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**Hint:**To solve these types of problems, we need to use some set theory. Let A be a set then N(A) represents the number of elements present in it. If there are two sets A and B then \[A\cup B\] is a set called the union of A and B it consists of the elements that belong to either A or B and \[A\cap B\] the set is called the intersection of A and B, which consists of elements that belong to both A and B.

We will use the formula for set theory that states that \[N\left( A\cap B \right)=N\left( A \right)+N\left( B \right)-N\left( A\cup B \right)\].

**Complete step by step solution:**

We are given that the total number of students that are studying is 150. We are given that 62 percent of total students study English, let A be the set of students who study English so \[N\left( A \right)=0.62\times 150=93\]. Also, we are given that 68 percent of students study Maths, let B is set of students who study Maths then \[N\left( B \right)=0.68\times 150=102\].

As the total number of students is 150, \[N\left( A\cup B \right)=150\]. We need to find the number of students who study both subjects, which means we need to find \[N\left( A\cap B \right)\].

Using the set theory formula which states \[N\left( A\cap B \right)=N\left( A \right)+N\left( B \right)-N\left( A\cup B \right)\]. Substituting the values, we get

\[\begin{align}

& \Rightarrow N\left( A\cap B \right)=N\left( A \right)+N\left( B \right)-N\left( A\cup B \right) \\

& \Rightarrow N\left( A\cap B \right)=93+102-150 \\

& \Rightarrow N\left( A\cap B \right)=45 \\

\end{align}\]

**Hence, the number of students studying both subjects is 45.**

**Note:**Using the set theory properties to solve these types of problems makes it easier to solve. One should know the properties and formulas of sets to use it. We used the formula \[N\left( A\cap B \right)=N\left( A \right)+N\left( B \right)-N\left( A\cup B \right)\]. We can also use it for more than two sets.

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