A student scored 85, 90, and 95 on three exams. If the final exam is worth twice as much as each of the others, what score is needed on the final exam to achieve an average of 90?

Sun, 03 May 2026 21:50:12 +0000

**Why exam weighting matters: Preparing for a balanced academic average** In today’s fast-paced, results-driven education landscape, students and parents alike are increasingly focused on how grades translate into long-term academic and career success. This awareness fuels curiosity about how weighted exam systems influence final performance—especially when one exam carries significant weight. A recent example shows a student scoring 85, 90, and 95 across three core exams, raising a key question: What final score is needed to reach a target average of 90, considering the final count double the value of each earlier exam? With exam structures evolving to reflect real-world demands, understanding these weighted averages is more relevant than ever. Navigating such calculations helps students set realistic, data-backed goals—without the pressure of vague or exaggerated claims. **The math behind weighted averages: Why final scores matter more when weighted** When calculating a geometric average based on exam weightings, each exam’s score is multiplied by its weight before summing and dividing by total weight. In this case, the regular three exams carry a weight of 1 point each, while the final exam counts as 2 points, making the total weight 4. To average 90 across all weighted points, the total weighted score must equal 90 × 4 = 360. Adding up the first three exams—85 + 90 + 95—gives 270. Subtracting this from 360 reveals the required final score must be 90. Since the final counts twice, dividing 90 by 2 yields 45. Thus, scoring 45 on the final is necessary to achieve a 90-point average. This not only clarifies the target but also helps students avoid common calculation errors tied to weighted scoring.