Finding the Value of m and k to Reach an Average of 7.3

[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!

 

[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.

 

[Andrew Mason]

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.

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

AM

 

[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.

 

[dextercioby]

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

AM

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

Daniel.

 

[Andrew Mason]

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

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

 

[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.

 

[Galileo]

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!

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.

 

[dextercioby]

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

Daniel.

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

 

[Galileo]

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

MUHAHAHAHWAHAHAHWHAWHAHWAWAAAAAAAAAAAAAAAAAAAAAAAAAA!

 

[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?

 

[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.

 

[Andrew Mason]

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?

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

 

[dextercioby]

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

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.

 

[Andrew Mason]

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...?

I'd call it 'trial and error'.

AM

 

[dextercioby]

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

Daniel.

 

[preet]

Thanks for the help