You are Here: Home >< A-levels

# Really hard maths question watch

1. 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?
2. (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.
3. (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?
4. It's just simulataneous equations. Model it algebraically, with x, y, z representingg how many people have 3, 4 or 5 people respectfully.
5. (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?
6. (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 and (say), one method is to solve for in one equation and substitute that into the other equation to get an equation only in . Another method is to multiply both equations by constants in such a way that when you subtract them, the s or s 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.
7. (Original post by IrrationalRoot)
If you have two simultaneous equations in and (say), one method is to solve for in one equation and substitute that into the other equation to get an equation only in . Another method is to multiply both equations by constants in such a way that when you subtract them, the s or s 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
8. (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.
9. (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
10. (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 . Then you're told that the amount of teams is equal to 27 so and then you are ALSO told there are twice as many teams of 4 as there are of 3 so 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.
11. (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
12. (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."

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 27, 2016
The home of Results and Clearing

### 3,579

people online now

### 1,567,000

students helped last year
Today on TSR

Hang on, have Edexcel's come out already?

### University open days

1. SAE Institute
Animation, Audio, Film, Games, Music, Business, Web Further education
Thu, 16 Aug '18
2. Bournemouth University
Fri, 17 Aug '18
3. University of Bolton
Fri, 17 Aug '18
Poll

## All the essentials

### Student life: what to expect

What it's really like going to uni

### Essay expert

Learn to write like a pro with our ultimate essay guide.

### Create a study plan

Get your head around what you need to do and when with the study planner tool.

### Resources by subject

Everything from mind maps to class notes.

### Study tips from A* students

Students who got top grades in their A-levels share their secrets