Grade 11 math puzzles — free practice & printable teacher quizzes

• Look at the three equally spaced inputs 0, 1, and 2. How does the middle value relate to the two outer values?
• For an exponential function, f(1)² = f(0)f(2). Use that with the given sum, then decide which outer value is larger.

Answer: E — 40

The outer values have sum 25 and product f(1)² = 100, so they are 5 and 20. Since f is increasing, f(0) = 5 and f(2) = 20. Each step therefore doubles the value, giving f(3) = 40. The tempting 5/2 comes from reversing 5 and 20, which would make f decrease.

• Rewrite each logarithm as an exponential statement. Remember that a logarithm’s base can lie between 0 and 1.
• Let t = log₂ x. Then x = 2^t, and the given equation also says x^t = 16.

Answer: D — 17/4

Let t = log₂ x. Since x = 2^t and x^t = 16, we have 2^(t²) = 2⁴. Thus t = ±2, giving x = 4 or x = 1/4. Their sum is 17/4. The tempting answer 4 misses the valid base 1/4, for which both logarithms are negative.

• How does a quadratic’s change in height behave across successive equal-width jumps?
• The height gains form an arithmetic sequence. Find its common difference, then count the gaps from the first dot to the twentieth.

Answer: B — 7508

The first two height gains are 110 − 41 = 69 and 215 − 110 = 105. Because the x-coordinates are equally spaced and the graph is quadratic, each successive gain increases by 36. There are 19 gains from dot 1 to dot 20, so the last gain is 69 + 18 × 36 = 717. The 20th height is 41 + 19(69 + 717)/2 = 7508. The tempting 8261 counts 20 gains, which reaches dot 21 instead.

• The product collects solutions from both quadratics. Compare when each can have two real roots before counting the solutions together.
• The two discriminants have opposite signs unless both are zero. In that special case, check whether the repeated roots coincide.

Answer: C — t = 1

The discriminants are 4(t² − 1) and 4(1 − t²). Unless t² = 1, one is positive, so the product has at least two distinct real solutions. At t = 1, both factors are (x − 1)², giving just x = 1. At t = −1, the factors are (x + 1)² and (x − 1)², giving two solutions. Thus t = 1; the tempting answer t = −1 or t = 1 overlooks that the repeated roots differ when t = −1.

• Find the total Ada marks and the total Ben marks separately, then think about the cups they both mark.
• The shared cups are numbered by multiples of 6. Each of the three groups has a geometric sum.

Answer: B — 16/9

The sums in Ada’s and Ben’s cups are 1 + 1/4 + ... = 4/3 and 1/2 + 1/16 + ... = 4/7. Cups they both mark have numbers divisible by 6 and sum to 1/16 + 1/1024 + ... = 4/63. To keep only cups marked by exactly one person, subtract that shared sum twice: 4/3 + 4/7 − 2(4/63) = 16/9. Subtracting it just once gives 116/63, which incorrectly includes the shared cups.

• Imagine continuing the draws until the bag is empty. How could the end of that complete sequence describe what was left when you stopped?
• Count reds after the last blue and blues after the last red separately. Think about the six gaps around the five marbles of the opposite color.

Answer: D — 5/3

In a complete draw order, the marbles left when you stop are either the reds after the last blue or the blues after the last red. The five blues create six gaps for the reds. By symmetry, each gap contains an expected 5/6 reds, so an expected 5/6 reds follow the last blue. Likewise, an expected 5/6 blues follow the last red. The expected number left is therefore 5/6 + 5/6 = 5/3. The tempting answer 5/6 counts only the cases where reds are left behind.

• First count arrangements on eight numbered seats. Then ask whether every valid arrangement produces eight different rotations.
• To count on numbered seats, use inclusion–exclusion: temporarily glue together any equal-letter pairs that are required to be adjacent.

Answer: A — 96

Choose k of the four equal-letter pairs to be adjacent. Gluing those pairs into blocks gives 8(7−k)!/2^(4−k) arrangements on numbered seats. Inclusion–exclusion therefore gives 2520−2880+1440−384+48=744 valid arrangements. Of these, 4!=24 repeat after a half-turn: their first four letters contain A, B, C and D once each. Every other valid arrangement has eight distinct rotations, while these 24 have only four. The answer is (744−24)/8+24/4=96. The tempting answer 93 comes from dividing all 744 by eight, overlooking the half-turn arrangements.

• Treat a², b², and c² as three new numbers, and look for what their sum and pairwise products tell you.
• To check whether a bound can actually be reached, set c=t and use (a−b)²≥0 to restrict t.

Answer: E — 66

Since a+b+c=0, we have ab+bc+ca=−3. Also, a²b²+b²c²+c²a²=(ab+bc+ca)²−2abc(a+b+c)=9. The sum-of-cubes identity applied to a², b², c² gives a⁶+b⁶+c⁶=6³−3·6·9+3a²b²c²=54+3a²b²c². Now let c=t. Then a+b=−t and ab=t²−3, so (a−b)²=12−3t²≥0 and t²≤4. Thus 4−a²b²c²=4−t²(t²−3)²=(4−t²)(t²−1)²≥0. The maximum is 54+3·4=66, reached at (a,b,c)=(−1,−1,2). The tempting 78 comes from treating a², b², c² as if all three could equal 2; their square roots cannot then sum to zero.

• The simplest formula fitting the five codes need not be the hidden program. What must every possible change to that formula have in common?
• Find a small polynomial that fits, then consider the roots of the difference between it and the hidden polynomial.

Answer: E — 218

The five codes fit B(x)=3x²−2x+2, which gives B(6)=98. For any permitted program P, the integer-coefficient polynomial P−B has roots 1, 2, 3, 4, 5. Hence P(x)−B(x)=(x−1)(x−2)(x−3)(x−4)(x−5)Q(x) for an integer-coefficient polynomial Q. At key 6, the code must therefore be 98+120Q(6). Only 218 fits, and taking Q(x)=1 shows it is possible. The tempting 158 differs from 98 by 60, which passes separate divisibility checks against the five earlier keys, but those checks miss the required factor of 120.

• Factor the increase since hour 4, then study its divisibility by 49 separately from the fourth-power condition.
• First check when powers of 2 equal 1 modulo 7. Then use 2³ = 1 + 7 to check which of those powers equal 1 modulo 49.

Answer: C — 67

At hour t, the count is 2^(t+1), so its increase since hour 4 is 32(2^(t−4)−1). Because 32 is coprime to 49, the increase is divisible by 49 exactly when 2^(t−4) ≡ 1 (mod 49). Powers of 2 repeat every three exponents modulo 7, so write t−4 = 3m. Then 2^(t−4) = (1+7)^m ≡ 1+7m (mod 49), which equals 1 modulo 49 exactly when 7 divides m. Thus t = 4+21j for a positive integer j. The count is a fourth power when t+1 = 5+21j is divisible by 4, or j ≡ 3 (mod 4). The smallest such j is 3, giving hour 67. Hour 25 is tempting: its increase is divisible by 49, but its count, 2^26, is not a fourth power.

• A move touches three positions. Can you group the lamps so that every move treats the groups alike?
• Try the groups {1, 4, 7}, {2, 5, 8}, and {3, 6, 9}. What happens to the parity of the number of lit lamps in each?

Answer: B — {3, 4, 6, 7}

Every move flips one lamp in each of the three groups, so their numbers of lit lamps must always have the same parity. For {3, 4, 6, 7}, those numbers are 2, 0, and 2; it is reachable by flipping {3, 4, 5} and then {5, 6, 7}. Every other choice has unequal group parities. In particular, {1, 2, 3, 4} has an even total of four lit lamps, but its group counts are 2, 1, and 1.

• Use AB and AC as coordinate directions; they need not be perpendicular.
• Find X, then compare areas with the whole triangle.

Answer: C — 2/7

Use coordinates A = (0,0), B = (1,0), and C = (0,1); this change of shape preserves area ratios. Then P = (2/3,0) and Q = (0,1/3). The lines CP and BQ have equations 3x + 2y = 2 and x + 3y = 1, so X = (4/7,1/7). Triangles ACX and ABX occupy 4/7 and 1/7 of ABC, respectively. The remaining triangle BXC therefore occupies 1 − 4/7 − 1/7 = 2/7. The tempting choice 5/7 adds the two triangles beside A instead of finding the requested triangle.