A circular road runs around a circular garden. If the sum of the circumferences of the inner and outer
circles is 88 metres and the radius of the inner circle is one-third of the outer circle, then the width of
the road is :
(A) 4 metres (B) 5 metres
(C) 6 metres (D) 7 metres

• let radius of inner circle=r1
  radius of outer circle=r2
  Also r1=1/3*r2
  3*r1=r2
  circumference of inner circle=2*pi*r1
  circumference of outer circle=2*pi*r2
  =6*pi*r1 (since r2=3*r1)
  2*pi*r1 + 6*pi*r1 =88 (as per question)
  r1=3.5 m
  r2=3*3.5
  =10.5 m
  width of road =10.5 - 3.5 =7 m Ans.
  
• 12 years agoHelpfull: Yes(27) No(0)
• D)7m
  
  let radius of inner circle is r1
  radius of outer circle is r2
  Circumference of circle=2*pi*r
  
  2*pi*r1 + 2*pi*r2=88
  2*pi(r1+r2)=88
  pi*(r1*r2)=44
  
  r1=1/3*r2
  
  pi*(r2/3 + r2)=44
  pi=22/7
  r2=10.5
  width of road=r2-r1=10.5-3.5=7m
  
• 12 years agoHelpfull: Yes(10) No(0)
• ans-> 7m
  
  2*pi*(1/3)*R + 2*pi*R = 88 ( where 1/3R = r)
  
  solving above we get R=10.5m
  r= 3.5m
• 8 years agoHelpfull: Yes(0) No(0)