# Brain Teasers are Back!

Well done Steve, well done Mulberry.

This one didn’t seem to phase either of you at all.

I found myself starring at it for quite a few minutes before I twigged what information Shape B and 6/7 was telling me !

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A shopping trip.

Mrs D went shopping this afternoon using cash to buy a few odds & ends.

In the first shop, she spent one third of the cash she had in her purse.

In the second shop, she spent one-third of the cash she had left.

Altogether she had spent £75 cash.

How much cash did she originally have in her purse ?

PS. I’ve only used ‘cash’ to eliminate credit-cards, cheques, phone apps etc. The ‘cash’ could be bank notes and/or coins. I appreciate that some of the younger members of the Forum might well be unfamiliar with cash, cheques, credit-cards etc

Hi Don,

now that’s a nice one. In hindsight I could have solved it in my head, instead of using a calculator to check my thought process. I’ll post my solution tomorrow instead of spoiling the fun for everybody else

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She had \$135, spent a third in the first shop - \$45, and had \$90 left. She then spend a third of the \$90, being \$30. So, spent \$45 + \$30 = \$75.

I worked this out as A being the initial amount, with A/3 spent first, leaving 2/3 A. Spend a third of 2/3 A in the second shop = 2/9 A. Total spend is 3/9 A + 2/9 A = 5/9 A.

As 5/9 A = \$75, A = \$75 x 9 / 5 = \$135. Also, a fifth of the \$75 is \$15, and the initial amount is nine fifths = 9 x \$15 = \$135.

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Hi Mulberry, Hi Mike,

Well done and a nice explanation Mike. I’m sure Mulberry has the correct solution and hopefully will post it, even if it is similar to Mike’s.

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Hi Don and Mike,

yes , 135 I was my solution as well. The road was quite different from Mikes clear and logical one, though. With math I’m somehow able to see things intuitively. In this case, it had to be something in the ballpark of 150. With an initial amount of £150, £100 would remain after the first shop and £66,67 after the second, giving a sum of £83,33 spent. To lower that amount to £75, the initial amount had to go down, likely to a number leading to both thirds having no decimal places. I simply tried £140 and £137 as the next two, before arriving at €135. Only then I realized, that stepping down multiples of 5 would have been smarter.
Only after writing the above, I saw that simply dropping the 0 from 150 and going with that would have been the easiest way.

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Although my solution was the same as Mike’s, I just love the intuitive thought process from Mulberry.

I’m a big fan of experimental mathematics, where anything goes to reach a solution, then sort out the formalities at the end.

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I think you all did very well.
And no doubt one or two others who might occasionally take a quick look at what’s going on.

Paddington – but not the Bear !

Yesterday evening, Mrs D and myself travelled by train from London Paddington to Didcot.

When we got to Paddington, Mrs D bought a cheap ticket and caught the 18:15 slow train to Didcot. This train travels at a steady 55 mph.

Unfortunately, I spotted a work colleague and spent a moment or two chatting. As a result, I missed the train that Mrs D had taken. However, I bought a slightly more expensive ticket and caught the faster 18:27 train which travels at a steady 70 mph.

The distance from Paddington to Didcot is 70 miles.

How far from Didcot did my train overtake Mrs D’s train ?

(assume these are non-stopping trains that run at the very steady speeds as quoted, they leave on time, they arrive on time, they are clean and comfortable … remember, it’s a Brain Teaser, not reality )

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Hi Revvie,
the odd thing is that, apart from math and a few things at work, my mind works precisely the way Mike used to tackle the teaser. The experimental way leads to a different level of understanding of concepts and relationships. I very much prefer understanding to memorizing, which made this particular teaser a joy to solve.

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Probably an unnecessarily long winded method, but my calculations suggest that your train overtook Mrs D’s train 18.67 miles from Didcot.
I’ll show my workings in due course.

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Steve, long winded is perfectly acceptable.
It delivered to correct solution.

With apologies for the delay (well it is a British Rail teaser!), here goes:-
The slow train has a 12 minutes “start” on the fast one. At a speed of 55mph, this means it has travelled 11 miles before the fast train sets off.
When the fast train starts, it will catch the slow train after T hours. It will travel a distance of 70T miles.
Meanwhile in the same time T, the slow train will travel 55T miles, which we know is 11 miles less than the fast train because of its “start”.
So, 70T-55T=11, and therefore T is 11/15 hours (or 44 minutes).
The fast train will have travelled 70 x 11/15 miles, equating to 51.33 miles from Paddington.
The distance between Paddington and Didcot is 70 miles, so both trains are 18.67 miles away (70-51.33) from Didcot when the fast train catches the slow one.
I have no doubt there is a more elegant solution.

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Well done Steve !

I don’t think that was long winded at all. It was certainly neat and clear. And, of course, correct.

My method was not dissimilar. I’ll post it later today or tomorrow. I worked in hours with 12 mins = 0.2 hrs and used T to represent the travel time of the fast train rather than the slow train.

I won’t be surprised if Ravvie, or Mrs R, has a one-line or two- Line solution … ?

I was going to guess Reading, mainly from gut feel and not being at all familiar with the stops on that line.

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Reading would be a superb guess. I think it’s within about half a mile or so of Steve’s figure.

I’m doing this on my iPad so …
Let T = time in hours that the slow train travels before it is overtaken.
Hence time travelled by the fast train to catch the slow train = T - 0.2 hours
Distance travelled = Speed x Time
Slow train ST = 55T
Fast train FT = 70(T - 0.2)
The distances are equal, hence
55T = 70(T - 0.2)
55T = 70T - 14
15T = 14
T = 14/15 hrs = 56 minutes

Distance D, travelled by slow train in 14/15 hrs
D = 55 x 14/15 = 51.33 miles
Distance to Didcot = 70 - 51.33 = 18.67 miles

I think that Steve and my method are more or less the same. I also noticed that I actually used T as the time of the slow train, not the fast train as I said above. Apologies

**This one is a bit cheesy **

Mrs D bought a Dutch Edam cheese the other day. It was a perfect sphere and its diameter was 12 inches. (Yes, I know, real ones are more like 5 inches, but you might thank me for choosing something like a foot)

She asked me to cut it into eight identical pieces, ie shape and volume. So I asked her which way would she prefer that I cut it.

This caused a bit of irritation, so I quickly added that there is more than one way to achieve this objective.

“Well, whichever way requires the least amount of cling film in which to wrap the the eight pieces individually”. I assumed she meant whichever way results in the minimum surface area of cheese, including the red wax, since I wasn’t going to allow for any overlaps in the cling film eg at edges or corners.

Into what shapes couldI have cut the cheese to give the required eight identical pieces ? And what is the ratio of surface area to volume of each of these shapes ?

Why did I say “into what shapes could I have cut the cheese…” because this wasn’t what Mrs D wanted, she wanted a more conventional set of eight pieces …