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Straightup

Posted: Tue Nov 08, 2011 9:42 pm 

Joined: Mon Nov 07, 2011 9:29 pm Posts: 3

Can anyone please shed some light on these two questions ?
Q1. At noon, the hour and minute hands of a clock point in the same direction. Determine, to the nearest second, the last time before noon that the hour and minute hands point in the same direction.
Also determine, to the nearest second, the last time before noon that the hour and minute hands point in exactly opposite directions.
Q2. A giant timeline is constructed, showing every year from 1AD until 2012. The company which has to make this needs to calculate how many of each digit 09 will be required. Assuming that no leading zeros are necessary, how many of each digit will be needed?
TIA


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Guest55

Posted: Tue Nov 08, 2011 10:14 pm 

Joined: Mon Feb 12, 2007 1:21 pm Posts: 13025

Where are these questions from?


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Blitz

Posted: Wed Nov 09, 2011 7:18 am 

Joined: Wed Aug 25, 2010 2:58 pm Posts: 550

Is the first question 11:59 an 59 seconds??


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Straightup

Posted: Wed Nov 09, 2011 9:15 am 

Joined: Mon Nov 07, 2011 9:29 pm Posts: 3

The questions are set by the Scottish Mathematical Council. My 11 year old son has to attempt them for homework.


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mystery

Posted: Wed Nov 09, 2011 9:40 am 

Joined: Tue Jul 21, 2009 9:56 pm Posts: 8418

The hands on a clock cross one another at 12:00, 1something, 2 something, etc ...... 10 something and 12:00 again. So each time the cross the hour hand has travelled one eleventh of a full circle.
There are 60 minutes in a clock circle. oneeleventh of this is 5mins and 5/11of a min. So the hands will last have crossed at 5and5/11of a min to 11.


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mystery

Posted: Wed Nov 09, 2011 9:45 am 

Joined: Tue Jul 21, 2009 9:56 pm Posts: 8418

I think they are good problem solving questions .............. unclear though which children will have done the questions themselves, and which with some help.
I think that with the first question playing with a watch or clock where you can move the hands round the full 12 hours will help ............. of course he'll see the approximate answer, but it might also help him work out how to work out the answer if you see what I mean.
The second question
1  2012 ........... well again it's a really good problem. If he thinks about it in terms of the units, tens, hundreds and thousands columns and how many times each digit will appear in each I think he will get there ........ eventually he might start to see some shortcuts.
Is he completely stuck?


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jd1

Posted: Wed Nov 09, 2011 12:58 pm 

Joined: Thu Oct 07, 2010 12:23 pm Posts: 4

This is quite tricky but may be solved by proceeding in a systematic way. For example, there are only 2 choices (1 and 2) for the thousand place, and therefore 1 will appear 1000 times (i.e. from 1000 to 1999) and 2 will appear 13 times (2000 to 2012) at this place.
Similarly, all the digits (0 to 9) may appear in the hundreds column. The digit 0 will appear 100 times where the thousand digit = 1, i.e. 1000 to 1099, and 13 times where the thousand digit = 2 (i.e. 2000  2012). The digits 1  9 will each appear 200 times, which is composed of 100 times for 3 digit numbers (e.g. 100199 for digit 1), and 100 times for 4 digit numbers (e.g. 11001199 for digit 1).
Similar analysis can be done for the tens and the units column.
The final solution after adding up all of the above is as follows: digit 0: 514 digit 1: 1605 digit 2: 615 digits 39: 601 times each.
This adds up to the total number of digits required = 6941. This also tallies with the total number of digits calculated by: number of 1digit numbers x 1 + # of 2digit numbers x 2 + # of 3digit numbers x3 + # of 4digit numbers x 4 = 9x1 + 90x2+ 900x3 + 1013x4 = 6941.
Hope this helps and is not too confusing!! There may be a simpler method to solve this but I don't know about that.
Regards


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Straightup

Posted: Thu Nov 10, 2011 8:37 am 

Joined: Mon Nov 07, 2011 9:29 pm Posts: 3

Many thanks for all the help. I'm trying to get my head round the answers...


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