M08-13

Question

A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

Show Answer

Official Answer

E

A. 60
B. 120
C. 240
D. 275
E. 300

Bunuel · Accepted Answer

Official Solution:

A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

Show Answer

Official Answer

E

A. 60
B. 120
C. 240
D. 275
E. 300

Let the number of chickens be  c  and the number of days the stock of feed is intended to last be  d . We can also assume that the amount of feed each chicken consumes per day is 1 unit (since it will be proportionally reduced anyway, the exact amount does not matter).
 
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned. This implies that 75 chickens in  d  days would have consumed the same amount of food as  (c - 75)  chickens in 20 days (since both represent the amount of food saved). Thus, we can write the equation:  75d = (c - 75)*20 .
 
If the farmer buys 100 more chickens, he will run out of feed 15 days earlier than planned. This implies that 100 chickens in  d-15  days would have consumed the same amount of food as  c  chickens in 15 days (since both represent the additional amount of food consumed). We can write another equation from this situation:  100(d-15) = 15c .
 
Solving the system of equations,  75d = (c - 75) *20  and  100(d-15) = 15c , gives us  c = 300 .
 
Therefore, the farmer has 300 chickens.

Answer: E

Gemmie · Answer

c: number of chickens
d: number of days

c * d = (c - 75) * (d + 20)
= (c + 100) * (d - 15)

=> cd = cd - 75d + 20c - 1500 (i)
= cd + 100d - 15c - 1500 (ii)

=> - 75d + 20c = 100d - 15c
=> 35c = 175d
=> c = 5d

From (i) => 100d - 15c = 1500
=> 20d - 3c = 300
=> 4c - 3c = 300 (coz c = 5d)
 => c = 300

heartbanger97 · Answer

I think this is a  high-quality  question and I agree with explanation.

Lampard42 · Answer

I think this is a  high-quality  question and I agree with explanation.

kaladin123 · Answer

I think this is a  high-quality  question and I agree with the explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

rickyric395 · Answer

Hey  Bunuel  , can you please elaborate the solution, I am having trouble understanding the solution.

Bunuel · Answer

Can you please tell me which part is unclear in the solution? Thank you!

P.S. Else you can check alternative solutions  HERE .

rickyric395 · Answer

Hey  Bunuel  the logic you have used i.e the increase in feed with reduced chicken consumption part. I know, even though you have clearly mentioned the logic, I still have some trouble understanding it. Can you please elaborate on the reasoning you have used, if that's possible.
Thanks in advance

Bunuel · Answer

1. The problem stipulates that if the farmer sells 75 chickens, his feed would last 20 more days than intended. This suggests that the amount of feed these 75 chickens would have eaten over the originally planned duration (d days) is equal to the amount of feed the remaining chickens (c - 75) would consume in the additional 20 days. We can equate these two quantities and form our first equation: 75d = (c - 75) * 20.

2. In the second scenario, the farmer buys 100 more chickens, leading to the feed running out 15 days earlier than initially planned. This implies that these additional 100 chickens' feed consumption over the adjusted period (d - 15 days) is equivalent to the amount of feed the original quantity of chickens (c) would consume in these 15 days saved. By equating these two quantities, we can form our second equation: 100(d - 15) = 15c.

Does this make sense?

rickyric395 · Answer

Yes, It makes sense now. Thanks a lot  Bunuel

tanishqgirotra · Answer

Hi bunuel, 
I tried solving the question using this method:
let x be the number of actual chicken and y be the number of days of food for these chicken and lets assum that these each chicken eats 1 unit of food each day
So, xy = (x-75)(y+20) and xy = (x+100)(y-15) but solving these equations the answer is not coming out actually. Could you pls explain why this concept is wrong?

Bunuel · Answer

xy = (x - 75)(y + 20) and xy = (x + 100)(y - 15) gives x = 300 and y = 60, which is correct.

adarsh2209 · Answer

I like the solution - it’s helpful.

sb995 · Answer

I did not quite understand the solution. Could you please further explain why 15c represents the additional amount of food consumed and is equal to 100(d-15)?

100(d-15) = amount of food consumed by 100 chickens in d-15 days but why is 15c equal to the additional food and therefore equal to 100(d-15)? Thank you.

Bunuel · Answer

When the farmer  adds 100 chickens , the feed runs out  15 days earlier  than the planned schedule. That means the  extra food consumed  by those 100 new chickens in the shortened period must equal the food that would have been eaten by the original flock in the missing 15 days.

Extra food eaten = food for 100 chickens * (d – 15) days = 100(d – 15). 
 Food that got “used up early” = food for the original flock of c chickens * 15 days = 15c.

These two quantities must be equal because they both represent the  same chunk of feed : the feed that disappeared 15 days too soon.

So:

100(d – 15) = 15c.

M08-13

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