Quizzes & Puzzles1 min ago
Grandson's Home Workquestion.
39 Answers
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 had reckoned there were some damned old stay-at-home children with average ages of 35 - if one is under 30 then the oldest is gonna be in his later thirties
but boys and girls there are five unknowns ( A-E) and six equations - and this can only mean there are too many equations and one is repeated
so there should be one solution - it could be possible there are none
( this follows from a result in linear equations - five unknown, five eqns etc will admit a unique solution so long as certain conditions are filled - five unknowns and SIX equations then one has to be redundant ( a linear multiple of one of the other five) because other wise there isnt a solution)
and if there is only one missing variable in each equation then one must be repeated. in FF example this is day 1 and 5 so one can be crossed out
but third and fourth have the same total - which means that there has to be twins - a repeated solution
but it could be the other way round dammit
OK baybayze.....
the general solution cd be given by .... first of all let the sum of five be ecks
say the first day as we have seen is A+B+C+D = 38 and is x-e
second day let us say ...................A +B+C + E = 35 and is x-d
third day let us say ................ A + B+ D+E= 36 and is x-c
fourth day let us say ........ A + C+D+E=36 and is x-b
so day 3 and 4 is the twinnzie day where we day that b and c have the same age
the fifith night is .......... A+B +C+D = 38 and is x-e
this is the repeated day that is we are saying on day one and five the same child goes out and so the age sum is the same
and the sixth night is ........ B+C+D+E=39 and is x-a
add them all up omitting the one we think is repeated day three in this case
and we get 184 and on the other side we get 5x minus a,b,c,d,e
which is x so that is 4x and the total sum is 184 so x is 46
and so we see that e is 8 ( 46 minus 38 ) d is 9 ( 46 minus35) c is 10 and so is b and the last one gives a as 7
7,10,10,9 and 8
and secondly ( o god it is getting late ) three and four are repeated equations and one and five are twinnzies
blah blah I ant get it together to do all the equations
but if you do that then you get a total for 4x to be 186 which is not divisible by 4 and so is not a solution to the problem
ha !
but boys and girls there are five unknowns ( A-E) and six equations - and this can only mean there are too many equations and one is repeated
so there should be one solution - it could be possible there are none
( this follows from a result in linear equations - five unknown, five eqns etc will admit a unique solution so long as certain conditions are filled - five unknowns and SIX equations then one has to be redundant ( a linear multiple of one of the other five) because other wise there isnt a solution)
and if there is only one missing variable in each equation then one must be repeated. in FF example this is day 1 and 5 so one can be crossed out
but third and fourth have the same total - which means that there has to be twins - a repeated solution
but it could be the other way round dammit
OK baybayze.....
the general solution cd be given by .... first of all let the sum of five be ecks
say the first day as we have seen is A+B+C+D = 38 and is x-e
second day let us say ...................A +B+C + E = 35 and is x-d
third day let us say ................ A + B+ D+E= 36 and is x-c
fourth day let us say ........ A + C+D+E=36 and is x-b
so day 3 and 4 is the twinnzie day where we day that b and c have the same age
the fifith night is .......... A+B +C+D = 38 and is x-e
this is the repeated day that is we are saying on day one and five the same child goes out and so the age sum is the same
and the sixth night is ........ B+C+D+E=39 and is x-a
add them all up omitting the one we think is repeated day three in this case
and we get 184 and on the other side we get 5x minus a,b,c,d,e
which is x so that is 4x and the total sum is 184 so x is 46
and so we see that e is 8 ( 46 minus 38 ) d is 9 ( 46 minus35) c is 10 and so is b and the last one gives a as 7
7,10,10,9 and 8
and secondly ( o god it is getting late ) three and four are repeated equations and one and five are twinnzies
blah blah I ant get it together to do all the equations
but if you do that then you get a total for 4x to be 186 which is not divisible by 4 and so is not a solution to the problem
ha !
tambo- I'm sure lordg didn't mean what you think he means. He can't gave been ASKING for the total age of the 4 - because there are 6 children, so asking for the total of 4 wouldn't make sense. He must have been TELLING us the the total age on each night was the figure given for each night- 38,35 etc. More importantly he didn't disagree with the solutions given- he just said he couldn't see the formula. Anyway maybe lordg can clarify so we can be sure.
togo has six kids: //That should have read. 7,8,9,10,11,12//
and not five - and that leaves one......
FF has the right answer - I can hardly believe he did it trial and error
I cant subtract as usual - 7,10,10,9 and 8 should be 7 10 10 11 and 8 - that is the same as FF, because 46 - 35 is eleven and not nine ( ouch!). Yikes I havent changed for fifty years
and that (FF's) is I think the only allowed answer - as there are six equations and five unknowns so there cant be more than one answer ( unless more than one kid goes out twice -we know one does already )
yeah good quezzie to get the kidz finking yeah.
no answer from the parents or grandparents who presumably are sitting there stunned ....
and not five - and that leaves one......
FF has the right answer - I can hardly believe he did it trial and error
I cant subtract as usual - 7,10,10,9 and 8 should be 7 10 10 11 and 8 - that is the same as FF, because 46 - 35 is eleven and not nine ( ouch!). Yikes I havent changed for fifty years
and that (FF's) is I think the only allowed answer - as there are six equations and five unknowns so there cant be more than one answer ( unless more than one kid goes out twice -we know one does already )
yeah good quezzie to get the kidz finking yeah.
no answer from the parents or grandparents who presumably are sitting there stunned ....
These unusual type of questions are good in my opinion as they encourage pupils to explore the problem rather than just stick the numbers into a formula teacher has written on the board. I think some of the class would have got it- and probably quicker than we did. I remember giving a top set a problem where they had to develop a formula for working out the angle at any time of day between the hour hand and minute hand- and a couple of tables in the class worked it out within the lesson (which was just as long as it took me)
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