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 Post subject: Mixed-up Clock
PostPosted: Sat Feb 23, 2013 10:01 pm 
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Joined: Mon Nov 26, 2007 4:14 am
Posts: 138
Location: Middlesex
Hi folks

DS3's teacher has given him this for homework. I can't make any sense of it. Can anybody help us solve this riddle please:

A Mixed-up Clock


There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from the ten statements below?

Here is a clock-face with letters to mark the position of the numbers so that the statements are easier to read and to follow.
clock face.

Image

No even number is between two odd numbers.
No consecutive numbers are next to each other.
The numbers on the vertical axis (a) and (g) add to 13.
The numbers on the horizontal axis (d) and (j) also add to 13.
The first set of 6 numbers [(a) - (f)] add to the same total as the second set of 6 numbers [(g) - (l)] .
The number at position (f) is in the correct position on the clock-face.
The number at position (d) is double the number at position (h).
There is a difference of 6 between the number at position (g) and the number preceding it (f).
The number at position (l) is twice the top number (a), one third of the number at position (d) and half of the number at position (e).
The number at position (d) is 4 times one of the numbers adjacent (next) to it.


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 Post subject: Re: Mixed-up Clock
PostPosted: Sat Feb 23, 2013 10:26 pm 
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Joined: Sat Mar 06, 2010 11:39 pm
Posts: 2080
Start with f.


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 Post subject: Re: Mixed-up Clock
PostPosted: Sat Feb 23, 2013 10:44 pm 
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Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
The number at position (f) is in the correct position on the clock-face.
So f is 5
There is a difference of 6 between the number at position (g) and the number preceding it (f).
We know f is 5 so g must be 11
The numbers on the vertical axis (a) and (g) add to 13.
We know g is 11 so a must be 2
The number at position (l) is twice the top number (a), one third of the number at position (d) and half of the number at position (e).
We know a is 2 so l must be 4, d must be 12 and e must be 8
The numbers on the horizontal axis (d) and (j) also add to 13.
We know d is 12 so j must be 1
The number at position (d) is double the number at position (h).
We know d is 12 so h must be 6
The number at position (d) is 4 times one of the numbers adjacent (next) to it.
We know d is 12 and has 8 on one side of it so c must be 3
So far we have
a=2
b=
c=3
d=12
e=8
f=5
g=11
h=6
i=
j=1
k=
l= 4
leaving 7,9,10 left to allocate


The first set of 6 numbers [(a) - (f)] add to the same total as the second set of 6 numbers [(g) - (l)] .
1+2+...+11+12 = 78 so a-f must be 39. The 5 numbers we have add to 30, so b must be 9

No even number is between two odd numbers.
So i can't be an odd number - meaning i is 10 and k is 7
a=2
b=9
c=3
d=12
e=8
f=5
g=11
h=6
i=10
j=1
k=7
l= 4


No consecutive numbers are next to each other.
But we didn't need to use this clue


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 Post subject: Re: Mixed-up Clock
PostPosted: Sat Feb 23, 2013 11:02 pm 
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Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
His teacher has lifted this question straight from nrich by the way - http://nrich.maths.org/primary-upper


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 Post subject: Re: Mixed-up Clock
PostPosted: Sun Feb 24, 2013 1:32 am 
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Joined: Sun Sep 23, 2012 6:58 pm
Posts: 208
That's what it's there for.....


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 Post subject: Re: Mixed-up Clock
PostPosted: Sun Feb 24, 2013 4:41 am 
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Joined: Mon Nov 26, 2007 4:14 am
Posts: 138
Location: Middlesex
Thanks all, for your help. I do feel that this question is quite difficult for a year 4 student (and the teacher messed it up by dividing question 10 into two like this: (10) The number at position (d) is 4 (11) times one of the numbers adjacent (next) to it.) This mistake made me suspicious and I Googled it up. It is here but without a solution:
http://nrich.maths.org/2127

Okanagan, I am much impressed. I am working towards completing my own GCSE maths (never felt the need to do it but now looking for something new) but could not make any headway dealing with this one as I was trying to start with 4 at d per teacher's mistake. Then I gave up and came here. I suppose I should stick with VR and a bit of NVR!


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 Post subject: Re: Mixed-up Clock
PostPosted: Sun Feb 24, 2013 8:11 am 
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Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
Drastic Dad wrote:
the teacher messed it up by dividing question 10 into two like this:
(10) The number at position (d) is 4
(11) times one of the numbers adjacent (next) to it.
Actually you could do it without that clue :) Like this:

The number at position (f) is in the correct position on the clock-face.
So f is 5
There is a difference of 6 between the number at position (g) and the number preceding it (f).
We know f is 5 so g must be 11
The numbers on the vertical axis (a) and (g) add to 13.
We know g is 11 so a must be 2
The number at position (l) is twice the top number (a), one third of the number at position (d) and half of the number at position (e).
We know a is 2 so l must be 4, d must be 12 and e must be 8
The numbers on the horizontal axis (d) and (j) also add to 13.
We know d is 12 so j must be 1
The number at position (d) is double the number at position (h).
We know d is 12 so h must be 6
So far we have
a=2
b=
c=
d=12
e=8
f=5
g=11
h=6
i=
j=1
k=
l= 4
leaving 3,7,9,10 left to allocate


The first set of 6 numbers [(a) - (f)] add to the same total as the second set of 6 numbers [(g) - (l)].
1+2+...+11+12 = 78 so a-f must be 39, as must g-l. The 4 numbers we have for a-g add to 27, so b+c must be 12.
The only way to make 12 from 3, 7 ,9 and 10 is 3+9 - therefore b and c are 3 and 9.
This leaves i and k to be 10 and 7

No consecutive numbers are next to each other.
So b can't be 3, meaning b is 9 and c is 3.
No even number is between two odd numbers.
So i can't be an odd number - meaning i is 10 and k is 7

a=2
b=9
c=3
d=12
e=8
f=5
g=11
h=6
i=10
j=1
k=7
l= 4


The number at position (d) is 4 times one of the numbers adjacent (next) to it.
Not needed as all the answers can be derived without this clue.

twelveminus wrote:
That's what it's there for.....
Exactly, but worth pointing out as a source of similar material for anyone who hadn't come across it. As it's a live question which means they haven't published the answers yet, I do wonder if the teacher has actually worked it out though!


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 Post subject: Re: Mixed-up Clock
PostPosted: Sun Feb 24, 2013 11:14 am 
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Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11931
I do think it is wrong to give full solutions - the initial hint to start with f should have been sufficient!

As a maths teacher I trust the parent will own up and say where the solution has come from!


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 Post subject: Re: Mixed-up Clock
PostPosted: Sun Feb 24, 2013 3:01 pm 
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Joined: Mon Nov 26, 2007 4:14 am
Posts: 138
Location: Middlesex
Okanagan wrote:
Drastic Dad wrote:
the teacher messed it up by dividing question 10 into two like this:
(10) The number at position (d) is 4
(11) times one of the numbers adjacent (next) to it.
Actually you could do it without that clue :) Like this:

The number at position (f) is in the correct position on the clock-face.
So f is 5
There is a difference of 6 between the number at position (g) and the number preceding it (f).
We know f is 5 so g must be 11
The numbers on the vertical axis (a) and (g) add to 13.
We know g is 11 so a must be 2
The number at position (l) is twice the top number (a), one third of the number at position (d) and half of the number at position (e).
We know a is 2 so l must be 4, d must be 12 and e must be 8
The numbers on the horizontal axis (d) and (j) also add to 13.
We know d is 12 so j must be 1
The number at position (d) is double the number at position (h).
We know d is 12 so h must be 6
So far we have
a=2
b=
c=
d=12
e=8
f=5
g=11
h=6
i=
j=1
k=
l= 4
leaving 3,7,9,10 left to allocate


The first set of 6 numbers [(a) - (f)] add to the same total as the second set of 6 numbers [(g) - (l)].
1+2+...+11+12 = 78 so a-f must be 39, as must g-l. The 4 numbers we have for a-g add to 27, so b+c must be 12.
The only way to make 12 from 3, 7 ,9 and 10 is 3+9 - therefore b and c are 3 and 9.
This leaves i and k to be 10 and 7

No consecutive numbers are next to each other.
So b can't be 3, meaning b is 9 and c is 3.
No even number is between two odd numbers.
So i can't be an odd number - meaning i is 10 and k is 7

a=2
b=9
c=3
d=12
e=8
f=5
g=11
h=6
i=10
j=1
k=7
l= 4


The number at position (d) is 4 times one of the numbers adjacent (next) to it.
Not needed as all the answers can be derived without this clue.

twelveminus wrote:
That's what it's there for.....
Exactly, but worth pointing out as a source of similar material for anyone who hadn't come across it. As it's a live question which means they haven't published the answers yet, I do wonder if the teacher has actually worked it out though!

Thanks once again Okanagan. As I said earlier, I am not confident with my maths. Just gave it to my boy after printing it from the website and he did it in less than 5 minutes without my assistance whatsoever. The addition of the 11th question by dividing question 10 in two had thrown both of us astray. Thanks once again.

Guest55, as Maths teacher you should know that teachers should learn to copy and paste first before handing out ready-made homework assignments without acknowledging their sources:

Image


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 Post subject: Re: Mixed-up Clock
PostPosted: Mon Feb 25, 2013 1:52 pm 
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Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8228
How annoying. Why didn't they just cut and paste from the nrich website or give you the web address?

The number of homework sheets that go out with errors on them have rocketed since schools have stopped using textbooks sometime during the last century. Textbooks also contained errors, but the teacher got to know them and asked children to correct the misprints.


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