A man has 1 million cow at the beginning of Year 1997. The numbers grow by x% (x > 0) during the year. A famine hits his village in the next year and many of his cow die. The cow population decreases by y% during 1998 and at the beginning of 1999 the man finds that he is left with 1 million cow. Which of the following is correct?
(a)x>y(b)y>x(c)x=y(d)can't determine

• in 1997 cows starting = 10^6;
  after x% growth , population = 10^6+ (x*10^4) = 10^4(100+x);
  after y% decrease , population in 1998 = 10^4(100+x)-10^4(100+x)*y/100;
  population becomes = 10^4(100+x)- 10^2(x+100)y = 10^6(in 1999)
  taking 10^2 common from LHS.
  10^2[10^2(100+x)-(x+100)y]=10^6
  10^2(100+x)-(x+100)y=10^4;
  10^4+100x-xy-100y=10^4;
  100x-100y=10^4-10^4+xy;
  100x-100y=xy;
  dividing both sides by xy, and taking 100 common lhs;
  100(1/y-1/x)=1
  1/y-1/x = 1/100;
  this means 1/y>1/x;
  this implies x>y;
  so a is the answer;
  
• 11 years agoHelpfull: Yes(1) No(0)
• Let us assume the value of x to be 10%.
  Therefore, the number of cows in the herd at the beginning of year 1998 (end of 1997) will be 1 million + 10% of 1 million = 1.1 million
  
  In 1998, the numbers decrease by y% and at the end of the year the number cows in the herd = 1 million.
  i.e., 0.1 million cows have died in 1998.
  
  In terms of the percentage of the number of cows alive at the beginning of 1998, it will be (0.1/1.1)*100 % = 9.09%.
  From the above illustration it is clear that x > y.
  
• 11 years agoHelpfull: Yes(1) No(0)