8.4Math Enrichment • Grade 5
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Chapter 8 • Growing and Shrinking Resources

Lesson 8.4 — When the Resource Shrinks

资源自然减少时

Sometimes the resource does not grow. It may dry out, leak away, spoil, or disappear while consumers are also using it. This lesson shows how to keep both losses in one clear model.

When nature and consumers both remove the resource, their rates add.
Grade 5 enrichment35–45 minutes10 interactive missionsNo this lesson required
Mission 1

Follow the arrows

Before calculating, identify what is entering the stock and what is leaving it.

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↓
Natural growth entersConsumers remove, but new resource arrives.
—
No natural changeOnly the consumers reduce the stock.
↑
Natural loss leavesNature and consumers both reduce the stock.
Interactive resource tankShrinking system

Both arrows point out of the stock, so the two removal rates add.

Check the direction of change

Mission 2

Build the shrinking-resource equation

One cow-day unit is the grass one cow eats in a day. Each cow eats at the same constant rate, so n cows consume n units per day. S is measured in units, d in units/day, and t in days. The equations below describe the exact emptying time, with S > 0 and n + d > 0.

Name the starting stock S, the natural loss per day d, the daily consumption rate n, and the time t.

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Consumers removen units/day
+
Nature removesd units/day
S − dt = nt   — after natural loss, the remaining resource is consumed.
S = nt + dt = (n+d)t
t = S ÷ (n+d)   — time equals starting stock divided by total depletion rate.
Compare: when the resource grows, the net rate is n-g. When it shrinks naturally, the net rate is n+d.

Choose the correct relationships

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Mission 3

Use two scenarios to reveal the natural loss

A pasture loses the same amount of grass each day. In separate trials with identical starting grass, it runs out after exactly 5 days with 20 cows or exactly 6 days with 16 cows.

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Two complete scenariosSame starting grass

The original grass is the same in both scenarios. The one-day difference isolates the natural loss.

Scenario A

20 × 5 = 100 cow-days

S − 5d = 100

Scenario B

16 × 6 = 96 cow-days

S − 6d = 96

  1. Subtract the consumption totals: 100−96=4.
  2. The scenarios differ by 6−5=1 day.
  3. Therefore one day of natural loss is 4÷1=4 cow-day units.

Enter the key quantities

Mission 4

Recover the original stock

Cows and natural drying remove grass throughout the same time interval. Add both losses back to recover the beginning.

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Cow use20 × 5 = 100
Natural loss4 × 5 = 20
Starting stock100 + 20 = 120
Second check96 + 24 = 120
AgreementBoth give 120
S=(20+4)×5=120
S=(16+4)×6=120
Reasonableness check: the starting stock must be greater than the amount the cows ate, because some grass also disappeared naturally.

Recover and verify the start

Mission 5

Find how long 11 cows can eat

Each day, the cows remove 11 units and nature removes 4 more. Together they reduce the original stock by 15 units per day.

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Change the herd size

11 cows
11cow use/day
4natural loss/day
15total depletion/day
8days
120 ÷ (11 + 4) = 8 days

Solve for 11 cows using S = 120 and d = 4

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Mission 6

Guided practice: a second drying pasture

A drying pasture feeds 20 cows for 5 days or 12 cows for 7 days. How long can 6 cows eat?

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Complete problem statement

The grass decreases naturally by the same amount each day. In separate trials with identical starting grass, it is completely used after either 20 cows eat for 5 days or 12 cows eat for 7 days. Find how many days it can feed 6 cows.

  1. Total use: 20×5=100 and 12×7=84.
  2. Natural loss: (100−84)÷(7−5)=8 units/day.
  3. Starting stock: (20+8)×5=140.
  4. Six cows plus natural loss deplete 6+8=14 units/day.
  5. Duration: 140÷14=10 days.

Enter every key result

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Mission 7

Compare growing, unchanged, and shrinking systems

The same starting stock and consumer rate can last very different lengths of time depending on what nature is doing.

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growth or loss per day
Shrinkingsystem
15effective depletion/day
8time until empty
fallsstock trend
Stock over timeS ÷ (n+d)

Compare the three models

Mission 8

Work backward from a shrinking-resource model

Once you know S=(n+d)t, you can solve for any missing quantity.

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Find time

t = S ÷ (n+d)

Find consumers

n = S ÷ t − d

Find natural loss

d = S ÷ t − n

Backward-solving laboratory

Mission 9

A new drying pasture — independent investigation

Try the eight steps yourself, using a small hint if needed. Check an attempt before opening worked review. Correct all eight to complete this investigation.

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A naturally shrinking pasture feeds 21 cows for 4 days or 13 cows for 6 days. These are separate trials with identical starting grass; each ends exactly when no grass remains. Natural loss and each cow’s eating rate stay constant. Use cow-day units.

Write quantities in one consistent unit. Enter decimals or fractions such as 30/7; a decimal within 0.001 is accepted.

Hint

Multiply rate by time.

Worked solution

21 × 4 = 84.

Hint

Compare totals only after using the same unit.

Worked solution

13 × 6 = 78.

Hint

Use the two trials and cancel their identical starting stock.

Worked solution

84 − 78 = 6. The longer trial loses more grass naturally.

Hint

Subtract the shorter trial’s 4 days from the longer trial’s 6 days.

Worked solution

6 − 4 = 2.

Hint

Divide the difference in grass eaten by the extra days of natural loss.

Worked solution

6 ÷ 2 = 3 cow-day units/day.

Hint

Add the grass eaten to all the grass lost naturally during the same trial.

Worked solution

84 + 4 × 3 = 96; check: 78 + 6 × 3 = 96.

Hint

Add the 9 cows’ daily use and the natural loss per day.

Worked solution

9 + 3 = 12. Both processes remove grass.

Hint

Divide remaining stock by its net depletion rate; check the time or capacity requested.

Worked solution

96 ÷ 12 = 8. Check: cows eat 72 and nature removes 24, totaling 96.

Mission 10

Exit ticket

Answer all five questions. A perfect score completes the exit ticket. The certificate also requires the other core missions and the independent investigation.

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days
✓

Certificate of Mastery

Shrinking-Resource Systems Engineer

This certifies that the learner can distinguish growth from natural loss, recover a hidden starting stock, and solve forward and backward resource problems using

S = (n+d)t

Lesson 8.4 • Grade 5 Math Enrichment

Teaching notes