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**Hint:**In this question we have to find the number of students who take both mathematics and physics in the class. For that, we are going to solve using the union and intersection method. Initially, we will assign the given values to certain variables and then we find the required result with the help of the following formula.

Formulas used:

\[{\rm{n(AUB) = n(A) + n(B) - n(A}} \cap {\rm{B)}}\]

**Complete step by step answer:**

It is given that the total number of students in the class is \[30\].

It is given that the number of students who have taken mathematics in the class is \[10\] and let it be \[{\rm{n(A)}}\].

Also given that the number of students taking physics in the class is \[15\] and let it be \[{\rm{n(B)}}\].

We also have information that the number of students taking neither mathematics nor physics in the class is 10.

We have to find the number of students who have taken both mathematics and physics in the class which is denoted as \[{\rm{n(A}} \cap {\rm{B)}}\].

Where \[{\rm{n(AUB)}}\] is found using the following fact,

\[{\text{n(AUB) = total number of students - students those who take neither subject}}{\rm{. }}\]

\[\Rightarrow {\rm{n(AUB) = 30 - 10}}\],

\[{\rm{n(AUB) = 20}}\]

Now let us substitute the known values in the formula which is used below,

\[\Rightarrow {\rm{n(AUB) = n(A) + n(B) - n(A}} \cap {\rm{B)}}\]

By substituting the known values in the above formula we get,

\[\Rightarrow 20 = 10 + 15 - {\rm{n(A}} \cap {\rm{B)}}\]

On simplification we get,

\[\Rightarrow 10 = 15 - {\rm{n(A}} \cap {\rm{B)}}\]

Let us subtract by 15 both sides we get,

\[10 - 15 = - {\rm{n(A}} \cap {\rm{B)}}\]

\[\Rightarrow - 5 = - {\rm{n(A}} \cap {\rm{B)}}\]

By multiplying with $-1$ on both sides we get,

\[\Rightarrow {\rm{n(A}} \cap {\rm{B) = 5}}\]

**\[\therefore \] The number of students has taken both mathematics and physics in the class is \[5\].**

**Note:**

Here we have to find \[{\rm{n(AUB)}}\]which is most important in finding the result. Also, we will subtract the total number of students to the number of students those who neither take both the subjects to find \[{\rm{n(AUB)}}\].

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