krishna kumar wrote:Can someone solve this please.
If a farmer sells 75 of his chickens, his stock of feed will last 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?
A. 60
B. 120
C. 240
D. 275
E. 300
OA E
The number of chickens is
inversely proportional to the number of days. As the number of chickens increases, the number of days must decrease, so that the same amount of feed is consumed. The product of the two values must remain constant. Thus, we can write the following equation:
(number of chickens)*(number of days) = (number of chickens)*(number of days)
Let d = number of days. Let's plug in the answer choices for the number of chickens.
Answer choice C: 240 chickens
Feed for 240 chickens lasts for d days, feed for 240-75=165 chickens lasts for d+20 days.
240d = 165(d+20)
d=44 days.
Thus feed = 240*44 = 10,560.
100 more chickens = 240+100=340 chickens. 10,560 is not divisible by 340.
Eliminate C.
Answer choice D: 275 chickens
Feed for 275 chickens lasts for d days, feed for 275-75=200 chickens lasts for d+20 days.
275d = 200(d+20)
d=53.33 days. Doesn't work.
Eliminate D.
Answer choice E: 300 chickens
Feed for 300 chickens lasts for d days, feed for 300-75=225 chickens lasts for d+20 days.
300d = 225(d+20)
d=60 days.
Thus feed = 300*60 = 18000.
100 more chickens = 300+100= 400 chickens. 18000/400 = 45 days.
60-45 = 15 fewer days. Success!
The correct answer is
E.
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