8.6Math Enrichment Lab
Mission progress0 of 10
Chapter 8 • Dynamic-resource models

Lesson 8.6 — Replenishing Inventory and Changing Rates

补充库存与变化的速率

A stock can be used while more is arriving. When the number of users or the incoming rate changes, pause the story, update what remains, and begin a new period.

Freeze the timeline. Update the stock. Restart with the new rates.
Grade 5 enrichment50–70 minutes8 core missions · 2 optional extensionsEverything needed is on this page.
Mission 1

See inventory as a changing stock

Use a steady-flow model: supplies arrive and are used continuously at constant rates within each period. Each employee uses the same amount per day. Alternative trials begin with identical stock, and the stated run-out time is the first time the stock reaches zero.

A replenishing inventory has an amount already present, an inflow that adds more, and an outflow that removes it.

Not complete
Stock-flow picture

The stock changes because incoming items are added and outgoing use removes items.

Three quantities

Starting stock, S: inventory already present at day 0.

Incoming rate, g: inventory added each day.

Outgoing rate, n: inventory used or removed each day.

Stock after t days = S + gt − nt
At the exact moment the stock reaches zero: S + gt = nt.

Identify each quantity

Mission 2

Use two run-out times to reveal daily replenishment

A vending machine receives the same number of drinks every day. Its stock lasts 30 days for 5 employees, but only 20 days for 6 employees.

Not complete

Measure in employee-days

One employee-day unit is the amount of drinks one employee uses in one day.

5 × 30 = 150 units used 6 × 20 = 120 units used

Both totals include the same starting stock. The 30-day story also includes ten extra days of replenishment.

g = (150 − 120) ÷ (30 − 20) = 3

The machine receives enough drinks each day for 3 employees for one day.

Two complete scenariossame starting stock

Subtracting the bars removes the same starting stock and isolates ten days of replenishment.

Build the comparison

employee-day units per day
Mission 3

Recover the stock that was there at the start

Once the replenishment rate is known, remove all newly added inventory from either complete scenario.

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Use the 30-day scenario

S = 5×30 − 3×30 = 60

Check with the 20-day scenario

S = 6×20 − 3×20 = 60

Both routes reveal the same starting stock:

S = 60 employee-day units
Meaning: without any daily replenishment, the starting inventory alone could support one employee for 60 days.
Starting stock + replenishment150 total units

The 30-day story contains 60 starting units and 90 units added during the 30 days.

Check the hidden stock

Mission 4

When demand changes, split the timeline

Start again with the 60-unit stock found in Mission 3. The company first has 4 employees. After 30 days, 2 more employees join. The replenishment remains 3 employee-day units per day.

Not complete

Period 1 — four employees

Net depletion = 4 − 3 = 1 per day 60 − 1×30 = 30 units remain

Period 2 — six employees

Net depletion = 6 − 3 = 3 per day 30 ÷ 3 = 10 more days 30 + 10 = 40 days altogether

Change the second-period employee count

6 employees
30stock at day 30
3net use per day
10extra days
40total days

Solve for four employees followed by six (regardless of the slider)

Mission 5

Apply the model to a replenished warehouse

In separate trials with identical starting stock and constant incoming deliveries, removing four truckloads per day empties a warehouse in 9 days, or five per day empties it in 6 days. Each truck can remove one truckload per day. Find the time one truck needs to carry only the original stock, ignoring later deliveries for that calculation.

Not complete

Reveal the daily delivery into the warehouse

4×9 = 36 truckload-units 5×6 = 30 truckload-units g = (36−30) ÷ (9−6) = 2

Recover the original stock

S = 36 − 2×9 = 18 truckloads

One truck can carry the original stock alone in:

18 ÷ 1 = 18 days
Read the wording carefully. If new goods kept arriving at 2 truckloads per day while only one truck removed 1, the whole warehouse would never empty.
Warehouse flow2 in each day

This question asks how long one truck needs for the original 18 truckloads, not for an endlessly replenished warehouse.

Check the warehouse model

truckloads per day
truckloads
Mission 6

Experiment with two periods

Change the starting stock, incoming rate, outgoing rate, and switch day. The graph will tell you whether the stock empties before the switch, after the switch, or never.

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60 units
3/day
4/day
30 days
3/day
6/day
30stock at switch
1period 1 net use
3period 2 net use
40finish day

State the strategy

Mission 7

When the incoming rate changes instead

A company has typing work already waiting. Five equally productive typists finish in 24 days or nine in 12 days, with the same constant inflow and starting work. How many typists, working at a constant rate, would finish in 40 days if inflow stays unchanged for 8 days and then halves for the remaining 32 days?

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Recover the original system

5×24 = S + 24g 9×12 = S + 12g g = (120−108) ÷ 12 = 1 S = 108−12 = 96

Now split the 40-day timeline

For the first 8 days, new work arrives at 1 unit per day. For the next 32 days, it arrives at half that rate:

40n = 96 + 8×1 + 32×0.5 40n = 120, so n = 3 typists
Typing-work timelineincoming rate halves after day 8

The outgoing rate stays the same; only the incoming work changes after day 8.

Complete the piecewise model

Mission 8

Recognize the same balance in pipes and escalators

Inventory is only one story. The same mathematics appears whenever a background flow adds to or works against a person’s action.

Not complete

Pipe challenge

An initially empty pool has an inlet running alone for x minutes, without overflowing. It then stays open. With one identical outlet pipe open, the pool empties 12 minutes later. With two identical outlet pipes open, it empties 4 minutes later. An outlet’s rate may differ from the inlet’s rate.

The two outlet arrangements are separate trials beginning with the same stored water. Let inlet rate be a and one outlet rate be b.

ax = 12(b−a) ax = 4(2b−a)

Equating the right sides gives:

12(b−a)=4(2b−a), so b=2a ax=12a, so x=12 minutes
Clarified model: the inlet remains open; the outlet pipes match one another, but an outlet need not match the inlet.
Pool flow balanceinlet stays open

The stored water is the starting stock; the inlet is continuing inflow; the outlets are continuing outflow.

Escalator moving downward

In separate trips on the same downward-moving escalator, a boy walks from bottom to top at 2 steps per second and takes 100 seconds. A girl walks from bottom to top at 3 steps every 2 seconds and takes 200 seconds. Walking rates are measured relative to the moving steps; all rates are constant, and each counted step advances one stair spacing.

Let the escalator move downward at e steps per second and let N be the number of visible steps when stopped.

N=(2−e)×100=(1.5−e)×200 e=1, and N=100

Two people on an upward escalator

A brother starts at the top of an upward-moving escalator and walks to the bottom, taking 100 steps. His sister starts at the bottom at the same time and walks to the top, taking 50 steps. Each walking step advances one stair spacing relative to the moving escalator. All rates are constant, and the brother’s walking rate is twice the sister’s.

The brother takes 100 ÷ (2r) = 50/r seconds, and the sister takes 50 ÷ r = 50/r seconds. Their journey times are equal. If the escalator contributes E steps of upward movement during their shared time:

N=100−E=50+E 2E=50, so E=25 and N=75
Walking rate and background motionnet motion

The escalator’s movement is a background rate: it can help or oppose the walker.

Check the transfer

steps per second
Mission 9

A new replenishing supply — 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.

Not complete

A supply receives the same amount each day. Six employees use it up in 12 days; eight use it up in 8 days. Each employee uses one unit per day. The separate trials begin with the same stock and end at the first run-out time. In a fresh trial, start again with this stock and unchanged replenishment: five employees use it for 6 days, followed by nine employees.

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

6 × 12 = 72.

Hint

Compare totals only after using the same unit.

Worked solution

8 × 8 = 64.

Hint

Use the two trials and cancel their identical starting stock.

Worked solution

(72 − 64) ÷ (12 − 8) = 2.

Hint

Track the extra time or remove all growth from total use.

Worked solution

72 − 12 × 2 = 48; also 64 − 8 × 2 = 48.

Hint

Subtract the daily replenishment from five employees’ daily use.

Worked solution

5 − 2 = 3 per day.

Hint

Subtract six days of net use from the starting stock.

Worked solution

48 − 6 × 3 = 30.

Hint

Subtract unchanged replenishment from nine employees’ daily use.

Worked solution

9 − 2 = 7 per day.

Hint

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

Worked solution

30 ÷ 7 = 30/7 ≈ 4.285714 additional days. Total time is 6 + 30/7 = 72/7 ≈ 10.285714 days.

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.

Not complete

Optional reflection — not automatically graded

Certificate of Mastery

Dynamic-Inventory Systems Engineer

This certifies that the learner can recover replenishment and starting stock, split a story when rates change, distinguish original stock from continuing inflow, and solve a fresh two-period inventory problem.

Freeze the timeline → update what remains → restart with the new rates

Lesson 8.6 • Grade 5 Math Enrichment

Teaching notes

The typist, pipe, and escalator investigations are based on the chapter exercises that follow. The pipe exercise is phrased here with the mathematically consistent interpretation made explicit: the inlet stays open, the outlet pipes are identical to one another, and an outlet’s rate may differ from the inlet’s rate.

Optional bonus: Balance the Water Grid