Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be

• ans:87.5%
  surface areas ratio:(r1/r2)^2=100/400 i.e r1/r2=1/2;
  volumes ratio is:(r1/r2)^3=(1/2)^3=1/8
  the percentage of A compared to B=(1/8)*100=12.5%
  so the volume of A is 87.5% less than volume of B
• 13 years agoHelpfull: Yes(32) No(4)
• Surface area of a sphere=4(pi)r^2
  Since Surf. area of B=300% of surf. area of A
  => rad. of B=2.*rad. of A
  Volume of sphere=4/3(pi)r^3
  =>Vol. of B=8* Vol of A
  => Vol. of A is (1-1/8) times lower than Vol. of B
  =>Vol of A=87.5% lower than Vol of B
• 13 years agoHelpfull: Yes(10) No(3)
• ANS: K=700
  EXP
  4*PI*R^2=300*4*PI*r^2/100 + 4PI*r^2
  =>R^2=4r^2
  =>R/r=2
  (4/3)PI*R^3=(4/3)PI*r^3[K/100+1]
  =>(R/r)^3=[K/100+1]
  =>8=[K/100+1]
  =>K=700 ANS..
• 13 years agoHelpfull: Yes(3) No(13)