Question 1204325
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The response from tutor @ikleyn shows a good typical formal algebraic solution.<br>
For problems like this involving averages of numbers that are all relatively close together, here is a VERY different way to solve the problem.<br>
The score on the last test was 12 points lower than on each of the other tests.<br>
"Distribute" those 12 points less over the 6 total tests.  That's 2 points per test.  That means the scores on his other 5 tests were each 97+2 = 99, and the score on his last test was 97-5(2) = 87.<br>
ANSWER: 87<br>
And for a more standard formal algebraic solution somewhat different than from the other tutor....<br>
let x = his score on each of the first 5 tests<br>
Then his score on the last test was x-12; and the average for the 6 tests was 97:<br>
{{{(5(x)+(x-12))/6=97}}}
{{{(6x-12)/6=97}}}
{{{x-2=97}}}
{{{x=99}}}<br>
ANSWER: his score on the last test was x-12 = 99-12 = 87<br>