Tag: math & puzzles

To solve this problem, let's assume that Farmer Charlie initially had "x" number of chickens on his farm.

According to the given information, if he sells 75 chickens, he can feed the remaining chickens for 20 days longer. This means that the amount of chicken food he has can last for (x - 75) chickens for (20 + 15) days.

Similarly, if he buys 100 extra chickens, he would run out of chicken food 15 days earlier. This means that the amount of chicken food he has can last for (x + 100) chickens for (20 - 15) days.

Now, let's set up the equations based on the above information:

Equation 1: (x - 75) * (20 + 15) = x * 20 Equation 2: (x + 100) * (20 - 15) = x * 20

Let's solve these equations to find the value of x, which represents the initial number of chickens.

Expanding the equations: Equation 1: (x - 75) * 35 = 20x Equation 2: (x + 100) * 5 = 20x

Simplifying the equations: Equation 1: 35x - 2625 = 20x Equation 2: 5x + 500 = 20x

Solving for x: Equation 1: 35x - 20x = 2625 15x = 2625 x = 175

Equation 2: 20x - 5x = 500 15x = 500 x = 33.33

Since the number of chickens cannot be fractional, we can conclude that Farmer Charlie initially had 175 chickens on his farm.

Therefore, the correct answer is option A) 350.

• Thanks
• Wrong explanation

To solve this problem, we can use a concept called the "Frobenius coin problem." In this problem, we want to find the largest number that cannot be expressed as a sum of multiples of given numbers.

In this case, the given numbers are 6, 9, and 20. We want to find the largest number of bitterballs that cannot be ordered in these portions.

Let's analyze the given options:

Option A) 43 - This option is the correct answer. We can express 43 as a sum of multiples of 6, 9, and 20. For example, we can order 6 bitterballs (6 * 7 = 42) and add an extra 1 bitterball to make a total of 43.

Option B) 46 - We can express 46 as a sum of multiples of 6, 9, and 20. For example, we can order 9 bitterballs (9 * 5 = 45) and add an extra bitterball to make a total of 46.

Option C) 76 - We can express 76 as a sum of multiples of 6, 9, and 20. For example, we can order 20 bitterballs (20 * 3 = 60), 9 bitterballs (9 * 1 = 9), and 6 bitterballs (6 * 1 = 6) to make a total of 76.

Option D) 79 - We can express 79 as a sum of multiples of 6, 9, and 20. For example, we can order 20 bitterballs (20 * 3 = 60), 9 bitterballs (9 * 1 = 9), and 6 bitterballs (6 * 1 = 6) to make a total of 79.

Therefore, the correct answer is option A) 43. This option is correct because it is the largest number that cannot be ordered in portions of 6, 9, or 20.

• Thanks
• Wrong explanation

To find the time it takes to fill the water-cask with all three taps together, we need to calculate the combined rate at which the taps fill the cask.

Let's denote the rates of the smallest, middle, and largest taps as R1, R2, and R3 respectively. The time it takes to fill the cask with each tap can be expressed as follows:

Time with smallest tap = 20 minutes (1/R1) Time with middle tap = 12 minutes (1/R2) Time with largest tap = 5 minutes (1/R3)

To find the combined rate, we can add the rates of the individual taps:

Combined rate = R1 + R2 + R3

Now, let's find the rates of the taps:

Rate with smallest tap (R1) = 1/20 cask per minute Rate with middle tap (R2) = 1/12 cask per minute Rate with largest tap (R3) = 1/5 cask per minute

To find the combined rate, we add the rates:

Combined rate = (1/20) + (1/12) + (1/5) = (3/60) + (5/60) + (12/60) = 20/60 = 1/3 cask per minute

The combined rate is 1/3 cask per minute, which means it takes 3 minutes to fill the water-cask with all three taps together.

Therefore, the correct answer is D) 3 min.

• Thanks
• Wrong explanation