A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 4 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
a) 800
b) 488
c) 900
d) 500

plz explain answer

• Size of big cube = 5cm
  Total volume of the big cube = 5*5*5 = 125cm^3
  
  Size of the hollow cube inside the big cube = 3cm
  Volume of the hollow space inside the big cube = 3*3*3 = 27cm^3
  
  Therefore, volume occupied by small cubes (or volume of
  thickness) = 125 - 27 = 98cm^3
  
  Size of each small cube = 1cm
  Volume of each small cube = 1*1*1 = 1cm^3
  Total number of small cubes in wall = 98 / 1 = 98
  
  In short, 98 small cubes make up the wall of the big cube.
  Each cube has 6 faces, so 98 cubes have = 98*6 faces = 588
  
  The four sides of the big cube have 100 painted faces.
  Because each big side has 25 faces of the small cubes.
  
  Therefore, total unpainted faces = 588 - 100 = 488
• 9 years agoHelpfull: Yes(20) No(3)
• 488
• 13 years agoHelpfull: Yes(16) No(6)
• 1. All 4 corner cubes(4*5) are painted on two sides, so remaining faces are 4= 4*5*4=80
  2. the remaining cubes on painted surfaces are (5-2)(5)*4, which are painted on one side, so no. of unpainted faces = 3*5*4*5=300
  3.now on top and bottom there are (5-2)(5-2) cubes unpainted, so =3*3*2*6=108
  total 108+80+300=488 :)
• 13 years agoHelpfull: Yes(11) No(3)
• plz explain your answer
• 13 years agoHelpfull: Yes(4) No(10)
• 588-100=488
  
• 9 years agoHelpfull: Yes(1) No(1)