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Cows on the Meadow

Newton's grazing problem, animated

Drag cows onto a meadow that regrows daily. Each day, +g tiles bloom, then N cows crop the grass — so the net change Δ = g − N becomes a meter you can watch breathe. Four levels build the cow grazing problem (牛吃草问题) from free play to a leaky water tank, the same model wearing a different skin.

What this game shows · Cow Grazing Problem (牛吃草问题)

Newton's grazing problem looks confusing because two things happen at once: grass grows while cows eat. This animated game runs both rates at the same time so you can read the daily net change Δ = g − N straight from the meter — no algebra required.

Daily growth (g)
tiles added to the meadow each day.
Cows (N)
tiles eaten each day, one per cow.
Net change (Δ)
Δ = g − N. Negative kills the field; zero sustains it forever.

Aligned with CCSS 6.EE.C.9 (two related quantities) and 6.RP.A.2 (unit rate). Recommended for Grade 6, with a level 4 transfer story for Grade 7.

Free Range

Drag cows and daily growth. Press ▶ Day +1 and watch the meadow rise or fall — the panel writes Δ = g − N for you.

Hint · Try cows = 0, growth = 5: the meadow fills up. Then cows = 10, growth = 5: it empties.

Meadow · 100 tiles

60/ 100

Day

0

Grass-o-meterΔ = 5 − 4 = +1
+5 grow/day−4 eaten/day Δ = +1

What the meadow says

  • Daily growth → +g = +5
  • Daily eaten → −N = −4
  • Net per day → Δ = g − N = 5 − 4 = +1
  • After 0 days → 60 + 0 × (+1) = 60

Olympiad thinking model

Who this demo helps, and where to practice next

Cows on the Meadow is built for students who need a visual way to decode multi-step puzzle structure. It gives the page a clear search purpose: learn the model, manipulate it, then continue into the matching grade-level practice.

Cows on the Meadow helps when a student can copy a procedure but cannot explain why it works. The demo slows the idea down into a visible model before sending the learner to guided missions.

Learning goals

  • The meadow has two simultaneous rates: grass grows by g tiles per day and N cows eat 1 tile each. Net change per day is Δ = g − N — every day the meter ticks by exactly that much.
  • Two clues give g and the original supply. From 27 cows × 6 days = 23 cows × 9 days exhausted, subtract: 3g = 45 → g = 15, original = 72. For 21 cows: 21t = 72 + 15t → t = 12 days.
  • The maximum sustainable cow count is N = g. Below that the meadow piles up to its cap (wasted capacity); above it the meadow dies on a predictable day. The same Δ = inflow − outflow rule re-skins as a leaky water tank in level 4.

How to play

  1. 1 Fix one quantity and watch which quantity changes.
  2. 2 Write the hidden relationship in words before writing an equation.
  3. 3 Use the related grade topics to transfer the puzzle move into standard word problems.
FAQ

Cow grazing, answered.

01 What is Newton's cow grazing problem (牛吃草问题)? Olympiad

A classic Olympiad puzzle where grass grows daily at rate g while N cows eat one tile each. Two rates run at once, so the meadow holds out for as long as the original supply plus accumulated growth covers consumption.

02 How do you find the daily growth rate from two scenarios? g = 15

Subtract cow-day totals. If 27 cows × 6 days and 23 cows × 9 days both finish the meadow, then 207 − 162 = 45 extra cow-days came from 3 extra days of growth, so g = 15. Original supply = 162 − 6 × 15 = 72.

03 What is the maximum sustainable cow count? N ≤ g

Exactly N = g. Below that, grass piles up against the cap; above it, the meadow eventually dies. Δ = g − N is the daily net change — set Δ ≥ 0 for sustainability.

04 Which grade is this game for? Grade 6

Grade 6, aligned with CCSS 6.EE.C.9 (two related quantities) and 6.RP.A.2 (unit rate). The leaky-tank transfer level reuses the same model for Grade 7 ratio applications.

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