# Generalizing with Variables

• Sep 3rd 2011, 07:28 PM
thamathkid1729
Generalizing with Variables
Generalize the average test grade. Suppose Jane has an average of G after n tests. What average score does she need on the next m tests to average H for all m + n tests?

t1 = test #1, t2 = test #2, ...

(t1 + t2 + ... + tn)/n = G

(t1 + t2 + ... + tn) = n*G

((t1 + t2 + ... + tn) + (tn+1 + tn+2 + ... + tm))/(m+n) = H

(n*G + (tn+1 + tn+2 + ... + tm))/(m+n) = H

n*G + (tn+1 + tn+2 + ... + tm) = (m+n)*H

(tn+1 + tn+2 + ... + tm) = (m+n)*H - n*G

(tn+1 + tn+2 + ... + tm)/m = ((m+n)*H - n*G)/m

((m+n)*H - n*G)/m is my answer but I'm a little unsure of this. Any help would be greatly appreciated!!!
• Sep 4th 2011, 05:59 AM
HallsofIvy
Re: Generalizing with Variables
Yes, that is correct. If she averaged "G" on n tests, then her total score is nG. If she averages "X" on the next m tests, her total score for those tests would be mX so here total score over all m+ n tests would be nG+ mX. Her average over all the tests would be $\frac{nG+ mX}{m+ n}$. Set that equal to H and solve for X:
$\frac{nG+ mX}{m+n}= H$
$nG+ mX= (m+ n)H$
$mX= (m+ n)H- nG$
$X= \frac{(m+n)H- nG}{m}$ which is what you have.

You could also write that as $X=\frac{mH+ n(H-G)}{m}= H+ \frac{n}{m}(H- G)$
showing just how the difference between "H", the desired average, and "G", the current average, influences the needed future average.