A man sold 12 candies in $10 had loss of b% then again sold 12 candies at $12 had profit of b% find the value of b.

• 12 candy sold at 10$ incurred b% loss, 12 candy sold at 12$ made a profit of b%
  
  let cost price be $x for 12 candy
  
  loss equation : x-10 = b% loss i.e., bx/100
  profit equation : 12-x = b% profit i.e., bx/100
  
  x-10=12-x
  x=11
  
  putting value of x in any equation gives b = 100/11 = 9.09%
  
  
• 9 years agoHelpfull: Yes(12) No(0)
• cp should be equal in this case
  so, when sp=$10 loss =b%
  CP=(100/(100-loss%) )*sp
  =>CP=(100/(100-b) )*10 -------------------- 1
  
  in 2nd type he sold at $12 , SP=12 and profit=b%
  then
  CP=(100/100+b)*12 ----------------2
  
  we equate 1 and 2
  12/100+b=10/100-b
  =>b=100/11
  => b=9.09%(Ans)
  
• 9 years agoHelpfull: Yes(2) No(0)
• 9.09 is the ans
• 9 years agoHelpfull: Yes(1) No(0)
• let to real cost of candies is =x
  b% loss so .. b% of x= (x*b)/100
  so sold price is : x-[ (x*b) / 100 ] =10 ----> as question say .............................(1)
  now b% profit.. b%of x= (x*b)/100
  so new sold price = x+[ (x*b)/100 ]=12 ......................................................(2)
  comparing these equation ...
  x-[ (x*b) / 100 ] / x+[ (x*b)/100 ]=10/12
  
  b=9.09 ans :)
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• (CP-10)*100/CP=b (LOSS%)
  (12-CP)*100/CP=b(profit %)
  
  Solving 1 and 2 we get CP=11
  therefore b=1/11=9.09%
• 5 years agoHelpfull: Yes(0) No(0)
• Here 12 candies is immaterial.
  Loss % =
  
  
  So Here SP = 10 and loss% = b%
  
  
  
  
  In the second case he got a profit of b%
  So Profit % =
  
  
  So Here SP = 12 and profit% = b%
  
  
  
  
  Solving 1 and 2 we get b = 1/11 or 9.09%
• 5 years agoHelpfull: Yes(0) No(0)