## Sunday, November 18, 2012

### 2007 State Harder Algebra Questions

Check out Mathcounts here, the best competition math program for middle school students.

#16 2007 Mathcounts State Sprint: A data set for a class of 25 sixth graders has their ages listed as
the integer values of either 10 or 11 years. The median age in the data set is 0.36 years greater than the mean. How many 10-year-olds are in the class?

#23: 2007 Mathcounts State Sprint:  A box contains some green marbles and exactly four red marbles.
The probability of selecting a red marble is x%. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x - 15)%. How many green marbles are in the box before the number of green marbles is doubled?

#16: Solution I:
Since there are 25 students, the median has to be either 10 or 11 (Why?)
Since the mean has to be 10 (if all the children are all 10 year old) or higher, the median has to be 11.
Let there be x 10 year old students, and there are (25 - x) 11 year old students.

Solution II:
Let there be x 10 year old and y 11 year old.
According to the given:  $\frac{10x + 11y}{25} = 11 - 0.36 = 10.64$
10x + 11 y = 266
For x to be integer, y has to either 6 or 16. Further checking, only when y = 16, x = 9 works.

#23: Let there be g numbers of Green marbles.
According to the given, we can set up two equations:

$\frac{4}{4 + g}= \frac{x}{100}$ $\rightarrow400 = 4x + xg$                                          equation 1

$\frac{4}{4 + 2g}= \frac{x-15}{100}$ $\rightarrow400 = 4x -60 + 2gx - 30g$           equation 2

Equation 1 minus equation 2 and you have: 0 = xg + 60 -2xg + 30g
Move all the variables to the other side:
xg - 30g = g (x - 30) = 60
If g = 4, x = 45 (doesn't work)
If g = 6, x = 40 (it works.)