Grade 7 math puzzles — free practice & printable teacher quizzes

• What must the screen have shown just before the very last digit swap?
• Work backward through the last round to find the number between the two rounds. Then work backward through the first round.

Answer: C — 17

Working backward, swap 53 to get 35, subtract 7 to get 28, and halve to get 14. But 14 was the input to the second round, not the first. Swap 14 to get 41, subtract 7 to get 34, and halve to get 17. Checking forward: 17 → 34 → 41 → 14 → 28 → 35 → 53.

• Write down the last digit each time you multiply a power of 2 by 2. What happens if you keep going?
• Once the last digits repeat, use the exponents to find where each power falls in the cycle.

Answer: E — 2

The last digits of powers of 2 repeat in a cycle: 2, 4, 8, 6. Since 2026 leaves remainder 2 when divided by 4, 2^2026 ends in 4. The next power ends in 8, and 4 + 8 = 12, so the sum ends in 2. The tempting 8 is the last digit of 2^2027 alone, not of the sum.

• After a full lap, which way is Maya facing compared with when she started? Think about her total change in direction.
• The three turns make one full revolution. Find the total of the first two, then use the 30° difference.

Answer: C — 95°

Maya makes one full turn, or 360°, during the lap. Her first two turns therefore add to 360° − 140° = 220°. Since the first is 30° larger, the second is (220° − 30°) ÷ 2 = 95°. The tempting 110° splits 220° equally but ignores the 30° difference.

• Try it with a $100 shirt.

Answer: D — 4% lower

Start at $100 → up 20% → $120 → down 20% of $120 → $96: 4% below the original. The −20% acts on a BIGGER base than the +20% did, so they don't cancel: 1.2 × 0.8 = 0.96. Percent changes compose by multiplication — a fact the SAT loves to test.

• Imagine adding a new corner between two neighboring vertices. What happens to the diagonal count?
• Once you know how many vertices there are, count the diagonals touching each blue vertex. Is any diagonal in both counts?

Answer: D — 11

A triangle has no diagonals. Adding a corner to make a k-vertex polygon creates k−3 diagonals from the new corner and turns the old side into a diagonal, adding k−2 in all. The totals for 4 through 8 vertices are therefore 2, 5, 9, 14, and 20, so this is an octagon. Each blue vertex touches 5 diagonals, but the diagonal joining them is counted twice. Thus 20−(5+5−1)=11 diagonals have no blue endpoint. The tempting answer 10 subtracts their shared diagonal twice.

• Pick a convenient distance, say 12 km each way, and compute the total time.

Answer: C — 4.8 km/h

Take 12 km each way: going takes 3 h, returning 2 h. Total: 24 km in 5 h → 4.8 km/h. The "obvious" 5 is wrong because Ala spends MORE TIME at the slower speed. Averages must be weighted by what actually accumulates (time) — a deep idea that reappears in statistics and physics.

• Think about which small prime factors must survive, even if you cross out a number that supplies several of them.
• In a group of nine consecutive numbers, count the evens, the multiples of 4, and the multiples of 3. How many of each can one crossed-out number take away?

Answer: C — 144

Nine consecutive numbers contain at least four evens, at least two multiples of 4, and exactly three multiples of 3. Crossing out one leaves enough factors for at least 2⁴ × 3² = 144. The trap is assuming an extra factor must survive too: starting with 1 through 9, crossing out 8, 9, 5, or 7 gives 144 × 315, 144 × 280, 144 × 504, or 144 × 360, respectively—ruling out 288, 432, 720, and 1008.

• Focus on one crossing and trace its two segments back to the dots that made it.
• Any four dots form a quadrilateral. How many pairs of its sides or diagonals cross inside the circle?

Answer: E — 126

Each crossing comes from four dots. For any four dots, exactly one pair of segments crosses inside the circle: the quadrilateral’s two diagonals. Since no three segments meet at one interior point, different groups of four give different crossing points. Counting the dots in order and dividing out their 4 × 3 × 2 × 1 possible orders gives (9 × 8 × 7 × 6) ÷ 24 = 126. The tempting 378 counts all three ways to pair four dots, but only one pairing crosses.

• Look at groups of three numbers. Can you find a relationship that works on both sides of the blank?
• Notice that 12 = 2 × 5 + 2 and 29 = 2 × 12 + 5. What would come after 29 if that relationship continues?

Answer: B — 70

Each number is twice the previous number plus the number before that. So the blank is 2 × 29 + 12 = 70. This checks in both later terms: 2 × 70 + 29 = 169 and 2 × 169 + 70 = 408. The tempting 63 comes from adding 5 instead of 12 to twice 29.

• The new line makes a triangular change in area. First decide whether each change makes the area grow or shrink.
• All three triangles span two days. Compare each dot’s height with the height of the new line at that day.

Answer: B — 18 degrees

Erasing Tuesday’s 14-degree dot shrinks the area: it is 3 degrees above the line joining 10 and 12. Erasing Thursday’s 18-degree dot shrinks it more: it is 4 degrees above the line joining 12 and 16. The triangles have equal bases, so the taller one removes more area. Wednesday’s 12-degree dot is also 4 degrees from its new line, but it lies below that line, so erasing it makes the area grow instead.

• Think about how few marbles two neighboring bowls could hold together if every bowl were fairly full. Then consider several pairs in the row.
• Pair bowls 1–2, 3–4, and 5–6, leaving bowl 7 on its own. Could every bowl hold at least 6 marbles?

Answer: E — 5

If every bowl held at least 6 marbles, each of the three neighboring pairs would hold at least 6 + 7 = 13. With at least 6 in the remaining bowl, that would require 45 marbles, so 6 is impossible. But the row 5, 12, 5, 6, 5, 6, 5 totals 44 and has different amounts in neighboring bowls. Thus the answer is 5. The tempting answer 6 comes from dividing 44 by 7 without allowing for the neighboring bowls.

• Can you split the cards into pairs whose numbers differ by 3?
• Find six cards that avoid such a pair.

Answer: B — 7

Pair the cards as (1,4), (2,5), (3,6), (7,10), (8,11), and (9,12). With 7 cards, Mila must have both cards from at least one pair. But 6 is not enough: she could draw 1, 2, 3, 7, 8, and 9, with no two numbers differing by 3. So the answer is 7; the trap in choosing 6 is thinking that one card from each pair is enough.