If a farmer sells 20 of his chickens, his stock of feed will last for 3 more days than planned, but ...
If a farmer sells 20 of his chickens, his stock of feed will last for 3 more days than planned, but if he buys 15 more chickens, he will run out of feed 4 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?
Answer/Solution
60
Steps/Work
Let x = total feed required for the planned period
n= number of chicken
t = total time of the planned feed
x = nt
1) x = (n-15) * (t+4)
2) x = (n+20) * (t-3)
equating 1&2
(n-15) * (t+4) = (n+20) * (t-3)
7n = 35t
n =5t
x= n * n/5
substituting this value in 1
n * n/5 = (n-15) * (n/5+4)
5n = 300
n =60
B
n= number of chicken
t = total time of the planned feed
x = nt
1) x = (n-15) * (t+4)
2) x = (n+20) * (t-3)
equating 1&2
(n-15) * (t+4) = (n+20) * (t-3)
7n = 35t
n =5t
x= n * n/5
substituting this value in 1
n * n/5 = (n-15) * (n/5+4)
5n = 300
n =60
B