**Question:**

Length of rectangle exceeds its breadth by 7 cm. If the length & breadth is decreased by 4 cm & 3 cm, then the area of the new rectangle is same as the area of the original. What will be the perimeter of the original rectangle? Plzz explain how to crack..

let b=X

thn l=x+7

A=x2+7x

second case

b=x-3

L=x+7-3

A=(x-3)(x-4)

Now A1=A2

x2+7x=x2+4x-3x+12

x=2

Original

b=2cm

l =9cm

P=2(11)=22cm

Even I think question is incorrect. Found the question in previous IBPS SO paper. Answer happends to be 50. I don't have any idea how the answer is 50...

Respectively lagaana bhool gya chore

Wrong question

Wrng ques

Question is wrong...length goes into minus...__lol__

@ Kulajeet bro, I think the qstn is wrng , because if both the length and breadth is reduced, area will not remain the same. It will be reduced.

Qstn should have been like this

1.if length is decreased by 4 & breadth is incrsd by 3.

Ans. l=16,b=9,p=50.

OR

2. If length is incrsd by 4 & breadth is decreased by 3.

Ans. l=40,b=33, p=146 .

Let the width is x and then length is x+7 then the area A=x(x+7)

Then reducing length and width by 4 and 3 i.e

(x+7)-4 and x-3 the new area becomes A=(x+3)(x-3)

Thn solve x(x+7)=(x+3)(x-3)

:__D__