Newton's Method ② The Spring and Water Pumping
Newton's method appears not only in "grass and cows" problems, but also in various situations such as "springs and water pumping" and "queues and service counters." Although the situations differ, the way of thinking is exactly the same.
Spring Problems
A spring initially contains water, and a small amount springs up every minute. Water is pumped out from it with pumps.
Example Problem
Water springs up in a spring, and pumps are used to pump it out. With 3 pumps, the spring is empty in 40 minutes; with 5 pumps, it is empty in 20 minutes. How many minutes will it take with 2 pumps?
Solution:Let the amount pumped by one pump in one minute be "1".
① 3 pumps×40 minutes＝120 (the amount pumped out)
② 5 pumps×20 minutes＝100 (the amount pumped out)
③ S＋r×40＝120 …㋐, S＋r×20＝100 …㋑
④ ㋐－㋑：r×20＝20 → r＝1
⑤ ㋑：S＋20＝100 → S＝80
⑥ With 2 pumps for N minutes: 80＋1×N＝2×N → N＝80 minutes
Water springs up in a spring, and pumps are used to pump it out. With 3 pumps, the spring is empty in 40 minutes; with 5 pumps, it is empty in 20 minutes. How many minutes will it take with 2 pumps?
Solution:Let the amount pumped by one pump in one minute be "1".
① 3 pumps×40 minutes＝120 (the amount pumped out)
② 5 pumps×20 minutes＝100 (the amount pumped out)
③ S＋r×40＝120 …㋐, S＋r×20＝100 …㋑
④ ㋐－㋑：r×20＝20 → r＝1
⑤ ㋑：S＋20＝100 → S＝80
⑥ With 2 pumps for N minutes: 80＋1×N＝2×N → N＝80 minutes
It Can Also Be Used for Queue Problems
Problems such as "○ people join the queue every minute" and "a service counter serves △ people every minute" can also be solved with the same equation. Think of "initial amount＝number of people initially in line," "increasing amount＝number of people joining every minute," and "decreasing amount＝number of people served at the counter."
Key Points
- Even when the situation changes, it is the same type of problem if you replace it with "initial amount, increasing amount, and decreasing amount."
- Set up "amount pumped out ＝ number of pumps × time" in two ways, then solve by subtraction.
- First find the "increasing amount r" by subtraction, and then find the "initial amount S".
Practice Problems
- There is a queue in which 4 new people join every minute. With 3 service counters, the queue disappears in 20 minutes, while with 2 counters it takes 40 minutes. Find the initial number of people in the queue.
- A spring initially contains 150 units of water and 5 units spring up every minute. If one pump pumps 20 units per minute, how many pumps are needed to empty the spring in 20 minutes?
- When 40 cows eat the grass in a pasture, it is gone in 20 days; with 25 cows, it is gone in 40 days. Find the amount of grass that grows in one day.
Answers and Explanations
- Answer:Cannot be determined from the conditions in the problem
Explanation: If one service counter serves 1 person per minute and 4 people join the line every minute, the number joining is greater than the number processed by 3 counters per minute, so the line will not become shorter. Therefore, the condition that the line disappears in 20 minutes is contradictory, and the initial number of people cannot be determined. - Answer:13 pumps are required
Explanation: The amount of water that must be pumped out by 20 minutes is 150＋5×20＝250 units. n pumps pump 20n units in 20 minutes, so 20n≧250, n≧12.5. Since the number of pumps must be an integer, at least 13 pumps are required. Twelve pumps are insufficient at 240 units, while 13 pumps can pump out 260 units. - Answer:Increase in grass (one day)＝equivalent to 10 cows
Explanation: Let the amount one cow eats in one day be 1. If the initial grass is S and the grass that grows in one day is r, then S＋20r＝40×20＝800, S＋40r＝25×40＝1000. The difference between the two equations gives 20r＝200, so r＝equivalent to 10 cows/day. Check: S＝600, 600＋10×40＝1000.