Question

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
[1] 85.5
[2] 92.5
[3] 90.5
[4] 87.5

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

OPTION - (D) 87.5