Question: When the length of a rectangular field is reduced by 5m and breadth is increased by 3m, it's area gets reduced by 9 sq.m. If the length is increased by 3m and breadth by 2m the area increases by 67 sq.m. Find the length and breadth of the room.

Solution:

Let the length of the rectangular field be 'l' m and the breadth of the rectangular field be 'b' m. Also, let the area of the rectangular field is 'lb' sq.m.

According to the question,

Condition I,
When the length of the rectangular field is reduced by 5m and breath is increased by 3m, it's area gets reduced by 9 sq.m.
$or, (l-5)(b+3) = lb -9$
$or, lb +3l -5b -15 = lb -9$
$or, lb - lb + 3l = 5b + 15 -9$
$or, 3l = 5b +6$
$or, l = \frac{5b +6}{3}$ - (i)

Condition II,
When the length is increased by 3m and breadth by 2m, the area increases by 67 sq.m.
$or, (l+3)(b+2) = lb +67$
$or, lb + 2l +3b +6 = lb +67$
$or, lb - lb +2l +3b = 67-6$
$or, 2l + 3b = 61$ - (ii)

Put value of l from equation (i) in equation (ii), we get,

$or, 2 \left ( \dfrac{5b +6}{3} \right ) + 3b = 61$

$or, \dfrac{2(5b +6)}{3} = 61 -3b$

$or, 10b +12 = 3(61-3b)$

$or, 10b +12 = 183 - 9b$

$or, 10b +9b = 183-12$

$or, 19b = 171$

$or, b = \dfrac{171}{19}$

$\therefore b = 9$

Put value of b in equation (i), we get,

$or, l = \dfrac{5×9 +6}{3}$

$or, l = \dfrac{51}{3}$

$\therefore l = 17$

So, (l,b) = (17,9)

Therefore, the required length of the rectangular field is 17m and the breadth is 9m.