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Really hard maths question

Heres the question;

One hundred and seven people get together in teams to take part in a quiz. Each team has three, four or
five members. There are twenty-seven teams altogether and there are twice as many teams with
four members as there are teams with three members. How many teams have five members?
Original post by CraigBackner
Heres the question;

One hundred and seven people get together in teams to take part in a quiz. Each team has three, four or
five members. There are twenty-seven teams altogether and there are twice as many teams with
four members as there are teams with three members. How many teams have five members?


Do you need help with this or is it just a challenge?
If you need help, try writing the information as two equations, assigning letters to mentioned unknown quantities.
Original post by IrrationalRoot
Do you need help with this or is it just a challenge?
If you need help, try writing the information as two equations, assigning letters to mentioned unknown quantities.


I've literally done that but I don't know how to solve it. Could you explain how please?
It's just simulataneous equations. Model it algebraically, with x, y, z representingg how many people have 3, 4 or 5 people respectfully.
Original post by BobBobson
It's just simulataneous equations. Model it algebraically, with x, y, z representingg how many people have 3, 4 or 5 people respectfully.


would the equation be
3x+4y+5z=170
and
3x+4y+5z=27?
Original post by CraigBackner
I've literally done that but I don't know how to solve it. Could you explain how please?


If you have two simultaneous equations in xx and yy (say), one method is to solve for yy in one equation and substitute that into the other equation to get an equation only in xx. Another method is to multiply both equations by constants in such a way that when you subtract them, the yys or xxs cancel out.
Surprised you haven't learnt simultaneous equations in school yet. If you still don't understand it'd probably be best to look them up for a more comprehensive explanation.
Original post by IrrationalRoot
If you have two simultaneous equations in xx and yy (say), one method is to solve for yy in one equation and substitute that into the other equation to get an equation only in xx. Another method is to multiply both equations by constants in such a way that when you subtract them, the yys or xxs cancel out.
Surprised you haven't learnt simultaneous equations in school yet. If you still don't understand it'd probably be best to look them up for a more comprehensive explanation.


I know how to solve simultaneous equations, I just dont know how to solve this problem which involes it , thats what I need explaning
Original post by CraigBackner
would the equation be
3x+4y+5z=170
and
3x+4y+5z=27?


The first ones right, but not the second one. The total number of Teams equals 27, so it's just the number of each type of team added together.
Original post by CraigBackner
would the equation be
3x+4y+5z=170
and
3x+4y+5z=27?


There's also a third equations that can be made with the later part of the question
Original post by CraigBackner
I know how to solve simultaneous equations, I just dont know how to solve this problem which involes it , thats what I need explaning


Simple. You have teams of 3, 4 and 5. You do not know how many of each team therefore you assign variables to each and it has to sum up to 107 so 3x+4y+5z=1073x+4y+5z=107. Then you're told that the amount of teams is equal to 27 so x+y+z=27x+y+z=27 and then you are ALSO told there are twice as many teams of 4 as there are of 3 so y=2xy=2x and you can simplify the two equations into 2 variables and solve it as normal.


Original post by CraigBackner
would the equation be
3x+4y+5z=170
and
3x+4y+5z=27?



That wouldn't make sense as this system of equations is inconsistent; you'd get 0=143. Geometrically, these are 2 parallel planes so no intersections obviously.
(edited 7 years ago)
Original post by BobBobson
There's also a third equations that can be made with the later part of the question


okay so the two equations are
3x+4y+5z=27
and
x+y+z = 27

and then I eliminate one of the varbiles, lets say x, which will create another equation, in which I use to solve
Original post by CraigBackner
okay so the two equations are
3x+4y+5z=27
and
x+y+z = 27

and then I eliminate one of the varbiles, lets say x, which will create another equation, in which I use to solve


The third equation. The third equation about "there are twice as many teams with four members as there are teams with three members."

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