a shopkeeper sold a certain no. of toys all ay a cetain price.the no. of toys that he sold is a three digit no.in which tens digit and units digit are same and are non-zero,and the price of each toy is a two-digit no. when expressed in rupees.by mistake he reversed the digits of both,the no. of items sold and the price of each item.in doing so,he found that his stock account at the end of the day showed 792 items more than what it actually was.if the faulty calculations show a total sale of rupees 5117,what was the actual selling price of each toy?pl. provide sol.

1) 43
2) 37
3) 75
4) 34

• number of toys= 100a+10b+b= 100a+11b
  Price= 10c+d
  Reverse number= 110b+a
  Reverse price= 10d+c
  Due to the mistake, stock showed 792 items more ie. number of items sold 792 less
  
  100a+11b - (110b+a) = 792
  a-b = 8
  Since, the digits are non zero it can only be 9,1. Thus the number is 911.
  
  With faulty calculation total sales =
  5117 = 119*x
  x = 43
  Thus actual price = 34
• 8 years agoHelpfull: Yes(0) No(0)
• Let the three 3 digit number be xyz.
  Such that 100x+10y+z-(100z+10y+x) = 792
  => 100x - x + z - 100z = 792.
  => 99(x-z) = 792.
  => (x-z) = 792/99 = 8.
  By mistake, Stock showed 792 items more ie. number of items sold 792 less.
  As mentioned 10's & 1's are same and non-zero i.e 9 & 1 only satisfy that So, number = 911.
  Reversing it - (119), Faulty Calculations = 5117 = 119*x
  x = 5117/119
  x = 43.
  Actual Selling price = 34.
• 3 years agoHelpfull: Yes(0) No(0)