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Postcards from the Moon
Look back
An answer is a beginning. Find out what it really tells you.
Li Wei tests a small tile frame while Rowan listens. The full broadcast and backdrop plans come next.
Imagine a radio studio inside a Moon base. Rowan and Li Wei are preparing audio postcards: a little mystery, a favourite tune, a memory worth sharing. CKSprout is their coach.
Rowan “The recordings are ready. Can we press play?”
CKSprout “What would you like to be sure of first?”
Li Wei “That the whole programme fits. Our listeners have saved us one slot.”
The studio and its plans are imagined. Each problem gives the rules for its own model.
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.
Does the whole programme fit?
The programme has four recorded clips, each 3 minutes long, and one 2-minute introduction.
Allow a 1-minute transition between each pair of consecutive clips. There is no transition before the first clip or after the last. Nothing overlaps, and there are no other parts.
Li Wei has added the introduction and clips: 14 minutes. The available slot is 16 minutes.
Is that calculation a complete check that the programme fits? Find the full running time and say how much time is spare or how much it overruns.
Introduction
2 minutes · once
Four clips
3 minutes each
Transitions
1 minute between consecutive clips only
Available slot
16 minutes
The programme has an introduction, four clips, transitions and an available slot.A question to help me begin
Compare the proposed calculation with everything the original plan says must happen.
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Is that calculation a complete check that the programme fits? Find the full running time and say how much time is spare or how much it overruns.
Does the whole programme fit?
The programme has four recorded clips, each 3 minutes long, and one 2-minute introduction.
Allow a 1-minute transition between each pair of consecutive clips. There is no transition before the first clip or after the last. Nothing overlaps, and there are no other parts.
Li Wei has added the introduction and clips: 14 minutes. The available slot is 16 minutes.
Introduction
2 minutes · once
Four clips
3 minutes each
Transitions
1 minute between consecutive clips only
Available slot
16 minutes
The programme has an introduction, four clips, transitions and an available slot.
Your answer
Enter the full running time and the nonnegative size of its gap from the slot. Also decide whether the programme fits and whether any gap is spare time or an overrun; that decision is unscored.
The quantities fit. Explain the fit decision, the direction of the gap and what every programme part contributes.
Break it into steps
Step 1 of 3
How many transitions occur between four clips?
Four clips have three intervening gaps. Those three minutes were missing from the proposed total.
Step 2 of 3
What is your answer for Full minutes?
Full minutes: 17. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
How large is the gap between the full running time and the slot?
Overrun minutes: 1. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 whole programme
Answer:No. The full programme takes 17 minutes, so it overruns by 1 minute.
Li Wei “Two plus four lots of three is fourteen. That arithmetic is right.”
CKSprout “Does it account for the whole programme?”
Rowan “We need a transition after the first, second and third clips. Those three minutes are missing.”
1. Introduction0 → 2 min
2. Clip 12 → 5 min
3. Transition5 → 6 min
4. Clip 26 → 9 min
5. Transition9 → 10 min
6. Clip 310 → 13 min
7. Transition13 → 14 min
8. Clip 414 → 17 min
The cards record successive stages, not a length scale for time. The last clip ends at minute seventeen; no transition follows it.
Count in minutes:
\[2+4\times3+3\times1=17\]
Check: all four clips and all three transitions appear, with the introduction counted once. There is no extra transition after the last clip. The partial total was correct; the claim that the whole plan fitted was not.
Look back
Return to the original question. Does the answer meet every condition? Does the argument establish what you claimed?
A useful check can uncover an error, confirm a good plan or show that you still need to prove something. A different method can make the reason clearer. Keep that reason: it may help with another problem.
Work forward follows what the givens lead to. Look back asks whether the result and its explanation settle the whole request. Both can belong in one solution.
Seven recordings
A different programme contains seven complete recordings lasting 38, 47, 49, 45, 62, 53 and 51 seconds, plus one 30-second introduction.
There are no transitions, pauses or other parts. The slot lasts 6 minutes.
Find the full running time. Does it fit? State the spare time or overrun, and give a way to check your recording total without repeating the same addition order.
The original recording lengths
Clip
Seconds
Clip 1
38
Clip 2
47
Clip 3
49
Clip 4
45
Clip 5
62
Clip 6
53
Clip 7
51
Separate introduction: 30 seconds, once
Each row identifies one complete recording. The introduction is separate; there are no other parts.
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Find the full running time. Does it fit? State the spare time or overrun, and give a way to check your recording total without repeating the same addition order.
Seven recordings
A different programme contains seven complete recordings lasting 38, 47, 49, 45, 62, 53 and 51 seconds, plus one 30-second introduction.
There are no transitions, pauses or other parts. The slot lasts 6 minutes.
The original recording lengths
Clip
Seconds
Clip 1
38
Clip 2
47
Clip 3
49
Clip 4
45
Clip 5
62
Clip 6
53
Clip 7
51
Separate introduction: 30 seconds, once
Each row identifies one complete recording. The introduction is separate; there are no other parts.
Your answer
Enter the full running time and the nonnegative size of its gap from the slot. Also decide whether the programme fits and whether any gap is spare time or an overrun; that decision is unscored.
The quantities fit. Explain the fit decision, the direction of the gap and what every programme part contributes.
Break it into steps
Step 1 of 3
How long are the recordings alone?
Three hundreds and forty-five make 345 seconds. The thirty-second introduction is still separate.
Step 2 of 3
What is your answer for Full seconds?
Full seconds: 375. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
How large is the gap between the full running time and the slot?
Overrun seconds: 15. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 recording total
Answer:No, it does not fit. The full programme takes 375 seconds, or 6 minutes 15 seconds, an overrun of 15 seconds. Check the recording total by pairing 38 with 62, 47 with 53, and 49 with 51, then keeping the remaining 45 seconds.
Grouping terms
Rowan “I added the recordings in playback order and got 345 seconds. I want to check it.”
CKSprout “What different arrangement could make that total easier to inspect?”
Li Wei “Thirty-eight with sixty-two makes a hundred. So do forty-seven with fifty-three, and forty-nine with fifty-one. Three hundreds, plus forty-five: 345 again.”
Regroup the seven recording lengths
Clips
Seconds
Subtotal
1 and 5
38 + 62
100
2 and 6
47 + 53
100
3 and 7
49 + 51
100
Recording 4 remains: 45 seconds
Separate introduction: 30 seconds
Three pairs use six recordings. Keep the unpaired recording and the introduction as well.
Each recording appears once in this new arrangement. The separate introduction still needs its own 30 seconds.
In seconds, the full running time is:
\[3\times100+45+30=375\]
Check: six minutes is 360 seconds, and \(375-360=15\). The groups use all seven original recordings; the introduction appears once. The playback-order running totals are 38, 85, 134, 179, 241, 294 and 345. The nonadjacent pairs reach the same recording total by a different route.
How the tools fit: Look back asks whether the total includes every part. Grouping terms supplies a different, easy-to-follow arrangement for checking it.
Two-clip plans
Six different clips have these lengths: A: 35, B: 45, C: 55, D: 65, E: 75 and F: 85 seconds.
Each possible plan plays exactly two different clips with at most 120 seconds in total. There is no other airtime. AB and BA are the same choice.
A proposed list is AB, AC and BC.
Is the proposed list complete? Find every allowed pair and explain how you know that none is missing.
Six different clips
Clip
Seconds
A
35
B
45
C
55
D
65
E
75
F
85
Choose two different clips. A reversed order is the same choice, and the complete pair must fit the two-minute limit.
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Is the proposed list complete? Find every allowed pair and explain how you know that none is missing.
Two-clip plans
Six different clips have these lengths: A: 35, B: 45, C: 55, D: 65, E: 75 and F: 85 seconds.
Each possible plan plays exactly two different clips with at most 120 seconds in total. There is no other airtime. AB and BA are the same choice.
A proposed list is AB, AC and BC.
Six different clips
Clip
Seconds
A
35
B
45
C
55
D
65
E
75
F
85
Choose two different clips. A reversed order is the same choice, and the complete pair must fit the two-minute limit.
Your answer
Enter the count. Also list every allowed pair and explain whether the proposed list was complete.
These quantities fit. Check the count. Also list every allowed pair and explain whether the proposed list was complete.
Break it into steps
Step 1 of 2
How many allowed later partners does clip A have?
All five fit with A. Continue systematically: B has three later partners and C has one.
Step 2 of 2
What is your answer for Allowed pair count?
Allowed pair count: 9. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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: every allowed pair
Answer:No. There are 9 allowed pairs: AB, AC, AD, AE, AF, BC, BD, BE and CD.
Li Wei “All three pairs on the list fit the limit.”
CKSprout “What has that check established?”
Rowan “That those three work. It has not shown that every working pair is there.”
Make an organized list
Fix the earlier letter and try every later partner. This covers each pair once.
Every distinct pair, once
Pair
Seconds
Fits?
AB
80
Yes
AC
90
Yes
AD
100
Yes
AE
110
Yes
AF
120
Yes
BC
100
Yes
BD
110
Yes
BE
120
Yes
BF
130
No
CD
120
Yes
CE
130
No
CF
140
No
DE
140
No
DF
150
No
EF
160
No
Thick lines start a new earlier-letter group. Checking all fifteen choices distinguishes nine allowed pairs from a merely working shortlist.
Check completeness: A has five allowed later partners, B has three, and C has one. C with E or F is already too long. D's shortest later partner is E, and \(65+75=140\), so no pair starting at D or later can fit. Thus \(5+3+1=9\).
Looking back revealed that “all these work” was weaker than “these are all.” The organised list closes that gap.
The light patch
A rectangular light patch has positive whole-number side lengths, measured in tile widths. Its perimeter is exactly 12 tile widths. One square tile fills each unit square, with no gaps or overlap. Squares are allowed rectangles.
A 2-by-4 patch uses 8 tiles. A note calls this “the greatest possible number.” Turning a rectangle does not make a different shape.
Is the greatest-possible claim correct? Find the greatest tile count, give its dimensions and explain why no allowed rectangle uses more.
Proposed patch: 2 by 4 · 8 tiles
The proposed patch has two rows of four tiles. The note makes a claim about its tile count.Try with the CKSTEM Thinking Compass
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Is the greatest-possible claim correct? Find the greatest tile count, give its dimensions and explain why no allowed rectangle uses more.
The light patch
A rectangular light patch has positive whole-number side lengths, measured in tile widths. Its perimeter is exactly 12 tile widths. One square tile fills each unit square, with no gaps or overlap. Squares are allowed rectangles.
A 2-by-4 patch uses 8 tiles. A note calls this “the greatest possible number.” Turning a rectangle does not make a different shape.
Proposed patch: 2 by 4 · 8 tiles
The proposed patch has two rows of four tiles. The note makes a claim about its tile count.
Your answer
Enter the greatest count and its dimensions. Explain why checking one allowed patch does not prove greatest.
These quantities fit. Check the greatest count and its dimensions. Explain why checking one allowed patch does not prove greatest.
Break it into steps
Step 1 of 4
What must one length plus one width equal?
Half of twelve is six. The shorter whole side can therefore be one, two or three.
Step 2 of 4
What is your answer for Greatest tiles?
Greatest tiles: 9. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 4
What is your answer for Shorter side?
Shorter side: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 4 of 4
What is your answer for Longer side?
Longer side: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 greatest patch
Answer:No. A 3-by-3 patch uses the greatest count, 9 tiles.
Rowan “The two-by-four patch has the right perimeter. It really does use eight tiles.”
CKSprout “What does that check tell us about the note?”
Li Wei “It proves this rectangle works, but not that it is greatest. I will test every possible shorter side.”
Casework
Look back asks whether the greatest claim has been justified. A legal candidate alone does not rule out a better one.
Two copies of the length and width make the perimeter, so one length plus one width is 6. The shorter side cannot exceed half of 6, so its positive whole-number cases are 1, 2 and 3.
1 by 5
5 tiles
2 by 4
8 tiles
3 by 3
9 tiles
The adjacent side lengths add to six. Up to turning, these are all the positive whole-number choices, so comparing them proves the maximum.
Up to turning, the complete possibilities are \(1+5\), \(2+4\) and \(3+3\). Their areas are 5, 8 and 9.
Check: the winning patch has perimeter \(3+3+3+3=12\) and area \(3\times3=9\). The complete list, rather than the one working example, proves the greatest claim.
The loose-tile backdrop
There are 20 trays of loose, identical square light tiles. Tray 1 contains 1 tile, tray 2 contains 3, tray 3 contains 5, and each successive tray contains 2 more. Tray 20 contains 39 tiles.
The backdrop has 20 rows of 20 unit squares. Tiles can be moved freely between trays and placed one per square. Every tile must be used, with no cutting, overlap or gaps.
A proposed total is \(20\times(1+39)=800\) tiles, so the note says the backdrop is too small.
Check the proposed total. Find the actual number of tiles and decide whether they fill the backdrop exactly. If the trays start with one tile and keep increasing by two, will they always fill a square whose side length is the number of trays? Justify your answer for any positive whole number of trays.
Loose tiles
20 trays: 1, 3, 5, …, 39
Total unknown
Backdrop
20 rows · 20 columns
The backdrop grid is reduced to fit. The tray rule is given, but no stock total or filled arrangement has been supplied.Try with the CKSTEM Thinking Compass
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Check the proposed total. Find the actual number of tiles and decide whether they fill the backdrop exactly. If the trays start with one tile and keep increasing by two, will they always fill a square whose side length is the number of trays? Justify your answer for any positive whole number of trays.
The loose-tile backdrop
There are 20 trays of loose, identical square light tiles. Tray 1 contains 1 tile, tray 2 contains 3, tray 3 contains 5, and each successive tray contains 2 more. Tray 20 contains 39 tiles.
The backdrop has 20 rows of 20 unit squares. Tiles can be moved freely between trays and placed one per square. Every tile must be used, with no cutting, overlap or gaps.
A proposed total is \(20\times(1+39)=800\) tiles, so the note says the backdrop is too small.
Loose tiles
20 trays: 1, 3, 5, …, 39
Total unknown
Backdrop
20 rows · 20 columns
The backdrop grid is reduced to fit. The tray rule is given, but no stock total or filled arrangement has been supplied.
Your answer
Enter the total. Also decide exact fit and justify the square rule for every positive tray count; this number does not grade that proof.
These quantities fit. Check the total. Also decide exact fit and justify the square rule for every positive tray count; this number does not grade that proof.
Break it into steps
Step 1 of 2
How many disjoint outside tray pairs can twenty trays form?
Twenty trays make ten pairs, not twenty. Each first-and-last pair totals forty tiles.
Step 2 of 2
What is your answer for Total tiles?
Total tiles: 400. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 full backdrop
Answer: The proposed total is wrong. There are 400 tiles, and they fill the 20-by-20 backdrop exactly. Yes, the rule always supplies exactly enough tiles for a square whose side length is the tray count.
Rowan “One plus thirty-nine makes forty. I multiplied that by the twenty trays.”
CKSprout “Which trays are inside one forty?”
Li Wei “Tray one and tray twenty. I can pair from the outside as we did with the recordings. Twenty trays make ten pairs, not twenty.”
The team reuses Grouping terms to inspect every tray once.
Twenty terms make ten outside pairs
Trays
Tiles
Total
1 and 20
1 + 39
40
2 and 19
3 + 37
40
3 and 18
5 + 35
40
4 and 17
7 + 33
40
5 and 16
9 + 31
40
6 and 15
11 + 29
40
7 and 14
13 + 27
40
8 and 13
15 + 25
40
9 and 12
17 + 23
40
10 and 11
19 + 21
40
Every tray appears once in the table. Each pair uses two original terms; ten groups of forty make four hundred tiles.
\[10\times40=400\]
Use a variable
Rowan “That settles twenty trays. Trying another number still would not explain every possible number.”
CKSprout “What could you name so that the same reasoning can follow the changing count?”
Li Wei “I will call the number of trays \(n\), and the tray we are adding \(k\). Tray \(k\) has had \(k-1\) increases of two since the first tile.”
The check settles the twenty-tray count, but it does not prove a rule for every tray count. Trying more totals could suggest a rule without explaining why it must hold.
Here \(n\) is a positive whole number, and \(k\) is a whole number with \(1\le k\le n\). The rule for tray \(k\) becomes:
\[1+2(k-1)=2k-1\]
The target is a square with side \(n\). The symbols keep the tray rule and the target clear, but they do not yet show how the tiles fit.
Draw a diagram
Li Wei “We have a rule for the count. I still want to see where each tray goes.”
CKSprout “What arrangement could you try with the first few trays?”
Rowan “The next tray is longer than one side. I can bend it around the corner as an L, growing both sides of the square.”
1234223433344444
Tray 1: 1 · tray 2: 3 · tray 3: 5 · tray 4: 7
Matching numbers and colours identify one complete tray. New bottom rows and right columns make disjoint L-shaped layers; this small picture illustrates the construction.
To grow a square of side \(k-1\) to side \(k\), add a bottom row of \(k\) tiles and a right column of \(k-1\) tiles above it. The shared corner belongs to the bottom row only. Together they use \(k+(k-1)=2k-1\), exactly tray \(k\).
Previous square
Side \(k-1\)
Amber bottom row
\(k\) new tiles, including the corner
Teal right column
\(k-1\) new tiles above that corner
This schematic labels the general step rather than measuring a particular drawing. The dashed corner is part of the bottom row only; the new row and column use the next tray exactly.
Start with tray 1. Each new layer fills the new row and column once, so there are no gaps or overlaps. After \(n\) trays, every position of the \(n\)-by-\(n\) square is filled, using \(n^2\) tiles. This construction explains the reusable rule; the four-tray picture alone is not a proof of every case.
Check the general total by grouping too
The first and last terms give \(1+(2n-1)=2n\). Moving inward adds two to one term and removes two from its partner, so each outside pair has that same total. For even \(n\), there are \(n/2\) pairs:
\[\frac{n}{2}(2n)=n^2\]
For odd \(n\), there are \((n-1)/2\) such pairs. The unpaired middle tray comes after \((n-1)/2\) increases of two from the first tile, so it contains \(n\) tiles:
\[\frac{n-1}{2}(2n)+n=n^2\]
Full check: the 20 trays contain all consecutive odd amounts from 1 through 39. Ten disjoint pairs give 400; the square holds \(20\times20=400\). The layer construction uses each complete tray once. We did not silently discard tiles or count a shared corner twice.
How the tools fit:Look back challenges what the calculation and its claim establish. Use a variable keeps the changing tray count and each layer’s meaning clear. Draw a diagram gives a structural explanation and a construction that works for every allowed tray count. The earlier grouping check provides a separate arithmetic confirmation.
Your turn
Decide what each check needs to establish before opening the answer. A plan may be correct.
A shorter programme
A programme has three 2-minute clips, one 1-minute introduction and a 30-second transition between consecutive clips only. There is no transition before the first clip or after the last. Nothing overlaps; there are no other parts.
Does the whole programme fit an 8-minute slot? State its running time and any spare time or overrun.
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Does the whole programme fit an 8-minute slot? State its running time and any spare time or overrun.
A shorter programme
A programme has three 2-minute clips, one 1-minute introduction and a 30-second transition between consecutive clips only. There is no transition before the first clip or after the last. Nothing overlaps; there are no other parts.
Your answer
Enter the full running time and the nonnegative size of its gap from the slot. Also decide whether the programme fits and whether any gap is spare time or an overrun; that decision is unscored.
The quantities fit. Explain the fit decision, the direction of the gap and what every programme part contributes.
Break it into steps
Step 1 of 3
How many seconds do the transitions take altogether?
Two transitions use sixty seconds, one minute. Add that once to the six clip minutes and one introduction minute.
Step 2 of 3
What is your answer for Full minutes?
Full minutes: 8. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
How large is the gap between the full running time and the slot?
Spare minutes: 0. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 shorter programme
Answer:Yes. It takes 8 minutes exactly, with no spare time or overrun.
Three clips make six minutes. The two transitions make one minute; the introduction adds one more. \(6+1+1=8\). The check confirms the plan instead of finding a mistake.
Another light patch
A spare rectangular light patch has perimeter exactly 20 tile widths and positive whole-number side lengths. Squares are allowed, and turning a rectangle does not make a different shape. One tile fills each unit square.
Rowan proposes a 5-by-5 patch with 25 tiles. His note says: “Its perimeter is 20, so 25 must be the greatest possible tile count.”
Does that argument prove the greatest claim? Decide whether 25 really is the greatest possible count, and justify your conclusion.
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Does that argument prove the greatest claim? Decide whether 25 really is the greatest possible count, and justify your conclusion.
Another light patch
A spare rectangular light patch has perimeter exactly 20 tile widths and positive whole-number side lengths. Squares are allowed, and turning a rectangle does not make a different shape. One tile fills each unit square.
Rowan proposes a 5-by-5 patch with 25 tiles. His note says: “Its perimeter is 20, so 25 must be the greatest possible tile count.”
Your answer
Enter the best count. Explain whether the original note proves it, then supply the missing greatest-value argument.
These quantities fit. Check the best count. Explain whether the original note proves it, then supply the missing greatest-value argument.
Break it into steps
Step 1 of 2
What must one length plus one width equal?
The side sum is ten. Test shorter whole sides one through five, including the square.
Step 2 of 2
What is your answer for Greatest tiles?
Greatest tiles: 25. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 spare patch
Answer:No, the perimeter check does not prove the greatest claim. Yes, 25 is the greatest possible tile count, from the 5-by-5 patch.
The note proves only that the proposed patch is allowed. To settle greatest, one length plus one width must be 10. Up to turning, the complete possibilities are 1 by 9, 2 by 8, 3 by 7, 4 by 6 and 5 by 5. Their tile counts are 9, 16, 21, 24 and 25. None is greater than 25.
Check: the 5-by-5 patch has perimeter \(5+5+5+5=20\) and uses \(5\times5=25\) tiles. A correct answer can still need a stronger argument.
When does the reply arrive?
In a simplified radio model, a short message takes 1.3 seconds from Moon to Earth and the same time back. As soon as Earth receives the whole message, the person there takes 3 seconds to prepare and send a short reply. Assume no other delay or transmission time.
A note gives 4.3 seconds from sending the Moon message until its reply reaches the Moon.
Is the note correct? Find the full elapsed time and explain what the check must include.
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Is the note correct? Find the full elapsed time and explain what the check must include.
When does the reply arrive?
In a simplified radio model, a short message takes 1.3 seconds from Moon to Earth and the same time back. As soon as Earth receives the whole message, the person there takes 3 seconds to prepare and send a short reply. Assume no other delay or transmission time.
A note gives 4.3 seconds from sending the Moon message until its reply reaches the Moon.
Your answer
Enter the full elapsed time in seconds. You may use a decimal or a fraction. Also decide whether the note is correct and explain your check using the complete journey.
The elapsed time fits the simplified model. Explain whether the note is correct and identify which stages your check includes.
Break it into steps
Step 1 of 3
How many one-way signal journeys occur before the reply reaches the Moon?
There is one Moon-to-Earth journey and one Earth-to-Moon journey. The stated endpoint needs both.
Step 2 of 3
How many seconds do the signal journeys take altogether, before adding preparation?
The two journeys take 1.3 + 1.3 = 2.6 seconds. Preparation is a separate three-second stage.
Step 3 of 3
How many seconds pass from sending the Moon message until the reply reaches the Moon?
The complete model gives 1.3 + 3 + 1.3 = 5.6 seconds. The proposed 4.3 seconds stops after only one signal journey and preparation. Explain why the return journey must also be included.
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 reply
Answer:No. The reply reaches the Moon after 5.6 seconds in this model.
Moon → Earth1.3 seconds
Prepare the answer3 seconds
Earth → Moon1.3 seconds
Follow the reply all the way back to the Moon. The model includes both signal journeys and the preparation time.
The outgoing journey, answer preparation and return journey give \(1.3+3+1.3=5.6\) seconds. The note’s 4.3 seconds counts only one journey. The model states what is included; a real communications system can have additional delays.
What will you check next time?
The best discovery was not that a plan could be wrong. It was that the team could ask a useful next question.
Does it meet the whole request? What does the argument prove? Can another route explain it? Which idea could help next time?
Draw a diagram gives the tile check a visible structure. Keep the habit of choosing a picture that exposes the relationship you need to understand.
Rowan “The recordings are ready. Can we press play?”
CKSprout “What would you like to be sure of first?”
Li Wei “That the whole programme fits. Our listeners have saved us one slot.”