Suppose you have a currency called miso in three denominations, 1 miso, 10misos and 50 misos. In how many ways can you pay a bill of 107 misos?
Option
A.17
B.16
C.18
D.15
E.19

• Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively.
  x+10y+50z=107
  
  Now the possible values of z could be 0, 1 and 2.
  
  For z=0: x+10y=107
  
  Number of integral pairs of values of x and y that satisfy the equation:
  x+10y=107 will be 11.
  
  These values of x and y in that order are:
  (7,10);(17,9);(27,8)…(107,0)
  
  For z=1: x+10y=57
  
  Number of integral pairs of values of x and y that satisfy the equation:
  x+10y=57 will be 6.
  
  These values of x and y in that order are:
  (7,5);(17,4);(27,3);(37,2);(47,1) and (57,0)
  
  For z=2: x+10y=7
  
  There is only one integer value of x and y that satisfies the equation:
  x+10y=7 in that order is (7,0)
  
  Therefore total number of ways in which you can pay a bill of 107 Misos:
  =11+6+1= 18
  
  Ans : (C)
• 10 years agoHelpfull: Yes(94) No(1)
• 18 is answer
  1.)1+1+1+......107 times=107
  2.)10+1+1......97 times=107
  3.)10+10+1+1......87 times=107
  ....
  ........
  11.)10+10+.....10 times + 1+1+...7 times=107
  12.)50+1+1+....57 times=107
  13.)50+10+1+1....47 times=107
  ....
  .......
  17.)50+10+10+...5 times + 1+1+....7 times=107
  18.)50+50+1+1....7 times=107
  so 18 ways
• 10 years agoHelpfull: Yes(7) No(0)
• ans is c
  18
• 10 years agoHelpfull: Yes(2) No(0)
• A)17 is right answer because
  1+1+1+...........=107
  10+1+1...........=107
  ....
  ........
  10+10+..........=107
  50+10+1+1.......=107
  50+50+1+1.......=107
  so 17 ways
• 10 years agoHelpfull: Yes(0) No(7)
• no of ways=maximum usage of each miso/2*minimum usage of each miso=(107 of 1 miso*10 of 10 miso*2 of 50 miso)/7 of 1 miso*1 0f 10 miso=2140/2*7=2140/14=15
• 9 years agoHelpfull: Yes(0) No(1)
• c
  obvious one can easily check.
  
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 18
• 2 years agoHelpfull: Yes(0) No(0)