A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager. According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and 1/5th of the remaining to second son and so on..... After giving coins to 5th son, divided the remaining coins among five daughters equally.All should get full coins. Find the minimum number of coins he has?

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Showing Answers 1 - 25 of 25 Answers

Sathiagiri.K

  • Jul 17th, 2006
 

The answer is 13

Let the no of gold coin be x

((17x+68)/25)+ ((8x+68)/125)=x

x=408/32

x=12.75

X=13

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DEEPU T S

  • Jul 11th, 2007
 

The answer 13 is wrong..since all his daughers (5) and all his sons (5 ) will get one coin when the total is divided the number will be 10 together and considering the coins the manager gets which is also 5 the total will be anyway greater than 15

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gaurav4067

  • Dec 5th, 2010
 

Answer is 3121 coins. Here is the breakup:
First son = 624 coins
Second son = 499 coins
Third son = 399 coins
Forth son = 319 coins
Fifth son = 255 coins
Daughters = 204 each
Manager = 5 coins

Shailesh Aher

  • Jul 20th, 2015
 

16406

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sharath kumar p

  • Dec 9th, 2015
 

In question he mentioned that daughters will get coins equally, 204 would not be divided by 5.

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Narutanks

  • Mar 16th, 2016
 

Since the manager gets only 5 coins, and all should get same no of coins.
Therefore, total no of coins is 5 * (5 sons + 5 daughters + 1 manager). i.e. 5*11 = 55.
Total coins is 55

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Tej

  • Jan 6th, 2018
 

The minimum can be 21

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