In an examination, there were two papers, A and B, and the maximum mark in each of the two papers was 10. However, the weights assigned to papers A and B were in the ratio 2: 1, respectively. Jonathan scored 8 out of 10 in Paper A and an overall 70% in the examination. How much did he score in Paper B out of 10?
This problem involves calculating a missing score in an examination where different papers have different weightages. We need to determine Jonathan's score in Paper B given his score in Paper A, the maximum marks for each paper, the weights of the papers, and his overall examination percentage.
In this examination, the scores from Paper A and Paper B do not contribute equally to the overall percentage. Paper A has a weightage twice that of Paper B. Let's denote the weights as $w_A$ and $w_B$. According to the given ratio, we can set:
The maximum marks for both papers are $M_A = 10$ and $M_B = 10$. Jonathan's score in Paper A is $S_A = 8$. Let his score in Paper B be $S_B$.
The total maximum possible marks for the examination, considering the weights, are calculated as follows:
Total Maximum Weighted Marks = $(w_A \times M_A) + (w_B \times M_B)$
Substituting the values:
Total Maximum Weighted Marks = $(2 \times 10) + (1 \times 10) = 20 + 10 = 30$.
Jonathan achieved an overall score of 70% in the examination. This percentage is calculated based on the total maximum weighted marks.
Total Weighted Score Obtained = 70% of Total Maximum Weighted Marks
Total Weighted Score Obtained = $\frac{70}{100} \times 30 = 0.70 \times 30 = 21$.
Jonathan's score in Paper A (8 out of 10) contributes to the total score based on its weight ($w_A = 2$).
Weighted Score for Paper A = $w_A \times S_A$
Weighted Score for Paper A = $2 \times 8 = 16$.
The total weighted score obtained (21) is the sum of the weighted scores from Paper A and Paper B.
Total Weighted Score Obtained = Weighted Score for Paper A + Weighted Score for Paper B
So, Weighted Score for Paper B = Total Weighted Score Obtained - Weighted Score for Paper A
Weighted Score for Paper B = $21 - 16 = 5$.
The weighted score for Paper B (which is 5) is calculated using its weight ($w_B = 1$) and Jonathan's actual score ($S_B$).
Weighted Score for Paper B = $w_B \times S_B$
$5 = 1 \times S_B$
$S_B = \frac{5}{1} = 5$.
Therefore, Jonathan scored 5 in Paper B.
Jonathan scored 5 out of 10 in Paper B.
The average height of 20 students of class 8 is 152 cm and the average height of 15 students of class 9 is 168 cm. What is the average height (to the nearest cm) of the students of both classes?
The average of 4, 6, 8, 12 and x is 7 and the average of x, 9, 13, 15 and y is 9. What is the value of 2x - 3y?
The average weight of 20 girls in a school was 52 kg. Two new students of weight 54 kg and 50 kg were admitted. The ratio of this new average to the old one is:
If the average of two numbers is 13 and the square root of their product is 12, then the difference between the numbers is:
If the average of 5 consecutive odd integers in increasing order is 11 , then the average of the last 3 of them is: