" What a disappointing article! From the headline I was expecting it to say how they were wiring up BC109s and logic gates. No chance of that I suppose ...

Regards, Mike"

Are you sure we can mention only BC109s without the opposite gender equivalent 108.

Gender discrimination in Electronics - go and cut'n'paste your coursework again, silly boy.

Re. the mineshaft maths question, from my back-of-the-envelope calculation (no calculator, I'm 46), I reckon

the circumference of the drum is pi.d where d = 0.7

it rotates at 10 rpm so the cage moves at 10.pi.d m/min = 7.pi m/min

To rescue 21 people 10 at a time it needs to go down and up 3 times = 6 journeys

each journey is 22m, so the total distance is 6.22 m

it needs to be loaded 3 times and unloaded 3 times = 6 times

each load/unload takes 10 seconds, so loading takes 6 x 10 seconds = 1 minute

total time is (6.22 m)/(7.pi m/min) + 1 minutes

22/7 = pi (give or take) so the answer's 'about seven minutes'.

Roger

I don't think so - I am 36, have a C grade O Level in maths

But if the drum is 0.7 and revolves 10 times per minute then it will take 3 minutes and nine seconds for ONE journey (0.7 x 10 = 7metres per minute)

3.9 X 6 +1 = 24minutes 40 seconds

Any corrections on my limited maths ?

Chris - the drums *diameter* is 0.7m - the distance across the drum. The cable would normally be wrapped around the drum, so we use 2*pi*r = pi * diameter to work out the length of cable reeled per revolution.

Given its a GCSE question, the fact that this works out at around 2.2m therefore each trip in the lift takes pretty much exactly 1 minute acts as a bit of a confirmation.

7 mins was what I got too (GCSE maths 1998, grade A*, but back from a pub lunch).

The drums DIAMETER is 0.7 but its' circumference is pi times that = 2.2m, it rotates 10 times per minute so suspiciously it is exactly one minute to get up and down the shaft, so seven minutes is correct.

50 and A-level maths (only an "E" mind)

Chris,

the diameter of the drum is .7m

the circumverence is pi x 0.7 meter.

So you 24.40 minutes is 3.14 times too large

24.40 / 3.14 ~ about 7 minutes

You've probably already realised your mistake Chris, and are cringing in anticipation of further comments, but it's the diameter that's 0.7, not the circumference.

So 10rpm x 22/7 (pi) x 7/10 (diameter) = 22m/min.

pi * diameter (3.1415 * 0.7) = 2.2m per revolution. 10 revolutions = the required distance which takes one minute. There are six journeys, = 6 minutes. Add the 3 lots of loading time, your total is 6 minutes 30 seconds.

No calculator! I took GCSE maths in 1995, and A level Maths in 1997.

its 7.

I'm a wee bairn at only 27, sat GCSEs 11 years back, and passed them. I managed it without a calculator.

Should probably mention I went on to get a Masters and a PhD in theoretical physics after my lowly state school education.

uhm, yeah forgot the unloading time. so 7 minutes it is then!

Ok, 38 no calc.

Circumference = pi.d = 22/7*7/10m = 2.2 m

So, with 10 rpm we need 1 minute to descend or ascend the shaft. We need 3 journeys to evacuate all of the people in the mine and each of these will take 2 minutes 20 seconds to complete. Total time taken is therefore 7 minutes.

No, I think 7 minutes is right. The _circumference_ of the drum is approximately 2.2m (pi*0.7 ... how many kids today would be aware of the 22/7 approximation for pi?) so at 10rpm it would raise or lower the cage the full 22m in 1 minute.

6 of these journeys plus 6 x 10 seconds at each end gives 7 minutes.

No calculator required - I'm 28.

24yo, Calculator used to double check...

I get 7 minutes too!

Without re-stating the calculations, I got 7 minutes too (I'm 27, GCSE in maths). Mind you, a girl in my office yesterday who's about my age asked for help working out postage for the franking machine because she couldn't add 1250g and 250g...

9 secs = 0.15 of a minute

Therefore using your formula = 3.15 * 6 + 1 = 19.9

Or 19 minutes 54 Seconds

But as Rodger pointed out it did say the diameter of the wheel. Therefore it moves: So 1 rotation is equal to the diameter * pi (3.14159)

Breaking down Rodgers explaination:

1 rotation = 0.7 * pi = 2.1991148575128552669238503682957 metres per rotation. So in a minute it moves 21.99 metres per minute. Say 22 metres per minute. Therefore it takes 1 minute per journey.

3 journeys to evacuate, 3 journeys to the bottom of the lift shaft. So 6 minutes for total journeys.

It will need to be loaded and unloaded 3 times each. 6 * 10 seconds = 1 minute.

So 6 + 1 = 7

Sorry Chris but my calcuator says that Rodger is correct. Nice one!

To evacuate 21 miners using a cage which holds 10 people, starting from the top, we will need to make 3 return journeys: down (load up miners 1-10), up (unload), down (load up miners 11-20), up (unload), down (load up miner 21), up (unload).

Speed of cage = circumference of drum * 10 revs per minute = 0.7 * pi * 10 = 7 * pi metres per minute.

(Noticing the figures "22" and "7" cropping up, we can infer that this question was taken from a paper which predates the use of calculators, which is how maths was done in the days when I did my O levels; so we'll use 22/7 as an approximation for pi.)

Therefore, speed of cage = 22m. / min.

Distance travelled = 22m. * 6 journeys = 22*6 m.

Time for which cage is moving = 22*6 m. / 22 m./min. = 6'.

Loading and unloading time = 10 sec. * 6 = 60 sec = 1 min.

So total evacuation time = 7 minutes.

I think somebody missed a pi out somewhere; but, fortunately, you at least shew your workings, so you will get some marks.

why I only just scraped a C !

Thanks for the explanations all, I suspect in a real GSCE paper, I would have got 8 out 10 for showing my workings and writing my name correctly on the paper.....

Everyone answering this has missed a trick: after the cage comes up the final time, everyone is out of the mine and you don't have to count the final unloading time. 6 minutes 50 seconds then.

Alternatively, just look at the countdown timer the terrorists put on their bomb, and subtract 1 second.

Given we are talking evacuation here, I would sort out the 11 smallest lightest miners and put them all in one cage together to get them out in 4:20….

Do I win?

I like the idea of squeezing in eleven miners - though it would actually take 4:40, not 4:20.

Alternatively, if it takes ten seconds to unload/load ten people, then it'll only take one second to load one person - right? So we can shave eighteen seconds off the seven minutes which I worked out in my head.

Oh, 52, degree in mathematical physics.

...as 21 people should give you the cluestick. 1 person at the top operating the lift so only 20 people to raise. therefore, you only need 2 loads to get them out

As most people these days seem to have been educated by TV, the answer to the mine question is obviously 47 minutes 20 seconds.

On TV all problems are solved before the end of the hour and allowing for adverts, the rescued miners will reach the surface in around 47 minutes. Except of course for poor George. It was Georges last day down the mine. After 35 years he was retiring. But when the support broke, George held up the beam while the others escaped to safety. As the last man entered the lift Georges strength failed and he was buried under a million tons of rock.

Right. The mineshaft is 22 metres deep. Is there a trick? What is a headstock?

Gosh, that cage must be crowded. The rope goes around the drum. What if the rope is very thick - won't the outside of the rope take longer to go around the drum than the inside of the rope, like an aeroplane's wing? Will the rope break, or take off? Perhaps it is imaginary maths-question rope, infinitely narrow. Like the monofilament rappelling fibre in "The Stainless Steel Rat", which was made out of a single chain of molocules. It could cut through skin.

22 metres. The rope goes around the drum. 22 divided by 0.7 is... something or other. Then divide that by 10, or multiply it by ten, and then multiply it by 60 until it looks like a reasonable length of time. Why are they in the mine? Is this a political question, a reference to the miner's strike?

Ten seconds times three is thirty seconds, so the time will be at least thirty seconds, unless this is one of those riddles whereby someone has to stay in the cage in order to operate the controls.

Oh, I give up. Next question.

The answer is: it'll be done when it's done.

Of if you want, answer like a plumber. "Next thursday mate if you're lucky."

You need to add another 45 seconds to the final time. That's how long it takes for Lassie to tell you that there are people trapped down the mine in the first place.

For the record, another 7 minutes here--well, either that or 6 minutes since I didn't take note of the lift's starting position or the number of tea-breaks required. Not reading the question properly has always been a speciality of mine.

I've an O-level grade B of 1984 vintage, and no calculator because I was too lazy to find one, or to fire up bc in my Unix window for that matter.

Ok

First we need to find out the circumference of the motorised drum.

2 * pi * 0.35 = 2.2 (rounded to 2DP, yes rounding in the middle of a question is bad but it fits nicely with the next part.)

22/2.2 Gives us 10 which shows us it takes 10 rotations of the drum from top to bottom, and since 10 RPM was given to us earlier we now know it take 1 minute to go from the top to the bottom.

From the top

60 second down

10 seconds loading

60 seconds up

10 seconds unloading

= 140 seconds

* 3 (because the lift only holds 10 so for 21 people three trips are needed.)

= 420 seconds (7 Minutes.)

Calculator was used because I was unsure whether the answer would become decimal or how accurately PI was expected to be :) (Well actually calculator is missing in action so i used google instead.)

Just for the record - 16 Year Old from Oxfordshire.

(And the reaction of the Physics teacher in the other attack will be praised by teachers and likeminded students. I was forced to endure double science with a load of wiffly waffly rubbish and not much real Physics. Biology and Chemistry was simply taught awfully and incredibly boring however for Physics I felt some hope. My teacher is just retiring and so his Physics education was bloody hard, he knew his stuff and although the spec was crap he often went of on interesting tandems that kept be interested. For the record I think we managed 2 practical sessions in two years (whoo,)I hope that AS Physics is a damn site better than this rubbish, I want to learn real Science.)

I worked this out on a pad and i promise I have not read the answers from above!

Regards

"he often went of on interesting tandems that kept be interested"

I imagine the tandem could be used to teach relativistic physics, by pedalling very hard with a pupil facing backwards in the second seat.

7 minutes.

Nobody else has mentioned coming up with the answer without recourse to paper & pencil or Notepad.

Surely apart from needing to know that c=2*pie*r, the question is solved using primary school arithmetic. Or it was in my day. I sat my Physics, Pure Maths and Applied Maths A levels in 1964.

One thing not noted lifting gear tested to 2.5 times safe working load (swl) .10x2.5=25 men per trip =one trip down ,one up ?

Surely the correct response is "What do you mean, there are people down a mine?"

>>if it takes ten seconds to unload/load ten people, then it'll only take one second to load one person - right?

Nope. Cos the 10 mins is to open the rusty cage doors which haven't been used in 20 years.

Tony Humphreys wrote "Are you sure we can mention only BC109s without the opposite gender equivalent 108."

108s and 109s are the same sex (npn). Draw your own conclusions.

No calc, no paper, 7 mins

Lloyd, you only need to know c=pi*d given that you are provided with the dia.

37, PhD in electrical engineering.

Re: other letter, hadn't even thought of a BC109 in so long...

I got:

(Trapped in a box / tug of war) * strong wind

The trick is that this question was from the 70s/80s and now there are no working mines left anymore, so the whole question is pointless. Of course, GCSE History or Modern Studies probably won't include 80's politics, so kids wouldn't know that either.

Relevant questions:

Does the cart start at the top, or the bottom of the shaft? Are there already people in the cart?

Irrelevant question:

From what location(s) is the cart controlled? (If an operator is needed IN the cart to bring it back down, then only 9 people can be evacuated per trip, except on the last trip. For the specific case given, it makes no difference. A proper question would have 37 people in the mine, which would require 4 trips ( 10-1+10-1+10-1+10=37) to evacuate everyone.)

(apologized for the long, (nested) parentehticals)

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Does the cart start at the top, or the bottom of the shaft? Are there already people in the cart?

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Exactly why I didn't bother to even try to do the calculations - not enough data in the first instance to give a definitive answer. Mind you, writing "I refuse to answer this question on the grounds that I have been given inadequate data to perform the calculations properly" is, perhaps, not the best way to pass your exams ... knew I went wrong somewhere ;)