The Lemon Fool

Bob decided to sell some of his bicycles so went to his local car boot sale and priced his bikes as follows, and he determined not to accept anything below the asking price:

Boardman         £860
Brompton         £425
Mercian          £425
Harry Quinn      £399
Claud Butler     £165
Dawes            £140
Raleigh          £132
Moulton          £124
Hercules          £68
Falcon            £53

In the day he did very well and only one bike remained unsold. Moreover the amount of money he collected in the afternoon was precisely twice the amount collected in the morning. Which bike was not sold and which bikes were sold in the morning and which in the afternoon?

Cinelli

Adding them up, M(orning) + A(fternoon) + U(nsold) = 2791

Since A=2M, that's 3M+U = 2791. Or M = (2791-U) / 3 . Which tells us U has remainder 1 when divided by 3. Only 124 fits. So U = 124, M = 889, A = 1778.

One more constraint: the two 425s can't be split up, or the problem as stated has no unique solution.

By simple inspection, you can't make 889 including either 860 or 2*425. So the most expensive three must be in A.

860 + 2*425 = 1710 = A - 68. So we have M=53+132+140+165+399, A=68+425+425+860.

I hope drinks are on Bob, with all that money!

Reply to solution.

This is a model answer. That the two £425 sums can't be split is a nice touch and one that I hadn't considered.

Cinelli

Sorry Johnny come lately.



If y is the unsold bike, then 3x+y=2791 (total of all prices) . Unsold bike must divisible by 3 with rem 1 the only one is the Moulton.

(2791 - 124)/3 = £889 sold in the morning, £1778 in the afternoon.

-The boardman couldn't sell in the morning, remaining takings would be only £39.
-Checking whether the same argument applies for the boardman, brompton and mercian in the afternoon, 1778-(860+425+425)=68 so by inspection these three bikes and the Hercules sold in the afternoon, the remainder in the morning.



GS
P.S. I thought the Raleigh was the unsold bike as it was massively overpriced but apparently not.

UncleEbenezer wrote:Adding them up, M(orning) + A(fternoon) + U(nsold) = 2791

Since A=2M, that's 3M+U = 2791. Or M = (2791-U) / 3 . Which tells us U has remainder 1 when divided by 3. Only 124 fits. So U = 124, M = 889, A = 1778.

I wondered who was designing those illegible pastel-on-pastel websites

V8

GoSeigen wrote:
GS
P.S. I thought the Raleigh was the unsold bike as it was massively overpriced but apparently not.

I wondered if Bob had any more Moultons he wanted rid of?

servodude wrote:

GoSeigen wrote:
GS
P.S. I thought the Raleigh was the unsold bike as it was massively overpriced but apparently not.

I wondered if Bob had any more Moultons he wanted rid of?

Many years ago when I did regular CTC club rides on a sunday, we had a member who came on a Moulton. A strange and wondrous beast it was, and from memory, much admired by other club members.

UncleEbenezer wrote:

servodude wrote:
I wondered if Bob had any more Moultons he wanted rid of?

Many years ago when I did regular CTC club rides on a sunday, we had a member who came on a Moulton. A strange and wondrous beast it was, and from memory, much admired by other club members.

The space frame is just a glorious design
even the F looks delightfully retro these dates

Please bear in mind that they were all second hand and some in pretty bad shape.

Cinelli

UncleEbenezer wrote:Many years ago when I did regular CTC club rides on a sunday, we had a member who came on a Moulton. A strange and wondrous beast it was, and from memory, much admired by other club members.

But what was the bike like?


Julian F. G. W.

jfgw wrote:

UncleEbenezer wrote:Many years ago when I did regular CTC club rides on a sunday, we had a member who came on a Moulton. A strange and wondrous beast it was, and from memory, much admired by other club members.

But what was the bike like?


Julian F. G. W.

Sticky??