Sorry guys!!!! both answers are wrong!.... please try again....

Best Regards,
Luke.

---

- Luke.

19. Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

A is the slowpoke, so give A the bicycle, he arrives at the end in 1 minute.
C arrives at the end in 2 minutes, takes the bicycle, rides back to B who is currently at the 2/5 mark so they have to cover 3/5 in total to intersect.
B covers 1/5 of the bridge per minute, C on bicycle covers 5/5 per minute, so they intersect in (3/5)/(1/5+5/5) = 1/2 minute. Time now 2 1/2 minutes.
B takes bicycle and goes to end; don't have to calculate that as C is now the bottleneck.
B was (2 1/2)/(5) or 1/2 of the way across the bridge leaving 1/2 left. C runs back to end in 1 minute and all have crossed.
Total time is 3 1/2 minutes or 3 mins. 30 seconds.

Logic falls down where I have bolded. In this scenario, both A and B are standing waiting for C to arrive for 1/2 a minute.

Once B is on the bicycle, he pedals like crazy for 24 seconds then drops it and continues on foot getting to the end in a further 30 seconds.

Meanwhile C has used 48 seconds from the halfway point to reach the bicycle dropped by B and needs 6 seconds to reach the end at exactly the same time as B.

So the cumulative time is 2.5 mins + 54 secs = 3 mins 24 secs.

F.

---

Sorry guys!!!! both answers are wrong!.... please try again....

Best Regards,
Luke.

Well, a cogent argument could be made that the first pirate really has no good proposal to make.

I will try to see if there is some other line of reasoning hough, I am fairly sure that if there is one then the first pirate takes the bulk of the loot.

regards,

DADDIO

How about this solution for the pirate problem

997
0
1
0
2

Bill

19. Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

A is the slowpoke, so give A the bicycle, he arrives at the end in 1 minute.
C arrives at the end in 2 minutes, takes the bicycle, rides back to B who is currently at the 2/5 mark so they have to cover 3/5 in total to intersect.
B covers 1/5 of the bridge per minute, C on bicycle covers 5/5 per minute, so they intersect in (3/5)/(1/5+5/5) = 1/2 minute. Time now 2 1/2 minutes.
B takes bicycle and goes to end; don't have to calculate that as C is now the bottleneck.
B was (2 1/2)/(5) or 1/2 of the way across the bridge leaving 1/2 left. C runs back to end in 1 minute and all have crossed.
Total time is 3 1/2 minutes or 3 mins. 30 seconds.

Logic falls down where I have bolded. In this scenario, both A and B are standing waiting for C to arrive for 1/2 a minute.

Once B is on the bicycle, he pedals like crazy for 24 seconds then drops it and continues on foot getting to the end in a further 30 seconds.

Meanwhile C has used 48 seconds from the halfway point to reach the bicycle dropped by B and needs 6 seconds to reach the end at exactly the same time as B.

So the cumulative time is 2.5 mins + 54 secs = 3 mins 24 secs.

F.

You are probably close or right on. Using only simple arithmetic and no calculator I have it down to 3:27 trick may be to have everyone including the bicycle finish at the same time. This would maximally utilize the most valuable resource.

19. Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

A is the slowpoke, so give A the bicycle, he arrives at the end in 1 minute.
C arrives at the end in 2 minutes, takes the bicycle, rides back to B who is currently at the 2/5 mark so they have to cover 3/5 in total to intersect.
B covers 1/5 of the bridge per minute, C on bicycle covers 5/5 per minute, so they intersect in (3/5)/(1/5+5/5) = 1/2 minute. Time now 2 1/2 minutes.
B takes bicycle and goes to end; don't have to calculate that as C is now the bottleneck.
B was (2 1/2)/(5) or 1/2 of the way across the bridge leaving 1/2 left. C runs back to end in 1 minute and all have crossed.
Total time is 3 1/2 minutes or 3 mins. 30 seconds.

Logic falls down where I have bolded. In this scenario, both A and B are standing waiting for C to arrive for 1/2 a minute.

Once B is on the bicycle, he pedals like crazy for 24 seconds then drops it and continues on foot getting to the end in a further 30 seconds.

Meanwhile C has used 48 seconds from the halfway point to reach the bicycle dropped by B and needs 6 seconds to reach the end at exactly the same time as B.

So the cumulative time is 2.5 mins + 54 secs = 3 mins 24 secs.

F.

You are probably close or right on. Using only simple arithmetic and no calculator I have it down to 3:27 trick may be to have everyone including the bicycle finish at the same time. This would maximally utilize the most valuable resource.

This ain't simple.

A's speed is 1/10 (in bridges per minute), B's speed is 1/5, C's speed is 1/2, and the bicycle's speed is 1.

I believe that the fastest way to get everyone across is for B and C to start out on foot and A to start out with the bicycle. At a point y A will get off the bicycle and walk the rest of the way. Eventually C will get to the bicycle abondoned by A, then ride backwards to a point x, leaving the bicycle there, then turning around and walk until he reaches the end. Person B will walk until he reaches the bicycle left by C and then ride the rest of the way.

Below are the times that each will take to cross, in terms of x and y:

A: 1*y + 10*(1-y)
B: 5*x + 1*(1-x)
C: 2*y + (y-x) + 2*(1-x)

Next equate these equations: 10 - 9y = -3x + 3y + 2 = 4x + 1.

To solve set up two linear equations:

10 - 9y = -3x + 3y + 2 -> 3x - 12y = -8
10 - 9y = 4x + 1 -> 4x + 9y = 9

Then solve for x and y:

x = 12/25, y=59/75.

Given these points it will take each person 73/25 = 2.92min.or 2min. 55.2sec. to cross.

---

They would have been simple if everyone had PM'd me for the hints.....

Bill wins Q18: A997 coins, B0 coins, C1 coin, D0 coins, E2 coins. Congrats again!
&
Dominique wins Q19: 2.92 Minutes / 2m 55s. Well Done!!!

Standings Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 1 1/2 Points
5. John McLeod VII - 1 Point
6. Dominique - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Simpler than ever before....

Q20 (Half-Point & No calcs) : (8x7)x(15x26) divided by 15 = ???

Best Regards,
Luke.

---

- Luke.

They would have been simple if everyone had PM'd me for the hints.....

Bill wins Q18: A997 coins, B0 coins, C1 coin, D0 coins, E2 coins. Congrats again!
&
Dominique wins Q19: 2.92 Minutes / 2m 55s. Well Done!!!

Standings Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 1 1/2 Points
5. John McLeod VII - 1 Point
6. Dominique - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Simpler than ever before....

Q20 (Half-Point & No calcs) : (8x7)x(15x26) divided by 15 = ???

Best Regards,
Luke.

1456

---

1456

too easy

They would have been simple if everyone had PM'd me for the hints.....

Bill wins Q18: A997 coins, B0 coins, C1 coin, D0 coins, E2 coins. Congrats again!
&
Dominique wins Q19: 2.92 Minutes / 2m 55s. Well Done!!!

Standings Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 1 1/2 Points
5. John McLeod VII - 1 Point
6. Dominique - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Simpler than ever before....

Q20 (Half-Point & No calcs) : (8x7)x(15x26) divided by 15 = ???

Best Regards,
Luke.

I'll point I've been a way and have not looked at all posts in this thread.
I've noticed with some a lack of explanation, just a statement of the answer.
Maybe Luke can start to award points based on answering versus answering with explanations.
Now, with this particular question, I'm sure it'll seem trivial, but ...

Obviously, the factor 15 will cancel, so we're left with 8 * 7 * 26. The associative property allows for dropping the parentheses. We can also factor and re-associate in various ways. (A sub-question some might find interesting is: how many ways could this be done for this multiplication problem.)
Here, I'm thinking of the prime factorization of 26, 2 * 13.
(8 * 7) * 26 = 56 * (2 * 13) = (56 * 2) * 13. 56 and 2 are relatively small, so getting to 112 * 13 is not too difficult for us.
Multiplication relies heavily on the distributive property and I do not want to go get paper and do this in my head late at night, so:
112 * 13 = (100 + 10 + 2) * (10 + 3) = 100 * 10 + 100 * 3 + 10 * 10 + 10 * 3 + 2 * 10 + 2 * 3 = 1000 + 300 + 100 + 30 + 20 + 6 = 1456.

---

Capitalize on this good fortune, one word can bring you round ... changes.

They would have been simple if everyone had PM'd me for the hints.....

Bill wins Q18: A997 coins, B0 coins, C1 coin, D0 coins, E2 coins. Congrats again!
&
Dominique wins Q19: 2.92 Minutes / 2m 55s. Well Done!!!

Standings Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 1 1/2 Points
5. John McLeod VII - 1 Point
6. Dominique - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Simpler than ever before....

Q20 (Half-Point & No calcs) : (8x7)x(15x26) divided by 15 = ???

Best Regards,
Luke.

I'll point I've been a way and have not looked at all posts in this thread.
I've noticed with some a lack of explanation, just a statement of the answer.
Maybe Luke can start to award points based on answering versus answering with explanations.
Now, with this particular question, I'm sure it'll seem trivial, but ...

Obviously, the factor 15 will cancel, so we're left with 8 * 7 * 26. The associative property allows for dropping the parentheses. We can also factor and re-associate in various ways. (A sub-question some might find interesting is: how many ways could this be done for this multiplication problem.)
Here, I'm thinking of the prime factorization of 26, 2 * 13.
(8 * 7) * 26 = 56 * (2 * 13) = (56 * 2) * 13. 56 and 2 are relatively small, so getting to 112 * 13 is not too difficult for us.
Multiplication relies heavily on the distributive property and I do not want to go get paper and do this in my head late at night, so:
112 * 13 = (100 + 10 + 2) * (10 + 3) = 100 * 10 + 100 * 3 + 10 * 10 + 10 * 3 + 2 * 10 + 2 * 3 = 1000 + 300 + 100 + 30 + 20 + 6 = 1456.

So, yeah, I know Dominique and William answered before me, so Dominique can get the points or whatever, but I just wanted to offer food for thought. Again, providing explanations and the related problem of how many ways could we have dealt with 8 * 7 * 26.

(How about (8 * 13) * (7 * 2) = 104 * 14? Is it an easier than the way I initially broke it down? Hmmm. Actually, yes!)

---

Capitalize on this good fortune, one word can bring you round ... changes.

Dominique wins.... 1/2 a point.... congrats!!!
Answer: 1456...
Sarge is also awarded 1/2 a point for his idea and explanation of the question... I will now desire explanation with every question....

Standings Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 2 Points
5. Dominique - 1 1/2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Ah Geez.... I hate to get onto this track... but anyone who maintains a thread know it is extremely hard to cater for everyone's needs....
Some people want hard questions... some people want easy ones... some people don't like probability or calculus... some people like probability and so on.....
I do my best... and it is horrible when people ask for different things all at once.... please try to be considerate and respectful of what I am doing....
I am also having trouble finding questions now... I am going to have to start internet searching soon enough.... (I have found a few questions with a specific site I like.... but I'm running low anyway)

And once again thank you for keeping this thread alive... It is your contribution that helps....

Here are Three More:

Q21: A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?

Q22: 5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?

Q23: Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

Best Regards,
Luke.

---

- Luke.

Quick answer to Q23, no explanation, 50 mins.

Now I got time explaination.
Wifes return journey is 20 mins shorter so therefore she meets hubby 10 mins earlier. As he got off train 60 mins early he must have met her 60 - 10 mins after getting off train.

Q21: A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?

The subtleties of question posing.
It seems as if the intake must have been occurring during the two instances of checking the outflow for full work and 3/5 the work.

---

Capitalize on this good fortune, one word can bring you round ... changes.

WinterKnight is correct.... Well Done! 1 Point for you
Q23 Answer: 50 Minutes
Q23 Explanation:

Wifes return journey is 20 mins shorter so therefore she meets hubby 10 mins earlier. As he got off train 60 mins early he must have met her 60 - 10 mins after getting off train.

And the Standings once again:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 2 Points
5. Dominique - 1 1/2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

Mr. Kevvy still holds a fairly strong lead.... Bill and WinterKnight trail... but they themselves seem to be making a gap between 3 & 4....

All I can say is that Q21 & Q22 are still going:

Q21: A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?

Q22: 5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?

Best Regards,
Luke.

---

- Luke.

Q21: A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?

Let x be the input rate in gallons per hour into the tank.
Let y be the output rate in gallons per hour from the tank for each value.
Let w be the number of gallons in the tank.

It's known that it takes 2.5 hours to empty the tank with 10 valves open. Over the 2.5 hours the sum of the water in the tank initially and the water going into the tank will equal the amount of water leaving the tank.

Set that up as the equation: w + 2.5x = 2.5*10y

It's also known that it takes 5.5 hours to empty the tank with 6 valves open.

Set that up as the equation: w + 5.5x = 5.5*6y

To make it easier by getting rid of the decimals just multiply both equations by 2:

2w+5x=50y
2w+11x=66y

To solve the problem you need to know the relationship between w and x, so lets solve for y in the first equation and subsitute in the the second:

y=(2w+5x)/50.

Substituting this in the second equation...

2w+11x=66*(2w+5x)/50

100w+550x=132w+330x

220x=32w

w=220x/32

w=6.875x

So 6.875 times the input rate equals the capacity of the tank. Thus it would take 6.875 hours to fill the tank.

---

Don't you have to know if the tank was full to begin with. Or are the valve timings with the input shut off and the tank initially full--hard to believe this is irrelevant. I will go back and reread the problem-once you fully understand the problem you are halfway to its solution.

Regards.

Bill

Well Done again there Dominique.... everyone is getting very good at this....
Q21 Answer: 6.875 Hours
I agree with the Explanation provided.... 1 Point to Dominique...

Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Dominique - 2 1/2 Points
5. Sarge - 2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

The Top 3 still hold a steady lead on users 4-7.... Although Dominique is steadily catching up.... Mr. Kevvy may soon be displaced at this rate.....

The Final Unanswered question so far:

Q22: 5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?

Have a go....
Luke.

---

- Luke.

Q22: 5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?

I think this one is easier than it sounds.

The cows will eat the initial grass plus total grass that regrows.

So, let me define some terms.

x=initial amount of grass in an acre.
y=amount of grass grown in one acre in one day.

Putting this info in the form of equations we get:

50=2x+20y
210=3x+90y

To put it in more simple English terms, 5 cows eating for 10 days
results in a consumption of 50 units of grass. This is equal to
the sum of 2x initial units of grass and 20y units of grass growth
(10 days times 2 acres).

So to solve for x and y. Rewriting the above equations I get:

150=6x+60y (multiplying by 3)
420=6x+180y (multiplying by 2)

Subtract the first equation from the second and you get 270=120y,
so y=9/4. Plug this into either equation and you get x=5/2.

Coming as no surprise let d be the number of days.

So setting this up as an equation we get.

16d=7x + 7dy

16d=35/2 + 63d/4

d/4=35/2

d=70.

So it will take 70 days for 16 cows to eat 7 acres of grass.

---