The rupee/coin changing machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100 rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10 hundred rupee notes when you put a one rupee coin!
Sivaji, after being ruined by his rivals in business is left with a one rupee coin and discovers the flaw in the machine by accident. By using the machine repeatedly, which of the following amounts is a valid amount that Sivaji can have when he gets tired and stops at some stage (assume that the machine has an infinite supply of notes and coins):
A. 26975
B. 53947
C. 18980
D. 33966

• initially sivaji had only one coin so he earns only 1000/-
  to earn more he must convert one 100 note to 10 notes and then one ten note to ten coins
  now he has Rs.990 and 10 coins
  after converting 10 coins he has 10990
  after converting another 10 coins he has 20980
  after converting another 10 coins he has 30970
  after converting another 10 coins he has 40960
  after converting another 10 coins he has 50950
  after converting another 10 rupee note to coins and using only 3
  he has seven coins and 53940
  so sum is 53940+7=53947
• 11 years agoHelpfull: Yes(55) No(10)
• first he gives 1rupee coin and gets 1000rs
  then 1000-1=999.with that 1rupee he again gets 1000
  it gets repeated as 999
  then amount in his hand will be 999
  +999
  +999...so on
  and when he gets tired he stops and then he has multiples of 999+another one rupee coin..max ans given is 53946(mul of 999)+1=53947.
  
• 11 years agoHelpfull: Yes(18) No(7)
• ans will be multiple of 999 so ans is 33966
• 11 years agoHelpfull: Yes(8) No(13)
• For every one rupee his gain is 999. Hence ans =multiple of 999+1
  By this 53947 which is your desire ans.
• 9 years agoHelpfull: Yes(5) No(1)
• why u people are using 54 here can any pls tell me clearly
• 9 years agoHelpfull: Yes(4) No(2)
• The process works like this:
  Rs.1 Coin ⇒ 10 × 100 = Rs.1000
  Rs.100 ⇒ 10 × 10
  Rs.10 ⇒ 1 × 10
  Sivaji gets more money when he inserts a rupee coin only. For each rupee coin he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000. So he has 1999 with him. Now if he inserts another coin, he has 1998 + 1000 = 2998.
  Now each of these numbers are in the form of 999n + 1. So option B can be written as 54 × 999 + 1.
  
• 8 years agoHelpfull: Yes(1) No(1)
• 32968
  is the open seesame solution
• 9 years agoHelpfull: Yes(0) No(8)
• 33966 will be the answer...
  Soln:
  
  The only point of increment is when he puts 1 re & gets 10 hundred rupee i.e. 1000 rs. Thus he gains
  1000 – 1 = 999. So the amount should be a multiple of 999. Only 33966 is a multiple of 999.
  
• 9 years agoHelpfull: Yes(0) No(6)
• initially sivaji had only one rupee coin,hence he put one rupee coin to machine and got 1->1000.he then change 1000 as 990rupees & 10 rupees as coins .we can find the ans from the gn ans
  now,
  10000(10 coins)+8000(8coins)+982==18982(but option is 18980),so this s wrong
  20000(20 coins)+6000(6coins)+974==26974(but option is 26975),so this s wrong
  30000(30 coins)+3000(3coins)+967==33967(but option is 33966),so this s wrong
  50000(50 coins)+3000(3coins)+947==53947( option is also 53947),so this s correct
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• The general term for the amount left would be 999n+1000 after each operation of the machine, hence the answer wold be 53947 because (53947-1000)%999=0 (multiple of 999)
  Ans: 53947
• 9 years agoHelpfull: Yes(0) No(0)
• Answer: B
  Explanation:
  The process works like this:
  Rs.1 Coin ⇒ 10 × 100 = Rs.1000
  Rs.100 ⇒ 10 × 10
  Rs.10 ⇒ 1 × 10
  Sivaji gets more money when he inserts a rupee coin only. For each rupee coin he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000. So he has 1999 with him. Now if he inserts another coin, he has 1998 + 1000 = 2998.
  Now each of these numbers are in the form of 999n + 1. So option B can be written as 54 × 999 + 1.
• 9 years agoHelpfull: Yes(0) No(0)
• ion:
  The process works like this:
  Rs.1 Coin ⇒ 10 × 100 = Rs.1000
  Rs.100 ⇒ 10 × 10
  Rs.10 ⇒ 1 × 10
  Sivaji gets more money when he inserts a rupee coin only. For each rupee coin he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000. So he has 1999 with him. Now if he inserts another coin, he has 1998 + 1000 = 2998.
  Now each of these numbers are in the form of 999n + 1. So option B can be written as 54 × 999 + 1.
  
• 9 years agoHelpfull: Yes(0) No(0)
• first he gives 1rupee coin and gets 1000rs
  then 1000-1=999.with that 1rupee he again gets 1000
  it gets repeated as 999
  then amount in his hand will be 999
  +999
  +999...so on
  and when he gets tired he stops and then he has multiples of 999+another one rupee coin..max ans given is 53946(mul of 999)+1=53947.
• 5 years agoHelpfull: Yes(0) No(0)
• Hi guys . This is simple problem , when one uses his logical thinking .
  If you failed to get the answer , dont worry . Here is the explanation for the solution to develop how to think .
  
  Given :
  
  100 ₹ note ---------> 10 * 10₹ note
  10₹ note ------------> 10 * 1₹ coin.
  1₹ coin --------------> 10* 100 ₹ note
  
  Question: what will be the possible answer from the following ?
  
  At first sivaji has 1₹ .
  So , he takes 10 * 100₹ note .
  and then using that 100₹ note , he takes 10* 10 ₹ notes .
  which finallises that he has 10 * 10 * 10 ₹ notes .
  
  and then using that 10 ₹ notes , he takes 10 * 1₹ coin .
  and it confirms that he has 10 * 10 * 10 * 1₹ coin = 1000 * 1₹ coin .
  
  
  
  Initially, Sivaji has 1* 1₹ coin and he now has 1000* 1₹ coin .
  which confirms that at each attempt he gain 1000 ₹ .
  
  
  
  In that case , from the 1000₹ , if he put 1₹ coin in machine and he will be having 999₹.
  
  
  After several steps ( as above) , finally he will gain another 1000 ₹ .
  And now he will be having 1999 .
  
  He will put 1₹ from that 1999 to get another 1000₹ ...
  By that time he will be having 1998₹
  and the gains will be increasing on........
  
  
  
  
  we can write the above in this form .....
  
  999 -> 1998 -> 2997 ............
  
  Hey ? Did you notice one thing ?
  
  Is the series looks like AP ?
  
  yeah !
  
  
  This is the series of the money he is having while he put the 1₹ from the previous gain in the machine .
  
  The question says that he became tierd and stops at some stage .
  This mean , he became tierd and stops putting 1₹ from the previous gain in the machine .
  
  I'M SAYING IT AGAIN .....
  
  
  999 -> 1998 -> 2997 ............
  This is the series of the money he is having while he put the 1₹ from the previous gain in the machine.
  
  So the possible answer would be something from this series + 1 ?
  
  The general form of this series is 999n .
  So the answer will be 999n+1 .
  
  Ryt ?
  
  check the options , which acccept the equation 999n+1.
  
  
  ( Which answer has respective n in whole number )
  
  By checking , we can find the answer in OPTION B .
  
  Cheers .
• 2 years agoHelpfull: Yes(0) No(0)