibility-by-3 subset-counting

A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3, or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

Show hint
Divisible by 3 iff digit-sum divisible by 3. The chosen 3 digits' sum is what matters; the order is irrelevant for divisibility.
Show solution
  1. Subsets of size 3 from {1, 2, 3, 4}: 4 total ({1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}).
  2. Digit sums: 6 ✓, 7, 8, 9 ✓ ⇒ 2 subsets give a multiple of 3.
  3. Probability: 2/4 = 1/2.
C 1/2.
Mark: · log in to save
Problem 25 · 1998 AJHSME Stretch
Algebra & Patterns work-backward invariant

Three generous friends redistribute their money as follows: Amy gives Jan and Toy enough to double each of their amounts; then Jan gives Amy and Toy enough to double theirs; finally Toy gives Amy and Jan enough to double theirs. Toy had $36 at the beginning and $36 at the end. What is the total amount the three friends have?

Show hint (soft nudge)
The total never changes — you only need to find it at one moment.
Show hint (sharpest)
Toy's money doubles in the first two rounds, so track it up to just before his own turn.
Show solution
  1. Toy's $36 doubles in each of the first two rounds: 36 → 72 → 144 just before his turn.
  2. He ends with $36, so he gave away 144 − 36 = 108, which exactly doubled Amy and Jan — meaning they held $108 then.
  3. The total, unchanged throughout, is 144 + 108 = $252.
D $252.
Mark: · log in to save