Newton's method involves problems in which there is an amount that exists at the beginning, an amount that keeps increasing, and an amount that is reduced, all at the same time. It originated from a problem considered by the famous scientist Newton.
The Three Elements of Newton's Method
- Initial amount (S):The amount of grass that already exists
- Increasing amount (r):The amount of grass that grows in one day
- Amount eaten:The amount one cow eats in one day × the number of cows
When the grass is gone after N days, the following equation holds: "initial amount ＋ increasing amount×number of days ＝ amount eaten×number of days".
Example Problem
A pasture has grass that grows by the same amount every day. If 10 cows are released, the grass is gone in 20 days; if 15 cows are released, it is gone in 10 days. How many days will it take for the grass to be gone if 6 cows are released?
Solution:Let the amount one cow eats in one day be "1".
① 10 cows×20 days＝200 (the amount eaten by all the cows)
② 15 cows×10 days＝150 (the amount eaten by all the cows)
③ Let the initial amount be S and the amount that grows in one day be r. Then,
S＋r×20＝200 …㋐
S＋r×10＝150 …㋑
④ ㋐－㋑：r×10＝50 → r＝5
⑤ Substitute into ㋑：S＋50＝150 → S＝100
⑥ If it takes N days with 6 cows, 100＋5×N＝6×N → N＝100 days
A pasture has grass that grows by the same amount every day. If 10 cows are released, the grass is gone in 20 days; if 15 cows are released, it is gone in 10 days. How many days will it take for the grass to be gone if 6 cows are released?
Solution:Let the amount one cow eats in one day be "1".
① 10 cows×20 days＝200 (the amount eaten by all the cows)
② 15 cows×10 days＝150 (the amount eaten by all the cows)
③ Let the initial amount be S and the amount that grows in one day be r. Then,
S＋r×20＝200 …㋐
S＋r×10＝150 …㋑
④ ㋐－㋑：r×10＝50 → r＝5
⑤ Substitute into ㋑：S＋50＝150 → S＝100
⑥ If it takes N days with 6 cows, 100＋5×N＝6×N → N＝100 days
Key Points
- The basic equation is "initial amount ＋ increasing amount×number of days ＝ amount eaten×number of days".
- Use the two conditions to find the initial amount and the increasing amount.
- Setting "the amount one cow eats in one day＝1" makes the calculations easier.
Practice Problems
- When 20 cows eat the grass in a pasture, it is gone in 30 days; with 30 cows, it is gone in 15 days. How many days will it take with 10 cows?
- A pasture has 200 units of grass initially, and the grass increases by 10 units every day. If one cow eats 15 units every day, how many days will it take for the grass to be gone when 20 cows are released?
- In a pasture, the grass is gone in 12 days with 25 cows and in 24 days with 15 cows. Find the amount of grass that grows in one day.
Answers and Explanations
- Answer:90 days
Explanation: Let the amount one cow eats in one day be 1. Condition ①20 cows×30 days＝600 (total consumption), condition ②30 cows×15 days＝450 (total consumption). S＋r×30＝600 …①, S＋r×15＝450 …②. ①－②：r×15＝150→r＝10. From ②：S＋150＝450→S＝300. With 10 cows for N days: 300＋10N＝10N→0N＝300→impossible (with 10 cows, consumption＝increase, so the grass will not disappear). Check: consumption by 10 cows per day＝increase of 10, so the grass does not decrease. → "There must be at least 11 cows for the grass to disappear." With 11 cows: 300＋10N＝11N→N＝300 days. The correct answer to the problem is "The grass will not disappear with 10 cows (infinite time)." - Answer:Approximately 13.6 days (200÷（20×15－10）＝200÷290≈0.69…Actually, 20 cows×15 units＝300 units/day consumed, while the increase is 10 units/day)
Explanation: Initial amount S＝200, increasing amount r＝10/day, and one cow eats 15 units every day, so 20 cows consume 300 units/day. When the grass is gone after N days: 200＋10N＝300N. 200＝290N→N＝200÷290≈0.69 days (less than one day). Since 20 cows×15＝300 units/day consumed greatly exceeds the 10-unit daily increase, the grass disappears very quickly. N＝200÷（300－10）＝200÷290≈0.69 days (approximately 17 hours). - Answer:Daily increase in grass＝15
Explanation: Let the amount one cow eats in one day be 1. Condition ①25 cows×12 days＝300 (total consumption), condition ②15 cows×24 days＝360 (total consumption). S＋r×12＝300 …①, S＋r×24＝360 …②. ②－①：r×12＝60→r＝5. Check: From ①, S＝300－60＝240. 15 cows×24 days＝360, and 240＋5×24＝240＋120＝360✓. r＝5 (based on one cow). If one cow eating one unit per day is 1, the increase is equivalent to 5 cows per day.