Question

3) Students are made to stand in rows. If one student is extra in a row there
would be 2 rows less. If one student is less in a row there would be 3 rows
more. Find the number of students in the class​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

Number of rows = x

and

Number of students in each row = y

So,

Total number of students = xy------ (1)

According to first condition

If one student is extra in a row there would be 2 rows less.

Number of rows = x - 2

Number of students in each row = y + 1

Total number of students = ( x - 2 ) × ( y + 1 )----(2)

From equation (1) and (2), we concluded that

$\rm :\longmapsto\:(x - 2)(y + 1) = xy$

$\rm :\longmapsto\:xy + x - 2y - 2 = xy$

$\rm :\longmapsto\:x - 2y - 2 = 0$

$\bf :\longmapsto\:x - 2y = 2 - - - - (3)$

According to second condition

If one student is less in a row there would be 3 rows

more.

Number of rows = x + 3

Number of students in each row = y - 1

Total number of students = ( x + 3 ) × ( y - 1 )----(4)

From equation (1) and (4), we concluded that

$\rm :\longmapsto\:(x + 3)(y - 1) = xy$

$\rm :\longmapsto\:xy - x + 3y - 3 = xy$

$\rm :\longmapsto\:- x + 3y - 3 = 0$

$\rm :\longmapsto\:- x + 3y = 3 - - - - (5)$

On adding equation (3) and (5), we get

$\rm :\longmapsto\:x - 2y - x + 3y = 2 + 3$

$\bf :\longmapsto\:y = 5$

On substituting y = 5, in equation (3), we get

$\rm :\longmapsto\:x - 2 \times 5 = 2$

$\rm :\longmapsto\:x - 10 = 2$

$\rm :\longmapsto\:x = 2 + 10$

$\bf :\longmapsto\:x = 12$

Hence,

Number of rows = 12

Number of students in each row = 5

Total number of students = 12 × 5 = 60

Answer 2

• Let  the original number of rows = x

• The original number of students in each row =y

So,

Total number of students in class = Number of students ×  Number of rows = xy

According to problem,

Case-1:

$\pink{ \boxed{\boxed{\begin{array}{cc}\bf\:Total \: no \: of \: students \: \\ \bf = > (y + 1)(x - 2) = xy \\ \\ \bf = > xy - 2y + x - 2= xy \\ \\ \bf = > x - 2y = 2 \: \: \: \: - - --(1)\end{array}}}}$

Case-2:

$\pink{ \boxed{\boxed{\begin{array}{cc}\bf \:Total \: no \: of \: students \: \\ = > \bf(y - 1)(x + 3) = xy \\ \\ \bf = > xy + 3y - x - 3 = xy \\ \\ \bf = > x - 3y = - 3 \: \: \: \: \: - - - -(2)\end{array}}}}$

$\blue{ \boxed{\boxed{\begin{array}{cc}   \bf \:Subtracting \: eqn.(1) \: from \: eqn.(2) \: \\ \bf \: we \: will \: get \: \\ \bf \: −y \: = \: −5 \\ = > \bf \: y=5 \\ \\ \bf \: Putting \: the \: value \: of \: y \: in \: eqn.(1) \\ \\ \bf \: x−2(5)=2 \\ \\ \bf \: x=12 \\ \\ \bf \: Hence \: we \: get \: x=12 \: \: and \: y=5 \\ \\ \bf \: Total \: no. \: of \: students  \: =xy \: =60 \end{array}}}}$