Grade 8 math puzzles — free practice & printable teacher quizzes

• The side lengths keep the same sum. Think about how far each one is from their shared midpoint before and after the change.
• If the original sides are 14 + d and 14 − d, the changed sides are 14 + (d − 2) and 14 − (d − 2). Compare their areas.

Answer: C — 160

The side lengths sum to 28 cm, so their midpoint is 14 cm. Write the original sides as 14 + d and 14 − d. Their area is 196 − d²; after the change it is 196 − (d − 2)². Thus d² − (d − 2)² = 20, giving 4d − 4 = 20 and d = 6. The original sides were 20 cm and 8 cm, so the area was 160 cm². The tempting 180 cm² is the area *after* the change.

• Draw the diagonal and look at the shape it makes with two sides of the rectangle. What kind of triangle is it?
• The width and height are the legs of a right triangle, and the diagonal is its hypotenuse. How are the squares of those three lengths related?

Answer: B — 10

The diagonal is the hypotenuse of a right triangle with legs 6 and 8. By the Pythagorean theorem, d² = 6² + 8² = 36 + 64 = 100, so d = 10. The choice 14 comes from adding 6 and 8, but that gives the length of a path along two sides of the rectangle, not the diagonal.

• Think about how the prime factors must be grouped for both roots to be whole numbers.
• The number must be a sixth power, k⁶. What do its two roots add up to in terms of k?

Answer: B — 729

Write the number as k⁶. Its square root is k³ and its cube root is k², so k³ + k² = 36. For k = 1, 2, and 3, the sums are 2, 12, and 36; larger k give larger sums. Thus k = 3 and the number is 3⁶ = 729. The tempting 1296 is 36², but its cube root is not a whole number.

• Traveling part of a straight segment gives you the same fraction of its total rise and its total horizontal distance. Think about which distance is still left.
• Let t be the fraction of the segment traveled. The climb so far is 4t, while the horizontal distance left is 2(1−t).

Answer: E — 1/3

Between the labeled points, the total rise is 4 and the total horizontal distance is 2. If t of the straight segment has been traveled, the climb is 4t and the horizontal distance left is 2(1−t). Thus 4t = 2(1−t), giving t = 1/3. The tempting 2/3 is the fraction still left, not the fraction traveled.

• Think about how the constant term limits the positive integer solutions of each equation. Which numbers could be a solution to both?
• Call the shared solution r. Find the other solution of each equation in terms of r, then check the possible values of r. Remember that each equation needs two distinct solutions.

Answer: A — 23

The shared solution r must divide both 24 and 36, so r is 1, 2, 3, 4, 6, or 12. The other solutions are 24/r and 36/r, giving a + b = 2r + 60/r. For r = 1, 2, 3, 4, and 12, the totals are 62, 34, 26, 23, and 29. The tempting total 22 comes from r = 6, but then the second equation has the repeated solution 6. Thus the minimum is 23, achieved when the solutions are 4 and 6 for one equation, and 4 and 9 for the other.

• Try sorting the three rolls. What successful patterns are possible, including when numbers match?
• For three distinct rolls, look for an equally spaced triple. Then count its arrangements on the colored dice separately from the all-equal rolls.

Answer: B — 7/36

The distinct triples that work are (1,2,3), (2,3,4), (3,4,5), (4,5,6), (1,3,5), and (2,4,6). Each has 6 arrangements on the colored dice, giving 36 outcomes. The 6 all-equal outcomes also work, for a total of 42 out of 216 equally likely outcomes. Thus the probability is 42/216 = 7/36. The tempting 1/6 counts the distinct triples but misses the all-equal rolls.

• Write N = 2ᵃ·3ᵇ and translate each condition into conditions on a and b.
• N/2 square → a−1 even, b even. N/3 cube → a divisible by 3, b−1 divisible by 3.

Answer: A — 648

Let N = 2ᵃ3ᵇ. N/2 = 2ᵃ⁻¹3ᵇ must be a square → a−1 and b even. N/3 = 2ᵃ3ᵇ⁻¹ must be a cube → a and b−1 divisible by 3. Smallest a: odd and divisible by 3 → 3. Smallest b: even with b−1 divisible by 3 → 4. N = 2³·3⁴ = 8 · 81 = 648 (check: 648/2 = 324 = 18², 648/3 = 216 = 6³). No other prime can help: its exponent would have to be even AND divisible by 3, and the smallest such exponent, 0, means "leave it out". Translating a number problem into bookkeeping on prime exponents is a signature olympiad technique.

• Counting squares gives a tempting limit, but can the dominoes actually reach it? Try coloring the grid like a checkerboard.
• The corners and center all have the same color. After finding an upper bound, look for a way to cut that many dominoes.

Answer: A — 20

A checkerboard coloring has 25 dark and 24 light squares. The five removed squares are all dark, leaving 20 dark and 24 light. Each domino covers one of each, so at most 20 can be cut; 22 is the trap from counting area alone. To reach 20, also leave unused the squares at (row, column) (2,1), (2,7), (6,1), and (6,7). The top and bottom two rows each become a tileable 2×5 rectangle. In the middle three rows, tile the 2×3 blocks in rows 3–4, columns 1–3 and rows 4–5, columns 5–7, and the 1×4 strips in row 3, columns 4–7 and row 5, columns 1–4.

• Rather than listing cards from the start, count how many come at or before a conveniently sized square. Remember that equal dot counts still belong to separate cards.
• A staircase with n rows has n(n+1)/2 dots. How many square cards and staircase cards have at most 40×40 dots?

Answer: B — 1764

Up to 1600 dots, there are 40 square cards. The 56th staircase has 56×57/2 = 1596 dots, while the 57th has 1653, so there are 56 staircase cards: 96 cards altogether. The next four dot counts are 1653, 1681, 1711 and 1764. Thus the 100th card has 1764 dots. The tempting choice 1849 results from merging tied cards, but ties must be kept separately.

• Reordering the increases changes which ones have happened by a given month, but not the final total. Think about what would make an intermediate total as small as possible.
• The four increases are $15, $10, $30, and $15. For each month, consider which increases would have to happen first to give the smallest possible total.

Answer: D — March

By March, the chart has used the $15 and $10 increases—the two smallest—so no reordering can lower March’s total. Putting the $30 increase before March could make it higher. February’s total could be lowered by putting $10 first, and April’s could be lowered by saving the $30 increase for last. January and May are tempting because their totals cannot be lowered, but reordering cannot make either one higher.

• Count all possible scarves first.
• Subtract those with no gold patches.

Answer: D — 37

Each of the three patches has 4 choices, giving 4 × 4 × 4 = 64 scarves. With no gold, each patch has 3 choices, giving 3 × 3 × 3 = 27 scarves. So the answer is 64 − 27 = 37. The tempting answer 48 comes from choosing which patch is gold and then coloring the other two, but it counts scarves with multiple gold patches more than once.

• How many times does an end number appear in the seven sums?
• Start by counting every number twice.

Answer: E — 1 and 2

The numbers 1 through 8 add to 36, so counting each twice gives 72. Each end number appears in only one gap sum instead of two, so the end numbers add to 72 − 69 = 3. The only two different numbers that work are 1 and 2. The tempting choice 1 and 3 would make the seven sums total 72 − 4 = 68.