The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area ?
A. $$37\frac{1}{2}\% $$
B. $$60\% $$
C. $$75\% $$
D. $$120\% $$
Answer: Option A
Solution (By Examveda Team)
Let original length = x and original breadth = yThen, original area = xy
New length :
$$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$
Let the new breadth = z
Then,
$$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$
∴ Decrease in breadth :
$$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$
This Question Belongs to Arithmetic Ability >> Area
Let length of rectangle =100 m
breadth =100 m
Original Area =100×100=10000m
2
New length =160 m
Let new breadth =x
New Area =160x
Since area of rectangle remains same
∴10000=160x
∴x=
2
125
Decrease in breadth =(100−
2
125
)%
=37.5%