Hint:

This is a classic economics problem that even economists get wrong!

Solution:

This is a terrible proposition for both you and me.

From my perspective, if I have a hundred people playing, I am losing 5% each round.

From your perspective, unless you are the lucky ten percent (~13.6%), you are losing 5% each round.

Here is the math behind it.

1. House perspective: expected value

Each round, the player’s bankroll is multiplied by:

Heads: ×1.5
Tails: ×0.6

Expected multiplier:

0.5 × 1.5 + 0.5 × 0.6

= 0.75 + 0.30

= 1.05

So the player gains 5% in expected value per round.

That means the house loses 5% in expected value per round.

After 100 rounds, expected player bankroll is:

$10,000 × 1.05^100 ≈ $1,315,013

So from the house’s perspective, this is a terrible game to offer.

2. Player perspective: compound growth

For the player, expected value is misleading because the whole bankroll is being compounded.

The compound-growth multiplier is the geometric mean:

sqrt(1.5 × 0.6)

= sqrt(0.9)

≈ 0.9487

So the typical player loses:

1 - 0.9487 = 0.0513

or about:

5.13% per round

That is why the player’s typical outcome is bad even though the expected value is positive.

3. After 100 rounds

If the player gets H heads and 100 - H tails, final bankroll is:

$10,000 × 1.5^H × 0.6^(100-H)

To finish ahead, they need:

1.5^H × 0.6^(100-H) > 1

Take logs:

H ln(1.5) + (100 - H) ln(0.6) > 0

Solve:

H > 55.749...

So the player needs at least:

56 heads out of 100

The probability of getting 56 or more heads with a fair coin is:

P(H ≥ 56) ≈ 13.6%

So:

About 13.6% of players win.
About 86.4% of players lose.

Hint:

Think like Kelly!

Solution:

For this you need to use the Kelly formula.

Let:

f = fraction of bankroll bet

Win probability:

p = 0.5

Lose probability:

q = 0.5

Win return on bet:

b = 0.5

Loss rate on bet:

a = 0.4

The bankroll multiplier is:

Win: 1 + bf = 1 + 0.5f

Loss: 1 - af = 1 - 0.4f

To maximize compound growth, maximize expected log growth:

G(f) = p ln(1 + bf) + q ln(1 - af)

Plug in the numbers:

G(f) = 0.5 ln(1 + 0.5f) + 0.5 ln(1 - 0.4f)

Take derivative and set to zero:

G'(f) = 0.5 × 0.5 / (1 + 0.5f) - 0.5 × 0.4 / (1 - 0.4f)

G'(f) = 0.25 / (1 + 0.5f) - 0.20 / (1 - 0.4f)

Set equal to zero:

0.25 / (1 + 0.5f) = 0.20 / (1 - 0.4f)

Cross-multiply:

0.25(1 - 0.4f) = 0.20(1 + 0.5f)

0.25 - 0.10f = 0.20 + 0.10f

0.05 = 0.20f

f = 0.25

So:

Optimal bet = 25% of current bankroll.

Solution:

The ages of the three children are 9, 2, and 2. The product of these ages is 36 (9 * 2 * 2 = 36), and their sum is 13. The address to the north must be 13, as it matches the sum of the ages. The mention of 'the oldest' indicates there is a distinct oldest child, ruling out other combinations like 6, 6, and 1, which would not have a single oldest child.

Solution:

The expression (a-x)(b-x)(c-x)(d-x)...(z-x) represents the product of linear factors where each factor is of the form (letter-x). If we let x equal any letter from a to z, one of the factors will be (x-x), which equals zero. Therefore, the entire product is zero, as the product of zero and any number is still zero.

Solution:

Fill the 13-gallon bucket completely and pour it into the 18-gallon bucket. Fill the 13-gallon bucket again and pour water into the 18-gallon bucket until it is full. Since the 18-gallon bucket already has 13 gallons, it can only take 5 more gallons. This will leave exactly 1 gallon of water in the 13-gallon bucket.

Solution:

The figures are the letter-counts of the Solar System’s bodies in order outward from the Sun: Sun (3), Mercury (7), Venus (5), Earth (5), Mars (4), Jupiter (7), Saturn (6), Uranus (6), Neptune (7). The next body is Pluto, which has 5 letters, so ? = 5.

Solution:

23

Solution:

The water-lily doubles in size every day. If the entire pool is covered on day 20, then half of the pool must have been covered the day before, which is day 19. Therefore, half of the pool will be covered by the water-lily after 19 days.

Solution:

500

Solution:

The formula for height, H = ut - 4.9t², is derived from the general equation of motion s = ut + 0.5at², where s is the distance traveled, u is the initial velocity, a is the acceleration (in this case, -9.8 m/s² due to gravity), and t is time. The term '4.9' represents half of the acceleration due to gravity, and the seconds are squared because acceleration is a change in velocity over time, making it meters per second per second.

Solution:

The smallest number that satisfies the condition of having a 2 at its unit place and, when doubled, results in the same digits with the 2 shifted to the front is 105263157894736842. When this number is multiplied by 2, it becomes 210526315789473684, which confirms the condition.

Hint:

This question requires calculus. If you don't know calculus, you probably shouldn't waste your time on this.

Solution:

The probability that a needle of length 1 will intersect a line on a floor with parallel lines 1 inch apart is 1/2. This is derived from considering the angle at which the needle falls and the distance from the center of the needle to the nearest line. The average probability of intersection can be calculated by integrating the cosine of the angle from 0 to π/2, leading to a result of 2/π, which is approximately 63.66%. Therefore, the correct probability is actually 2/π, not 1/2.

Hint:

Think about the worst-case order in which the colors could appear, and apply the pigeonhole principle.

Solution:

To guarantee two gloves of every color, consider the worst case: try to delay completing one of the color pairs as long as possible.

Suppose blue is the color that is delayed. She could first draw every red and yellow glove plus one blue glove without yet having two blues. That is

25 (red) + 45 (yellow) + 1 (blue) = 71 gloves.

The very next glove she draws must be blue (only blues are left), giving her the second blue she needs. Therefore she must take

71 + 1 = 72 gloves

to be certain of having a pair of each color. Answer: A) 72.

Solution:

The confusion arises from the incorrect addition of amounts. The ladies initially paid £30, and after receiving £3 back, they effectively paid £27 for the TV. Out of this £27, £25 went to the shop and £2 went to the salesman. The error in reasoning comes from trying to add the £2 in the salesman's pocket to the £27, which already includes that amount. Therefore, there is no missing £1; the £27 accounts for both the £25 and the £2.

Solution:

To determine the width of the river without crossing it, first measure a specific distance along the bank using the meter stick. Then, look at a tree or rock directly across the river and use the protractor to measure the angle from your new position to the tree/rock. Using the tangent of the angle, you can calculate the width of the river with the formula: width = distance traveled along the bank * tan(angle).

Solution:

Split 271 into 100 equal terms:

271 = 2.71 + 2.71 + … + 2.71 (100 times)

This yields the maximal product, (2.71)^100.

Solution:

The open lunch boxes correspond to the perfect squares among the numbers 1 to 1000. This is because a lunch box is toggled (opened or closed) for every divisor it has, and only perfect squares have an odd number of divisors. The perfect squares up to 1000 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, totaling 31 open lunch boxes.

Hint:

Ask yourself: which visible cards could possibly violate the rule? Those are the only cards you need to check.

Solution:

The rule can be violated in two ways:

1. A card shows 2 on the face we see, but the hidden side is not 5.
2. A card shows something other than 5 on the face we see, but the hidden side is 2.

Therefore we must turn over:

• The card showing 2, to confirm that its reverse side is 5.
• Every card that does not show 5 — that is, the cards showing 1, 3 and 4 — to be sure none of them hide a 2.

We do not need to turn over the card showing 5, because regardless of what is on its back it cannot contradict the rule.

Thus the minimum is four cards: 2, 1, 3, and 4.

Solution:

Multiply 5×2 = 10, then 5×3 = 15, then 5² − 3 = 22; write the three results consecutively: 10 15 22 → 101522.

Solution:

Rs. 31

Solution:

Let the number of stickers in a full sheet be 'x'. Winston has 3x + 4 stickers, while Wendy has 2x + 12 stickers. Since they have the same total number of stickers, we can set up the equation 3x + 4 = 2x + 12. Solving for x gives x = 8, so there are 8 stickers in a full sheet.

Solution:

The pricing policy of the shop is based on the number of digits in the quantity picked rather than the quantity itself. Each digit costs $2, so for 246, which has three digits, the cost is $6 (3 digits x $2 per digit). The shopkeeper charged $6 for the three digits in 246, which falls within the range of 100-999 items, where the price is $6.

Solution:

To find the person's weight on Earth based on the weights given for the moon, Mars, and Jupiter, we first need to understand that weight is the product of mass and gravitational acceleration. The weight on the moon is 12 kg, which corresponds to a gravitational acceleration of about 1.62 N/kg. This implies a mass of approximately 7.41 kg (12 kg / 1.62 N/kg). On Earth, with a gravitational acceleration of 9.81 N/kg, this mass would weigh about 72.7 N (7.41 kg * 9.81 N/kg). The weight on Jupiter is given as 189.982 kg, which corresponds to a gravitational acceleration of about 24.79 N/kg, leading to a mass of approximately 7.66 kg (189.982 kg / 24.79 N/kg). Therefore, the person's weight on Earth is approximately 75.18 N (7.66 kg * 9.81 N/kg). Thus, the person's weight on Earth is around 75.18 N or approximately 75 kg.

Hint:

Let x be the number of apples each person picked at the first tree. Translate each bullet into an equation, keeping track of the running total.

Solution:

Let x be the number of apples each friend picked at the first tree.

First tree: total = 3x.

Second tree: each picks 3x more, so total second-tree harvest = 3 × 3x = 9x.
Cumulative total after second tree = 3x + 9x = 12x.

Third tree: when they finish, they have 5 times what they had when they started this tree, i.e. 5 × 12x = 60x.
They therefore picked 60x − 12x = 48x at the third tree.

Fourth tree: they add 5 apples, giving a final total of 60x + 5.

This final total is given as 65 apples, so
60x + 5 = 65 ⇒ 60x = 60 ⇒ x = 1.

Therefore each person picked 1 apple at the first tree.

Solution:

To determine the probability of getting a head on the fourth toss after tossing three heads, we need to consider the coins in the jar: two coins are heads and tails (A and B), and one coin is heads and heads (C). The probability of selecting each coin is 1/3.

Given that we have already tossed three heads, we can use conditional probability. The only way to get three heads is if we have selected coin C (the heads and heads coin) or one of the other coins (A or B) and happened to get heads all three times. The probability of getting three heads with coin A or B is (1/2)^3 = 1/8 for each, and since there are two such coins, the total probability for A and B is 2 * (1/3) * (1/8) = 1/12.

The total probability of getting three heads is the sum of the probabilities for all coins: P(3 heads) = P(3 heads | C) + P(3 heads | A or B) = 1/3 + 1/12 = 5/12.

Now, we want to find the probability of having coin C given that we got three heads: P(C | 3 heads) = P(3 heads | C) * P(C) / P(3 heads) = (1/3) * (1) / (5/12) = 4/5.

Thus, the probability of getting a head on the fourth toss is: P(head on 4th toss | 3 heads) = P(head | C) * P(C | 3 heads) + P(head | A or B) * P(A or B | 3 heads) = 1 * (4/5) + (1/2) * (1/5) = 4/5 + 1/10 = 9/10.

Therefore, the probability of getting a head on the fourth toss is 9/10.

Solution:

The number that satisfies all the conditions is 381654729. It is a 9-digit number using each digit from 1 to 9 exactly once, is divisible by 9, and removing digits from the right results in numbers that are divisible by 8, 7, 6, 5, 4, 3, 2, and 1 respectively.

Solution:

The biologist can use the capture-recapture method to estimate the total elk population. If 125 elk were initially tagged and 34 out of 920 sampled elk have transmitters, the estimated total elk population is calculated as (920 * 125) / 34, which equals approximately 3,382 elk in the preserve.

Solution:

In a round robin format, the number of matches played is given by the formula n(n-1)/2, where n is the number of teams. Setting this equal to 45 gives the equation n(n-1) = 90. Solving this quadratic equation, we find that n = 10, so there are 10 teams participating.

Solution:

The confusion arises from misadding the amounts. The three men originally paid $30, and after receiving $3 back, they effectively paid $27 for the room. This $27 includes the $25 for the room and the $2 kept by the bellboy. Therefore, there is no missing dollar; the total should not be calculated by adding the bellboy's $2 to the $27, as that amount already includes it.

Solution:

He arrives 1 minute early.

Solution:

The maximum number of inhabitants can be 518 if one person is bald (having 0 hairs), allowing for 517 others to have unique hair counts from 1 to 517. If all inhabitants have hair, the maximum is 517, as no one can have exactly 518 hairs.

Solution:

The shopkeeper lost $800 in cash given to the lady and $200 worth of goods, totaling a loss of $1000. The $1000 returned to the neighboring shopkeeper is not an additional loss, as it is a reimbursement for the fake note.

Solution:

The game involves two players alternately stating numbers, with each number needing to be within 10 of the previous number. The key to winning is to start with the number 1, allowing the first player to control the game. No matter what number the second player chooses, the first player can always respond with a number that is 11 more than the second player's choice, ultimately leading to 100 and securing the win.

Solution:

Initially, the 100 lbs of potatoes consist of 99% water, meaning they have 1 lb of solid content. When the water content decreases to 98%, the solid content remains the same at 1 lb. To find the new weight, we set up the equation: 1 lb (solid) / (new weight) = 0.02 (2% solids), leading to a new weight of 50 lbs.

Solution:

The dealership has 20 Hondas, 20 Subarus, and 40 Fords, totaling 80 vehicles. However, the wording does not specify that these are the only cars, nor does it clarify if they include trucks or other types of vehicles.

Solution:

Let the cost of the stick of gum be x dollars. Then the cost of the can of soda is x + $1. According to the puzzle, the total cost is x + (x + $1) = $1.10. Solving this gives 2x + $1 = $1.10, which simplifies to 2x = $0.10, so x = $0.05. Therefore, the stick of gum costs $0.05.

Solution:

You should always switch to the other door. Initially, you have a 1/3 chance of picking the car and a 2/3 chance of picking a goat. When the host reveals a goat behind one of the other doors, switching gives you the 2/3 chance of winning the car, while sticking with your original choice keeps you at 1/3. If you originally picked a goat and then switch, you will get the car, which is why switching is the better strategy.

Solution:

Initially, 15% of 27 goats got pregnant, which is 4.05, rounded down to 4 goats. Each had 7 kids, so 4 x 7 = 28 kids. In total, there are now 27 + 28 = 55 goats. However, 62% of the 28 kids died, which is 17.36, rounded down to 17 kids. Therefore, 28 - 17 = 11 kids survived. The total number of goats left is 27 + 11 = 38 goats.

Solution:

Sea x el costo de cada martillo y y el costo de cada brocha. Se tienen las siguientes ecuaciones: 7x + 3y = 555 y 5x + 10y = 750. Resolviendo este sistema de ecuaciones, se encuentra que x = 75 y y = 30. Por lo tanto, cada martillo costó $75.00.

Solution:

Para resolver esto, asignamos valores a las letras A, B, C, y D. La ecuación se puede reescribir como: 1000A + 100B + 10C + D - D = 1000D + 100C + 10B + A. Simplificando, obtenemos: 999A + 90B - 90C - 999D = 0. Esto implica que A y D deben ser iguales y que B y C deben ser iguales. Una solución válida es A=1, B=0, C=0, D=1, lo que satisface la ecuación.

Solution:

One possible solution is: T=7, H=8, I=9, S=6, A=1, R=0, D=5. This gives us 7816 + 19 = 7835, which is correct.

Solution:

The shopkeeper faced a loss of Rs. 1000. Here's the breakdown: The shopkeeper gave away goods worth Rs. 200 and Rs. 800 in cash, totaling Rs. 1000. Since the 1000 rs note was fake, he had to return Rs. 1000 to the neighboring shopkeeper, resulting in a total loss of Rs. 1000.

Solution:

The ‘51 rupees’ comes from an invalid addition of overlapping leftovers. After each spend you have less money than before, but those leftover amounts are not disjoint portions of your original 50 rupees—they are successive remainders of the same pool. You cannot add them together. The correct total spent is 20 + 15 + 9 + 6 = 50 rupees, and the final remainder is 0. The 51 rupees arises only if you mistakenly sum the successive remainders (30 + 15 + 6 + 0), which is a logical error, not a real extra rupee.

Solution:

The man binds 200 books on his days, the helper 50 books (one-quarter of 200). They alternate days, so in each two-day cycle they bind 200 + 50 = 250 books. To reach 1000 books they need 1000 ÷ 250 = 4 cycles, which is 4 × 2 = 8 days.

Solution:

Step 1: Fill the 3-gallon jug and pour it into the 5-gallon jug.
Step 2: Fill the 3-gallon jug again. Pour from it into the 5-gallon jug until the 5-gallon jug is full. Since it already had 3 gallons, you pour in 2 more gallons, leaving 1 gallon in the 3-gallon jug.
Step 3: Empty the 5-gallon jug and pour the remaining 1 gallon from the 3-gallon jug into the 5-gallon jug.
Step 4: Fill the 3-gallon jug once more and pour all 3 gallons into the 5-gallon jug, which already contains 1 gallon. You now have exactly 4 gallons in the 5-gallon jug.

Solution:

In a group of four people, each person shakes hands with three others. Since each handshake involves two people, the total number of unique handshakes can be calculated using the formula n(n-1)/2, where n is the number of people. Thus, the total number of handshakes is 4(4-1)/2 = 6.

Solution:

14

If the first egg never breaks, the test floors are 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100, which is fourteen total attempts. If the first egg breaks on the first drop (the 14th floor), use the second egg and test: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which is fourteen total attempts. If the first egg breaks on the fifth drop (the 60th floor), use the second egg and test: 51, 52, 53, 54, 55, 56, 57, 58, 59, which is fourteen total attempts.

Hint:

Let b be his speed in still water and c the speed of the current. Write two equations for the time taken in each direction, then solve for b.

Solution:

Let the one-way distance be d km, the paddling speed in still water be b km/h, and the current speed be c km/h.

Against the current:   d / (b - c) = 4.
With the current:     d / (b + c) = 3.

Equating the expressions for d:

4(b - c) = 3(b + c)
4b - 4c = 3b + 3c
b = 7c.

Substitute back to find the distance:
d = 4(b - c) = 4(7c - c) = 24c.

Time in still water:
t = d / b = 24c / 7c = 24 / 7 hours ≈ 3 h 26 min.

Therefore, the trip would take 24⁄7 hours (about 3 hours 26 minutes) on still water.

Hint:

Consider the timing of his birthday in relation to the current date.

Solution:

He was born on December 31st.

Hint:

Look at the shapes of the digits. How many closed loops (holes) does each digit contain?

Solution:

The rule is to count the number of closed loops in the digits.

• 9 9 9 9: each 9 has one loop → 4 loops.
• 8 8 8 8: each 8 has two loops → 8 loops.
• 1 8 1 6: 8 has two loops, 6 has one → 3 loops.
• 1 2 1 2: digits 1 and 2 have no loops → 0 loops.

For 1 9 1 9, each 9 contributes one loop and the 1s contribute none, giving 2 loops in total.

Therefore the missing digit is 2.

Hint:

Work in stages: (1) find how many adults are pregnant, (2) work out how many kids are born, (3) subtract the kids that die, then add the survivors to the original adults.

Solution:

1. Pregnant adults: 15 % of 27 = 4.05 → 4 goats (rounded).
2. Kids born: 4 × 7 = 28 kids.
3. Kids that die: 62 % of 28 = 17.36 → 17 kids (rounded).
4. Kids that survive: 28 − 17 = 11 kids.
5. Total goats left: original 27 adults + 11 surviving kids = 38 goats.

Hint:

You can set up a system of equations to solve the problem.

Solution:

The cost of each hammer is $75.00 pesos.

Solution:

9

Hint:

Think about the people who consume turkey and chicken in relation to those who do not consume either of the two.

Solution:

15 people eat turkey and chicken.

Hint:

Think about the least common multiple of the packet counts 1, 2, 3, 4, 5 and 6.

Solution:

Let T be the common total number of candies each daughter receives. For the totals to be whole numbers of candies per packet we need T to be divisible by each daughter’s packet count 1, 2, 3, 4, 5 and 6. The least common multiple of those numbers is 60, so the smallest possible total is T = 60 candies.

Then the candies in each packet are:

• 1st daughter: 60 candies in her single packet (60 ÷ 1)
• 2nd daughter: 30 candies in each of her two packets (60 ÷ 2)
• 3rd daughter: 20 candies in each of her three packets (60 ÷ 3)
• 4th daughter: 15 candies in each of her four packets (60 ÷ 4)
• 5th daughter: 12 candies in each of her five packets (60 ÷ 5)
• 6th daughter: 10 candies in each of her six packets (60 ÷ 6)

Thus the smallest feasible distribution is 60 candies each, with packet sizes of 60, 30, 20, 15, 12 and 10 candies respectively.

Solution:

270 minutes, 1 pill left

Solution:

There were 33 sheep in the group.

Hint:

Calculate the filling rate of each dispenser and add them up.

Solution:

The first fountain fills 1/30 of the pool per hour, the second 1/40 and the third 1/120 (5 days = 120 hours). Together they fill 1/30 + 1/40 + 1/120 = 1/24. Therefore, it takes 24 hours to fill the pool.

Hint:

Set up two equations: let x be the number on the first tree and y on the second. Translate each statement directly into an equation.

Solution:

Let x = parrots on the first tree, y = parrots on the second tree.

1. If one parrot moves from the second to the first tree:
(x + 1) = (y − 1)

2. If one parrot moves from the first to the second tree:
(y + 1) = 2 (x − 1)

From (1): y = x + 2.

Substitute into (2):
(x + 2) + 1 = 2(x − 1)
x + 3 = 2x − 2
5 = x

Hence y = x + 2 = 7.

Answer: 5 parrots on the first tree and 7 parrots on the second tree.

Hint:

Consider how many books each person can bind in a two-day cycle.

Solution:

It will take them 5 days to bind 1000 books.

Hint:

Consider using mathematical operations such as addition, multiplication, and parentheses.

Solution:

(4 * 4) + (4 / 4) = 20

Hint:

Convert the total distance from metres to millimetres to make the calculation easier.

Solution:

Mary walks 18 metres, which is 18000 mm. She plucks a rose every 225 mm. To find the number of roses, divide 18000 mm by 225 mm: 18000 / 225 = 80. Therefore, Mary plucks 80 roses.

Hint:

Consider the total distance and the time taken at different speeds to find the time difference.

Solution:

The man reaches the office 5 minutes early when traveling at 25 km/hr.

Hint:

Consider the equations formed by the number of men, women, and children based on the bananas given.

Solution:

Let M be the number of men, W be the number of women, and C be the number of children. The equations are:
3M + 2W + 0.5C = 25
M + W + C = 25.
Solving these equations, one possible solution is M = 5, W = 10, C = 10.

Hint:

Find each product separately, then add the five results together.

Solution:

The paired products are:
1 × 6 = 6
2 × 12 = 24
3 × 18 = 54
4 × 24 = 96
5 × 30 = 150

Sum: 6 + 24 + 54 + 96 + 150 = 330.

Therefore, the required sum is 330.

Hint:

Let the cost of the pencil be x rupees. Write an equation for the total cost, using the fact that the pen is Rs 1.00 more expensive.

Solution:

Let the pencil cost x rupees.

Then the pen costs x + 1.00 rupees.

Total cost:
(x) + (x + 1.00) = 1.10

2x + 1.00 = 1.10
2x = 0.10
x = 0.05

Therefore, the pencil costs Rs 0.05 and the pen costs Rs 1.05.

Hint:

Let the number of oranges be x. Then the number of apples is 2x.

Solution:

Let the number of oranges be x. Then the number of apples is 2x. The equation is x + 2x = 12, which simplifies to 3x = 12. Therefore, x = 4. So, there are 4 oranges and 8 apples.

Solution:

Assuming he answered all 108 questions:

Let:

Correct answers = x

Wrong answers = 108−x

Score:

x−
3
1
​

(108−x)=0

Multiply by 3:

3x−108+x=0
4x=108
x=27

So he got:

108−27=81

wrong answers.

Answer: 81 wrong, 27 correct.

Hint:

Consider how many 45-second intervals fit into 1 hour.

Solution:

In 1 hour (3600 seconds), there are 80 intervals of 45 seconds (3600 / 45 = 80). Since the hose puts out half a litre every 45 seconds, in 1 hour it will put out 80 * 0.5 = 40 litres.

Hint:

This elevator is in the UK, therefore there are six other floors above the ground floor.

Solution:

Each person can choose from 6 floors (1 to 6). Therefore, the total number of ways they can exit is 6^4 = 1296.

Hint:

Let the age of Radha's daughter be x. Then Radha's age is 5x. Set up an equation based on their ages in four years.

Solution:

Radha's daughter is 8 years old now.

Hint:

Set up two equations: let the smaller number be x. Then the larger is x + 3 and their sum is 21.

Solution:

Let the smaller number be x. Then the larger is x + 3.

x + (x + 3) = 21 → 2x + 3 = 21 → 2x = 18 → x = 9.

So the numbers are 9 and 12.

Hint:

Consider the relationship between their hourly wages and the total earnings.

Solution:

Sister's hourly pay is $19.50 and Sushi's hourly pay is $25.35.

Hint:

Let h be the number of hits and m the number of misses. You know both h + m and h − m.

Solution:

Let h = number of hits and m = number of misses.

1. Total shots: h + m = 100
2. Total score: (+1)×h + (–1)×m = 30, so h − m = 30

Add the two equations: (h + m) + (h − m) = 100 + 30 → 2h = 130 → h = 65.

Therefore the marksman hit the target 65 times (and missed 35 times).

Hint:

Consider the time taken to walk each segment and the speed of the escalator.

Solution:

The number of steps on the escalator is 48 steps.

Solution:

After all students have gone through the lockers, only the lockers that are perfect squares will remain open. This is because a locker is toggled (opened or closed) for each of its divisors, and only perfect squares have an odd number of divisors. Therefore, the open lockers correspond to the perfect squares from 1 to 1000, which are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, totaling 31 open lockers.

Solution:

Originally stolen coins: 765.

Night divisions: 1) First thief discards 1, takes 191, leaves 573. 2) Second thief discards 1, takes 143, leaves 429. 3) Third thief discards 1, takes 107, leaves 321. 4) Fourth thief discards 1, takes 80, leaves 240.

Morning division: 240 coins ÷ 4 = 60 each.

Hint:

Think of the problem backwards: if you are allowed at most k drops and still have b balls left, how many floors can you distinguish between? Work out a recurrence for that quantity.

Solution:

Let (T(b,r)) be the maximum number of floors that can be tested with (b) balls and (r) drops remaining.

On any drop, either the ball breaks or it does not.

If it breaks, we have (b-1) balls and (r-1) drops left.

If it survives, we still have (b) balls and (r-1) drops left.

So:

[
T(b,r)=1+T(b-1,r-1)+T(b,r-1)
]

with:

[
T(0,r)=0,\qquad T(b,0)=0
]

Solving this gives:

[
T(b,r)=\sum_{i=1}^{b}\binom{r}{i}
]

Therefore, the minimum worst-case number of drops is the smallest integer (k) such that:

[
\sum_{i=1}^{m}\binom{k}{i}\ge n
]

The strategy is to choose the next floor so that, whether the ball breaks or survives, the remaining floors can still be handled with the remaining balls and drops.

For example, with 2 balls and 100 floors, we need the smallest (k) such that:

[
\binom{k}{1}+\binom{k}{2}\ge 100
]

For (k=14):

[
14+91=105
]

So 14 drops are sufficient, and 13 drops are not. Thus the optimal worst-case answer for 2 balls and 100 floors is 14 drops.

Solution:

Step 1. Form a way to order the boxes. Potentially, for example, have the box closest to the door be labeled 1, the next closest 2, etc.
Step 2. Sort by names alphabetically, and assign each name a corresponding number
Step 3. As each prisoner walks into the room, go to the box that would correspond with his number - which corresponds to his name. Open the box. Read the name inside, recall what number is associated with that name, go to the box associated with that number. Repeat.

Following this path, there is a chain of boxes that would eventually lead back to the box he started at. If he were to return to his starting box, it will be because he found a box containing a name that has a number corresponding to the box he started at. But - big but - his name was the one that corresponded to the box he started at. So if the chain brings him back to his starting box, he must've found the box with his name. If that chain is shorter than 50 boxes, he will have found the box with his name within his 50 box limit. These chains are random, and if names are placed randomly in the 100 boxes, there is a 31% chance that no chains are longer than 50, and everybody survives.
As for proof of that number - each box has a 1/(100 - #boxes in chain so far) probability of leading back to the original box. Over fifty boxes in a chain, the overall probability becomes 31.8% for each chain.

Solution contributed by Zyxion.

Hint:

Try making the first orb jumps of decreasing size so that, no matter where it breaks, the total number of drops (first orb so far + second-orb search) is the same.

Solution:

The key is to balance the number of drops left with the number gained.

So, first you drop at X, and if it breaks, you need X-1 drops to fill from 1.
Then at X + (X-1), you have X-2 drops to fill from X.
Then at X + (X-1) + (X-2), you have X-3 drops to fill from 2X-1.
Then at X + (X-1) + (X-2) + (X-3), etc.
Until you top out at X + (X-1) + (X-2) + ... + (X-(X-1)) + (X-X), which, when reversed is Sum(0: X), or X/2*(X+1).

Then we just need to find out the smallest integral value of X where X/2*(X+1) > 100.

It turns out that 13/2*14 = 91, and 14/2*15 = 105, so we need at least 14 drops in the worst case.

First orb drops are at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, and 100.
Whenever the first bulb breaks, you start at the next lower number and count up by one. It is impossible to do with a worst case of 13 drops.

Solution contributed by chrisfortytwo.

Solution:

53.
Working it through:

"Ninety are attending my ball" → the cachers' ball, so C = 90.
"One third of all cachers are sane" → sane cachers = 90/3 = 30 (and since all sane are cachers, that's all sane people).
"Eight whackos are cachers" → whacko cachers = 8.
"Only one whacko that is sane" → the sane‑and‑whacko overlap = 1 (and since that person is sane, they're already a cacher).

Inclusion‑exclusion on the cachers:
cachers that are sane or whacko = 30 + 8 − 1 = 37
cachers that are neither = 90 − 37 = 53

Hint:

Let g be the tree’s yearly growth (in metres). Express the heights after 6 and 7 years in terms of g and set up the given ratio.

Solution:

Let the constant yearly growth be g metres.

Height after 6 years: 2 + 6g
Height after 7 years: 2 + 7g

The statement “7 years height is one-eighth taller than 6 years height” means

(2 + 7g) = (2 + 6g) \(1 + \tfrac{1}{8}) = \tfrac{9}{8}(2 + 6g).

Multiply both sides by 8:

8(2 + 7g) = 9(2 + 6g)
16 + 56g = 18 + 54g

Rearrange:

56g – 54g = 18 – 16
2g = 2
g = 1 m per year.

Height after 10 years:

2 + 10g = 2 + 10(1) = 12 m.

Answer: 12 metres.

Solution:

2100010006

The positions mean:

1st digit says number of 1s: there are 2 ones.
2nd digit says number of 2s: there is 1 two.
3rd digit says number of 3s: there are 0 threes.
4th digit says number of 4s: there are 0 fours.
5th digit says number of 5s: there are 0 fives.
6th digit says number of 6s: there is 1 six.
7th digit says number of 7s: there are 0 sevens.
8th digit says number of 8s: there are 0 eights.
9th digit says number of 9s: there are 0 nines.
10th digit says number of 0s: there are 6 zeros.

Hint:

List all sets of three positive integers whose product is 72 and check which sums coincide. Then use the clue about an oldest daughter.

Solution:

First, enumerate all triples of positive integers whose product is 72:

(1,1,72), (1,2,36), (1,3,24), (1,4,18), (1,6,12), (1,8,9), (2,2,18), (2,3,12), (2,4,9), (2,6,6), (3,3,8), (3,4,6).

Compute their sums. Among these, only two triples share the same sum:

(2,6,6) and (3,3,8) both sum to 14.

Because the census taker knew the house number yet still couldn’t decide, that number must be 14, leaving those two possibilities.

The father then mentions an oldest daughter. The triple (2,6,6) has no single oldest child (the two six-year-olds tie), whereas (3,3,8) does. Therefore the daughters’ ages are 3, 3 and 8.

Hint:

Let b be the paddler’s speed in still water and c the speed of the current. Write two equations for the distance using the times given, then solve for b.

Solution:

Let the width of the river be d.

Against the current: \(\dfrac{d}{4}=b-c\).

With the current: \(\dfrac{d}{3}=b+c\).

Add the two equations:
\(\dfrac{d}{4}+\dfrac{d}{3}=2b \;\Rightarrow\; \dfrac{7d}{12}=2b \;\Rightarrow\; b=\dfrac{7d}{24}.\)

The time required in still water is
\(\displaystyle \text{time}=\frac{d}{b}=\frac{d}{\tfrac{7d}{24}}=\frac{24}{7}\text{ hours}\approx3\text{ h }25\text{ min}.\)

Solution:

He originally collected 1,026 mangoes.

After the 10 gates he retained 3 mangoes in his basket.

Solution:

983

Hint:

Write equations: the number of black tiles is ab; the number of white tiles equals the area of the outer rectangle minus the inner one. Then use the fact that there are equally many black and white tiles.

Solution:

Let the inner black rectangle have dimensions a × b (with a, b > 0). The outer rectangle, including the white border of one-tile thickness, has dimensions (a+2) × (b+2).

Black tiles: ab
White tiles: (a+2)(b+2) − ab = 2a + 2b + 4

Because there are equally many black and white tiles,

ab = 2a + 2b + 4.

Rearranging gives

(a − 2)(b − 2) = 8.

The positive integer factor pairs of 8 are (1, 8), (2, 4), (4, 2) and (8, 1), yielding

• a = 3, b = 10 or a = 10, b = 3 → ab = 30 black, 30 white ⇒ 60 tiles in total;
• a = 4, b = 6 or a = 6, b = 4 → ab = 24 black, 24 white ⇒ 48 tiles in total.

Therefore the arrangement is possible with either 48 tiles or 60 tiles altogether.