# Brain Teasers are Back!

There are eight combinations of engine situation.
I’ll denote a Live engine as L and a failed engine as F

LLL .8 x .8 x .8 = 0.512 = 51.2%
LLF .8 x .8 x .2 = 0.128 = 12.8%
LFL .8 x .2 x .8 = 0.128 = 12.8%
FLL .2 x .8 x .8 = 0.128 = 12.8%
LFF .8 x .2 x .2 = 0.032 = 3.2%
FLF .2 x. 8 x .2 = 0.032 = 3.2%
FFL .2 x .2 x .8 = 0.032 = 3.2%
FFF .2 x .2 x .2 = 0.008 = 0.8%

By adding up the various options :-
Zero engine failures LLL= 51.2%
One engine failing LLF + LFL + FLL = 12.8% + 12.8% + 12.8% = 38.4%
Two engines failing LFF + FLF + FFL = 3.2% + 3.2 % + 3.2% = 9.6%
Three engines failing FFF = 0.8%

There are other ways of setting out the explanation. I hope this one was clear.

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Thanks Don.
I remembered using Pascal’s triangle to solve probability problems and used that, but your explanation is much clearer.

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I often resort to a tree diagram which helps me keep track of the various options as well as the associated probabilities.

Good call Mike !

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Mr Pascal has a lot to answer for in my current field of employment !

Altitude (and height) are pretty crucial measurements that need to be got right. The precision altimeter is pretty accurate. It’s basically a barometer calibrated in feet (or metres in some aircraft). It has a sub-scale that is calibrated in units of atmospheric pressure. Someone on the ground can radio that surface pressure (or the associated sea-level pressure) to the flight crew who can reset the altimeter sub-scale to match that pressure.

We used to measure atmospheric pressure in units of Inches of Mercury - and still do so in the USA and Canada. Some while back, here in the UK we changed to millibars (mb). One Bar being roughly atmospheric pressure at sea level. The International Standard Atmosphere (ISA) uses 1013.25 mb as the standard sea-level pressure. In N America it’s 29.92 inches of Mercury, but they only quote the last three digits eg 992 when the pressure is below 30 inches.

Then a few years ago, some bright spark in Europe decided it would be even better to use Pascals, or rather - hecto pascals. The good thing being that 1,000mb is the same pressure as 1,000 hPa. Non-the-less, I still find it a mouthful to readback (say) “998 hecto-pascals” to make it clear that i’m not using inches of mercury. It was much easier to say “998 millibar”

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It must be a great feeling to have one’s name attached to something for posterity, eg Pascal’s Triangle, Fibonacci Series, Archimedes’ Principle, Newton’s Laws of Motion etc etc. I suspect though that most of them have been awarded posthumously.
Perhaps you may eventually become famous for Don’s Devilish Teasers

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If it’s got to be dome posthumously, believe me - i’m in no rush !!

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The Airport !

You know what it’s like, you sat in the terminal’s waiting room but got bored looking at the screen displaying departure gates and times until … yes, you hear “Final call for Flight No xxxx …” and you realise it’s your flight and the gate is soon to close !

It’s a ten minute walk to the departure gate. But fortunately there are moving walkways for half the journey. You have your carry-on bag with you, but decide that you can actually run for about a minute without collapsing and needing medical attention. You really do not want to miss your flight.

Should you do your running on the moving walkways or should you do it on the bits in between ?

Hmmm!
Intuitively, one might think you need to know the speed of the walkway, and your walking and running speeds.
Also, intuitively, one might think your running should be on the walkway (if that’s not an oxymoron!).
However, playing about in Excel with a few “reasonable” assumptions, it appears that the quickest overall journey occurs when you run on the bits in between.
Haven’t worked out why yet though…

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I believe that you cover the same additional distance over that minute on either platform

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I don’t think it quite works out that way JR . Try the following.

If I double my walking speed (S) at my running speed (2S) I will halve the time to cover a given distance between the walkways.
If we assume the walkway also moves at my walking speed (S) then my overall speed when walking on the walkway will be (2S) and my overall speed when running on the walkway will be (3S). I will only reduce the time to cover a given distance to 2/3 on the walkway.

A quick run through of a few scenarios on Excel, as Steve found out, is pretty convincing.

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Well done Steve. I agree, it does seem somewhat counter-intuitive.

What would we all do without Excel ?

L’Hopital paid an impecunious stranger for the mathematical rule that bears his name.

I thought a “reasonable” example might help browsers to follow Steve and my own reasoning about doing the running on the bit between the moving walkways.

Assume I walk at 1m/sec, I run at 2m/sec and the moving walkway (travelator) moves at 1m/sec. This means my combined speed when walking on the travelator will be 2m/sec and when running on the travelator it will be 3m/sec.

The normal 10 minute journey would comprise 400m of pathway at 1m/sec = 400 secs, plus 400m of walking along the travelator at a combined speed 2m/sec = 200 secs. Total journey time 400 + 200 = 600 secs = 10 minutes

So far, so good ?

If I now do the journey again but incorporate 1 minute (60 secs) of running :-

First; running on the travelator.

1 minute running (60 secs) along the travelator at a combined speed 3m/sec covers 180m. This leaves 220m of travelator walking at 2m/sec ie 110 secs ie 110 + 60 = 170 secs of time on the travelator.

I also have 400 secs of walking on the 400m of path between the travelator sections.

Total journey time of 170 + 400 = 570 secs. Which is less than the basic 600 sec/10 minute journey.

Second; running on the path between the travelator sections.

1 minute running (60 secs) along the pathway between the travelator sections at 2m/sec covers 120m. This leaves 280 m of path walking at 1m/s ie 280 secs ie 280 + 60 = 340 secs of time on the bits of path between the travelators.

I also have 200 secs of walking along the 400m of travelator(s) at a combined speed of 2m/s.

Total journey time of 340 + 200 = 540 secs. Which less than the 570 sec journey time if I do the running on the travelator.

You can substitute other values for walking/running/travelator speed but will always come to the same conclusion.

Electric v Diesel

Diane has an electric car which has a range of 210 miles at 70 mph and which takes 45 mins to recharge. Donald has a diesel car with a range in excess of 420 miles providing it is driven at a steady 60 mph.

They set off at the same time on a 420 mile journey. Who arrives first ?

PS you can assume that Donald has a good bladder and doesn’t need to stop en-route !

It looks easy, but I thought that about the last one. Their actual speeds are not stated, but I will assume that Diane drives at 70 mph and Don at 60 mph. Assuming that Diane’s car starts out fully charged and has she a charger available 210 miles out, Diane drives 210 miles in three hours, recharges for 45 minutes, and drives for another three hours to get there in 6 hours and 45 minutes. Don drives straight through and takes 7 hours to go 420 miles at 60 mph.

I will sit back and wait for the Excel lords to tell me otherwise.

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Your assumption is correct - well done. My apologies, I should have made it clearer that the journey was driven at the quoted speeds, no other comfort breaks etc, etc.

It was straight forward, uncomplicated arithmetic. But somehow, when I first saw this teaser, I had an intuitive feeling that the diesel car would get there first. I had to do the arithmetic twice, just to be sure the intuition was wrong.

A bit like the travellator !

Not just any old Tom, D*ck and Harry storey

Tom, D*ck and Harry have been given £500 to share between them.

They agree that Tom will get £10 more than D*ck,

and D*ck will get £20 more than Harry.

How much will each of them receive ?

PS Yikes ! I’ve had to put an ‘*’ in to allow a pretty common name to be used !!. - Take care with this aspect of your answers !!. Should have used Ann, Anne and Annie

Tom (T) gets £180, D*ck (D) gets £170, Harry (H) gets £150.

Workings

T + D + H = 500

T = D + 10

D = H + 20, so H = D - 20

(D + 10) + D + (D - 20) = 500

3D - 10 = 500, so 3D = 510

D = 510/3 = 170

T = 170 + 10 = 180

H = 170 - 20 = 150

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Well done Mike. And very nicely explained.

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