[email protected] wrote: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
A. 85.5
B. 92.5
C. 90.5
D. 87.5
formula surafce area =4pir^2
volume 4/3pi r^3
let the radius of sphere A be r1
and radius of sphere B be r2
now given surface area of B is 300% higher than the surface area of A. The
(4ir2^2-4pir1^2)/4pir1^2)*100=300
r2^2-r1^2/ri^2=3
r2^2=4r1^2
r2=+ - 2r1 equation 1(we will take r2=2r1 as r can not be -ve)
volume of A=4/3pi r1^3
volume of B=4/3pi r2^3
4/3pi r2^3=4/3pi r1^3 +(k/100*4/3pi r2^3) (cancelling 4/3 pi both side)
r2^3=r1^3+kr2^3/100 multiplying throughout by 100
100 r2^3=100r1^3+kr2^3
r2^3(100-k)= 100r1^3
(2r1)^3(100 -k)= 100r1^3 (r2=+ - 2r1 equation 1)
8(100-k)=100
700=8k
k=87.5
go step by step
