0
$\begingroup$

Johnny had to take a test a day late. His 96 raised the class average from 71 to 72. How many students, including Johnny, took the test?

I tried to do trial and error to see how many students there were but I couldn't figure it out.

$\endgroup$

1 Answer 1

1
$\begingroup$

Hint: Letting $n$ be the number of students excluding Johnny... $$\text{Old average} = \frac{71n}{n}$$ $$\text{New average} = \frac{71n + 96}{n+1}$$

Note that $\text{New average} - \text{Old average} = 1$.

$\endgroup$
5
  • $\begingroup$ I set the equations equal to each other and I got n^2 +25n. So would the number of students be 25? Without Johnny? Because when I plug in 25 for n the equations do not equal eachother $\endgroup$
    – Ella
    Feb 8, 2014 at 22:27
  • $\begingroup$ @Layla The expressions should not be equal, but, rather, they should differ by $1$. $\endgroup$
    – apnorton
    Feb 8, 2014 at 22:51
  • $\begingroup$ How do I solve for n if they need to differ by 1? $\endgroup$
    – Ella
    Feb 8, 2014 at 23:03
  • $\begingroup$ @Layla Subtract the old average from the new average and set that equal to $1$. $\endgroup$
    – apnorton
    Feb 8, 2014 at 23:17
  • $\begingroup$ By subtracting I get (71n^2 + 96n - 71n^2 - 71n)/ (n(n+1) and I get 25/(n+1)??? So I set that equal to 1 and get 25=n+1? So it's 25 total students? $\endgroup$
    – Ella
    Feb 8, 2014 at 23:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .