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