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Hint – In this question use the property that the number of students in a class is equal to multiplication of number of rows and number of students in each row ,also the number of students in a class remains the same.

Let number of rows $ = x$

And number of students in each row $ = y$

So, the total number of students in class = number of rows $ \times $ number of students in each row

Therefore total number of students in class $ = xy$.

Case 1

If one student is extra in a row i.e.$\left( {y + 1} \right)$, there would be two rows less i.e. $\left( {x - 2} \right)$.

But the total number of students in the class remains the same.

$

\Rightarrow xy = \left( {x - 2} \right)\left( {y + 1} \right) \\

\Rightarrow xy = xy + x - 2y - 2 \\

\Rightarrow x - 2y - 2 = 0.................\left( 1 \right) \\

$

Case 2

If one student is less in a row i.e. $\left( {y - 1} \right)$ there would be 3 rows more i.e. $\left( {x + 3} \right)$.

But the total number of students in the class remains the same.

$

\Rightarrow xy = \left( {x + 3} \right)\left( {y - 1} \right) \\

\Rightarrow xy = xy - x + 3y - 3 \\

\Rightarrow x - 3y + 3 = 0.................\left( 2 \right) \\

$

Now subtract equation (1) from equation (2)

$

x - 3y + 3 - x + 2y + 2 = 0 \\

- y + 5 = 0 \\

\Rightarrow y = 5 \\

$

Therefore from equation (1)

$

x - 2y - 2 = 0 \\

x - 2\left( 5 \right) - 2 = 0 \\

x = 10 + 2 = 12 \\

$

Therefore total number of students in the class

$ = xy = \left( 5 \right)\left( {12} \right) = 60$

So, there are 60 students in the class.

Note – In such types of questions first construct the equations according to given conditions as above then simplify these equations and calculate the values of number of rows and number of students in a particular row, then multiply these two values we will get the required number of students in the class.

Let number of rows $ = x$

And number of students in each row $ = y$

So, the total number of students in class = number of rows $ \times $ number of students in each row

Therefore total number of students in class $ = xy$.

Case 1

If one student is extra in a row i.e.$\left( {y + 1} \right)$, there would be two rows less i.e. $\left( {x - 2} \right)$.

But the total number of students in the class remains the same.

$

\Rightarrow xy = \left( {x - 2} \right)\left( {y + 1} \right) \\

\Rightarrow xy = xy + x - 2y - 2 \\

\Rightarrow x - 2y - 2 = 0.................\left( 1 \right) \\

$

Case 2

If one student is less in a row i.e. $\left( {y - 1} \right)$ there would be 3 rows more i.e. $\left( {x + 3} \right)$.

But the total number of students in the class remains the same.

$

\Rightarrow xy = \left( {x + 3} \right)\left( {y - 1} \right) \\

\Rightarrow xy = xy - x + 3y - 3 \\

\Rightarrow x - 3y + 3 = 0.................\left( 2 \right) \\

$

Now subtract equation (1) from equation (2)

$

x - 3y + 3 - x + 2y + 2 = 0 \\

- y + 5 = 0 \\

\Rightarrow y = 5 \\

$

Therefore from equation (1)

$

x - 2y - 2 = 0 \\

x - 2\left( 5 \right) - 2 = 0 \\

x = 10 + 2 = 12 \\

$

Therefore total number of students in the class

$ = xy = \left( 5 \right)\left( {12} \right) = 60$

So, there are 60 students in the class.

Note – In such types of questions first construct the equations according to given conditions as above then simplify these equations and calculate the values of number of rows and number of students in a particular row, then multiply these two values we will get the required number of students in the class.

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