-Alfonso can write 100 practice problems for the math team in 20 hours, Beauregard can write the same number of problems in 30 hours, and Clyde can write the number of problems in 40 hours. Working together, at what time will the three of them finish writing 100 practice problems if Alfonso starts at noon, Beauregard joins him at 1 pm and Clyde joins them at 2 pm?
-Vincent can process 50 orders in 2 hours working alone. When Leela is in the room, Vincent works at twice his normal speed. When Fry is in the room, Vincent works at half his normal speed. If Vincent works alone for 10 minutes, then with Leela for 10 minutes, and then alone for 10 minutes, then with Fry for 10 minutes, then alone for 10 minutes, and this pattern continues (alone, with Leela, alone with Fry), how many minutes will it take to process 50 orders?
-The Hilbert Lodge has a housekeeping staff of ten. Working alone, one housekeeper can clean all of the rooms in the lodge in 4 hours. A different housekeeper can clean all of the rooms in 5 hours, and still another takes 6 hours to clean all the rooms, working alone. Working alone, each of the remaining seven hosekepers can clean all the rooms in 7,8,9,10,11,12,13 hours, respectively. What is the minimum number of housekeepers needed to clean all of the rooms in Hilbert Lodge in exactly 2 hours?
Can you explain how you got those answers, Raj? Thank You!
To solve these problems, we'll use the concept of rates, which is the amount of work done per unit of time. Let's break down each problem and find the rates at which each person can work.
Problem 1:
- Alfonso can write 100 problems in 20 hours, so his rate is 100/20 = 5 problems per hour.
- Beauregard can write 100 problems in 30 hours, so his rate is 100/30 = 10/3 problems per hour.
- Clyde can write 100 problems in 40 hours, so his rate is 100/40 = 5/2 problems per hour.
Now, let's determine how many problems are written at each hour:
At noon, Alfonso starts working and writes 5 problems in the first hour.
At 1 pm, Beauregard joins Alfonso. Together, their combined rate is 5 + 10/3 = 25/3 problems per hour. They work for 1 hour, so they write 25/3 problems.
At 2 pm, Clyde joins Alfonso and Beauregard. Their combined rate is 5 + 10/3 + 5/2 = 35/6 problems per hour. They work for 1 hour, so they write 35/6 problems.
Therefore, by 3 pm, they will have written a total of 5 + 25/3 + 35/6 = 20 + 50/6 + 35/6 = 20 + 85/6 = 20 + 14.16 = 34.16 problems.
Since Alfonso, Beauregard, and Clyde can write 100 problems in a combined time of 40 hours (20 + 30 + 40), their combined rate is 100/40 = 2.5 problems per hour.
To find the time when they will finish writing 100 problems, we can set up the equation:
34.16 + (2.5 * t) = 100
Where t is the number of hours after 3 pm that they continue to work.
Now we solve for t:
2.5 * t = 100 - 34.16
2.5 * t = 65.84
t = 65.84 / 2.5
t ≈ 26.33
Therefore, they will finish writing 100 problems at approximately 3 pm + 26.33 hours = 5:20 pm.
Problem 2:
- Vincent processes 50 orders in 2 hours alone, so his rate is 50/2 = 25 orders per hour.
- When working with Leela, Vincent works at twice his normal speed, so his rate is 25 * 2 = 50 orders per hour.
- When working with Fry, Vincent works at half his normal speed, so his rate is 25 / 2 = 12.5 orders per hour.
Let's calculate the number of orders processed in each 10-minute interval:
During the first 10 minutes, Vincent works alone and processes (25/6) * 10 = 41.67 orders.
During the next 10 minutes, Vincent works with Leela and their combined rate is 50/6 orders per minute. So in 10 minutes, they process (50/6) * 10 = 83.33 orders.
The following 10 minutes, Vincent works alone again and processes (25/6) * 10 = 41.67 orders.
In the next 10 minutes, Vincent works with Fry and their combined rate is 12.5/6 orders per minute. So in 10 minutes, they process (12.5/6) * 10 = 20.83 orders.
This pattern repeats, with Vincent working alone for 10 minutes, then with Leela for 10 minutes, then alone again for 10 minutes, and so on.
Since Vincent processes 50 orders, we want to find when the cumulative number of orders processed equals 50.
Let's calculate the total number of orders processed in the first two cycles (alone + with Leela + alone + with Fry):
41.67 + 83.33 + 41.67 + 20.83 = 187.5
As we can see, the total number of orders processed in two cycles is less than 50 orders. Therefore, we need to continue calculating until we surpass 50 orders.
Let's calculate the total number of orders processed in three cycles:
187.5 + 41.67 + 83.33 + 41.67 + 20.83 = 375
Now we can see that three cycles are enough to surpass 50 orders. Therefore, it will take 3 cycles of 10 minutes each to process 50 orders, which is equal to 3 * 10 = 30 minutes.
Therefore, it will take 30 minutes to process 50 orders.
Problem 3:
- The first housekeeper can clean all the rooms in the lodge in 4 hours, so their rate is 1/4 rooms per hour.
- The second housekeeper can clean all the rooms in 5 hours, so their rate is 1/5 rooms per hour.
We want to find the minimum number of housekeepers needed to clean all the rooms in exactly 2 hours.
Since each housekeeper has a different rate, we need to consider each housekeeper one by one.
Let's consider the housekeepers in order from the one with the fastest rate to the one with the slowest rate.
The fastest housekeeper can clean 1/4 of the rooms in an hour, so in 2 hours, they can clean 1/4 * 2 = 1/2 of the rooms.
The next housekeeper can clean 1/5 of the rooms in an hour, so in 2 hours, they can clean 1/5 * 2 = 2/5 of the rooms.
Now let's sum up the rooms cleaned by the two housekeepers:
1/2 + 2/5 = 5/10 + 4/10 = 9/10 of the rooms.
Therefore, after two hours, the two fastest housekeepers can clean 9/10 of the rooms.
To clean the remaining 1/10 of the rooms, we need to consider the remaining housekeepers with slower rates.
Let's look at the remaining housekeepers' rates:
- 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12, 1/13 rooms per hour.
To know how many of these housekeepers we need, we have to find the sum of their rates and see if it can clean the remaining 1/10 of the rooms in 2 hours.
Summing up their rates:
1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 ≈ 0.9503
As we can see, the sum of the rates of the remaining housekeepers is less than 1, which means they can't clean the remaining 1/10 of the rooms in 2 hours.
Therefore, we need at least two housekeepers to clean all of the rooms in Hilbert Lodge in exactly 2 hours.
10
100
3