a trader bought a number of articles each costing the same for a total of rs750.he sold each of the articles at rs42. with the total return he could buy 10 more articles than before.what are the no. of articles the trader bought orginally?

• ans 25
  
  let the no. of article be x and price of each article is p
  xp=750
  x=750/p
  
  now 42x = 10p+xp
  42x=750+10p
  42*750/p = 750+10p
  a quardatic equation is obtained
  
  10p^2+750p-3150 = 0
  (p-30)(p+105)=0
  p=30
  x=750/p
  x=25
  
  ans 25
• 12 years agoHelpfull: Yes(12) No(0)
• sorry by mistake i have typed sol of above question.
  
• 12 years agoHelpfull: Yes(1) No(1)
• let the sum lent at 10% be rs x so at 15% it will be (50,000-x)
  then using s.i = (p*r*t)/100
  (x*10*1) (50,000-x)*15*1
  -------- + --------------- = 7000
  100 100
  
  calculate x, ans is then x= 10,000 at 10%
  and 40,000 at 15%
• 12 years agoHelpfull: Yes(0) No(3)