Answer
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Hint: Assume that there are 100 students in the examination. Use the formula in set theory to find out the number of students who failed in at least one of the subject. Then subtract this number from 100 to get the number of students who passed in both the subjects.
Complete step-by-step answer:
Let us assume that there are 100 students in the examination. So, we have 42 students who failed in Hindi, 52 students who failed in English and 17 students who failed in both. Let us denote the number of students who failed in Hindi and English by ‘$n(H)$’ and ‘$n(E)$’ respectively. Therefore, the number of students who failed in both can be denoted by ‘$n(H\cap E)$’. Applying formula for the union of sets, we have,
Total number of students who failed in at least one subject
$\begin{align}
& =n(H\cup E) \\
& =n(H)+n(E)-n(H\cap E) \\
& =42+52-17 \\
& =77 \\
\end{align}$
So, number of students who passed in both the subjects
$\begin{align}
& =100-\text{number of students who failed in at least one subject} \\
& \text{=100-77} \\
& \text{=23} \\
\end{align}$
Therefore, percentage of students who passed in both the subjects
$\begin{align}
& =\dfrac{23}{100}\times 100\% \\
& =23\% \\
\end{align}$
Hence, option (a) is the correct answer.
Note:We have assumed the total number of students 100 because if we choose it as any variable then we may get confused. At last we have to find the percentage, that is, a ratio multiplied by 100, so it doesn’t matter what we choose as the total number of students. This problem is purely based on set theory, so we have applied the formula of union of two sets to determine the number of students who failed in at least one subject.
Complete step-by-step answer:
Let us assume that there are 100 students in the examination. So, we have 42 students who failed in Hindi, 52 students who failed in English and 17 students who failed in both. Let us denote the number of students who failed in Hindi and English by ‘$n(H)$’ and ‘$n(E)$’ respectively. Therefore, the number of students who failed in both can be denoted by ‘$n(H\cap E)$’. Applying formula for the union of sets, we have,
Total number of students who failed in at least one subject
$\begin{align}
& =n(H\cup E) \\
& =n(H)+n(E)-n(H\cap E) \\
& =42+52-17 \\
& =77 \\
\end{align}$
So, number of students who passed in both the subjects
$\begin{align}
& =100-\text{number of students who failed in at least one subject} \\
& \text{=100-77} \\
& \text{=23} \\
\end{align}$
Therefore, percentage of students who passed in both the subjects
$\begin{align}
& =\dfrac{23}{100}\times 100\% \\
& =23\% \\
\end{align}$
Hence, option (a) is the correct answer.
Note:We have assumed the total number of students 100 because if we choose it as any variable then we may get confused. At last we have to find the percentage, that is, a ratio multiplied by 100, so it doesn’t matter what we choose as the total number of students. This problem is purely based on set theory, so we have applied the formula of union of two sets to determine the number of students who failed in at least one subject.
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