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06 Nov 2009, 06:30
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Q.
Adam,Bik and Cherian have a total of 80 coins among them.Adam triples the number of coins with the others by giving them some coins from his own collection. Next, Bik repeats the same process. After this Bik now has 20 coins. Find the no of coins he had in the beginning.
1.11
2.10
3.9
4.12
Adam,Bik and Cherian have a total of 80 coins among them.Adam triples the number of coins with the others by giving them some coins from his own collection. Next, Bik repeats the same process. After this Bik now has 20 coins. Find the no of coins he had in the beginning.
1.11
2.10
3.9
4.12
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06 Nov 2009, 07:10
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Hmm am I reading this wrong or is the question stated incorrectly?
Let \(A,B,C\) be the guys and let \(a,b,c\) be their respective number of coins with \(a+b+c = 80\)
If \(B\) has 20 coins in the end, \(a+c = 60\)
\(B\) has just tripled \((a+c)\) (since he has done the "same process").
So prior to \(B\)'s donation, \(a+c = 60/3 = 20\). Which means \(b = 80-20 = 60\).
This is right after \(A\) has done his donation and tripled the amount with \(B\) and \(C\).
So prior to this, \(B\)'s original sum should \(= 60/3 = 20\).
So the answer is 20 - but it's not in the options.
Let \(A,B,C\) be the guys and let \(a,b,c\) be their respective number of coins with \(a+b+c = 80\)
If \(B\) has 20 coins in the end, \(a+c = 60\)
\(B\) has just tripled \((a+c)\) (since he has done the "same process").
So prior to \(B\)'s donation, \(a+c = 60/3 = 20\). Which means \(b = 80-20 = 60\).
This is right after \(A\) has done his donation and tripled the amount with \(B\) and \(C\).
So prior to this, \(B\)'s original sum should \(= 60/3 = 20\).
So the answer is 20 - but it's not in the options.
06 Nov 2009, 12:48
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4test1
Answer given in the book is 10 coins...are we missing a point smwhr coz your solution seems apt....just one clarification needed...is it individual's coins which is getting tripled or is it (B+C) together...then we will need to do divide that amount by 3
