The following table presents the details about the number of candidates who appeared in an entrance exam and percentage (%) of candidates who qualified the exam from two states P and Q during the years 2017-2021. Some values are missing in the table, and have to be computed, if required. Year-wise Distribution of Candidates in an Entrance ExamYear State P State Q Appeared Qualified Appeared Qualified 2017 1800 60% 1520 30% 2018 2400 43% - 40% 2019 - 60% 1040 60% 2020 1920 70% 800 70% 2021 1520 - 1320 -
Let A and B represent the number of candidates appearing and qualifying from State Q in 2018 respectively. Let C and D represent the number of candidates appearing and qualifying from State Q in 2021 respectively. If A is 33⅓% more than C and B : D = 11 : 12, then what is B + D?
1472
Given the problem, we need to determine the combined number of candidates who qualified from State Q in the years 2018 and 2021, respectively.
- First, we focus on the information provided for 2018. Let \(A\) represent the candidates appearing from State Q in 2018.
- According to the problem, the percentage of candidates who qualified in 2018 from State Q is 40%. Thus, the number of candidates qualifying, \(B\), is calculated as:
\(B = \frac{40}{100} \times A\)
- We are given that \(A\) is 33⅓% more than \(C\) (candidates appearing in 2021). Thus:
\(A = C + \frac{1}{3}C = \frac{4}{3}C\)
- Next, focus on the year 2021. Here, the number of candidates appearing, \(C\), is given as 1320.
- Substituting the available values, we calculate:
\(A = \frac{4}{3} \times 1320 = 1760\)
- Thus, the number of candidates who qualified in 2018:
\(B = \frac{40}{100} \times 1760 = 704\)
- The problem states that \(B : D\) (ratio of candidates qualifying in 2018 to those in 2021) is 11 : 12. Solving for \(D\):
\(\frac{704}{D} = \frac{11}{12}\)
\(D = \frac{704 \times 12}{11} = 768\)
- Finally, calculate \(B + D\):
\(B + D = 704 + 768 = 1472\)
The correct answer is 1472, matching the question's provided option.
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