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 Post subject: examplus paper 9 ( question 19)Posted: Tue Jun 30, 2020 3:22 pm

Joined: Wed Jun 12, 2019 3:26 pm
Posts: 21
256 children compete in knock-out tennis tounament. The children are numbered from 1to 256.each child plays one match in the first round : child 1 plays child 129, 3 plays 131 and so on.
the loser of each match is knocked out of tournment and the winner goes through to the next round.
this continuesuntil only one child is left, having won all of their matches. How many rounds need to be played to find this one winner?

Answer = 8 rounds

2) A farmer wants to make a square or rectangular paddock for her horse. She has 80m of fence panels.each fence is 1m long and cannot be cut.if the farmer uses all the fence panels what is the difference between the areas of the smallest paddock and largest paddock that she can make?

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 Post subject: Re: examplus paper 9 ( question 19)Posted: Tue Jun 30, 2020 4:34 pm

Joined: Mon Feb 12, 2007 2:21 pm
Posts: 16254
There's no point in doing these for you.

What have you tried for the first one?

256 in round 1, how many in round 2?

2) Try some shapes - long and thin will have small area ...

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 Post subject: Re: examplus paper 9 ( question 19)Posted: Wed Jul 01, 2020 6:57 am

Joined: Wed Jan 18, 2012 12:41 pm
Posts: 10559
Location: Essex
One thing re the first one is that the labelling of the competitors as No.1 / No.129 etc is irrelevant. It's a 'knockout', therefore only the winner of each match 'survives' to the next round.

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 Post subject: Re: examplus paper 9 ( question 19)Posted: Fri Aug 28, 2020 12:55 pm

Joined: Wed Sep 18, 2019 2:57 pm
Posts: 30
On the first one teh slow way to do it is to say
round 1 - 256
round 2 - 128
.
.
.
Round 8 - 2

but that will eat up precious time and the will be looking for your child to think fast i.e
quickly work out that 256 =2^8 ... hence theer will be two left on the eighth round.

On the second,
its a "play" between perimeter and area.
if he uses all fences then the perimeter must be 80. there are many way to make the perimeter 80 e.g 1 and 39 fences on either side which gives an area of ...you guessed it 39 (39*1). The maximum are is the one the is achieve when both sides are equal i.e. ......17*23<18*22<19*21<20*20

Hence 39 is the minimum area and 400 is the maximum so difference is 361.

Note that square or rectangle shape make no difference to the outcome.

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