Answer
Verified
325.6k+ views
Hint: In this question, first draw the Venn diagram with the given data it will give us a clear picture of what we have to find out. The number of students who had taken none of the subjects can be found by subtracting the number of students taken at least one subject from the total number of students in the survey. So, use this concept to reach the solution of the question.
Complete step-by-step answer:
Given,
Total number of students in the survey \[n\left( U \right) = 25\]
The set of students who have taken mathematics \[n\left( M \right) = 15\]
The set of students who have taken Physics \[n\left( P \right) = 12\]
The set of students who have taken chemistry \[n\left( C \right) = 11\]
The set of students taken mathematics and chemistry \[n\left( {M \cap C} \right) = 5\]
The set of students taken physics and chemistry \[n\left( {P \cap C} \right) = 4\]
The set of students taken mathematics and physics \[n\left( {M \cap P} \right) = 9\]
The set of students taken all the 3 subjects \[n\left( {M \cap P \cap C} \right) = 3\]
Here number of students taking at least one subject \[ = n\left( {M \cup P \cup C} \right)\]
From the Venn diagram, we have
\[
\Rightarrow n\left( {M \cup P \cup C} \right) = n\left( M \right) + n\left( P \right) + n\left( C \right) - n\left( {M \cap P} \right) - n\left( {P \cap C} \right) - n\left( {M \cap C} \right) + n\left( {M \cap P \cap C} \right) \\
\Rightarrow n\left( {M \cup P \cup C} \right) = 15 + 12 + 11 - 9 - 4 - 5 + 3 \\
\Rightarrow n\left( {M \cup P \cup C} \right) = 41 - 18 \\
\therefore n\left( {M \cup P \cup C} \right) = 23 \\
\]
So, the number of students taking none of the subjects \[ = n\left( U \right) - n\left( {M \cup P \cup C} \right) = 25 - 23 = 2\]
Thus, the correct option is B i.e, 2
Note: The intersection of two sets \[A\] and \[B\], denoted by \[A \cap B\], is the set containing all elements of \[A\] that also belong to \[B\] (or equivalently, all elements of \[B\] that also belong to \[A\]). The union of two sets \[A\] and \[B\], denoted by \[A \cup B\] is the set of all elements that are found in \[A\] OR \[B\] (or both)
Complete step-by-step answer:
Given,
Total number of students in the survey \[n\left( U \right) = 25\]
The set of students who have taken mathematics \[n\left( M \right) = 15\]
The set of students who have taken Physics \[n\left( P \right) = 12\]
The set of students who have taken chemistry \[n\left( C \right) = 11\]
The set of students taken mathematics and chemistry \[n\left( {M \cap C} \right) = 5\]
The set of students taken physics and chemistry \[n\left( {P \cap C} \right) = 4\]
The set of students taken mathematics and physics \[n\left( {M \cap P} \right) = 9\]
The set of students taken all the 3 subjects \[n\left( {M \cap P \cap C} \right) = 3\]
Here number of students taking at least one subject \[ = n\left( {M \cup P \cup C} \right)\]
From the Venn diagram, we have
\[
\Rightarrow n\left( {M \cup P \cup C} \right) = n\left( M \right) + n\left( P \right) + n\left( C \right) - n\left( {M \cap P} \right) - n\left( {P \cap C} \right) - n\left( {M \cap C} \right) + n\left( {M \cap P \cap C} \right) \\
\Rightarrow n\left( {M \cup P \cup C} \right) = 15 + 12 + 11 - 9 - 4 - 5 + 3 \\
\Rightarrow n\left( {M \cup P \cup C} \right) = 41 - 18 \\
\therefore n\left( {M \cup P \cup C} \right) = 23 \\
\]
So, the number of students taking none of the subjects \[ = n\left( U \right) - n\left( {M \cup P \cup C} \right) = 25 - 23 = 2\]
Thus, the correct option is B i.e, 2
Note: The intersection of two sets \[A\] and \[B\], denoted by \[A \cap B\], is the set containing all elements of \[A\] that also belong to \[B\] (or equivalently, all elements of \[B\] that also belong to \[A\]). The union of two sets \[A\] and \[B\], denoted by \[A \cup B\] is the set of all elements that are found in \[A\] OR \[B\] (or both)
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE