Grade 10 math puzzles — free practice & printable teacher quizzes

• Think of the expression as a graph. What feature of an upward-opening parabola marks its minimum, and how could you locate that feature?
• For y = x² + bx + c, the axis of symmetry is x = −b/2. Use it to find the vertex, then evaluate the expression there.

Answer: E — 2

The coefficient of x² is positive, so the parabola opens upward and its vertex gives the minimum value. Its axis of symmetry is x = −(−6)/2 = 3. At x = 3, the expression is 3² − 6(3) + 11 = 2, so the correct choice is 2. The choice 3 is the x-value where the minimum occurs, not the minimum value.

• Imagine multiplying every side by the same unknown factor. How would the perimeter and area respond?
• Let t be the fractional increase in side length. Express both percent increases in terms of t, then compare them.

Answer: A — 20%

Let t be the fractional increase in each side. The perimeter increases by 100t%, while the area increases by 100((1+t)² − 1)%. Their difference gives t² + t = 0.24, or (t − 0.2)(t + 1.2) = 0. Since the side increased, t = 0.2, so the answer is 20%. The 44% choice is the area’s increase, not the side’s.

• The region between the triangles is only part of the outer triangle. What is the area of the whole outer triangle compared with the inner one?
• For similar triangles, the area ratio is the square of the perimeter ratio.

Answer: A — 78 cm

The outer triangle has 1 + 5/4 = 9/4 times the inner triangle’s area. Its perimeter is therefore 3/2 times the inner perimeter, so the two perimeters are in the ratio 3:2. The larger perimeter is (3/5) × 130 = 78 cm. The tempting 90 cm comes from incorrectly using the area ratio 9:4 as the perimeter ratio.

• That side could belong to a square on either side of the line. How can the origin help you choose between them?
• The side from (0, 1) to (1, 3) moves right 1 and up 2. For the next side, swap the step sizes and reverse one direction; then decide which turn points toward the origin’s side.

Answer: A — (3, 2)

From (0, 1) to (1, 3), the side moves (1, 2). An equal-length perpendicular move toward the origin’s side is (2, −1), which has slope −1/2. Adding it to (1, 3) gives the diagonally opposite corner (3, 2). The tempting (−1, 4) uses the opposite turn, placing the square on the wrong side of the line.

• Look for complementary-angle twins inside each sum before calculating any exact values.
• After simplifying A and B, combine the fractions and use a double-angle identity.

Answer: C — 4

Because 15° and 75° are complementary, A = 2 sin 15° and B = 2 cos 15°. Thus A/B + B/A = (sin² 15° + cos² 15°)/(sin 15° cos 15°) = 1/(sin 15° cos 15°) = 2/sin 30° = 4. The tempting 2 comes from treating A and B as equal, but sin 15° and cos 15° are not equal.

• Think about what the triangles tell you about the polygon’s original boundary, not just the total number of triangles.
• Let x be the number of triangles with two boundary sides. Express both the total number of triangles and the number of polygon sides in terms of x.

Answer: B — 17

Let x triangles have two boundary sides. There are 4 + 5 + x = 9 + x triangles, so the polygon has 11 + x sides. Counting its boundary sides through the triangles instead gives 5 + 2x. Thus 11 + x = 5 + 2x, so x = 6 and the polygon has 17 sides. The tempting answer 15 is the number of triangles, not sides.

• A number is a square when every prime in its factorization occurs an even number of times. Think about both divisor counts at once.
• For each prime with exponent a in d, its contributions to the two divisor counts are a+1 and 6−a.

Answer: B — 96

For j = 1, 2, …, 6, let n_j count the primes whose exponent in d is j−1. Each contributes j to the divisor count of d and 7−j to that of N/d. For both counts to be squares, the prime factor 5 forces n₅ and n₂ to be even; the factor 2 then forces n₆ and n₁ to be even; and the factor 3 forces n₃ and n₄ to be even. Conversely, these conditions make both counts squares. Since the six counts n_j add to 4, either one j occurs four times (6 ways), or two different j values occur twice each (C(6,2)·C(4,2) = 90 ways). The answer is 96; the tempting 90 forgets the all-four-equal cases.

• Moving seven places keeps the base the same modulo 7. Does it also keep the exponent at the same place in its cycle?
• Consider 42 consecutive terms. For each fixed base modulo 7, what exponent remainders occur?

Answer: B — 2

For a base not divisible by 7, powers repeat with a period dividing 6. In any 42 consecutive terms, each base modulo 7 occurs with all six exponent remainders modulo 6. A base of 0 contributes 0; a base of 1 contributes 6; and each base a from 2 through 6 contributes 0 because 1 + a + ⋯ + a⁵ ≡ 0 (mod 7). Thus each 42-term block sums to 6 modulo 7. Since 2020 = 48·42 + 4, the remaining four terms have remainders 1, 4, 6, and 4. The total is 48·6 + 1 + 4 + 6 + 4 = 303, leaving remainder 2. The tempting answer 6 comes from treating seven terms as a full cycle: after seven places the base repeats modulo 7, but the exponent has shifted in its cycle.

• Think about what the product condition tells you about ratios of consecutive terms, rather than calculating many terms.
• Find the first two ratios from the given terms. Then check the exponents of 2 and 3 at even and odd positions separately.

Answer: D — 204

Let r and s be the first two ratios of consecutive terms. The product condition makes the ratios repeat as r, s, r, s, ... . The 9th term gives 2(rs)^4 = 2592, so rs = 6; the 4th gives 2r²s = 36, so r = 3 and s = 2. Thus the term at position 2k is 6^k, while the term at position 2k+1 is 2^(k+1)3^k. An odd-position term cannot be a cube because its two exponents differ by 1. At an even position, k must be at least 101 for divisibility and a multiple of 3 for the term to be a cube. The least such k is 102, giving position 204. The tempting choice 202 gives 6^101, which is divisible by the required number but is not a cube.

• Three different slices leave exactly one slice out. Are the four possible missing slices equally likely?
• For each possible missing slice, multiply the percentages of the three slices that remain. Each group of three can occur in the same number of orders.

Answer: A — 88%

The four groups of three slices have relative weights 40×30×20, 40×30×10, 40×20×10, and 30×20×10. Their total is 50,000; the group without books has weight 6,000. So the probability is 1 − 6,000/50,000 = 88%. The tempting 75% treats the four possible missing slices as equally likely, but their weights differ.

• Instead of following one card, look at pairs of cards: which pairs have reversed their original left-to-right order?
• A rotation of three cards can be made with two swaps. Think about what each swap does to the parity of the number of reversed pairs.

Answer: B — 2, 3, 4, 1, 5

Each rotation is two swaps, so it preserves the parity of the number of reversed pairs. In 2, 3, 4, 1, 5, the card 1 has passed three cards, giving an odd number; that row is impossible. The other rows can all be made in two turns: 2, 3, 4, 5, 1 by rotating positions 1, 2, 3 left, then 3, 4, 5 left; 2, 1, 4, 3, 5 by rotating 1, 2, 3 right, then 1, 3, 4 left; and 3, 1, 2, 4, 5 by rotating 1, 2, 3 left twice. Even the original row 1, 2, 3, 4, 5 is possible: rotate the same three positions left, then right.

• What happens if a faster printer gets a shorter-lasting battery than a slower printer?
• Start with the best assignment, then look for the smallest loss from a swap.

Answer: D — 92

If a faster printer has a shorter-lasting battery than a slower printer, swapping their batteries increases the total. So the largest total pairs the printers and batteries in increasing order: 1×2 + 2×5 + 3×9 + 4×14 = 95. In any other assignment, such a swap can increase the total by at least (2−1)(5−2) = 3, so no other total exceeds 92. Swapping the 2- and 5-minute batteries in the best assignment loses exactly 3, giving 92. The tempting choice 95 is the largest total, not the second-largest.