# Math question

1. Feb 8, 2005

### preet

The integers from 1 to 9 are listed on a blackboard. If an additional m eights and k nines are added to the list, the average of all of the number is the list is 7.3. The value of k+m is

a)24 b)21 c)11 d)31

I got as far as (45+8m+9k) / (9+m+k) = 7.3
What do I do now?
TiA!

2. Feb 8, 2005

### dextercioby

Cross multiply and use the fact that the "old" numerator must be a whole/integer number.That should tell u something about the "old" denominator.
U'll have to solve simple 2-2 algebraic systems...In N.

Daniel.

3. Feb 8, 2005

### Andrew Mason

You mean to say that there is a method for solving this other than trial and error?

AM

4. Feb 8, 2005

### uranium_235

I am doing a unit on such systems in math. I do not see a solution that does not involve simplifying the equation you have and solving with the equations the answers suggest and seeing which one works out.

5. Feb 8, 2005

### dextercioby

In this case it's a finite sequence of 2-2 systems of (simple) algebraic equations.If you wanna call it trial & error,b my guest. To me those words mean something like "guessing"... :tongue2: It's not the case here.

Daniel.

6. Feb 9, 2005

### Andrew Mason

One way of doing this without guessing is to draw a solution of the equation (in this case: .7m + 1.7k = 20.7) on graph paper (with each graph line representing one unit of k or m). The solution lies where m > 0, k>0 and the line passes through an intersection of vertical and horizontal graph lines (which represent points where m and k are natural numbers).

AM

7. Feb 9, 2005

### dextercioby

Sure,Andrew,you're right.Such method would work.But i'm sure the OP's initial idea was to solve this problem by an algebraic method...

Daniel.

8. Feb 9, 2005

### Galileo

Well, only four choices. Just enter them into your calc. and see which one is correct. You couldn't get an easier question than this.

That is: Check which answer gives allowable values for m and k.

Last edited: Feb 9, 2005
9. Feb 9, 2005

### dextercioby

That's an EVIL answer... :tongue2: 666 posts...

Daniel.

P.S.You should do something about it,pretty fast... :tongue2:

10. Feb 9, 2005

### Galileo

MUHAHAHAHWAHAHAHWHAWHAHWAWAAAAAAAAAAAAAAAAAAAAAAAAAA!!!!!!!!!!!!!!!

11. Feb 10, 2005

### preet

I still don't understand what to do.... I don't really care for the calculation method (checking the answers), but I don't know how to do that either. How do I solve it algebraically?

12. Feb 10, 2005

### dextercioby

I've already told u.See post #2.Basically,the denominator has to be a multiple of 10,because,when multiplied with 7.3,it would yield a natural number,as the numerator is.So you have to set the denominator equal to 10,20,30,...For each equation,u'll get one for the numerator as well,since the ratio must be 7.3
In this way you generate a series of simple systems 2 (eq.)-2 (unknowns).

Daniel.

13. Feb 10, 2005

### Andrew Mason

I don't think there is a way of doing it algebraically. The solution is the set of values of k,m belonging to the set of natural numbers, that satisfy the relation: 7m + 17k = 207 Algebra will give you the set of all real numbers - ie. all points on the line. The solution is a subset of the points on that line that belong to the set of natural numbers.

The only way I can see of doing it is to draw the relation on paper (with m as the horizontal and k as the vertical axes, say). It is a line with slope k/m = -7/17. If you do it on graph paper, the intersection of the line with the intersection of any of the graph lines, will provide a solution.

AM

14. Feb 10, 2005

### dextercioby

What???

1.Denominator 10,numerator 73.System:

$$m+k=1;8m+9k=28$$
Doesn't have solution:

2.Denominator 20,numerator 146.Sytem:
$$m+k=11;8m+9k=101$$

k=13,m=-2.Doesn't work.

3.Denominator 30,numerator 219.Sytem:
$$m+k=21;8m+9k=174$$

k=6;m=15.Does work."21" is a valid solution.

4.Denominator 40,numerator 292.Sytem:
$$m+k=31;8m+9k=247$$

Doesn't have solution in N.

If this ain't algebraic solution,then WHAT IS...?

Daniel.

15. Feb 10, 2005

### Andrew Mason

I'd call it 'trial and error'.

AM

16. Feb 10, 2005

### dextercioby

Your call.I'd call it the most rigurous nongeometric solution one might get.

Daniel.

17. Feb 10, 2005

### preet

Thanks for the help