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The weather-lab dispatch
Eliminate the shared part
You may know enough to decide—even when some numbers are missing.
Amara prepares the cases while Li Wei checks the records. CKRobi helps them think through a reliable dispatch plan.
Rain taps against the weather lab’s windows. Amara and Li Wei are preparing equipment for a field team. Some mass labels have rubbed away, but the packing records survived.
Amara “We have the totals, but not every separate mass.”
CKRobi “Which decision do the records need to help you make?”
Li Wei “First, which supply bag is heavier. Then we can plan the delivery.”
CKSTEM Thinking CompassStuck? Find your next move.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
Two records, one decision
Each sealed pack contains one empty case, one logger and one supply bag. The empty cases are identical, and the loggers are identical. Every item has positive mass. These exact records include all the contents:
Pack A
Empty caseLoggerBag A
14 kg
Pack B
Empty caseLoggerBag B
11 kg
Each record includes one case, one logger and its named supply bag.
Which supply bag is heavier, and by how much? Do the records determine either bag’s individual mass?
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Which supply bag is heavier, and by how much? Do the records determine either bag’s individual mass?
Two records, one decision
Each sealed pack contains one empty case, one logger and one supply bag. The empty cases are identical, and the loggers are identical. Every item has positive mass. These exact records include all the contents:
Pack A
Empty caseLoggerBag A
14 kg
Pack B
Empty caseLoggerBag B
11 kg
Each record includes one case, one logger and its named supply bag.
Your answer
Enter the mass difference. Identify which bag is heavier and explain whether either individual mass is determined.
These quantities fit. Explain your choice of parts, how they recombine and why every original rule is met; reasoning stays unscored.
Break it into steps
Step 1 of 2
What is the difference between the complete pack totals?
Fourteen minus eleven is three kilograms.
Step 2 of 2
How much heavier is the heavier supply bag?
Bag A is three kilograms heavier; neither separate bag mass is fixed.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the two supply bags
Answer: Bag A is 3 kg heavier. Neither bag’s individual mass is determined.
Li Wei “I thought we needed the mass of the case and logger first.”
Amara “They match in both packs. Whatever they weigh, they add the same amount to each total.”
CKRobi “Could two different packings still fit these totals?”
Li Wei “The bags differ by three kilograms. It still doesn’t tell us what either bag weighs.”
Pack A
Empty caseSet asideLoggerSet asideBag A
14 kg
Pack B
Empty caseSet asideLoggerSet asideBag B
11 kg
Set aside an equal case and logger from each record; compare the bags that remain.
Set aside the equal case-and-logger mass from each record. It cannot account for the difference:
\[14-11=3\text{ kg}.\]
To check the limit of that conclusion, compare two possible explanations of the same records:
Possible explanation 1
Each empty case
1 kg
Each logger
7 kg
Bag A
6 kg
Bag B
3 kg
Pack A: 1 + 7 + 6 = 14 kg Pack B: 1 + 7 + 3 = 11 kg
Possible explanation 2
Each empty case
2 kg
Each logger
7 kg
Bag A
5 kg
Bag B
2 kg
Pack A: 2 + 7 + 5 = 14 kg Pack B: 2 + 7 + 2 = 11 kg
These are alternative explanations of the same records. Both leave the bags 3 kg apart.
Both explain 14 kg and 11 kg. The bag masses change, but their difference stays 3 kg. One consistent example checks a possibility; two different examples show why the individual masses are not fixed.
Eliminate the shared part
Choose two records that contain an equal part. Compare them in a way that removes that part and keeps the relationship you need. You do not always need to solve for every unknown.
Look for a reason that a part is equal: identical equipment, the same fixed charge, or the same positive test duration. A similar picture or a matching colour is not enough.
If you feel stuck, arrange the records side by side and name what is known to match. That small action can reveal a useful next move.
Optional mathematics: why subtracting the records works
Elimination
Let \(c\) be the mass of one case and logger together, and \(a,b\) the bag masses, in kilograms. Then \(c+a=14\) and \(c+b=11\). Subtract the complete second relation from the first:
\[(c+a)-(c+b)=14-11\]\[a-b=3.\]
The same \(c\) is removed twice, once from each record. We have found the requested difference without determining \(c\), \(a\) or \(b\) separately.
A match that needs checking
Another pair of packing cards looks convenient. This time, one important fact has changed.
This is a separate pair of packs. Each contains one empty case, one logger and one supply bag. The empty cases are identical, but the two loggers are different models with unknown masses. Every item has positive whole-kilogram mass.
Pack A
Empty caseLogger XBag A
14 kg
Pack B
Empty caseLogger YBag B
11 kg
The cases match, but Logger X and Logger Y are different models with unknown masses.
Do these records prove which supply bag is heavier? If not, give two possible packings that fit both totals but make opposite bags heavier.
CKSTEM Thinking CompassPlan and check your reasoning.
Do these records prove which supply bag is heavier? If not, give two possible packings that fit both totals but make opposite bags heavier.
A match that needs checking
Another pair of packing cards looks convenient. This time, one important fact has changed.
This is a separate pair of packs. Each contains one empty case, one logger and one supply bag. The empty cases are identical, but the two loggers are different models with unknown masses. Every item has positive whole-kilogram mass.
Pack A
Empty caseLogger XBag A
14 kg
Pack B
Empty caseLogger YBag B
11 kg
The cases match, but Logger X and Logger Y are different models with unknown masses.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
Show the answer and reasoning: the changed logger
Answer:No. Either bag could be heavier. The two packings below both fit the records.
Amara “The total difference is still three. Can we put that on the bag label?”
CKRobi “What made the earlier comparison reliable?”
Li Wei “Both removed parts had equal mass. Here the loggers could account for some—or all—of the difference.”
Possible explanation 1
Each empty case
1 kg
Logger X
9 kg
Logger Y
2 kg
Bag A
4 kg
Bag B
8 kg
Pack A: 1 + 9 + 4 = 14 kg Pack B: 1 + 2 + 8 = 11 kg
Bag B is heavier.
Possible explanation 2
Each empty case
1 kg
Logger X
2 kg
Logger Y
1 kg
Bag A
11 kg
Bag B
9 kg
Pack A: 1 + 2 + 11 = 14 kg Pack B: 1 + 1 + 9 = 11 kg
Bag A is heavier.
Both explanations fit 14 kg and 11 kg, but they give opposite comparisons of the bags.
In the first explanation, bag B is heavier; in the second, bag A is heavier. Every original total still works. Identical cases can be removed, but we have no right to remove the two unknown logger masses as though they were equal.
The useful next information would concern the logger masses, rather than another guess about the bags.
Two recorders, one run
Two recorders run for the same unknown positive whole number of complete minutes. Each records at its own constant positive whole-number rate.
Recorder A
Run time: unknown
Same positive duration
24 readings
Recorder B
Run time: unknown
Same positive duration
16 readings
Each total comes from its recorder’s constant rate over the same positive run time.
What is the ratio of A’s rate to B’s rate? Must we know the duration, and are the individual rates determined?
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What is the ratio of A’s rate to B’s rate? Must we know the duration, and are the individual rates determined?
Two recorders, one run
Two recorders run for the same unknown positive whole number of complete minutes. Each records at its own constant positive whole-number rate.
Recorder A
Run time: unknown
Same positive duration
24 readings
Recorder B
Run time: unknown
Same positive duration
16 readings
Each total comes from its recorder’s constant rate over the same positive run time.
Your answer
Enter A’s rate compared with B’s, for example as a:b or a/b. Equivalent ratios are accepted. Explain separately what the records tell you about the duration and the individual rates.
The ratio fits. Explain why positive equal run time cancels and why these totals do not determine the two individual rates.
Break it into steps
Step 1 of 2
What is the greatest whole factor shared by 24 and 16?
Eight divides both counts: 24 = 3 × 8 and 16 = 2 × 8.
Step 2 of 2
What is the ratio of A’s rate to B’s rate?
A:B=24:16=3:2. Multiplying both ratio parts by the same positive factor gives the same ratio.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: compare the rates
Answer: The rate ratio is 3:2. We do not need the duration, and the individual rates are not determined.
Li Wei “Without the timer, can we still compare them?”
CKRobi “Which comparison could these records settle?”
Amara “Each total repeats its per-minute rate for the same number of minutes. That common multiplier can leave the ratio unchanged.”
The totals are in the ratio \(24:16=3:2\). Because both accumulated over the same positive duration, their per-minute rates have that ratio too.
Possible run 1
Run time
8 minutes
A each minute
3 readings
B each minute
2 readings
A: 3 × 8 = 24 B: 2 × 8 = 16
Possible run 2
Run time
4 minutes
A each minute
6 readings
B each minute
4 readings
A: 6 × 4 = 24 B: 4 × 4 = 16
Different possible runs produce the same totals and the same rate ratio.
Both runs produce exactly the given totals, so neither the duration nor either individual rate is fixed. The ratio is fixed.
Optional mathematics: remove the shared multiplier
For positive duration \(t\), the rates are \(24/t\) and \(16/t\). Therefore
\[\frac{24/t}{16/t}=\frac{24}{16}=\frac{3}{2}.\]
The shared part is a multiplier here. Subtracting the totals would give a difference in readings, not the rate ratio. Dividing out the multiplier requires \(t\ne0\).
A test that never ran
In a different test, the recorders are stopped before starting. The duration is 0 minutes, and each records 0 readings. Their constant positive rates are unknown.
Do those equal zero totals prove that the rates are equal? Give a check that settles the claim.
CKSTEM Thinking CompassPlan and check your reasoning.
Do those equal zero totals prove that the rates are equal? Give a check that settles the claim.
A test that never ran
In a different test, the recorders are stopped before starting. The duration is 0 minutes, and each records 0 readings. Their constant positive rates are unknown.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
Show the answer and reasoning: the zero-minute test
Answer:No. Rates of 2 and 5 readings per minute both give 0 readings in 0 minutes.
Amara “I’ll test rates of two and five readings per minute.”
Li Wei “In zero minutes, both still record nothing. The test cannot distinguish those rates.”
Multiplication by zero erased the information. We cannot divide by that shared factor. This is why the earlier test required a positive duration.
What the missing labels still allow
The dispatch team now needs complete field kits, each containing one wind unit and one rain unit. The individual mass labels are missing.
All wind units have the same mass, all rain units have the same mass, and all batteries have the same mass. Each component mass is a positive whole number of kilograms. These exact records list every weighed item:
Record 1
Wind unitWind unitRain unitBattery
30 kg
Record 2
Wind unitBattery
21 kg
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit, with no battery or other item.
What does one complete kit weigh? Find every possible triple of wind, rain and battery masses, and explain why no possibility is missing.
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What does one complete kit weigh? Find every possible triple of wind, rain and battery masses, and explain why no possibility is missing.
What the missing labels still allow
The dispatch team now needs complete field kits, each containing one wind unit and one rain unit. The individual mass labels are missing.
All wind units have the same mass, all rain units have the same mass, and all batteries have the same mass. Each component mass is a positive whole number of kilograms. These exact records list every weighed item:
Record 1
Wind unitWind unitRain unitBattery
30 kg
Record 2
Wind unitBattery
21 kg
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit, with no battery or other item.
Your answer
Add every allowed wind/rain/battery mass triple, one per row. Also enter one complete kit’s mass. The order of your rows does not matter.
The complete set and kit mass fit. Explain how your positive whole-number bounds cover every possible mass triple exactly once.
Break it into steps
Help step 1
After removing the shared wind unit and battery from each record, how many wind units remain in the first?
One wind unit remains in the first record after equal shared contents are removed.
Help step 2
How many rain units remain alongside it?
One rain unit remains, so the difference is exactly one complete wind-and-rain kit.
Help step 3
What does one complete wind-and-rain kit weigh?
Each full kit weighs nine kilograms.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the kit and every possible mass
Answer: Each kit weighs 9 kg. There are 8 possible mass triples, shown below. The individual masses are not uniquely determined.
Eliminate the shared part
Li Wei “We could try masses for all three parts.”
Amara “Or compare what one record has beyond the other.”
CKRobi “Could you defend that comparison using the original records?”
One wind unit and one battery occur in both records. Removing that equal group leaves one wind unit and one rain unit, so a complete kit weighs \(30-21=9\) kg. This settles the kit mass, but not the separate component masses.
Removing one wind unit and one battery from each record leaves exactly one complete kit.
Make an organized list
A fixed kit mass does not identify every allowed packing. List the wind mass in order. It must be at least 1 kg and at most 8 kg, because the rain unit must also have positive whole mass. For each wind mass, the kit fixes the rain mass, and the 21 kg record fixes the battery mass.
Every possible component mass, in kg
Wind
Rain
Battery
1
8
20
2
7
19
3
6
18
4
5
17
5
4
16
6
3
15
7
2
14
8
1
13
The wind mass increases by one each row; each choice fixes the other two masses.
Why the list is complete: wind masses 1 through 8 are every allowed whole value. Each gives exactly one rain mass and one battery mass. Each row satisfies both original records. There is no second choice for either remaining mass once its wind mass is fixed.
For example, wind 7 kg, rain 2 kg and battery 14 kg give \(2\times7+2+14=30\) and \(7+14=21\). The other rows can be checked in the same original records.
The comparison cut a three-unknown search down to one ordered list. Trialling all three masses is valid too, but makes completeness harder to see.
A promise before the call
The team must choose a kit count now. A colleague will soon call with the wind-unit mass, allowing one final choice of extras.
Every wind unit, rain unit and battery matches its own measured type. Their masses are positive whole numbers of kilograms. These are earlier lab records for these same equipment types:
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit, with nothing else. There are 20 complete kits available.
There are also two separate wind spares and one separate battery. Taking these extras does not break up the kit stock.
The depot requires exactly one extras bundle: either one battery or two whole wind spares. Either bundle is accepted.
After vehicle and packaging mass have been reserved, the complete kits and chosen extras may weigh at most 96 kg. No other payload is carried.
The team must fix a whole kit count now. After learning the wind-unit mass, they may choose the extras bundle, but cannot change the kit count.
What is the largest kit count they can guarantee for every mass triple allowed by the records? Give a rule for choosing the extras after the call, and prove that a larger promise can fail.
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What is the largest kit count they can guarantee for every mass triple allowed by the records? Give a rule for choosing the extras after the call, and prove that a larger promise can fail.
A promise before the call
The team must choose a kit count now. A colleague will soon call with the wind-unit mass, allowing one final choice of extras.
Every wind unit, rain unit and battery matches its own measured type. Their masses are positive whole numbers of kilograms. These are earlier lab records for these same equipment types:
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit, with nothing else. There are 20 complete kits available.
There are also two separate wind spares and one separate battery. Taking these extras does not break up the kit stock.
The depot requires exactly one extras bundle: either one battery or two whole wind spares. Either bundle is accepted.
After vehicle and packaging mass have been reserved, the complete kits and chosen extras may weigh at most 96 kg. No other payload is carried.
The team must fix a whole kit count now. After learning the wind-unit mass, they may choose the extras bundle, but cannot change the kit count.
Your answer
Enter the guaranteed count. Supply a valid after-report extras rule and an allowed counterexample to any larger promise; those quantifier and proof parts are unscored.
These quantities fit. Explain your choice of parts, how they recombine and why every original rule is met; reasoning stays unscored.
Break it into steps
Step 1 of 3
What is the mass of one complete kit from the fresh lab records?
The fresh records again force a nine-kilogram kit.
Step 2 of 3
If ten kits were promised, how much mass allowance would remain for the required extras?
Ten kits weigh ninety kilograms, leaving six. To test the guarantee, find an allowed report under which neither extras bundle fits that allowance.
Step 3 of 3
What is the greatest whole kit count guaranteed for every allowed report?
Nine always works with a report-dependent bundle. An allowed report defeats every promise of ten or more.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
A reliable promise must survive every mass report the records allow. Distinguish finding one working load from giving a plan for all of them.
Show the answer and full reasoning: a reliable largest promise
Answer: They can guarantee 9 complete kits. One valid rule is: choose the two wind spares for wind masses 1–6 kg, and the battery for wind masses 7–8 kg. A promise of 10 or more can fail.
Eliminate the shared part
Li Wei “One copy of the smaller record does not leave just a kit. I’ll compare two copies with the larger record.”
CKRobi “Can you justify that comparison using the whole records?”
Amara “Two copies mean two wind units and two batteries, weighing twice twenty-one. Those match inside the larger record.”
The kit mass is not supplied by these new records. Compare the 51 kg record with two complete copies of the 21 kg record. Double every item and the total: two wind units and two batteries weigh 42 kg. Removing that equal group leaves one wind unit and one rain unit—a complete kit weighing \(51-2\times21=9\) kg. Positive whole masses then give wind masses 1–8 kg; each battery weighs 21 minus its wind unit’s mass.
Doubling the complete smaller record makes the removable group; the total doubles too.
Bound, then build the best
Amara “Heavier wind spares mean a lighter battery. A report where the two bundles weigh the same might leave no cheap escape.”
Li Wei “The wind-and-battery record is twenty-one. If the battery weighs two wind units, that is three wind units making twenty-one: seven each, with a fourteen-kilogram battery.”
This suggests trying wind 7 kg, battery 14 kg and rain 2 kg. Check the new records: \(3\times7+2+2\times14=51\) and \(7+14=21\). All masses are positive whole numbers, so this is an allowed report. A single allowed report that defeats ten is enough to rule out a ten-kit guarantee.
A load that works for one mass report does not establish a guarantee. Consider the allowed report wind 7 kg, rain 2 kg and battery 14 kg. Either accepted extras bundle then weighs 14 kg. Ten kits need \(10\times9+14=104\) kg, exceeding 96 kg. Any larger kit count is heavier. No plan can guarantee ten or more.
One allowed report: wind 7 kg · rain 2 kg · battery 14 kg
Battery bundle
10 kits: 90 kg
One battery: 14 kg
90 + 14 = 104 kg
104 kg > 96 kg
Wind-spares bundle
10 kits: 90 kg
Two wind spares: 7 + 7 = 14 kg
90 + 14 = 104 kg
104 kg > 96 kg
This one allowed report defeats ten kits with either extras choice. A larger kit load is heavier.
That bounds the promise at nine. It does not yet show that nine always works.
Li Wei “I’ll group the allowed wind masses into one to six, and seven or eight, with one safe extras choice for each group.”
Amara “Then we can keep the nine kits fixed and check every possible report.”
Casework
The bound leaves a construction to prove for every allowed mass report. The two cases below are disjoint and cover all wind masses 1–8. Nine kits weigh \(9\times9=81\) kg in both cases.
Wind masses 1–6
Wind mass in kg
123456
Take two wind spares. Each is at most 6 kg.
Nine kits: 81 kg
Payload ≤ 93 kg
81 + 12 = 93 ≤ 96
Wind masses 7–8
Wind mass in kg
78
Take one battery. The 21 kg record makes it at most 14 kg.
Nine kits: 81 kg
Payload ≤ 95 kg
81 + 14 = 95 ≤ 96
Every allowed report belongs to one case. Keep the nine kits fixed; choose only the extras.
For wind masses 1–6, the wind spares weigh at most \(2\times6=12\) kg, so the payload is at most \(81+12=93\) kg. For wind masses 7–8, the 21 kg record makes the battery at most \(21-7=14\) kg, so the payload is at most \(81+14=95\) kg. Both fit 96 kg.
Full check: nine whole kits are available out of twenty. Each extras choice uses only its separate stock, and exactly one permitted bundle is included. The kit count is fixed before the call; only the bundle changes afterwards. Every allowed report is covered, and the reserved packaging allowance is respected.
The upper bound and the complete construction meet at nine, proving the largest guarantee. The rule is not unique: at wind masses 6 and 7 kg, either extras bundle fits. Our rule is one valid plan.
A full table testing both bundles for each of the eight reports is also valid. Casework turns that table into two short arguments and a practical dispatch rule.
Li Wei sends the plan to the field team. When the call arrives, they will know which extras to take without changing their promise.
Your turn
Choose what helps in each new situation. The answers remain closed until you want to check.
Two ticket receipts
Adult fares are fixed and positive, and child fares are fixed and positive. There are no other charges.
Receipt 1
Adult fares
1
Child fares
1
$14
Receipt 2
Adult fares
1
Child fares
3
$26
The same adult and child fares are used on both receipts.
What is one child fare, and what is one adult fare? Check both receipts.
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What is one child fare, and what is one adult fare? Check both receipts.
Two ticket receipts
Adult fares are fixed and positive, and child fares are fixed and positive. There are no other charges.
Receipt 1
Adult fares
1
Child fares
1
$14
Receipt 2
Adult fares
1
Child fares
3
$26
The same adult and child fares are used on both receipts.
Your answer
Enter both fares in their labelled roles and check both complete receipts.
These quantities fit. Explain your choice of parts, how they recombine and why every original rule is met; reasoning stays unscored.
Break it into steps
Step 1 of 3
What do the two extra child fares cost together?
Twenty-six minus fourteen is twelve.
Step 2 of 3
What is one child fare?
Twelve divided by two is six.
Step 3 of 3
What is one adult fare?
Fourteen minus six is eight.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the fares
Answer: One child fare is $6 and one adult fare is $8.
The second receipt includes the first receipt’s adult and child, plus two more children. Those extra fares cost \(26-14=12\) dollars, so each costs \(12\div2=6\) dollars. The adult fare is \(14-6=8\) dollars. Check: \(8+6=14\) and \(8+3\times6=26\).
Different run times
Two recorders run at constant positive rates for these exact times:
Recorder A
Run time
4 minutes
Total
20 readings
Recorder B
Run time
6 minutes
Total
30 readings
Both rates are constant, but these two run times differ.
A note says their rate ratio is 2:3 because their totals are 20:30.
Is the note correct? Find the actual rates and explain whether the duration can be cancelled.
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Is the note correct? Find the actual rates and explain whether the duration can be cancelled.
Different run times
Two recorders run at constant positive rates for these exact times:
Recorder A
Run time
4 minutes
Total
20 readings
Recorder B
Run time
6 minutes
Total
30 readings
Both rates are constant, but these two run times differ.
A note says their rate ratio is 2:3 because their totals are 20:30.
Your answer
Enter both actual rates. Also evaluate the claimed ratio and explain why the different durations cannot be cancelled.
These quantities fit. Explain your choice of parts, how they recombine and why every original rule is met; reasoning stays unscored.
Break it into steps
Step 1 of 2
What is recorder A’s actual rate?
Twenty divided by four is five.
Step 2 of 2
What is recorder B’s actual rate?
Thirty divided by six is five.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the run times
Answer:No. Both rates are 5 readings per minute, so the ratio is 1:1. The different durations cannot be cancelled as one common factor.
A records \(20\div4=5\) per minute; B records \(30\div6=5\). Check: \(5\times4=20\) and \(5\times6=30\). The totals differ because the run times differ.
Choose the extras earlier
Every wind unit, rain unit and battery matches its own type. Their masses are positive whole kilograms:
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit only. There are 20 complete kits, plus two wind spares and one battery in separate stock. The depot accepts exactly one extras bundle: one whole battery or two whole wind spares. After vehicle and packaging mass have been reserved, the kits and extras may weigh at most 96 kg; nothing else is carried.
This time, both the whole kit count and the extras bundle must be fixed before the wind-mass report. Neither may change afterwards.
What is the largest kit count they can guarantee with a single fixed bundle? Give a working choice and explain why neither fixed bundle can guarantee nine kits.
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What is the largest kit count they can guarantee with a single fixed bundle? Give a working choice and explain why neither fixed bundle can guarantee nine kits.
Choose the extras earlier
Every wind unit, rain unit and battery matches its own type. Their masses are positive whole kilograms:
All three component types have fixed positive whole-kilogram masses in these two records.
A complete kit contains one wind unit and one rain unit only. There are 20 complete kits, plus two wind spares and one battery in separate stock. The depot accepts exactly one extras bundle: one whole battery or two whole wind spares. After vehicle and packaging mass have been reserved, the kits and extras may weigh at most 96 kg; nothing else is carried.
This time, both the whole kit count and the extras bundle must be fixed before the wind-mass report. Neither may change afterwards.
Your answer
Enter the count. Choose a fixed bundle and justify both its guarantee and the failure of a nine-kit promise.
These quantities fit. Explain your choice of parts, how they recombine and why every original rule is met; reasoning stays unscored.
Break it into steps
Step 1 of 3
What is the largest battery mass allowed by the records?
At wind mass one kilogram, the battery weighs twenty kilograms.
Step 2 of 3
What is the largest total mass of two wind spares?
Two spares of eight kilograms each weigh sixteen.
Step 3 of 3
What is the largest guaranteed kit count with the extras bundle fixed before the report?
Either fixed bundle guarantees eight; each has an allowed report defeating nine.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: before the report
Answer: They can guarantee 8 kits with either fixed bundle. Neither fixed bundle guarantees nine.
The records still give a 9 kg kit and wind masses 1–8 kg. With a fixed battery, its largest possible mass is 20 kg: eight kits plus it weigh at most \(8\times9+20=92\) kg. With fixed wind spares, their largest total is 16 kg: \(8\times9+16=88\) kg. Both fit 96 kg.
A different allowed report defeats each fixed bundle. Neither can guarantee nine kits.
The first allowed report defeats nine kits with a fixed battery; the second defeats nine with fixed wind spares. Any larger kit count is heavier. Eight works and nine cannot be guaranteed, so eight is largest.
This does not contradict the earlier nine-kit plan. That plan could choose the bundle after the report; this task removes that choice. The timing of information changes what a reliable plan can do.
What makes the comparison valid?
Recognise the cue: two records contain a quantity known to be equal. Choose an operation that removes it while preserving the relationship you need. A determined difference or ratio does not always determine the individual values.
Use a variable helps you express unknown relationships. Eliminate the shared part helps you combine comparable relationships. They can work together; neither name makes every calculation useful.
An organised list can expose all possibilities. Casework can cover a whole group with one argument. A largest guarantee needs both an excluding bound and a plan that always attains it.
When stuck, put the records side by side and identify one genuinely shared quantity. Then check what your next step would establish—and what it would still leave unknown.
Amara “We have the totals, but not every separate mass.”
CKRobi “Which decision do the records need to help you make?”
Li Wei “First, which supply bag is heavier. Then we can plan the delivery.”