population of city increases by 10 and 15 successively in two years.find the average annual percentage increase in two years

• starting population as 100.
  when it increases,then it increement by 10.ans=110,then 10% of 100=10
  Sequence-->it increements by 15.so 15% of 110=16.5
  110+16.5=126.5
  average=126.5-100=26.5
  
  
  
• 10 years agoHelpfull: Yes(24) No(2)
• let the initial population be 100.
  when it inc. first by 10, then 10% of 100 is 10 therefore it becomes 110.
  now since successively inc. by 15 then 15% of 110 is:16.5
  therefore new amount becomes 110 + 16.5= 126.5
  now initial was 100 and now we have 126.5
  therefore a diff. of 26.5%
  
• 10 years agoHelpfull: Yes(4) No(3)
• 15% of 110=16.5
  110+16.5=126.5
  average=126.5-100=26.5
  
  26.5/2=13.25 per year
  the average annual percentage increase in two years
• 10 years agoHelpfull: Yes(2) No(0)
• population of a city for two years is a successive increase.
  so for successive increase formula is a+b+ab/100
  first year=a% and second year=b%
• 10 years agoHelpfull: Yes(1) No(0)
• we can solve it by successive % change :
  10+15+((10*15)/100)=26.5%
• 10 years agoHelpfull: Yes(0) No(0)
• since calculator is allow in tcs its not a big deal to solve dis
  10% i.e. 100-110 --- (1)
  15% i.e. 100-115 ----(2)
  multiply (1) amd (2)
  10000-12650
  it is 25.5%
• 10 years agoHelpfull: Yes(0) No(0)