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 Posted: Fri Sep 12, 2008 8:47 pm

Joined: Thu Jul 17, 2008 11:09 am
Posts: 10
Location: lewisham
A computer checks emails and news updates on a regular basis. In the 24-hour period the computer checks emails every 5 minutes and news updates every 7 minutes. Neither work when both checks are being done at the same time and this causes an error message on the screen. There is 1 error message after an hour and three error messages after 2 hours. How many error messages are there after 24 hours?

Thanks in advance. I am puzzled. I thought it was supposed to be 48?

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 Post subject: Posted: Fri Sep 12, 2008 8:56 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 13994
There will be an error message every 35 minutes so we need to know how many in 24 hours.

ie 24 x 60 divided by (5 x 7)

ie 24 x 12 divided by 7 which is 41. ...... ie 41 error messages.

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 Post subject: Posted: Fri Sep 12, 2008 9:42 pm

Joined: Thu Jul 17, 2008 11:09 am
Posts: 10
Location: lewisham
Thanks.

I understand the first bit but why 24x12 ? and then divided by 7? And what do we do with the information 1 error after 1 hour and 3 errors after 2 hours?

Sorry for being slow.

And does anyone think that that's too complicated and can't be in the tests?

Thanks to all.

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 Post subject: Posted: Fri Sep 12, 2008 10:15 pm

Joined: Mon Jan 30, 2006 4:07 pm
Posts: 2668
Dear Irene

We know errors arrive every 35 minutes, need to know how many 35s are in 24 hours.

24 hours converted into minutes = 24 x 60 = 1440 [or 24 x 6 add a nought]

1440 divided by 35

To make it simple without the help of a calculator or if the child cannot divide large numbers. We can break it down, as we know...

10 x 35 =350

20 x 35 is 700

40 x 35 = 1400

1400 to 1440 would produce 1 more 35, therefore answer is 41.

Patricia

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 Post subject: Posted: Fri Sep 12, 2008 11:43 pm

Joined: Tue Jan 01, 2008 1:05 pm
Posts: 515
irene wrote:
Thanks.

I understand the first bit but why 24x12 ? and then divided by 7? And what do we do with the information 1 error after 1 hour and 3 errors after 2 hours?

Sorry for being slow.

And does anyone think that that's too complicated and can't be in the tests?

Thanks to all.

As already indicated above, lowest common multiple question. 7x5 = 35, to give you the error message frequency. The 1 error after 1hr etc. information is actually superfluous in this example, and you can get the correct answer by default. But in practice gives you an envelope in which the first error occurs, so that you know the start point, and which can affect how many whole error periods fit into 24hrs.

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 Post subject: Posted: Sat Sep 13, 2008 9:07 pm

Joined: Thu Jul 17, 2008 11:09 am
Posts: 10
Location: lewisham
Wow.

I was wondering what type of question could that be. I only hope my D won't get something like that in the test next week.

Do they teach this in the primary school? I don't think they do in ours. Not untill the start of year 6.

Many thanks to all

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 Post subject: Posted: Sun Sep 14, 2008 7:34 pm

Joined: Sat Sep 13, 2008 11:56 pm
Posts: 5
Yes, I think it is a common type of question

Another example

John paints ahouse in 4 hrs and Jim paints a house in 6 hrs.
How long does it take both working together to paint 10 houses?

-----------------

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 Post subject: Posted: Mon Sep 15, 2008 9:06 am

Joined: Wed Jan 02, 2008 12:18 pm
Posts: 490
Location: kent
Irene,
I don't think it is supposed to be something that is "taught" in schools. This looks like a good question for the 11+ to me as it requires some innate mathematical reasoning ability. In my view, the questions that require a lot of "teaching" in order for any child to be able to answer them are not tests of ability, but tests of how well the child has been taught a particular syllabus in school or by a tutor.

When I read through NFER maths practise questions I would not be surprised by a question like the one you have quoted, but by ones that require specific mathematical "knowledge".

I don't know whether your 11+ would contain this type of question or not. Are you in an NFER area or not? If you are, read through the nfer practise papers and see if this type of question is in there or not.

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 Post subject: Posted: Tue Sep 16, 2008 2:31 pm

Joined: Thu Jul 17, 2008 11:09 am
Posts: 10
Location: lewisham
Quote:
John paints ahouse in 4 hrs and Jim paints a house in 6 hrs.
How long does it take both working together to paint 10 houses?

Is it 50 hours (6+4)/2x10?

Could not explain very well the computer&updates Q to my D.
Just hope she'll figure something out.

Thanks.

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 Post subject: Posted: Tue Sep 16, 2008 4:21 pm

Joined: Tue Jan 01, 2008 1:05 pm
Posts: 515
irene wrote:
Quote:
John paints ahouse in 4 hrs and Jim paints a house in 6 hrs.
How long does it take both working together to paint 10 houses?

Is it 50 hours (6+4)/2x10?

Could not explain very well the computer&updates Q to my D.
Just hope she'll figure something out.

Thanks.

Lowest common multiple is 12. By which time they would, between them have painted 5 houses. Therefore 10 houses would take 24 hrs.

Cannot be 50 hrs if you think about it, because john could paint more than 10 houses on his own.

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