Too techie.. is there an easier method?

11 Plus Maths – Preparation and Information

Moderators: Section Moderators, Forum Moderators

11 Plus Platform - Online Practice Makes Perfect - Try Now
srk13
Posts: 44
Joined: Mon Feb 09, 2009 2:12 pm

Too techie.. is there an easier method?

Post by srk13 »

Can someone tell me how to explain my DD questions like this?

All bridges must have 2 section A's (in the start and end) and nowhere else. Length of individual sections is as follows:

Section A - 8.19m, B - 5.21m, C=2.91m, D=3.64m, E=7.52m, F=2.40m

Find the best combination of sections that will make the following bridges.

a) 44.18m
b) 65m
c) 94m

I know the following:

1) 2 section A's = 8.19 + 8.19 = 16.38
2) For (a) then the length required becomes 44.18 - 16.38 = 27.80

Now how to get the other sections add upto 27.8?
BarnetDad
Posts: 395
Joined: Thu Apr 24, 2008 10:51 pm

Post by BarnetDad »

I think I'd suggest mine skipped it!
yoyo123
Posts: 8099
Joined: Mon Jun 18, 2007 3:32 pm
Location: East Kent

Post by yoyo123 »

could you repeat the question?
All bridges must have 2 section A's and nowhere else

:?: :?:


possibly best way to get it is estimation, you need something to add up to approx 28

either that or trial and error
WP
Posts: 1331
Joined: Thu Jan 03, 2008 9:26 am
Location: Watford, Herts

Post by WP »

This is an example of what's known in the computing literature as a Knapsack Problem. It's one of a large class of problems (the NP-complete problems, also including the Travelling Salesman Problem) that are known to be equally hard, but no-one knows whether there is an efficient way to solve them. That question is one of the Millenium Prize Problems. Most people think the answer is no, but no-one has proved it.

In short, I think BarnetDad's advice is sound.

But then the one thing I can't resist is temptation, so, getting a computer to try all the possibilities:
a) 5D + 4F (unique solution)
b) 3B + C + 4E (shortest solution)
c) 2B + 4D + 7E or 6C + 8E (shortest solutions)
(It's typical of NP-complete problems that it's easy to check solutions, but the number of potential solutions to check is very large.)
Crazydad
Posts: 42
Joined: Mon Jun 30, 2008 4:08 pm

Post by Crazydad »

This kind of question should be kept until your DD finishes all others. Here is what I think from what you have done for a)

If you take a look at 27.80 m, you will see that you require the last digit as 0.

Now look at each section

B = 5.21m - you need 10B to make the last digit = 0 ... No (52.10m)
C = 2.91m - you need 10C to make the last digit = 0 ... No (29.10m)
D = 3.64m - you need 5D to make the last digit = 0 ... Yes - 3.64x5 = 18.20m
E = 7.52m - you need 5E to make the last digit = 0 ... No (37.6m)
F = 2.40m - you always have the last digit = 0 ... Yes

So 27.80 - 18.20 = 9.60m
Then 2.40m x 4 = 9.60m

Answer 5D + 4F

For b) and c) you have to start from the biggest section, E.
WP
Posts: 1331
Joined: Thu Jan 03, 2008 9:26 am
Location: Watford, Herts

Post by WP »

Crazydad wrote:If you take a look at 27.80 m, you will see that you require the last digit as 0.
It's a lot harder than that. You've assumed that two numbers ending in 0 were added to get the final 0, but it could have been one ending in 6 and another ending in 4, etc. And it could have been adding three, four or five numbers.
SunlampVexesEel
Posts: 1245
Joined: Fri Jul 06, 2007 9:31 pm

Post by SunlampVexesEel »

I have a heuristic method that meant I got at least some of the answers without trying all possibilities...

Here's how it works.. Assume 2A

Take all the other lengths and sort in descending length... giving order.. 7.52, 5.21, 3.64, 2.91, 2.4

Now choose the most of the biggest length that fits the remaining gap and then successively allocate the remainder using the next size down, when there is a remainder reduce the larger size and use a multiple of a smaller size...

I'm not sure such an approach works but I have just deduced that
44.18 is 2x8.19 + 5x3.64 and 4x2.4 without enumerating all the possibilities....
Animis opibusque parati
SunlampVexesEel
Posts: 1245
Joined: Fri Jul 06, 2007 9:31 pm

Post by SunlampVexesEel »

and...

65 = 2x8.19 + 4x7.52 + 3x5.21 and 2.91
Animis opibusque parati
SunlampVexesEel
Posts: 1245
Joined: Fri Jul 06, 2007 9:31 pm

Post by SunlampVexesEel »

94 = 2x8.19 + 7x7.52 + 2x5.21+4x3.64
Animis opibusque parati
SunlampVexesEel
Posts: 1245
Joined: Fri Jul 06, 2007 9:31 pm

Post by SunlampVexesEel »

The moral of this story is that when packing your rucksac... start with the biggest things first. :lol:
Animis opibusque parati
Post Reply
11 Plus Mocks - Practise the real exam experience - Book Now