A) 2 : 1
B) 1 : 2
C) 1 : 4
D) 4 : 1
Correct Answer: C
Solution :
Surface area of the inner cubical box \[=4(4.5\times 4)+4\times 4=72+16=88\,c{{m}^{2}}\] \[\therefore \] Total surface area \[=125+88+(18\times 0.5)=222\,c{{m}^{2}}\] Volume of solid cube \[={{(4)}^{3}}=64\,c{{m}^{3}}\] Volume of recast cube \[={{(1)}^{3}}=1\,c{{m}^{3}}\] \[\therefore \] Total surface area of cube : total surface area of recast cube = x : y \[\Rightarrow \] \[x:y=6{{(4)}^{2}}:6{{(1)}^{2}}\times 64=1:4\]You need to login to perform this action.
You will be redirected in
3 sec