
In an examination, 42% students failed in Hindi and 52% failed in English. If 17% failed in both of these subjects, the percentage of students who passed in both the subjects are
(a) 23
(b) 27
(c) 34
(d) 40
Answer
595.5k+ views
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

