Question

Mariana averaged 72 on her first three history exams. The first score was 78. The second score was 6 less than the third score. What did she score on the second and third exams?

Answer 1

F multiply both sides by 3, combine like terms
93 - 44 + x + x = 3(83)
49 + 2x = 249
2x = 249 - 49
2x = 200
x = 100 her score on the 3rd exam
then
100 - 44 = 56 on the 2nd exam
:
;
Check
%2893%2B56%2B100%29%2F3 = 83

Answer 2

Okay, let's assign variables:
second exam: a
third exam: b
Because the second score is 6 less than the third, we know that
a = b - 6
knowing this, the numbers now become:
second exam: b - 6
third exam: b
now that we have these numbers. We have to find the average by adding all the numbers and dividing:
(72 + b + b - 6) / 3 = 78
(72 + 2b - 6) / 3 = 78
(2b + 66) / 3 = 78
To get rid of the 3, we multiply both sides by 3:
2b + 66 = 234
2b = 168
b = 84
Mariana scored 84 points on her third exam and 78 on her second exam.