if length, breadth and height of a cuboid is increased by x%,y% and z% respectively then its volume is increased by

Anwesha Barman
38 Points
7 years ago
Actual length = l
New length = l + x/100 = l’ = (100+x) l /100
New breadth =  b’ = (100+y) b /100
Actual height = h
New height = h’ = (100+z) h /100
Actual volume = lbh
New volume = l’b’h’
={(100+x) x (100+y) b (100+z) h} 100 x100 x 100
= {1000000 + 10000y + 10000x +100xy +10000z + 100yz + 100xz +xyz} lbh / 1003
Therefore, volume increased = lbh –  {1000000 + 10000y + 10000x +100xy +10000z + 100yz + 100xz +xyz} lbh / 1003
= 1000000lbh –  [{1000000 + 10000y + 10000x +100xy +10000z + 100yz + 100xz +xyz} lbh ]/ 1003
= lbh [ 1000000 – 1000000 + 10000y + 10000x +100xy +10000z + 100yz + 100xz +xyz] / 1003
= lbh [ 10000y + 10000x +100xy +10000z + 100yz + 100xz +xyz] / 1003
= lbh[ (x/100) +(y/100) +(z/100) +(xy+yz+zx)/10000 +xyz/1003]
Volume increase in % = {(lbh [ (x/100) +(y/100) +(z/100) +(xy+yz+zx)/10000 +xyz/103] ) / lbh } 100
=x +y +z +(xy+yz+zx)/100 +xyz/100%
:) Hope it helps.....