The length of a rectangle is
increased by 109 and its breadth
is decreased by 10%. Then the
area of the new rectangle is :
ISSC, 2007]
a neither increased nor
decreased
by increased by 1%
i decreased by 1%
decreased by 10%
Answers
Step-by-step explanation:
Correction :-
length increased by 10% not 109
Given :-
Let the length of the rectangle be l units
Let the breadth of the rectangle be b units
Area of the original rectangle = lb sq.units
If the length is Increased by 10%then
The new length = l+10%of l
=> l+(10/100)(l)
=> l+(1/10)(l)
=>>(10l+l)/10
=> 11l/10 units
If the breadth is decreased by 10% then
The new breadth = b-10% of b
=> b -(10/100)×b
=> b - (1/10)b
=> b - (b/10)
=> (10b-b)/10
=> 9b/10 units
The area of the new rectangle
=> (11l/10×(9b/10)
=> (11l×9b)/(10×10)
=> (99lb)/100 sq.units
Area of the Original rectangle > Area of the new rectangle
decreased in the area =Original area-New area
=> lb - (99lb/100)
=> (100lb-99lb)/100
=> lb/100 sq.units
Increased Percentage in the area
=> [increased area/Original area]×100
=> [lb/100)/lb]×100
=> (lb/100lb)×100
=> 1%
Answer :-
The new area is decreased by the original area is 1%
Used formulae:-
Area of a rectangle = lb sq.units
Where, l = length ,b = breadth of a rectangle
Answer:
I THINK instead of 109 , it is 10 % .
Lemme Try : -
Let ,
Length of Rectangle is X & Breadth is Y
So , Area = X × Y
A / Q : -
NEW Length = X + 109 ,
NEW Breadth = Y - Y of 10 %
= Y - Y × 10 / 100
= Y - Y / 10
= 9Y / 10
SO , NEW Area = ( X + 109 ) × ( 9 Y / 10 )
= 9 XY / 10 + 981 Y / 10
= 0.9 XY + 98.1 Y
∆ Area
= 0.9 XY + 98.1 Y - XY
= - 0.1 XY + 98.1 Y
As we can See , NEW AREA > AREA
SO , IT'S INCREASED & There is Single Option Regarding Increasing
So , Increased by 1 %