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by nikhilkatira » Tue Jul 27, 2010 8:57 am
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?

#60

# 120

# 240

# 275

# 300
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by selango » Tue Jul 27, 2010 9:22 am
C - No of chickens
D - No of days

CD=(C-75)(D+20)--Eqn 1

(C-75)(D+20)=(C+100)(D-15).

C=5D,Sub this in Eqn 1

C=300.

I am not sure if this approach is correct.
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by nikhilkatira » Tue Jul 27, 2010 9:28 am
selango wrote:C - No of chickens
D - No of days

CD=(C-75)(D+20)--Eqn 1

(C-75)(D+20)=(C+100)(D-15).

C=5D,Sub this in Eqn 1

C=300.

I am not sure if this approach is correct.
Anand your method is right...Do you know any other approach ?
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