Question

Length of rectangular blackboard is 8m more than its breadth. If its length is increased by 7m and its breadth is decreased by 4m, its area remains unchanged. What is the length of the blackboard?

Answer 1

Let the length be L and breadth be B

Let the length be L and breadth be BGiven, L=8+B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60B=20 cm

Let the length be L and breadth be BGiven, L=8+BArea of the board, L×B=(8+B)×B Given, If its length is increased by 7 cm and breadth is decreased by 4 cm, its area remains unchanged.⇒(L+7)(B−4)=L×BBut L=8+B⇒(8+B+7)(B−4)=(8+B)×B⇒(15+B)(B−4)=(8+B)×B⇒15B−60+B 2 −4B=8B+B 2 3B=60B=20 cmSo, Length =20+8=28 cm