Solve your puzzle and present the possible solution in the Comments While you tried to solve the puzzle, what mathematical knowledge and skills did you apply to solve the problem?

One of the solution i discovered is: 1. Move 1 and 3 ( Their speed num) to the left (6,8,12 at right) 27 minutes 2. Move 3 back to the right (3,6,8,12 at right) 24 minutes 3. Move 8 and 12 to the left (3,6 at right) 12 minutes 4. Move 1 back to the right (1,3,6 at right) 11 minutes 5. Move 1 and 6 to the left (3 at right) 5 minutes 6. Move 1 back to the right(1,3 at right) 4 minutes 7. Move 1 and 3 to the left ( None at right) 1 minute :D

The puzzle requires us to move all the people from the right to the left in 30 minutes (lamp life). By moving 2 people at the same time, the time used would be the person that has more minutes. (Eg 1 and 12 move at the same time, 12 minutes would be used) So by moving 2 people, we would have "profit". The most profited would be 8 + 12 as we would have a "profit" of 8 minutes. Another thing is to add up all the people's minutes when they walk back and fro which is (1+3+6+8+12)2 = 40 :O so by using "profit" we could decrease our 40 to (below)30.

Move 1 and 3 to the left. Bring 3 back. Move 8 and 12 to the left. Bring 1 back. Move 1 and 6 to the left. Then move 1 back again And then move 1 and 3 to the left.

TAADAAA~~~

This puzzle actually make me find out all the speeds of the people and move them according to their speeds. So by moving the 1st 2 fastest people first, time will not be wasted to bring them back. Then moving the 2 slowest people, they will be "cleared". Always leaving one of the 2 fastest people also helped in transporting the other slower people. By using their speeds to calculate the time used by all people, they can be shifted according to their advantage.

Move 1 and 3 to the left ->30-3=27min left Move 1 back ->27-1=26min left Move 8 and 12 to the left->26-12=14min left Move 3 back ->12-3=11min left Move 6 and 1 to the left ->11-6=5min left Move 1 back ->5-1=4min left Move 1 and 3 to the left ->4-3=1min left => Win!

The puzzle tells us to move the 2 slowest persons to the left together and bring the fastest person back for the next move. This will solve the puzzle. Nice game!

Through some experimentations, i seem to find that the only solution is: Move 1 and 3 to the left Move 3 back to the right Move 8 and 12 to the left Move 1 back to the right Move1 and 6 to the left Move 1 back to the right Lastly,move 1 and 3 to the left.

I applied a little of LCM or HCF. As if u move 8 and 12 or 1 and 3 the time/ lamp life will be of the bigger number. So if you move 12 the biggest number, you should also move the second biggest number as the time taken will still be the same. And you can also save time for bringing the lamp back by using the smallest number as it is just transporting themselves back and the time taken will be of their number alone.

Move 1 and 3 to the left. Move 3 back to the right. Move 8 and 12 to the right. Move 1 back to the right. Move 1 and 3 to the left. Move 1 back to the right. Move 1 and 6 to the left. There's your ans :D/

This puzzle teaches me to move the people with the slowest speed to the left as soon as possible so that they would not get in the way. And the time is usually followed by the person with the slower speed crossing the bridge.

1 and 3 to LHS of the bridge 1 to RHS of the bridge(3 still at LHS) 8 and 12 to LHS of the bridge 3 to RHS of the bridge(8 and 12 at LHS) 1 and 6 to LHS of the bridge 1 to RHS of the bridge(6,8,12 at LHS) 1 and 3 to LHS of the bridge Finish. LHS= left hand side, RHS= right hand side.

1 and 3 are moved immediately to the LHS of the bridge so that one of them is able to deliver the lamp back when 8 and 12 crossed the bridge.

Move 1 and 3 to the left (3 mins) Move 3 back to the right (3 mins) Move 8 and 12 to the left (12 mins) Move 1 back to the right (1 min) Move 1 and 6 to the left (6 mins) Move 1 back to the right (1 min) Move 1 and 3 to the left (3 mins) ________ 29 mins (1 min left)

This puzzle required some logic to solve. When moving the person that takes 12 minutes to cross the bridge, whichever person moving with her will still take 12 minutes to cross the bridge, so it is only logical that we move the person that takes 8 minutes to cross the bridge with her, so as to save time. We should also use the person that only takes 1 minute to cross the bridge to transport the lantern as we can also save time that way.

The solution i found is: Move 1 and 3 to the left. Bring 3 back to the right. Move 8 and 12 to left. Bring 1 back to the right. Move 1 and 6 to the left. Move 1 back to the right Move 1 and 3 to the left.

As the time taken for the pair to cross the bridge is the time taken for the slower person to cross the bridge, therefore, to lessen the time taken for everyone to cross the bridge, the slowest people should go together in order to lessen the time taken. Done :D

OK. to be truthful i was so fed up with the game i went to find the tips.

But i discovered that the rationales behind the trick to the game, was to move the 2 fastest ppl to the left first because we would want to save the most time in moving the lamp back to the front, after that we move the fatest guy back to the right, and we let 8 walk with 12 back to the right since they're both so slow and it'll be a waste to ask 1 to stroll with 12 and then get another fast guy to stroll with 8 also, and plus there is still 3 to fetch the lamp back who is moderately fast. so 3 will come back to the right, and let 1 and 6 go the left, so that 1 can bring the lamp back to the right the fastest. then when 1 come back with the lamp, he will bring 3 back, HALLELUJAH.

Move 1 and 3 to left, Move 1 back. Move 8 and 12 to left, Move 3 back. Move 1 and 6 to left, Move 1 back. Move 1 and 3 to left. All are at the left, and 1 minute left!

First move the fastest people over (1and3), they will be "transporters" so that moving back to right will be faster. So move 1 back. Currently: 3 at left | 1,6,8,12 at right Then now you can first get rid of the slowest people. Which is to move (8and12). Currently: 3,8,12 at left | 1,6 at right Move the fastest at the left back to right, which is (3). Currently: 8,12 at left | 1,3,6 at right Now, move 1 and 6 to the left. Currently: 1,6,8,12 at left | 3 at right Move 1 back and move (1and3) to left All at left! and you still have 1 minute left.

Move 1 and 3 to the left (27 minutes left) Move 3 back to the right (24 minutes left) Move 8 and 12 to the left (12 minutes left) Move 1 back to the right (11 minutes left) Move1 and 6 to the left (5 minutes left) Move 1 back to the right (4 minutes left) Move 1 and 3 to the left. (1 minutes left) And everybody travelled across the bridge

To solve this I basically used trial and error. As the lamp can only last for 30 minutes, it is only smart to make the faster people 1 and 3 the ones that would be travelling over the bridge back and forth while the slowest people 8 and 12 just move from the right to let and stay put there. And you just need to keep trying till you get everyone over within 30 minutes.

Move 1 and 3 to the left side. Bring 3 to the right side. Move 8 and 12 to the left side. Move 1 to right side. Move 1 and 6 to the left side. Then move 1 back again to the right side Lastly,move 1 and 3 to the left.

To solve this,first I use logic,since they would take the slower person time,thus of course 1 and 3 would be the one traveling back and forth to transport the lamp while the slowest 8 and 12 would stay after being transported to the left.After that,I use the trial and error method to solve the question.

The method i found was to; 1st: Bring 1 and 3 to LHS 2nd: Bring 1 back to RHS 3rd: Bring 8 and 12 to LHS 4th: Bring 3 back to RHS 5th: Bring 1 and 6 to LHS 6th: Bring 1 back to RHS 7th: Finally, bring 1 and 3 to LHS

This puzzle requires us to think methodically, such as to minimize the time difference between the 2 people traversing and not requiring them to cross back such as 12 and 8, which are very time consuming. Also, we need to take into account which people should traverse back and forth, with the longer time-taking people not traversing such as 12, 8 and 6, leaving us with 1 and 3. This is basically a logical test where we must decide the proper order of people moving.

Solution: 1) Bring 1 and 3 to the left 2) Bring 1 back to the right 3) Bring 1 and 6 to the left 4) Bring 1 back to the right 5) Bring 8 and 12 to the left 6) Bring 3 back to the right 7) Bring 1 and 3 to the left

At first, I used trial and error to determine the possible combinations and see the time for the different combinations( e.g. Person 1+Person 3=3min). I found out that 1 and 3 moves the fastest, while 8 and 12 moves the slowest. Thus, to save time, which is what the game is all about, 1 and 3 would do most of the moving of the lantern. Another conclusion from the trial and error is that once 8 and 12 is on the left side, they must not move anymore. Finally, I tried moving 1 and 3, which is the most important, to the left side, and did trial and error for the rest. To sum it up, I used trial and error.

Move 1 and 3 to the left side of the bridge Send 1 back to the right side of the bridge Move 12 and 8 to the left side of the bridge Send 3 back to the right side of the bridge Move 6 and 1 to the left side of the bridge Send 1 back to the right side of the bridge Move 3 and 1 to the right side of the bridge

This puzzle require us to plan and calculate the speed and the time the light still lasts.

One of the solution i discovered is:

ReplyDelete1. Move 1 and 3 ( Their speed num) to the left (6,8,12 at right) 27 minutes

2. Move 3 back to the right (3,6,8,12 at right) 24 minutes

3. Move 8 and 12 to the left (3,6 at right) 12 minutes

4. Move 1 back to the right (1,3,6 at right) 11 minutes

5. Move 1 and 6 to the left (3 at right) 5 minutes

6. Move 1 back to the right(1,3 at right) 4 minutes

7. Move 1 and 3 to the left ( None at right) 1 minute :D

The puzzle requires us to move all the people from the right to the left in 30 minutes (lamp life). By moving 2 people at the same time, the time used would be the person that has more minutes. (Eg 1 and 12 move at the same time, 12 minutes would be used) So by moving 2 people, we would have "profit". The most profited would be 8 + 12 as we would have a "profit" of 8 minutes. Another thing is to add up all the people's minutes when they walk back and fro which is (1+3+6+8+12)2 = 40 :O

so by using "profit" we could decrease our 40 to (below)30.

Owen here.

The solution that I discovered is:

ReplyDeleteMove 1 and 3 to the left.

Bring 3 back.

Move 8 and 12 to the left.

Bring 1 back.

Move 1 and 6 to the left.

Then move 1 back again

And then move 1 and 3 to the left.

TAADAAA~~~

This puzzle actually make me find out all the speeds of the people and move them according to their speeds. So by moving the 1st 2 fastest people first, time will not be wasted to bring them back. Then moving the 2 slowest people, they will be "cleared". Always leaving one of the 2 fastest people also helped in transporting the other slower people. By using their speeds to calculate the time used by all people, they can be shifted according to their advantage.

My solution:

ReplyDeleteMove 1 and 3 to the left ->30-3=27min left

Move 1 back ->27-1=26min left

Move 8 and 12 to the left->26-12=14min left

Move 3 back ->12-3=11min left

Move 6 and 1 to the left ->11-6=5min left

Move 1 back ->5-1=4min left

Move 1 and 3 to the left ->4-3=1min left => Win!

The puzzle tells us to move the 2 slowest persons to the left together and bring the fastest person back for the next move. This will solve the puzzle. Nice game!

Through some experimentations, i seem to find that the only solution is:

ReplyDeleteMove 1 and 3 to the left

Move 3 back to the right

Move 8 and 12 to the left

Move 1 back to the right

Move1 and 6 to the left

Move 1 back to the right

Lastly,move 1 and 3 to the left.

I applied a little of LCM or HCF. As if u move 8 and 12 or 1 and 3 the time/ lamp life will be of the bigger number. So if you move 12 the biggest number, you should also move the second biggest number as the time taken will still be the same. And you can also save time for bringing the lamp back by using the smallest number as it is just transporting themselves back and the time taken will be of their number alone.

Move 1 and 3 to the left.

ReplyDeleteMove 3 back to the right.

Move 8 and 12 to the right.

Move 1 back to the right.

Move 1 and 3 to the left.

Move 1 back to the right.

Move 1 and 6 to the left.

There's your ans :D/

This puzzle teaches me to move the people with the slowest speed to the left as soon as possible so that they would not get in the way. And the time is usually followed by the person with the slower speed crossing the bridge.

1 and 3 to LHS of the bridge

ReplyDelete1 to RHS of the bridge(3 still at LHS)

8 and 12 to LHS of the bridge

3 to RHS of the bridge(8 and 12 at LHS)

1 and 6 to LHS of the bridge

1 to RHS of the bridge(6,8,12 at LHS)

1 and 3 to LHS of the bridge

Finish.

LHS= left hand side, RHS= right hand side.

1 and 3 are moved immediately to the LHS of the bridge so that one of them is able to deliver the lamp back when 8 and 12 crossed the bridge.

The solution that I discovered is:

ReplyDeleteMove 1 and 3 to the left (3 mins)

Move 3 back to the right (3 mins)

Move 8 and 12 to the left (12 mins)

Move 1 back to the right (1 min)

Move 1 and 6 to the left (6 mins)

Move 1 back to the right (1 min)

Move 1 and 3 to the left (3 mins)

________

29 mins (1 min left)

This puzzle required some logic to solve. When moving the person that takes 12 minutes to cross the bridge, whichever person moving with her will still take 12 minutes to cross the bridge, so it is only logical that we move the person that takes 8 minutes to cross the bridge with her, so as to save time. We should also use the person that only takes 1 minute to cross the bridge to transport the lantern as we can also save time that way.

The solution i found is:

ReplyDeleteMove 1 and 3 to the left.

Bring 3 back to the right.

Move 8 and 12 to left.

Bring 1 back to the right.

Move 1 and 6 to the left.

Move 1 back to the right

Move 1 and 3 to the left.

As the time taken for the pair to cross the bridge is the time taken for the slower person to cross the bridge, therefore, to lessen the time taken for everyone to cross the bridge, the slowest people should go together in order to lessen the time taken.

Done :D

OK. to be truthful i was so fed up with the game i went to find the tips.

ReplyDeleteBut i discovered that the rationales behind the trick to the game, was to move the 2 fastest ppl to the left first because we would want to save the most time in moving the lamp back to the front, after that we move the fatest guy back to the right, and we let 8 walk with 12 back to the right since they're both so slow and it'll be a waste to ask 1 to stroll with 12 and then get another fast guy to stroll with 8 also, and plus there is still 3 to fetch the lamp back who is moderately fast. so 3 will come back to the right, and let 1 and 6 go the left, so that 1 can bring the lamp back to the right the fastest. then when 1 come back with the lamp, he will bring 3 back, HALLELUJAH.

Move 1 and 3 to left,

ReplyDeleteMove 1 back.

Move 8 and 12 to left,

Move 3 back.

Move 1 and 6 to left,

Move 1 back.

Move 1 and 3 to left.

All are at the left, and 1 minute left!

First move the fastest people over (1and3), they will be "transporters" so that moving back to right will be faster. So move 1 back.

Currently: 3 at left | 1,6,8,12 at right

Then now you can first get rid of the slowest people. Which is to move (8and12).

Currently: 3,8,12 at left | 1,6 at right

Move the fastest at the left back to right, which is (3).

Currently: 8,12 at left | 1,3,6 at right

Now, move 1 and 6 to the left.

Currently: 1,6,8,12 at left | 3 at right

Move 1 back and move (1and3) to left

All at left! and you still have 1 minute left.

I kinda used trial-and-error.

Move 1 and 3 to the left (27 minutes left)

ReplyDeleteMove 3 back to the right (24 minutes left)

Move 8 and 12 to the left (12 minutes left)

Move 1 back to the right (11 minutes left)

Move1 and 6 to the left (5 minutes left)

Move 1 back to the right (4 minutes left)

Move 1 and 3 to the left. (1 minutes left)

And everybody travelled across the bridge

To solve this I basically used trial and error. As the lamp can only last for 30 minutes, it is only smart to make the faster people 1 and 3 the ones that would be travelling over the bridge back and forth while the slowest people 8 and 12 just move from the right to let and stay put there.

And you just need to keep trying till you get everyone over within 30 minutes.

Move 1 and 3 to the left side.

ReplyDeleteBring 3 to the right side.

Move 8 and 12 to the left side.

Move 1 to right side.

Move 1 and 6 to the left side.

Then move 1 back again to the right side

Lastly,move 1 and 3 to the left.

To solve this,first I use logic,since they would take the slower person time,thus of course 1 and 3 would be the one traveling back and forth to transport the lamp while the slowest 8 and 12 would stay after being transported to the left.After that,I use the trial and error method to solve the question.

The method i found was to;

ReplyDelete1st: Bring 1 and 3 to LHS

2nd: Bring 1 back to RHS

3rd: Bring 8 and 12 to LHS

4th: Bring 3 back to RHS

5th: Bring 1 and 6 to LHS

6th: Bring 1 back to RHS

7th: Finally, bring 1 and 3 to LHS

This puzzle requires us to think methodically, such as to minimize the time difference between the 2 people traversing and not requiring them to cross back such as 12 and 8, which are very time consuming. Also, we need to take into account which people should traverse back and forth, with the longer time-taking people not traversing such as 12, 8 and 6, leaving us with 1 and 3. This is basically a logical test where we must decide the proper order of people moving.

Solution:

ReplyDelete1) Bring 1 and 3 to the left

2) Bring 1 back to the right

3) Bring 1 and 6 to the left

4) Bring 1 back to the right

5) Bring 8 and 12 to the left

6) Bring 3 back to the right

7) Bring 1 and 3 to the left

At first, I used trial and error to determine the possible combinations and see the time for the different combinations( e.g. Person 1+Person 3=3min). I found out that 1 and 3 moves the fastest, while 8 and 12 moves the slowest. Thus, to save time, which is what the game is all about, 1 and 3 would do most of the moving of the lantern. Another conclusion from the trial and error is that once 8 and 12 is on the left side, they must not move anymore. Finally, I tried moving 1 and 3, which is the most important, to the left side, and did trial and error for the rest.

To sum it up, I used trial and error.

Move 1 and 3 to the left side of the bridge

ReplyDeleteSend 1 back to the right side of the bridge

Move 12 and 8 to the left side of the bridge

Send 3 back to the right side of the bridge

Move 6 and 1 to the left side of the bridge

Send 1 back to the right side of the bridge

Move 3 and 1 to the right side of the bridge

This puzzle require us to plan and calculate the speed and the time the light still lasts.

Move 1 and 3 to left,

ReplyDeleteMove 1 back.

Move 8 and 12 to left,

Move 3 back.

Move 1 and 6 to left,

Move 1 back.

Move 1 and 3 to left.

After 25 tries >.<

Actually, I found out that this activity uses a bit of logical thinking and using the timing of each person.