A student scored 85, 90, and 88 on three exams. If the final exam is worth double the points of a regular exam, what score is needed on the final to achieve an average of 90?

["Achieving an Average of 90: How a Final Exam Score Impacts Your Overall Grade", "Students often find themselves wondering how specific exam scores influence their overall academic performance—especially when important assessments count more than regular quizzes or midterms. Let’s explore a realistic scenario: a student scored 85, 90, and 88 on three regular exams, and now faces a final exam worth double the points. How high must the student score on the final to achieve an average of 90 across all exams?", "### Understanding the Weight of Exam Formats", "In most grading systems, regular exams are weighted more heavily than smaller quizzes or homework, but some courses assign greater points to fewer, more comprehensive assessments. Here, we assume the final exam counts double a regular exam’s worth. This matters because a single high final score can pull up the average significantly.", "Given:\n- Three regular exams: 85, 90, 88\n- The final exam is worth equivalent to two regular exams combined (i.e., a weighted 2-exam score out of 4 total weighted units)\n- Target average grade: 90", "To find the needed final score, let’s break it down:", "---", "### Step 1: Calculate the total points earned so far", "Sum of regular exam scores:\n85 + 90 + 88 = 263", "Since the regular exams likely constitute 2 normalized weighted units (each worth 1 point), total current weighted points:\n263", "Let ( x ) be the score on the final exam, which counts as 2 units. So, weighted points from final: ( 2x )", "Total weighted points after including final:\n( 263 + 2x )", "---", "### Step 2: Define the total number of weighted units", "With one regular exam (1 unit) and three zero-or one-weight exams:\nTotal weight = 1 (regular) + 3 + 1 doubled = 5 total weighted units", "Wait — correction: The problem states the final exam is worth double a regular exam, but we don’t know how many regular exams. Since only three are mentioned, assume 3 regular exams + final counted double.\nSo total weighted units = 3×1 + 1×2 = 5 units", "We want the overall weighted average to be 90, so total weighted points should be:", "[\n90 \ imes 5 = 450\n]", "---", "### Step 3: Set up the equation", "[\n263 + 2x = 450\n]", "Solve for ( x ):", "[\n2x = 450 - 263 = 187\n\quad \Rightarrow \quad\nx = \frac{187}{2} = 93.5\n]", "---", "### Conclusion", "To achieve a 90 average when:\n- Three regular exams score 85, 90, and 88\n- The final exam counts double (equivalent to two regular exams)", "The student must score at least 93.5 on the final exam. Since exam scores are typically reported as whole numbers, scoring 94 or higher will round to an average above 90 depending on rounding edges, but strictly mathematically, a 93.5 is required.", "---", "Key Takeaways:\n- Assigning double weight to the final amplifies its impact on the average.\n- Understanding required performance on key assessments helps strategic studying.\n- Balancing weighted and regular exam scores ensures a targeted final push.", "For students aiming for a 90 average in weighted systems, never underestimate the final—your final might just decide your grade."]