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# Mathematics Brain Teasers.

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On 20/10/2018 at 4:11 PM, Ozymandias said:

Assume that a bee flies directly back and forth at a constant 5m/s between its hive situated beside a straight road and a honey pot being carried towards the hive by a beekeeper walking at a steady 3km/hr along the same road. When the bee starts to fly the honey pot is initially 1 kilometre from its hive. How far will the bee have flown in total when the beekeeper reaches its hive? [Admittedly, the poor bee will be knackered!!!]

This teaser has had nearly 100 views over the past few days with no takers so I will post the solution later today.

Edited by Ozymandias

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On 10/20/2018 at 4:11 PM, Ozymandias said:

Assume that a bee flies directly back and forth at a constant 5m/s between its hive situated beside a straight road and a honey pot being carried towards the hive by a beekeeper walking at a steady 3km/hr along the same road. When the bee starts to fly the honey pot is initially 1 kilometre from its hive. How far will the bee have flown in total when the beekeeper reaches its hive? [Admittedly, the poor bee will be knackered!!!]

OK. I'm somewhat disappointed that there have now been over 100 views of this teaser without any attempt to solve it, so here is the solution:

The bee will fly continuously at 5m/s while the beekeeper walks the1 kilometre distance to its hive. Walking at a steady 3km/hr that will take the beekeeper 20 minutes.

Therefore, the bee flies for 20 minutes at 5m/s: i.e. 5m/s x 60sec/min x 20mins = 6000m = 6km. The answer is 6 kilometres.

To continue this thread someone else can set a question.

Edited by Ozymandias
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3 hours ago, Ozymandias said:

OK. I'm somewhat disappointed that there have now been over 100 views of this teaser without any attempt to solve it, so here is the solution:

The bee will fly continuously at 5m/s while the beekeeper walks the1 kilometre distance to its hive. Walking at a steady 3km/hr that will take the beekeeper 20 minutes.

Therefore, the bee flies for 20 minutes at 5m/s: i.e. 5m/s x 60sec/min x 20mins = 6000m = 6km. The answer is 6 kilometres.

To continue this thread someone else can set a question.

Ah shyt I forgot to post my answer, was meant to do it before physiotherapy. I was wondering why I didn't get a response.

I'll post a simple one tomorrow, hopefully to entice the masses.

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Screw it. Easy this one ........is.

1+1+1+1+1

1+1+1+1+1

1+1+1*0+1=X

Solve for X.

Edited by danydandan
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On 23/10/2018 at 10:40 PM, danydandan said:

Screw it. Easy this one ........is.

1+1+1+1+1

1+1+1+1+1

1+1+1*0+1=X

Solve for X.

X = 3.

Can't figure out what the first and second lines are there for. The sum of each is 5. How are they supposed to relate to the equation in X on the third line?

Edited by Ozymandias
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7 hours ago, Ozymandias said:

X = 3.

Can't figure out what the first and second lines are there for. The sum of each is 5. How are they supposed to relate to the equation in X on the third line?

Yeap, it's a coding riddle, kinda. Some people continue the string and add up all the ones.

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Another easy one:-

0 = (2*X^2 - 5*X) / (X^3 + X^2 - 7*X + 1)

Solve for X.

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On 27/10/2018 at 1:09 AM, Ozymandias said:

Another easy one:-

0 = (2*X^2 - 5*X) / (X^3 + X^2 - 7*X + 1)

Solve for X.

Seriously no one is going to answer.

There are theoretically two answers. The obvious one is x=0. The not so obvious one 2.5 if you use roots and single variable equation techniques!

Ozy I suspect you were looking for just zero?

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21 minutes ago, danydandan said:

Seriously no one is going to answer.

There are theoretically two answers. The obvious one is x=0. The not so obvious one 2.5 if you use roots and single variable equation techniques!

Ozy I suspect you were looking for just zero?

Spot on, Dan.

The problem was in the form of a fraction equalling zero. That can only be true if the numerator on top of the fraction equals zero.

Inspection, and native wit, quickly identifies X = 0 as a solution. That is the simple answer I hoped people would get.

But the numerator is a quadratic expression that equals zero for the statement to be true. Quadratic equations, as you point out, have two roots. In this case one is X = 0 and the other is X = 2.5. I expected you, at least, to give both results.

If you want to continue this thread then set another teaser. I'd be happy to let it finish here.

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13 minutes ago, Ozymandias said:

Spot on, Dan.

The problem was in the form of a fraction equalling zero. That can only be true if the numerator on top of the fraction equals zero.

Inspection, and native wit, quickly identifies X = 0 as a solution. That is the simple answer I hoped people would get.

But the numerator is a quadratic expression that equals zero for the statement to be true. Quadratic equations, as you point out, have two roots. In this case one is X = 0 and the other is X = 2.5. I expected you, at least, to give both results.

If you want to continue this thread then set another teaser. I'd be happy to let it finish here.

I'll think of one tomorrow.

Hopefully get a month out if this thread lol.

Edited by danydandan
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1+1=1.

Is this a true statement? Whatever your answer, why?

Edited by danydandan
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13 hours ago, danydandan said:

1+1=1.

Is this a true statement? Whatever your answer, why?

Depends on the specific definitions of 1, + and =, and the context. Most people seeing that equation will view it through the narrow prism of base 10 arithmetic using natural numbers and say it is false, but in other contexts it is quite true.

If 1 symbolises 'a colour', + means 'mixed with', and = signifies 'results in', then 1 + 1 = 1.

I could say more but what the hell. The interest is overwhelming.

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21 minutes ago, Ozymandias said:

Depends on the specific definitions of 1, + and =, and the context. Most people seeing that equation will view it through the narrow prism of base 10 arithmetic using natural numbers and say it is false, but in other contexts it is quite true.

If 1 symbolises 'a colour', + means 'mixed with', and = signifies 'results in', then 1 + 1 = 1.

I could say more but what the hell. The interest is overwhelming.

Indeed. Think I'll leave it for awhile.

Revive it in a few weeks.

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If I parachute out of a plane at midnight on a moonless night, and land in a roughly circular lake, will I be closer to the shore, or the middle of the lake, on the balance of probabilities ?

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3 hours ago, Habitat said:

If I parachute out of a plane at midnight on a moonless night, and land in a roughly circular lake, will I be closer to the shore, or the middle of the lake, on the balance of probabilities ?

Pretty simple one. I'm assuming lake R=2 thus closer to centre of lake R=1. Just assuming working that out in my head, you have a P=0.333 odds of being closer to the centre and 0.666 odds of being closer to the edge. So your more likely to randomly hit a point closer to the edge

Edited by danydandan
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3 hours ago, Habitat said:

If I parachute out of a plane at midnight on a moonless night, and land in a roughly circular lake, will I be closer to the shore, or the middle of the lake, on the balance of probabilities ?

Thanks for joining the thread, Habitat. And with an interesting question. My take on it is this:

Landing nearer the centre means dropping into the central circular area whose radius is no greater than half the lakes radius. Landing nearer the edge means dropping into the annular area whose radius is greater than half that of the lake. The probability distribution is the ratio of these areas.

Area of Annulus : Area Centre Circle

(pi)(1^2 - 0.5^2) : (pi)0.5^2

0.75 : 0.25

75% chance of being closer to the lake's edge.

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1 hour ago, Ozymandias said:

Thanks for joining the thread, Habitat. And with an interesting question. My take on it is this:

Landing nearer the centre means dropping into the central circular area whose radius is no greater than half the lakes radius. Landing nearer the edge means dropping into the annular area whose radius is greater than half that of the lake. The probability distribution is the ratio of these areas.

Area of Annulus : Area Centre Circle

(pi)(1^2 - 0.5^2) : (pi)0.5^2

0.75 : 0.25

75% chance of being closer to the lake's edge.

Yeah that's more correct then my response I think.

Getting all fancy with your equations n stuff.

Edited by danydandan
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Yeah it's definitely 1/4 at being closer to the centre.

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Have Ozy or myself given the correct answer?

If nah or yah, please let us know.

Also you can set the next teaser if you wish.

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1 minute ago, danydandan said:

Have Ozy or myself given the correct answer?

If nah or yah, please let us know.

Also you can set the next teaser if you wish.

Obviously an exact answer would depend on the lake, but clearly there is greater probability of being closer to the shore, than the centre of the lake, by virtue of the outer annulus having greater area than the bullseye centre.

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52 minutes ago, Habitat said:

Obviously an exact answer would depend on the lake, but clearly there is greater probability of being closer to the shore, than the centre of the lake, by virtue of the outer annulus having greater area than the bullseye centre.

Do you want to set another teaser?

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8 hours ago, danydandan said:

Do you want to set another teaser?

OK, we sometimes hear that the Second Coming ( of Jesus Christ) is soon to be upon us, but I have seen an interesting observation made about this possibility, and it concerns what would have become of a sum of money that the Lord may have left in the bank, at a modest interest rate of say 2% compounding annually, at the time of his untimely demise, and upon his return, now wants to distribute to the poor, and in excess of what would be available, were the world's richest man today, be giving away all his money. So, how much, at today's \$ values, would he have needed on deposit in that completely "safe" bank, 2018 years ago, at 2%, to be able to now distribute that much money ? The answer may shock !

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12 minutes ago, Habitat said:

OK, we sometimes hear that the Second Coming ( of Jesus Christ) is soon to be upon us, but I have seen an interesting observation made about this possibility, and it concerns what would have become of a sum of money that the Lord may have left in the bank, at a modest interest rate of say 2% compounding annually, at the time of his untimely demise, and upon his return, now wants to distribute to the poor, and in excess of what would be available, were the world's richest man today, be giving away all his money. So, how much, at today's \$ values, would he have needed on deposit in that completely "safe" bank, 2018 years ago, at 2%, to be able to now distribute that much money ? The answer may shock !

You will need to specify how much money he will be giving away today in order to calculate the amount put on deposit 2018 years ago.

Edited by Ozymandias
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1 minute ago, Ozymandias said:

You will need to specify how much money he will be giving away today in order to calculate the amount put on deposit 2018 years ago.

Whatever the nett worth of the world's richest man is. Plus a little more. Let's be very generous, and say \$200 Billion

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19 minutes ago, Habitat said:

Whatever the nett worth of the world's richest man is. Plus a little more. Let's be very generous, and say \$200 Billion

Thanks. Nice question. I'll withhold my ridiculously tiny answer until tomorrow to let others have a go.

Edited by Ozymandias

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