in a town,the population was 8000.in one year,male polpulatin increased by 10% and female population increased by 8% but the total population increased by 9%.the number of males in the town was:

• M+F= 8000 ....(1)
  1.1M+ 1.08F = 1.09(8000) ...(2)
  
  multiply 1 eqtn with 1.08 so that 1.08 F factor will be removed from eqtn and only M factor will remain nd solved male number.
  eqtn 2 - eqtn (1)*1.08===>> 1.02M = (8720-8640) ===> M = 4000
  
  thanks
  
• 11 years agoHelpfull: Yes(4) No(2)
• Answer
  4000
  No of males=A
  No of females=B
  totalpopulation=8000
  A+B=8000..........(1)
  ((10A/100)+A)+((8B/100)+B)=(9(A+B)/100)+A+B
  Solving both sides, we get
  A-B=0.............(2)
  Now solving eqn 1 and 2,
  we get,
  A=B=4000.
  therefore no of males=4000.
• 11 years agoHelpfull: Yes(4) No(2)
• Could you please explain the solution step by step
• 8 years agoHelpfull: Yes(0) No(0)