#### Answer

H) 98

#### Work Step by Step

You already have the value of the first four exams $78, 85, 97, 92$. The fifth exam is $x$, a variable, as you don't know what the value will be. In this problem, you find the average of the five exams, and what will be needed to make the average 90.
$(79+85+97+92+x)\div5=90$
$(x+352)\div5=90$
$((x+352)\div5)\times5=90\times5$
$x+352=450$
$(x+352)-352=(450)-352$
$x=98$
To check your answer, plug $98$ into the original equation and see if it works.
$((98)+352)\div5=90$
$450\div5=90$
$90=90$