Practice Questions¶

Send the two slowest together, and use the fastest to shuttle the torch back.

• C + D cross: 2 min
• D returns: 1 min
• A + B cross: 8 min
• C returns: 2 min
• C + D cross: 2 min

Total = 2 + 1 + 8 + 2 + 2 = 15 min.

The key insight is that the two slowest (A, B) should cross together so their 8 min is paid only once, and the fastest (D) does the torch shuttling.

This is the classic Cheryl's Birthday problem.

1. You (month) says C doesn't know. This rules out months that contain a unique day — a day that appears only once across all dates. Jun 18 and May 19 are unique days, so the month cannot be May or June.
2. C (day) now knows. Among the remaining months (Jul, Aug), the day must be unique. Jul 14 and Aug 14 share the day 14, so it cannot be the 14th. The remaining candidates are Jul 16, Aug 15, Aug 17.
3. You (month) now knows. If the month were August, there would still be two options (15, 17), so it must be a month with a single remaining candidate: July.

Answer: July 16.

By symmetry, the number of red pairs you collect equals the number of black pairs the dealer collects in expectation — the game is fair in terms of card counts. However, ties count as a loss for you.

The probability of a strict win is less than 1/2 (there is a positive probability of a tie), so the expected value is negative. You should not pay anything to play — the fair price is $0 (or you simply decline).

The house edge comes entirely from the tie rule, not from the card mechanics.

1. Light rope 1 at both ends and rope 2 at one end simultaneously.
2. Rope 1 burns completely in 30 min (both ends). At that moment, rope 2 has 30 min of burn left.
3. Immediately light the other end of rope 2. It now burns from both ends and finishes in 15 min.

Total = 30 + 15 = 45 min.

Split the 9 balls into three groups of 3: A, B, C.

1. Weigh A vs B.
2. If A > B: the heavy ball is in A.
3. If A < B: the heavy ball is in B.
4. If A = B: the heavy ball is in C.
5. From the identified group of 3, weigh any two balls against each other.
6. If one is heavier, that's the heavy ball.
7. If they balance, the third ball is heavy.

Each weighing has 3 outcomes, so \(3^2 = 9\) outcomes cover exactly 9 cases.

Trailing zeros come from factors of 10 = 2 × 5. Since 2s are far more abundant than 5s, count the number of factors of 5.

• Multiples of 5: \(\lfloor 500/5 \rfloor = 100\)
• Multiples of 25: \(\lfloor 500/25 \rfloor = 20\)
• Multiples of 125: \(\lfloor 500/125 \rfloor = 4\)
• Multiples of 625: \(\lfloor 500/625 \rfloor = 0\)

Total = 100 + 20 + 4 = 124 trailing zeros.

7 races.

1. 5 races: split into 5 groups of 5, race each group. Record the order within each group.
2. 1 race: race the 5 group winners. This gives an overall ranking of the groups.

Let the winner of the winners' race be group A, second be group B, third be group C. Eliminate: - Groups D and E entirely (their winners are already slower than 3 horses). - In group C, only its winner can still be in the top 3 (its 2nd and 3rd are slower than its winner, which is already 3rd overall). - In group B, only its winner and 2nd can still qualify. - In group A, its winner is already #1; its 2nd and 3rd can still qualify.

Candidates: A2, A3, B1, B2, C1 (5 horses). 3. 1 race: race these 5. The top 2, combined with A1, give the 3 fastest.

Total = 5 + 1 + 1 = 7 races.

Let the infinite expression equal \(x\). Then the expression inside the first square root is also \(x\):

Squaring both sides: \(x^2 = 2 + x \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0\).

Since \(x > 0\), we take \(x = 2\).

Answer: \(x = 2\).

Let \(D\) be the total number of visible steps, and \(x\) the number of steps the escalator carries a stationary rider.

Alice's speed is \(3/2\) times Bob's. Let Bob's climbing rate be \(v\), Alice's be \(1.5v\).

• Alice: \(D = 30 + x \cdot \frac{t_A}{t_{total}}\)... more directly, the escalator moves the same number of steps \(x\) for both, but in different times.

Time ratio: Alice counts 30 steps at rate \(1.5v\), so \(t_A = 30/(1.5v) = 20/v\). Bob counts 20 steps at rate \(v\), so \(t_B = 20/v\). They take the same time!

So the escalator contributes the same \(x\) steps to both: - \(D = 30 + x\) - \(D = 20 + x\)

These are inconsistent unless... let's redo. The escalator moves \(x\) steps in the time it takes to climb. Since both take the same time, \(x\) is the same.

\(D = 30 + x\) and \(D = 20 + x\) cannot both hold. The resolution: the escalator's contribution differs because the climb times differ. Let \(k\) be the ratio of Bob's time to Alice's time.

Alice: \(D = 30 + x_A\), Bob: \(D = 20 + x_B\), where \(x_B = k \cdot x_A\).

Alice time \(t_A = 30/(1.5v) = 20/v\). Bob time \(t_B = 20/v\). So \(k = 1\) — same time, same escalator contribution \(x\).

\(D = 30 + x = 20 + x\) is impossible, meaning the given numbers are inconsistent. This is a trick: with these speeds and counts, no consistent \(D\) exists. The intended setup (as in the original) uses different climb times. With Alice at 3 steps per Bob's 2, and counts 30 vs 20, both take equal time, so the problem as stated has no solution.

(For the original-style numbers — 25 and 20 with a 3:2 speed ratio — the answer is 50.)

Take any 10 coins into a separate pile, and flip all 10 of them.

Let the 10-coin pile contain \(m\) heads (and \(10 - m\) tails). After flipping, it contains \(10 - m\) heads. The remaining 90 coins contain \(10 - m\) heads (the rest of the original 10 heads). Both piles now have exactly \(10 - m\) heads.

Answer: separate 10 coins and flip them.

Draw one fruit from the bag labelled "Mixed".

Since all labels are wrong, the "Mixed" bag is actually all apples or all oranges. - If you draw an apple, the "Mixed" bag is the Apples bag. - If you draw an orange, the "Mixed" bag is the Oranges bag.

Now the bag labelled "Apples" cannot be apples (label wrong) and cannot be the bag you just identified, so it must be the Mixed bag. The remaining bag is the last fruit type.

One draw from the "Mixed" bag is sufficient.

Designate one counter (say man #1). The other 49 men each flip the glass from upside-down to upright exactly once — the first time they see it upside down — and otherwise do nothing.

The counter, whenever called, flips the glass from upright to upside down (resetting it) and increments a mental count. When the counter has seen the glass upright 49 times, all 49 other men have been called at least once, so the counter announces everyone has been called.

The counter must also handle the initial state: if the glass starts upside down, the counter's first flip-up doesn't count toward the 49.

There are 3 colours, so with 3 socks you could have one of each colour. The 4th sock must match one of them.

Answer: 4 socks (3 pigeonholes, so \(3 + 1\)).

Split the square into a \(5 \times 5\) grid of 25 smaller squares, each of side \(1/5\). A circle of radius \(1/5\) can cover one such small square (its circumradius is \(\frac{1/5}{\sqrt2} < 1/5\)).

By the pigeonhole principle, 26 ants in 25 squares means at least one square contains 2 ants. Place the circle over that square.

Answer: guaranteed by pigeonhole.

1. Fill the 5L jug.
2. Pour from the 5L jug into the 3L jug until the 3L jug is full. Now the 5L jug has 2L.
3. Empty the 3L jug.
4. Pour the 2L from the 5L jug into the 3L jug. The 3L jug now has 2L, with 1L of space.
5. Fill the 5L jug again.
6. Pour from the 5L jug into the 3L jug until the 3L jug is full (adds 1L). The 5L jug now has 4L.

Answer: 4L in the 5L jug.

Let A's extra two coins be considered separately. First compare A's first \(n\) coins against B's \(n\) coins.

By symmetry, the probability A's \(n\) coins beat B's \(n\) coins equals the probability they lose, and there is a tie with probability \(p_{tie}\).

• If A's \(n\) coins already beat B: A wins regardless of the 2 extra coins.
• If A's \(n\) coins tie B: A wins if the 2 extra coins produce at least 1 head, i.e. with probability \(3/4\).
• If A's \(n\) coins lose to B: A can still win if the 2 extra coins produce 2 heads and the deficit is exactly 1. This is more involved.

A cleaner approach: pair each of A's \(n\) coins with one of B's. For the \(n\) pairs, A wins a pair with prob \(1/4\), loses with prob \(1/4\), ties with prob \(1/2\). A's two extra coins add 0, 1, or 2 heads.

The full computation is tedious; the elegant result is that with \(n+2\) vs \(n\) coins, the probability A has strictly more heads is \(1/2\) (the extra two coins give A a strict advantage only through ties, and the symmetry argument generalises: \(P(\text{A more}) = P(\text{B more})\), and ties are broken in A's favour with probability \(3/4\) of the tie mass, but the net is still \(1/2\)).

Answer: \(1/2\).

Let \(X\) = you win, \(Y\) = tie, \(Z\) = dealer wins. By symmetry \(P(X) = P(Z)\).

The probability of a tie: after you draw a card, there are 3 cards of the same value among the remaining 51, so \(P(Y) = 3/51 = 1/17\).

Since \(P(X) + P(Y) + P(Z) = 1\) and \(P(X) = P(Z)\):

Answer: \(8/17\).

Consider only seats #1 and #100. The process is decided by which of these two seats gets taken first by a "displaced" passenger.

Whenever a passenger is forced to choose randomly, they are equally likely to pick seat #1 or seat #100. If seat #100 is taken first, passenger 100 loses; if seat #1 is taken first, passenger 100 wins (because then everyone else can sit properly and seat #100 remains).

By symmetry, the two outcomes are equally likely.

Answer: \(1/2\).

Fix one point. For each of the other \(N-1\) points, the probability it lies within the semicircle starting at the fixed point is \(1/2\). The events are disjoint for the \(N\) choices of the "starting" point, so:

Answer: \(\dfrac{N}{2^{N-1}}\).

• Probability A and B are in different quarters: A is placed in one quarter; B must be in one of the other 3 quarters out of 7 remaining spots: \(3/7\).
• Probability A wins their quarterfinal: \(1/2\).
• Probability B wins their quarterfinal: \(1/2\).

Answer: \(3/28\).

Let \(P_i\) be the probability A wins given A is \(i\) games ahead (so \(P_{-3}=0\), \(P_3=1\)).

For \(i \in \{-2,-1,0,1,2\}\): $\(P_i = \frac{2}{3}P_{i+1} + \frac{1}{3}P_{i-1}\)$

Solve the recurrence. Writing the equations: - \(P_2 = \frac{2}{3}(1) + \frac{1}{3}P_1 = \frac{2}{3} + \frac{1}{3}P_1\) - \(P_1 = \frac{2}{3}P_2 + \frac{1}{3}P_0\) - \(P_0 = \frac{2}{3}P_1 + \frac{1}{3}P_{-1}\) - \(P_{-1} = \frac{2}{3}P_0 + \frac{1}{3}P_{-2}\) - \(P_{-2} = \frac{2}{3}P_{-1} + \frac{1}{3}(0) = \frac{2}{3}P_{-1}\)

Solving this system gives \(P_0 = \frac{16}{27} \approx 0.593\).

Answer: \(16/27\).

Answer: \(1/221\).

Optimal strategy: keep the first roll if it is 4, 5, or 6; re-roll if it is 1, 2, or 3.

• Probability of rolling 4, 5, or 6: \(1/2\). Expected value if you keep: \((4+5+6)/3 = 5\).
• Probability of re-rolling: \(1/2\). Expected value of the second roll: \(3.5\).

Answer: \(4.25\).

Total hands: \(\binom{52}{5} = 2{,}598{,}960\).

A straight flush: choose the suit (4 ways) and the starting value of the run. Runs can start at A, 2, ..., 10 (10 possible runs, where A can be high or low). So \(4 \times 10 = 40\) straight flushes.

Answer: \(40 / \binom{52}{5} \approx 1.54 \times 10^{-5}\).

Let \(f(n)\) be the number of ways. The last hop is either 1 stair (from \(n-1\)) or 2 stairs (from \(n-2\)):

This is the Fibonacci sequence. For example, \(f(10) = 89\).

Answer: Fibonacci numbers.

For every group of 3 pirates (who must NOT be able to open the safe), there must be a lock that none of them has a key to, but that every one of the other 4 pirates does have a key to.

Number of 3-pirate groups: \(\binom{7}{3} = 35\) locks.

Each lock is opened by the 4 pirates not in its associated group, so each lock has 4 keys. Total keys = \(35 \times 4 = 140\). Each pirate carries \(140/7 = 20\) keys.

Answer: 35 locks, 20 keys per pirate.

1. Arrange the 5 men around the circle: \((5-1)! = 24\) ways.
2. There are 5 gaps between the men. Choose 3 gaps for the women: \(\binom{5}{3} = 10\).
3. Arrange the 3 women in those gaps: \(3! = 6\).

Total = \(24 \times 10 \times 6 = 1440\).

Answer: 1440 ways.

Sample space (equally likely): BB, BG, GB, GG. Given at least one boy, we restrict to {BB, BG, GB}. Only BB has two boys.

Answer: \(1/3\).

Let \(A\) = double-headed coin, \(B\) = 10 heads in a row.

• \(P(A) = 1/1000\)
• \(P(B \mid A) = 1\)
• \(P(B \mid A^c) = (1/2)^{10} = 1/1024\)

By Bayes: $\(P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A^c)P(A^c)} = \frac{1 \cdot \frac{1}{1000}}{1 \cdot \frac{1}{1000} + \frac{1}{1024} \cdot \frac{999}{1000}}\)$

Answer: \(\approx 0.506\).

Let \(p\) = probability the first player loses. The first player loses if the gun fires on their first pull (prob \(1/6\)), or if it doesn't (prob \(5/6\)) and then the roles reverse — the first player now faces the same situation as the second player, whose loss probability is \(1 - p\).

The first player loses with probability \(6/11 > 1/2\), so you should go second, with loss probability \(5/11\).

Answer: go second; loss probability \(5/11\).

Deal the aces first. The first ace can go to any player; the second ace must go to a different player (39 of the remaining 51 positions); the third to a different player (26 of 50); the fourth to the last player (13 of 49).

Answer: \(\approx 0.105\).