Question 158017
Let {{{x=Midterm}}}
{{{y=FinalExam}}}
Let's start with an average of 70 (MIN) first:
Given {{{x=48}}}
{{{(2/3)x+(1/3)y=70}}}
{{{(2/3)(48)+(1/3)y=70}}}
{{{(96/3)+(1/3)y=70}}}
{{{(1/3)y=70-(96/3)=70-32=38}}}, cross multiply
{{{y=(38)(3)=114}}}, MINIMUM SCORE on Final Exam
Next with an average of 79 (MAX): Given {{{x=48}}}
{{{(2/3)x+(1/3)y=79}}}
{{{(2/3)(48)(1/3)y=79}}}
{{{(1/3)y=79-(96/3)=79-32=47}}}, cross multiply
{{{y=3*47=141}}}, MAX SCORE  on Final Exam
Therefore:To have an average of {{{70-79}}} inclusive, he should have a final score of {{{114-141}}} inclusive.
Check it out by substituting 1st {{{y=114}}}, 2nd {{{y=141}}} to the above eqn and you'll get an average of 70 and 79 inclusive.
thank you,
jojo