Question

- if length and breadth of a cuboid is increased by 20%, then by how much percent the height should be reduced to keep the volume same?

Answer 1

Let the length be l, breadth be b and height be h
So,
Volume of cuboid = lbh
Now,
As per question the length and breadth are increased by 20%
So, new length =
$l + \frac{20}{100} l = l + \frac{1}{5} l = \frac{6}{5} l$
Similarly,
New breadth =
$b + \frac{20}{100} b = \frac{6}{5} b$
Now,
New length *new breath * reduced height = previous volume.
So,
$\frac{6}{5} l \times \frac{6}{5} b \times height= lbh \\ \frac{36}{25} lb \times height = lbh \\ height = \frac{25}{36lb} \times lbh = \frac{25}{36} h$
Therefore, new height should be 25/36 of previous height.
Percentage of reduction = difference /initial *100
So,
Change in height = h- 25/36h = 9/36 h = 1/4 h
So,
Percentage =
$( \frac{ \frac{1}{4} h}{h} \times 100)\% = (\frac{1}{4} \times 100)\% = 25\%$

Therefore, height should be reduced by 25%

Hope this will be helping you ✌️