ChatterBank1 min ago
Grandson's Home Workquestion.
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My 11 year old grandson recently started life at grammar school. This is his first maths homework question. I know you can get to the answer by trial and error but surely there is a mathematical/logical way to approach it. Parents and grandparents are stumped. Any suggestions?
There are five children in the house.
On each of the next six nights four of them go out.
No one child goes out on all six nights.
On the fjrst night the average age of the four is 38; second night 35; third night 36; fourth night 36; fifth night 38; sixth night 39.
What is the age pf the children?
Put us out of our misery please
There are five children in the house.
On each of the next six nights four of them go out.
No one child goes out on all six nights.
On the fjrst night the average age of the four is 38; second night 35; third night 36; fourth night 36; fifth night 38; sixth night 39.
What is the age pf the children?
Put us out of our misery please
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I get the average age of the children as 9.22 years.
1 Take the total ages on the 6 nights ie 38, 35, 36,36,38,& 39 and divide it by the number of children who went out which is 4.
2 you now end up with the average ages of the children who went out on any given night, which gives you the figures 9.2,8.75,9,9,9.4 & 9.75
3 Add them up and you get 55.3 and divide by the number of nights which the children went out, which was 6.
4 55.3 divided by 6 is 9.22. The average age of the children was 9.22yrs old.
1 Take the total ages on the 6 nights ie 38, 35, 36,36,38,& 39 and divide it by the number of children who went out which is 4.
2 you now end up with the average ages of the children who went out on any given night, which gives you the figures 9.2,8.75,9,9,9.4 & 9.75
3 Add them up and you get 55.3 and divide by the number of nights which the children went out, which was 6.
4 55.3 divided by 6 is 9.22. The average age of the children was 9.22yrs old.
I don't think he is expected to use a formula. It looks as if here is more than one answer depending on whether you assume they are all different ages or some can be the same age. if we knew which people went on each day you could solve it by algebra (solving simultaneous equations) but we don't know the combinations for each day. So I felt the best approach was using trial and improvement. You could narrow down the range of numbers by noting that the mean of the ages would be 9-10 and that the difference between highest and lowest could only be small as the sums of any four all fell in the range 35-39.
Then just try some numbers and see if you make the 6 totals, and if it doesn't work just tweak some numbers until it works.
It's a good puzzle.
Then just try some numbers and see if you make the 6 totals, and if it doesn't work just tweak some numbers until it works.
It's a good puzzle.
After messing about on a spreadsheet for a while I came up with 7 8 10 10 11 too.
But I think it may change if you assume night 1 and night 5 are the same group, or night 3 and night 4.
I suspect there isn't enough information for a unique solution, but I've spent too much time on this one already to check.
IMO, unless how to get the required answer has been spelt out in class, this seems rather complex for an 11 year old. Unless it's a school for genii.
But I think it may change if you assume night 1 and night 5 are the same group, or night 3 and night 4.
I suspect there isn't enough information for a unique solution, but I've spent too much time on this one already to check.
IMO, unless how to get the required answer has been spelt out in class, this seems rather complex for an 11 year old. Unless it's a school for genii.
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