Data Interpretation - Line Charts - Discussion
Discussion Forum : Line Charts - Line Chart 10 (Q.No. 4)
Directions to Solve
In a school the periodical examination are held every second month. In a session during April 2001 - March 2002, a student of Class IX appeared for each of the periodical exams. The aggregate marks obtained by him in each perodical exam are represented in the line-graph given below.
Marks Obtained by student in Six Periodical Held in Every Two Months During the Year in the Session 2001 - 2002.
Maximum Total Marks in each Periodical Exam = 500
4.
What are the average marks obtained by the student in all the periodical exams during the last session ?
Answer: Option
Explanation:
Average marks obtained in all the periodical exams
| = | 1 | x [360 + 365 + 370 + 385 + 400 + 405] = 380.83  381. |
| 6 |
Discussion:
4 comments Page 1 of 1.
Samil said:
4 years ago
Why divide 1 by 6 at first?
Shouldn't we compute directly with 6/ (Sum of All). I'm so confused, Anyone, please clarify this.
Shouldn't we compute directly with 6/ (Sum of All). I'm so confused, Anyone, please clarify this.
Pema said:
7 years ago
Why dividing by 6 instead of average Mark 500? anyone help me to know the reason.
Pema said:
7 years ago
Why dividing by 6 instead of average Mark 500? anyone help me to know the reason.
(1)
Vatshal said:
8 years ago
The Last session means previous year, so why is feb02 included in it?
(1)
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