Math, asked by pokelife128, 3 months ago

if the length and breadth of a rectangle are increased by 10% then the percentage of area increases by

Answers

Answered by manpreetkaur77
3

Step-by-step explanation:

If the length of a rectangle is increased by 10%, by what percentage should the breadth be decreased so the area remains the same?

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K S Narayanan

Answered 3 years ago

For Aptitude / competitive exams :

Increase in length by 10% or 1/10th ,will require 1/(10 + 1 ) or 9 1/11% decrease in width to kep area constant.

This is a rule worth remembering.

Alternate quick method :

Net percentage change if two variables are multiplied together is given by :

A + B + AB/100

Here, A= 10% , let B = -x, net change = 0

10 +(-x) + (10 * -x)/100 = 0

=> 10 - x - x/10 = 0

10 = 11x/10

X = 100/11 = 9 1/11%

Example :

Original length = 120

Original width = 100

Original area = 120 x 100 = 12000

Let length increase by 10% or 1/10th

New length = 120 + 12 = 132

New width = 12000/132 = 90.9090

Hence, width decreases by :

(100 - 90.9090)/100 * 100 = 9.0909% or

9 1/11% or 1/11 part.

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Purav Parekh

Updated 3 years ago

By 9.09

Let us assume the length of the rectangle is x and breadth is y . This makes the area of the rectangle x⋅y . Now, as mentioned, you have increased the length by 10 which makes it 1.1⋅x .

You want to maintain the area at x⋅y .

Length is 1.1⋅x .

So the breadth has to be 11.1 so that it maintains the area

=> (1.1⋅x)⋅(11.1⋅y)=x⋅y

11.1 is 0.909090 ...

so the breadth is reduced by 1−0.909090 ...) which is 0.090909 ...

this is percentage terms is 9.0909 ...

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