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The rupee/coin changing machine at a bank had a flaw. It gives 10 ten rupees note 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 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 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)
Read Solution (Total 13)
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- 53947 ( As a ten rupee note will give exactly 10 one rupee coins and a hundred rupee note will be exactly 10 ten rupee notes , only thing which multiplies money is the one rupee coin --> gives thousand at once . Means the Magnification will be in terms of 1000 . And any amount at a certain time will be of the form ((1000-X) + (X*1000) ) , means once you get thousand , you can take out a certain amount and multiply it with 1000 ...and so on .... it will be recursive !! Here 53947 = 947 + (1000-947)*1000 , once shiva gets 1000 he keeps 947 apart and inserts the 53 coins in the changing machine ... other options are not possible
- 11 years agoHelpfull: Yes(7) No(1)
- DUMMY QUESTION
- 11 years agoHelpfull: Yes(6) No(4)
- options are required
- 11 years agoHelpfull: Yes(5) No(4)
- 1 rupee coin => 10*100 rupees notes
10*100 rupees notes => 100*10 rupees notes
100*10 rupees notes => 1000*1 rupees coins
1000*1 rupee coin => 10000*100 rupee notes
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this will go on forever...
there is no answer to this question - 11 years agoHelpfull: Yes(4) No(6)
- 1 rupee coin => 10*100 rupees notes
10*100 rupees notes => 100*10 rupees notes
100*10 rupees notes => 1000*1 rupees coins
1000*1 rupee coin => 10000*100 rupee notes
now he will have 10^3*1rupee notes.from this we can say that,he can earn 10^3*1 rupees for 1 rupee...
so he can earn=10^3*10^3*10^3...*10^3*1rupees=((10^3)^n)*1 rupees where n=1,2,3,4,..... - 11 years agoHelpfull: Yes(3) No(4)
- Answer formula is>>>>
1st he will get 1000(10*100) rs from 1 rs coin
then he has 1000 rs with him,
now what ever exchange he does,,
the sum of last 3 digit and 1st 2 digit will be 1000,
[[ as example:--
if he exchanges 1 rs with 1000 rs. he has total 999+1000=1999 rs.
notice 1+999=1000;
if he exchanges 33 nos of 1 rs coin then he total has 33000+967=33967
notice 33+967=1000;
SO ANSWER WILL BE 53947 - 9 years agoHelpfull: Yes(2) No(0)
- 10 times one rupee coin to get 1 10 rupee note then ten times one rupee note to get 1 100 rupee note then ten times hundred rupee note to get one rupee coin
- 11 years agoHelpfull: Yes(1) No(2)
- i thnk ans is 1
- 11 years agoHelpfull: Yes(1) No(5)
- no option,, dummy question i guess so
- 11 years agoHelpfull: Yes(1) No(0)
- guys check the options
the amt shuld be of the form 999n , n>0 , n is and integer - 11 years agoHelpfull: Yes(0) No(2)
- he gets infnte rupees...
- 11 years agoHelpfull: Yes(0) No(1)
- the ans is 33966
- 8 years agoHelpfull: Yes(0) No(1)
- 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.
- 8 years agoHelpfull: Yes(0) No(0)
TCS Other Question
find the last two digits of 583^512?
1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4....... what is the 2888th number in this sequence???